NIMCET Mathematics Previous Year Question Paper and Solution with PDF

Solution

Qus : 5NIMCET PYQ

If $\vec{a}=4\hat{j}$ and $\vec{b}=3\hat{j}+4\hat{k}$ , then the vector form of the component of $\vec{a}$ alond $\vec{b}$ is

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Solution

Qus : 6NIMCET PYQ

If $\vec{a}=\hat{i}-\hat{k}$, $\vec{b}=x\hat{i}+\hat{j}+(1-x)\hat{k}$ and $\vec{c}=y\hat{i}+x\hat{j}+(1+x-y)\hat{k}$, then $\begin{bmatrix}{\vec{a}} & {\vec{b}} & {\vec{c}}\end{bmatrix}$ depends on

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Solution

Quick Solution

Given:

$ \vec{a} = \hat{i} - \hat{k}, \quad \vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}, \quad \vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k} $

Form the matrix:

$ M = \begin{bmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{bmatrix} $

Find the determinant:

$ \det(M) = \begin{vmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{vmatrix} = 1 $

Since the determinant is constant and non-zero, the vectors are linearly independent.

$ \boxed{\text{The matrix does not depend on } x \text{ or } y} $

Qus : 7NIMCET PYQ

Angle of elevation of the top of the tower from 3 points (collinear) A, B and C on a road leading to the foot of the tower are 30°, 45° and 60°, respectively. The ratio of AB and BC is

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Solution

According to the given information, the figure should be as follows.

Let the height of tower = h

Qus : 8NIMCET PYQ

If $\vec{a}$ and $\vec{b}$ in space, given by $\vec{a}=\frac{\hat{i}-2\hat{j}}{\sqrt{5}}$ and $\vec{b}=\frac{2\hat{i}+\hat{j}+3\hat{k}}{\sqrt{14}}$ , then the value of $(2\vec{a}+\vec{b}).[(\vec{a} \times \vec{b}) \times (\vec{a}-2\vec{b})]$ is

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Solution

Qus : 9NIMCET PYQ

If $\vec{a}, \vec{b}$ are unit vectors such that $2\vec{a}+\vec{b} =3$ then which of the following statement is true?

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Solution

Quick Solution

Given: $ \vec{a}, \vec{b} $ are unit vectors and

$ 2\vec{a} + \vec{b} = 3 $

Take magnitude on both sides:

$ |2\vec{a} + \vec{b}| = 3 \Rightarrow |2\vec{a} + \vec{b}|^2 = 9 $

Use identity:

\[ |2\vec{a} + \vec{b}|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 + 4(\vec{a} \cdot \vec{b}) = 4 + 1 + 4(\vec{a} \cdot \vec{b}) = 5 + 4(\vec{a} \cdot \vec{b}) \]

Set equal to 9:

\[ 5 + 4(\vec{a} \cdot \vec{b}) = 9 \Rightarrow \vec{a} \cdot \vec{b} = 1 \Rightarrow \cos\theta = 1 \Rightarrow \theta = 0^\circ \]

Qus : 10NIMCET PYQ

The area enclosed between the curves y² = x and y = |x| is

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Solution

Solving y² = x and y = x,

we get, y = 0, x = 0, y = 1, x = 1 Therefore,

Qus : 11NIMCET PYQ

$\int f(x)\mathrm{d}x=g(x)$, then $\int {x}^5f({x}^3)\mathrm{d}x$

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Solution

Quick Solution

Given:

$ \int f(x)\, dx = g(x) $

Required: $ \int x^5 f(x^3)\, dx $

Use substitution:

Let $ u = x^3 \Rightarrow du = 3x^2\, dx \Rightarrow dx = \frac{du}{3x^2} $

Now rewrite the integral:

\[ \int x^5 f(x^3)\, dx = \int x^5 f(u) \cdot \frac{du}{3x^2} = \frac{1}{3} \int x^3 f(u)\, du \]

But $ x^3 = u $, so:

\[ \frac{1}{3} \int u f(u)\, du \]

Now integrate by parts or use the identity:

\[ \int u f(u)\, du = u g(u) - \int g(u)\, du \]

Final answer:

\[ \int x^5 f(x^3)\, dx = \frac{1}{3} \left[ x^3 g(x^3) - \int g(x^3) \cdot 3x^2\, dx \right] = x^3 g(x^3) - \int x^2 g(x^3)\, dx \]

\[ \boxed{ \int x^5 f(x^3)\, dx = x^3 g(x^3) - \int x^2 g(x^3)\, dx } \]

Qus : 12NIMCET PYQ

Test the continuity of the function at x = 2

$f(x)= \begin{cases} \frac{5}{2}-x & \text{ if } x<2 \\ 1 & \text{ if } x=2 \\ x-\frac{3}{2}& \text{ if } x>2 \end{cases}$

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Solution

LHL ≠ f(2)

Qus : 13NIMCET PYQ

$\lim _{{x}\rightarrow1}\frac{{x}^4-1}{x-1}=\lim _{{x}\rightarrow k}\frac{{x}^3-{k}^2}{{x}^2-{k}^2}=$, then find k

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Solution

Quick DL Method Solution

Given:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \]

LHS using derivative:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \left.\frac{d}{dx}(x^4)\right|_{x=1} = 4x^3|_{x=1} = 4 \]

RHS using DL logic:

\[ \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \approx \frac{3k^2(x - k)}{2k(x - k)} = \frac{3k}{2} \]

Equating both sides:

\[ \frac{3k}{2} = 4 \Rightarrow k = \frac{8}{3} \]

\[ \boxed{k = \frac{8}{3}} \]

Qus : 14NIMCET PYQ

The value of

2tan^-1[cosec(tan^-1x) - tan(cot^-1x)]

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Solution

Qus : 15NIMCET PYQ

The graph of function $f(x)=\log _e({x}^3+\sqrt[]{{x}^6+1})$ is symmetric about:

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Solution

Qus : 16NIMCET PYQ

If $3 sin x + 4 cos x = 5$, then $6tan\frac{x}{2}-9tan^2\frac{x}{2}$

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Solution

Qus : 17NIMCET PYQ

If the equation $|x^2 – 6x + 8| = a$ has four real solution then find the value of $a$?

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Solution

Qus : 18NIMCET PYQ

If A is a subset of B and B is a subset of C, then cardinality of A ∪ B ∪ C is equal to

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Solution

Qus : 19NIMCET PYQ

Largest value of $cos^2\theta -6sin\theta cos\theta+3sin^2\theta+2 $ is

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Solution

Qus : 20NIMCET PYQ

Given to events A and B such that odd in favour A are 2 : 1 and odd in favour of $A \cup B$ are 3 : 1. Consistent with this information the smallest and largest value for the probability of event B are given by

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Solution

Qus : 21NIMCET PYQ

If A and B are square matrices such that $B=-A^{-1} BA$, then $(A + B)^2$ is

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Solution

Qus : 22NIMCET PYQ

A bag contain different kind of balls in which 5 yellow, 4 black & 3 green balls. If 3 balls are drawn at random then find the probability that no black ball is chosen

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Solution

Probability — No Black Ball is Chosen

Given:

Yellow balls = 5
Black balls = 4
Green balls = 3
Total balls = 5 + 4 + 3 = 12

We are to find:

Probability that no black ball is selected when 3 balls are drawn at random.

Step 1: Total number of ways to choose any 3 balls from 12:

\[ \text{Total ways} = \binom{12}{3} = 220 \]

Step 2: Ways to choose 3 balls such that no black ball is chosen:

Only yellow and green balls are allowed ⇒ Total = 5 (yellow) + 3 (green) = 8
\[ \text{Favorable ways} = \binom{8}{3} = 56 \]

Step 3: Probability

\[ P(\text{no black ball}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{56}{220} = \frac{14}{55} \]

\[ \boxed{\text{Probability} = \frac{14}{55}} \]

Qus : 23NIMCET PYQ

Between any two real roots of the equation $e^x sin x = 1$, the equation $e^x cos x = –1$ has

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Solution

Number of Roots

Given:

$ e^x \sin x = 1 $ has two real roots → say $ x_1 $ and $ x_2 $

Apply Rolle’s Theorem:

Since $ f(x) = e^x \sin x $ is continuous and differentiable, and $ f(x_1) = f(x_2) $, ⇒ There exists $ c \in (x_1, x_2) $ such that $ f'(c) = 0 $

Compute:

\[ f'(x) = e^x(\sin x + \cos x) = 0 \Rightarrow \tan x = -1 \] At this point, \[ e^x \cos x = -1 \]

\[ \boxed{\text{At least one root}} \]

Qus : 24NIMCET PYQ

If f(x) is a polynomial of degree 4, f(n) = n + 1 & f(0) = 25, then find f(5) = ?

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Solution

Correct Shortcut Method — Find $ f(5) $

Step 1: Define a helper polynomial:

\[ g(x) = f(x) - (x + 1) \]

Given: $ f(1) = 2, f(2) = 3, f(3) = 4, f(4) = 5 \Rightarrow g(1) = g(2) = g(3) = g(4) = 0 $

So, \[ g(x) = A(x - 1)(x - 2)(x - 3)(x - 4) \quad \Rightarrow \quad f(x) = A(x - 1)(x - 2)(x - 3)(x - 4) + (x + 1) \]

Step 2: Use $ f(0) = 25 $ to find A:

\[ f(0) = A(-1)(-2)(-3)(-4) + (0 + 1) = 24A + 1 = 25 \Rightarrow A = 1 \]

Step 3: Compute $ f(5) $:

\[ f(5) = (5 - 1)(5 - 2)(5 - 3)(5 - 4) + (5 + 1) = 4 \cdot 3 \cdot 2 \cdot 1 + 6 = 24 + 6 = \boxed{30} \]

✅ Final Answer: $ \boxed{f(5) = 30} $

Qus : 25NIMCET PYQ

The maximum value of $f(x) = (x – 1)^2 (x + 1)^3$ is equal to $\frac{2^p3^q}{3125}$ then the ordered pair of (p, q) will be

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Solution

Maximum Value of $ f(x) = (x - 1)^2(x + 1)^3 $

Step 1: Let’s define the function:

\[ f(x) = (x - 1)^2 (x + 1)^3 \]

Step 2: Take derivative to find critical points

Use product rule:
Let $ u = (x - 1)^2 $, $ v = (x + 1)^3 $
\[ f'(x) = u'v + uv' = 2(x - 1)(x + 1)^3 + (x - 1)^2 \cdot 3(x + 1)^2 \] \[ f'(x) = (x - 1)(x + 1)^2 [2(x + 1) + 3(x - 1)] \] \[ f'(x) = (x - 1)(x + 1)^2 (5x - 1) \]

Step 3: Find critical points

Set $ f'(x) = 0 $: \[ (x - 1)(x + 1)^2 (5x - 1) = 0 \Rightarrow x = 1,\ -1,\ \frac{1}{5} \]

Step 4: Evaluate $ f(x) $ at these points

$ f(1) = 0 $
$ f(-1) = 0 $
$ f\left(\frac{1}{5}\right) = \left(\frac{1}{5} - 1\right)^2 \left(\frac{1}{5} + 1\right)^3 = \left(-\frac{4}{5}\right)^2 \left(\frac{6}{5}\right)^3 $

\[ f\left(\frac{1}{5}\right) = \frac{16}{25} \cdot \frac{216}{125} = \frac{3456}{3125} \]

Step 5: Compare with given form:

It is given that maximum value is $ \frac{3456}{3125} = 2^p \cdot 3^q / 3125 $

Factor 3456: \[ 3456 = 2^7 \cdot 3^3 \Rightarrow \text{So } p = 7, \quad q = 3 \]

✅ Final Answer: $ \boxed{(p, q) = (7,\ 3)} $

Qus : 26NIMCET PYQ

The coefficient of $x^{50}$ in the expression of ${(1 + x)^{1000} + 2x(1 + x)^{999} + 3x^2(1 + x)^{998} + ...... + 1001x^{1000}}$

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Solution

Easiest Method — Coefficient of $ x^{50} $

Given Expression:

\[ (1 + x)^{1000} + 2x(1 + x)^{999} + 3x^2(1 + x)^{998} + \cdots + 1001x^{1000} \]

This follows a known identity that simplifies the full expression to:

\[ f(x) = (1 + x)^{1002} \]

Now: The coefficient of $ x^{50} $ in $ f(x) $ is:

\[ \boxed{\binom{1002}{50}} \]

✅ Final Answer: $ \boxed{\binom{1002}{50}} $

Qus : 27NIMCET PYQ

If ${{x}}_k=\cos \Bigg{(}\frac{2\pi k}{n}\Bigg{)}+i\sin \Bigg{(}\frac{2\pi k}{n}\Bigg{)}$ , then $\sum ^n_{k=1}({{x}}_k)=?$

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Solution

Sum of Complex Roots of Unity

Given:

\[ x_k = \cos\left(\frac{2\pi k}{n}\right) + i \sin\left(\frac{2\pi k}{n}\right) = e^{2\pi i k/n} \]

Required: Find: \[ \sum_{k=1}^{n} x_k \]

This is the sum of all $ n^\text{th} $ roots of unity (from $ k = 1 $ to $ n $).

We know: \[ \sum_{k=0}^{n-1} e^{2\pi i k/n} = 0 \] So shifting index from $ k = 1 $ to $ n $ just cycles the same roots: \[ \sum_{k=1}^{n} e^{2\pi i k/n} = 0 \]

✅ Final Answer: $ \boxed{0} $

Qus : 28NIMCET PYQ

Number of point of which f(x) is not differentiable $f(x)=|cosx|+3$ in $[-\pi, \pi]$

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Solution

Points of Non-Differentiability of $ f(x) = |\cos x| + 3 $

Step 1: $ \cos x $ is differentiable everywhere, but $ |\cos x| $ is not differentiable where $ \cos x = 0 $.

Step 2: In the interval $ [-\pi, \pi] $, we have:

\[ \cos x = 0 \Rightarrow x = -\frac{\pi}{2},\ \frac{\pi}{2} \]

So $ f(x) = |\cos x| + 3 $ is not differentiable at these two points due to sharp turns.

✅ Final Answer: $ \boxed{2 \text{ points}} $

Qus : 29NIMCET PYQ

If $n_1$ and $n_2$ are the number of real valued solutions $x = | sin^{–1} x |$ & $x = sin (x)$ respectively, then the value of $n_2– n_1$ is

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Solution

Qus : 30NIMCET PYQ

The negation of $\sim S\vee(\sim R\wedge S)$ is equivalent to

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Solution

Qus : 31NIMCET PYQ

A point P in the first quadrant, lies on $y^2 = 4ax$, a > 0, and keeps a distance of 5a units from its focus. Which of the following points lies on the locus of P?

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Solution

Locus of Point on Parabola

Given: Point on parabola $ y^2 = 4a x $ is at distance $ 5a $ from focus $ (a, 0) $.

Distance Equation:

\[ (x - a)^2 + y^2 = 25a^2 \Rightarrow (x - a)^2 + 4a x = 25a^2 \Rightarrow x^2 + 2a x - 24a^2 = 0 \]

Solving gives: $ x = 4a $, $ y = 4a $

✅ Final Answer: $ \boxed{(4a,\ 4a)} $

Qus : 32NIMCET PYQ

If $\int x\, \sin x\, sec^3x\, dx=\frac{1}{2}\Bigg{[}f(x){se}c^2x+g(x)\Bigg{(}\frac{\tan x}{x}\Bigg{)}\Bigg{]}+C$, then which of the following is true?

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Solution

Qus : 33NIMCET PYQ

Let a, b, c, d be no zero numbers. If the point of intersection of the line 4ax + 2ay + c = 0 & 5bx + 2by + d=0 lies in the fourth quadrant and is equidistance from the two are then

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Solution

Qus : 34NIMCET PYQ

$\theta={\cos }^{-1}\Bigg{(}\frac{3}{\sqrt[]{10}}\Bigg{)}$ is the angle between $\vec{a}=\hat{i}-2x\hat{j}+2y\hat{k}$ & $\vec{b}=x\hat{i}+\hat{j}+y\hat{k}$ then possible values of (x,y) that lie on the locus

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Solution

Qus : 35NIMCET PYQ

Let R be reflexive relation on the finite set a having 10 elements and if m is the number of ordered pair in R, then

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Solution

Qus : 36NIMCET PYQ

If $| x - 6|= | x - 4x | -| x^2- 5x +6 |$ , where x is a real variable

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Solution

Qus : 37NIMCET PYQ

The range of values of $\theta$ in the interval $(0,\pi)$ such that the points (3, 2) and $(cos\theta ,sin\theta)$ lie on the samesides of the line x + y – 1 = 0, is

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Solution

Same Side of a Line — Geometric Condition

Line: $ x + y - 1 = 0 $
Point 1: (3, 2) → Lies on the side where value is positive:
\[ f(3, 2) = 3 + 2 - 1 = 4 > 0 \]

Point 2: $ (\cos\theta, \sin\theta) $ lies on same side if: \[ \cos\theta + \sin\theta > 1 \] Using identity: \[ \cos\theta + \sin\theta = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) \Rightarrow \sin\left(\theta + \frac{\pi}{4}\right) > \frac{1}{\sqrt{2}} \] So: \[ \theta \in \left(0,\ \frac{\pi}{2}\right) \]

✅ Final Answer: $ \boxed{\theta \in \left(0,\ \frac{\pi}{2}\right)} $

Qus : 38NIMCET PYQ

Which of the following number is the coefficient of $x^{100}$ in the expansion of $\log _e\Bigg{(}\frac{1+x}{1+{x}^2}\Bigg{)},\, |x|{\lt}1$ ?

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Solution

Qus : 39NIMCET PYQ

A real valued function f is defined as $f(x)=\begin{cases}{-1} & {-2\leq x\leq0} \\ {x-1} & {0\leq x\leq2}\end{cases}$. Which of the following statement is FALSE?

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Solution

Qus : 40NIMCET PYQ

A line segment AB of length 10 meters is passing through the foot of the perpendicular of a pillar, which is standing at right angle to the ground. Top of the pillar subtends angles $tan^{–1}$ 3 and $tan^{–1} 2$ at A and B respectively. Which of the following choice represents the height of the pillar?

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Solution

Qus : 41NIMCET PYQ

If a vector having magnitude of 5 units, makes equal angle with each of the three mutually perpendicular axes,then the sum of the magnitude of the projections on each of the axis is

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Solution

Vector Projection Problem

Given: A vector of magnitude 5 makes equal angles with x, y, and z axes.
To Find: Sum of magnitudes of projections on each axis.

Let angle with each axis be $ \alpha $. Then, from direction cosine identity: \[ \cos^2\alpha + \cos^2\alpha + \cos^2\alpha = 1 \Rightarrow 3\cos^2\alpha = 1 \Rightarrow \cos\alpha = \frac{1}{\sqrt{3}} \]

Projection on each axis: $ 5 \cdot \frac{1}{\sqrt{3}} $
Sum = $ 3 \cdot \frac{5}{\sqrt{3}} = \frac{15}{\sqrt{3}} = \boxed{5\sqrt{3}} $

✅ Final Answer: $ \boxed{5\sqrt{3}} $

Qus : 42NIMCET PYQ

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ballsis transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred is red, is:

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Solution

Conditional Probability – Bayes' Theorem

Given: One ball is transferred from Bag I to Bag II, and then a ball is drawn from Bag II and is black.

Goal: Find the probability that the transferred ball was red, given that a black ball was drawn.

Using Bayes' theorem: \[ P(R|A) = \frac{P(R \cap A)}{P(A)} = \frac{\frac{3}{10} \cdot \frac{5}{10}}{\frac{3}{10} \cdot \frac{5}{10} + \frac{4}{10} \cdot \frac{6}{10} + \frac{3}{10} \cdot \frac{5}{10}} = \frac{15}{54} = \boxed{\frac{5}{18}} \]

✅ Final Answer: $ \boxed{\frac{5}{18}} $

Qus : 43NIMCET PYQ

Let $f(x)=\frac{x^2-1}{|x|-1}$. Then the value of $lim_{x\to-1} f(x)$ is

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Solution

Qus : 44NIMCET PYQ

The mean of 5 observation is 5 and their variance is 12.4. If three of the observations are 1,2 and 6; then the mean deviation from the mean of the data is:

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Solution

Qus : 45NIMCET PYQ

In a beauty contest, half the number of experts voted Mr. A and two thirds voted for Mr. B 10 voted for both and 6 did not for either. How may experts were there in all.

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Solution

Let the total number of experts be

N

E

is the set of experts who voted for miss

A

F

is the set of experts who voted for miss

B

.
Since

6

did not vote for either, n

(E \cup F) = N - 6

n (E) = \frac{N}{2}, n (F) = \frac{2}{3} N

and

n (E \cap F) = 10

.
So,

N - 6 = \frac{N}{2} + \frac{2}{3} N - 10

Solving the above equation gives

\frac{N}{6} = 4 \Rightarrow N = 24

Qus : 46NIMCET PYQ

A circle touches the x–axis and also touches the circle with centre (0, 3) and radius 2. The locus of the centre of the circle is

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Solution

Qus : 47NIMCET PYQ

The value of non-zero scalars α and β such that for all vectors

and

such that

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Solution

Qus : 48NIMCET PYQ

A force of 78 grams acts at the point (2,3,5). The direction ratios of the line of action being 2,2,1 . The magnitude of its moment about the line joining the origin to the point (12,3,4) is

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Solution

Qus : 49NIMCET PYQ

Number of real solutions of the equation

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Solution

Qus : 50NIMCET PYQ

The sum of infinite terms of a decreasing GP is equal to the greatest value of the function

in the interval [-2,3] and the difference between the first two terms is f'(0). Then the common ratio of GP is

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Solution

Qus : 51NIMCET PYQ

Number of onto (surjective) functions from A to B if n(A)=6 and n(B)=3, is

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Solution

Qus : 52NIMCET PYQ

If $|z|<\sqrt{3}-1$, then $|z^{2}+2z cos \alpha|$ is

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Solution

Qus : 53NIMCET PYQ

A computer producing factory has only two plants T₁ and T₂ produces 20% and plant T₂ produces 80% of the total computers produced. 7% of the computers produced in the factory turn out to be defective. It is known that P (computer turns out to be defective given that it is produced in plant T₁ 10P(computer turns out to be defective given that it is produced in plant T₂ ). A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant T₂ is

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Solution

Qus : 54NIMCET PYQ

If A > 0, B > 0 and A + B = $\frac{\pi}{6}$ , then the minimum value of $tanA + tanB$

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Solution

On differentiating

x= tanA + tan(π/6-A)

we get :

dx/dA = sec²A-sec²(π/6-A)

now putting

dx/dA=0

we get

cos²(A) = cos²(π/6-A) so 0≤A≤π/6

therefore

A=π/6-A from here we get A = π/12 = B

so minimum value of that function is

2tanπ/12 which is equal to 2(2-√3)

Qus : 55NIMCET PYQ

Inverse of the function $f(x)=\frac{10^x-10^{-x}}{10^{x}+10^{-x}}$ is

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Solution

Let $f(x)=y$, then

⇒ $\frac{10^x-10^-x}{10^x+10^-x}=y$

⇒ $\frac{10^{2x}-1}{10^{2x}+1}=y$

⇒ $10^{2x}=\frac{1+y}{1-y}$ By Componendo Dividendo Rule

⇒ $x=\frac{1}{2}\log _{10}\Bigg{(}\frac{1+y}{1-y}\Bigg{)}$

⇒ ${f}^{-1}(y)=\frac{1}{2}\log _{10}\Bigg{(}\frac{1+y}{1-y}\Bigg{)}$

⇒ ${f}^{-1}(x)=\frac{1}{2}\log _{10}\Bigg{(}\frac{1+x}{1-x}\Bigg{)}$

Qus : 56NIMCET PYQ

The tangent at the point (2, -2) to the curve $x^2 y^2-2x=4(1-y)$ does not passes through the point

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Solution

Qus : 57NIMCET PYQ

The integral $\int \sqrt{1+2 cot x(cosec x+cotx)} dx$ , $(0<x<\frac{\pi}{2})$ (where C is a constant of integration) is equal to

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Solution

Qus : 58NIMCET PYQ

If all the words, with or without meaning, are written using the letters of the word QUEEN add are arranged as in English Dictionary, then the position of the word QUEEN is

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Solution

Letters of the word QUEEN are E,E,N,Q,U

Words beginning with E (4!) = 24

Words beginning with N (4!/2!)=12

Words beginning with QE (3!) = 6

Words beginning with QN (3!/2!)= 3

Total words = 24+12+6+9=45

QUEEN is the next word and has rank 46th.

Qus : 59NIMCET PYQ

The curve satisfying the differential equation

and passing through the point (1,1) also passes through the point __________

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Solution

Qus : 60NIMCET PYQ

is equal to

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Solution

Qus : 61NIMCET PYQ

If S and S' are foci of the ellipse

, B is the end of the minor axis and BSS' is an equilateral triangle, then the eccentricity of the ellipse is

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Solution

Qus : 62NIMCET PYQ

The equation of the circle passing through the point (4,6) and whose diameters are along x + 2y - 5 =0 and 3x - y - 1=0 is

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Solution

Qus : 63NIMCET PYQ

In a parallelogram ABCD, P is the midpoint of AD. Also, BP and AC intersect at Q. Then AQ : QC =

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Solution

Qus : 64NIMCET PYQ

The median AD of ΔABC is bisected at E and BE is extended to meet the side AC in F. The AF : FC =

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Solution

Qus : 65NIMCET PYQ

Let $X_i, i = 1,2,.. , n$ be n observations and $w_i = px_i +k, i = 1,2, ,n$ where p and k are constants. If the mean of $x_i 's$ is 48 and the standard deviation is 12, whereas the mean of $w_i 's$ is 55 and the standard deviation is 15, then the value of p and k should be

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Solution

Qus : 66NIMCET PYQ

Not Available

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Solution

Qus : 67NIMCET PYQ

, then the values of A₁, A₂, A₃, A₄ are

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Solution

Qus : 68NIMCET PYQ

then x =

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Solution

Qus : 69NIMCET PYQ

Equation of the tangent from the point (3,−1) to the ellipse 2x² + 9y² = 3 is

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Solution

Qus : 70NIMCET PYQ

The position vectors of the vertices

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Solution

Qus : 71NIMCET PYQ

The $sin^2 x tanx + cos^2 x cot x-sin2x=1+tanx+cotx $, $x \in (0 , \pi)$, then x

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Solution

Qus : 72NIMCET PYQ

Not Available right now

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Solution

Qus : 73NIMCET PYQ

In a chess tournament, n men and 2 women players participated. Each player plays 2 games against every other player. Also, the total number of games played by the men among themselves exceeded by 66 the number of games that the men played against the women. Then the total number of players in the tournament is

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Solution

Qus : 74NIMCET PYQ

Suppose A₁ , A₂ , A₃ , …..A₃₀ are thirty sets each having 5 elements with no common elements across the sets and B₁ , B₂ , B₃ , ..... , B_n are

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Solution

Qus : 75NIMCET PYQ

Let

,where [x]denotes the greatest integer

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Solution

Qus : 76NIMCET PYQ

Let U and V be two events of a sample space S and P(A) denote the probability of an event A. Which of the following statements is true?

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Solution

Qus : 77NIMCET PYQ

If a man purchases a raffle ticket, he can win a first prize of Rs.5,000 or a second prize of Rs.2,000 with probabilities 0.001 and 0.003 respectively. What should be a fair price to pay for the ticket?

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Solution

Qus : 78NIMCET PYQ

If the mean deviation 1, 1+d, 1+2d, … , 1+100d from their mean is 255, then d is equal to

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Solution

Qus : 79NIMCET PYQ

and

, then a possible value of n is among the following is

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Solution

Qus : 80NIMCET PYQ

Let S be the set $\{a\in Z^+:a\leq100\}$.If the equation $[tan^2 x]-tan x - a = 0$ has real roots (where [ . ] is the greatest integer function), then the number of elements is S is

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Solution

Qus : 81NIMCET PYQ

The solution set of the inequality

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Solution

Qus : 82NIMCET PYQ

If $a, b, c$ are in GP and $log a - log 2b$, $log 2b - log 3c$ and $log 3c - log a$ are in AP, then $a, b, c$are the lengths of the sides of a triangle which is

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Solution

Qus : 83NIMCET PYQ

Not Available

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Solution

Qus : 84NIMCET PYQ

If (1 + x – 2x²)⁶= 1 + a₁x + a₂x²+ ... + a₁₂x¹², then the value a₂+ a₄+ a₆+ ... + a₁₂

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Solution

Qus : 85NIMCET PYQ

Not Available

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Solution

Qus : 86NIMCET PYQ

A man takes a step forward with probability 0.4 and backward with probability 0.6. The probability that at the end of eleven steps, he is one step away from the starting point is

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Solution

Qus : 87NIMCET PYQ

Not Available

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Solution

Qus : 88NIMCET PYQ

If x, y, z are distinct real numbers then

= 0, then xyz=

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Solution

Qus : 89NIMCET PYQ

Not Available

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Solution

Qus : 90NIMCET PYQ

If $a, a, a_2, ., a_{2n-1},b$ are in AP, $a, b_1, b_2,...b_{2n-1}, b $are in GP and $a, c_1, c_2,... c_{2n-1}, b $ are in HP, where a, b are positive, then the equation $a_n x^2-b_n+c_n$ has its roots

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Solution

Qus : 91NIMCET PYQ

For the two circles $x^2+y^2=16$ and $x^2+y^2-2y=0$, there is/are

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Solution

Qus : 92NIMCET PYQ

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Solution

Qus : 93NIMCET PYQ

Let $\vec{a}, \vec{b}, \vec{c} $ be distinct non-negative numbers. If the vectors $a\hat{i}+a\hat{j}+c\hat{k}$ , $\hat{i}+\hat{k}$ and $c\hat{i}+c\hat{j}+b\hat{k}$ lie in a plane, then c is

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Solution

$\vec{a}=a\hat{i}+a\hat{j}+c\hat{k}\, ,\, \vec{b}=\hat{i}+\hat{k}\, \&\, \vec{c}=c\hat{i}+c\hat{j}+b\hat{k}$ are coplanar.

$\Rightarrow\begin{vmatrix}{a} & {a} & {c} \\ {1} & {0} & {1} \\ {c} & {c} & {b}\end{vmatrix}=0$

$\Rightarrow-ac-ab+ac+{c}^2=0$

$\Rightarrow{c}^2=ab$

Qus : 94NIMCET PYQ

A particle P starts from the point

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Solution

Qus : 95NIMCET PYQ

The correct expression for $cos^{-1} (-x)$ is

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Solution

You should learn it as an important formula.

Qus : 96NIMCET PYQ

If (4, 3) and (12, 5) are the two foci of an ellipse passing through the origin, then the eccentricity of the ellipse is

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Solution

Given: Foci are (4, 3) and (12, 5), and the ellipse passes through the origin (0, 0).

Step 1: Use ellipse definition

$PF_1 = \sqrt{(0 - 4)^2 + (0 - 3)^2}

= \sqrt{25}

= 5$

$PF_2 = \sqrt{(0 - 12)^2 + (0 - 5)^2}

= \sqrt{169}

= 13$

Total distance = $5 + 13 = 18 \Rightarrow 2a = 18 \Rightarrow a = 9$

Step 2: Distance between the foci

$2c = \sqrt{(12 - 4)^2 + (5 - 3)^2} = \sqrt{64 + 4} = \sqrt{68} \Rightarrow c = \sqrt{17}$

Step 3: Find eccentricity

$e = \dfrac{c}{a} = \dfrac{\sqrt{17}}{9}$

✅ Final Answer: $\boxed{\dfrac{\sqrt{17}}{9}}$

Qus : 97NIMCET PYQ

If $\Delta=a^2-(b-c)^2$, where $\Delta$ is the are of the triangle ABC, then $tanA=$

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Solution

Qus : 98NIMCET PYQ

The number of one - one functions f: {1,2,3} → {a,b,c,d,e} is

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Solution

Given: A one-one function from set $\{1,2,3\}$ to set $\{a,b,c,d,e\}$

Step 1: One-one (injective) function means no two elements map to the same output.

We choose 3 different elements from 5 and assign them to 3 inputs in order.

So, total one-one functions = $P(5,3) = 5 \times 4 \times 3 = 60$

✅ Final Answer: $\boxed{60}$

Qus : 99NIMCET PYQ

Two numbers $a$ and $b$ are chosen are random from a set of the first 30 natural numbers, then the probability that $a^2 - b^2$ is divisible by 3 is

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Solution

Qus : 100NIMCET PYQ

Suppose that the temperature at a point (x,y), on a metal plate is $T(x,y)=4x^2-4xy+y^2$, An ant, walking on the plate, traverses a circle of radius 5 centered at the origin. What is the highest temperature encountered by the ant?

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Solution

Qus : 101NIMCET PYQ

The value of the limit $$\lim _{{x}\rightarrow0}\Bigg{(}\frac{{1}^x+{2}^x+{3}^x+{4}^x}{4}{\Bigg{)}}^{1/x}$$ is

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Solution

Qus : 102NIMCET PYQ

The 10th and 50th percentiles of the observation 32, 49, 23, 29, 118 respectively are

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Solution

Qus : 103NIMCET PYQ

The value of m for which volume of the parallelepiped is 4 cubic units whose three edges are represented by a = mi + j + k, b = i – j + k, c = i + 2j –k is

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Solution

Given: Volume of a parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ is 4 cubic units.

Vectors:

$\vec{a} = m\hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} - \hat{j} + \hat{k}$
$\vec{c} = \hat{i} + 2\hat{j} - \hat{k}$

Step 1: Volume = $|\vec{a} \cdot (\vec{b} \times \vec{c})|$

First compute $\vec{b} \times \vec{c}$:

$ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = \hat{i}((-1)(-1) - (1)(2)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(2) - (-1)(1)) \\ = \hat{i}(1 - 2) - \hat{j}(-1 - 1) + \hat{k}(2 + 1) = -\hat{i} + 2\hat{j} + 3\hat{k} $

Step 2: Compute dot product with $\vec{a}$:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = (m)(-1) + (1)(2) + (1)(3) = -m + 2 + 3 = -m + 5$

Step 3: Volume = $| -m + 5 | = 4$

So, $|-m + 5| = 4 \Rightarrow -m + 5 = \pm 4$

Case 1: $-m + 5 = 4 \Rightarrow m = 1$
Case 2: $-m + 5 = -4 \Rightarrow m = 9$

✅ Final Answer: $\boxed{m = 1 \text{ or } 9}$

Qus : 104NIMCET PYQ

Angles of elevation of the top of a tower from three points (collinear) A, B and C on a road leading to the foot of the tower are 30°, 45° and 60° respectively. The ratio of AB and BC is

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Solution

Qus : 105NIMCET PYQ

The number of distinct real values of $\lambda$ for which the vectors ${\lambda}^2\hat{i}+\hat{j}+\hat{k},\, \hat{i}+{\lambda}^2\hat{j}+j$ and $\hat{i}+\hat{j}+{\lambda}^2\hat{k}$ are coplanar is

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Solution

Given: Vectors:

$\vec{a} = \lambda^2 \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} + \lambda^2 \hat{j} + \hat{k}$
$\vec{c} = \hat{i} + \hat{j} + \lambda^2 \hat{k}$

Condition: Vectors are coplanar ⟹ Scalar triple product = 0

$\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$

Step 1: Use determinant:

$ \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} \lambda^2 & 1 & 1 \\ 1 & \lambda^2 & 1 \\ 1 & 1 & \lambda^2 \end{vmatrix} $

Step 2: Expand the determinant:

$ = \lambda^2(\lambda^2 \cdot \lambda^2 - 1 \cdot 1) - 1(1 \cdot \lambda^2 - 1 \cdot 1) + 1(1 \cdot 1 - \lambda^2 \cdot 1) \\ = \lambda^2(\lambda^4 - 1) - (\lambda^2 - 1) + (1 - \lambda^2) $

Simplify:

$= \lambda^6 - \lambda^2 - \lambda^2 + 1 + 1 - \lambda^2 = \lambda^6 - 3\lambda^2 + 2$

Step 3: Set scalar triple product to 0:

$\lambda^6 - 3\lambda^2 + 2 = 0$

Step 4: Let $x = \lambda^2$, then:

$x^3 - 3x + 2 = 0$

Factor:

$x^3 - 3x + 2 = (x - 1)^2(x + 2)$

So, $\lambda^2 = 1$ (double root), or $\lambda^2 = -2$ (discard as it's not real)

Thus, real values of $\lambda$ are: $\lambda = \pm1$

✅ Final Answer: $\boxed{2}$ distinct real values

Qus : 106NIMCET PYQ

If the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ are coincide, then the value of $b^2$

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Solution

Qus : 107NIMCET PYQ

There are 9 bottle labelled 1, 2, 3, ... , 9 and 9 boxes labelled 1, 2, 3,....9. The number of ways one can put these bottles in the boxes so that each box gets one bottle and exactly 5 bottles go in their corresponding numbered boxes is

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Solution

Total bottles and boxes: 9 each, labeled 1 to 9.

We are asked to count permutations of bottles such that exactly 5 bottles go into their own numbered boxes.

Step 1: Choose 5 positions to be fixed points (i.e., bottle number matches box number).

Number of ways = $\binom{9}{5}$

Step 2: Remaining 4 positions must be a derangement (no bottle goes into its matching box).

Let $D_4$ be the number of derangements of 4 items.

$D_4 = 9$

Step 3: Total ways = $\binom{9}{5} \times D_4 = 126 \times 9 = 1134$

✅ Final Answer: $\boxed{1134}$

Qus : 108NIMCET PYQ

A particle is at rest at the origin. It moves along the x −axis with an acceleration $x-x^2$ , where x is the distance of the particle at time t. The particle next comes to rest after it has covered a distance

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Solution

Qus : 109NIMCET PYQ

If the perpendicular bisector of the line segment joining p(1,4) and q(k,3) has yintercept -4, then the possible values of k are

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Solution

Given: Points: $ P(1, 4) $, $ Q(k, 3) $

Step 1: Find midpoint of PQ

Midpoint = $ \left( \dfrac{1 + k}{2}, \dfrac{4 + 3}{2} \right) = \left( \dfrac{1 + k}{2}, \dfrac{7}{2} \right) $

Step 2: Find slope of PQ

Slope of PQ = $ \dfrac{3 - 4}{k - 1} = \dfrac{-1}{k - 1} $

Step 3: Slope of perpendicular bisector = negative reciprocal = $ k - 1 $

Step 4: Use point-slope form for perpendicular bisector:

$ y - \dfrac{7}{2} = (k - 1)\left(x - \dfrac{1 + k}{2}\right) $

Step 5: Find y-intercept (put $ x = 0 $)

$ y = \dfrac{7}{2} + (k - 1)\left( -\dfrac{1 + k}{2} \right) $

$ y = \dfrac{7}{2} - (k - 1)\left( \dfrac{1 + k}{2} \right) $

Given: y-intercept = -4, so:

$ \dfrac{7}{2} - \dfrac{(k - 1)(k + 1)}{2} = -4 $

Multiply both sides by 2:

$ 7 - (k^2 - 1) = -8 \Rightarrow 7 - k^2 + 1 = -8 \Rightarrow 8 - k^2 = -8 $

$ \Rightarrow k^2 = 16 \Rightarrow k = \pm4 $

✅ Final Answer: $\boxed{k = -4 \text{ or } 4}$

Qus : 110NIMCET PYQ

If $a{\lt}b$ then $\int ^b_a\Bigg{(}|x-a|+|x-b|\Bigg{)}dx$ is equal to

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Solution

Qus : 111NIMCET PYQ

Let C denote the set of all tuples (x,y) which satisfy $x^2 -2^y=0$ where x and y are natural numbers. What is the cardinality of C?

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Solution

Qus : 112NIMCET PYQ

The domain of the function $f(x)=\frac{{\cos }^{-1}x}{[x]}$ is

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Solution

Qus : 113NIMCET PYQ

If $x=1+\sqrt[{6}]{2}+\sqrt[{6}]{4}+\sqrt[{6}]{8}+\sqrt[{6}]{16}+\sqrt[{6}]{32}$ then ${\Bigg{(}1+\frac{1}{x}\Bigg{)}}^{24}$ =

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Solution

Given:

\[ x = 1 + 2^{1/6} + 4^{1/6} + 8^{1/6} + 16^{1/6} + 32^{1/6} \]

Step 1: Write in powers of $ a = 2^{1/6} $

\[ x = 1 + a + a^2 + a^3 + a^4 + a^5 = 1 + \frac{a(a^5 - 1)}{a - 1} \]

Step 2: Use identity $ a^6 = 2 \Rightarrow a^5 = \frac{2}{a} $

\[ x = 1 + \frac{2 - a}{a - 1} = \frac{1}{a - 1} \Rightarrow 1 + \frac{1}{x} = a \Rightarrow \left(1 + \frac{1}{x} \right)^{24} = a^{24} \]

Step 3: Final calculation

\[ a = 2^{1/6} \Rightarrow a^{24} = (2^{1/6})^{24} = 2^4 = \boxed{16} \]

✅ Final Answer: $\boxed{16}$

Qus : 114NIMCET PYQ

If the volume of the parallelepiped whose adjacent edges are $\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$, $\vec{b}=\hat{i}+\alpha \hat{j}+2\hat{k}$ and $\vec{c}=\hat{i}+2\hat{j}+\alpha \hat{k}$ is 15, then $\alpha$ is equal to

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Solution

Qus : 115NIMCET PYQ

The number of solutions of ${5}^{1+|\sin x|+|\sin x{|}^2+\ldots}=25$ for $x\in(-\mathrm{\pi},\mathrm{\pi})$ is

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Solution

Step 1: Recognize the series

This is a geometric series with first term $ a = 1 $, common ratio $ r = |\sin x| \in [0,1] $, so: $$ \text{Sum} = \frac{1}{1 - |\sin x|} $$

Step 2: Rewrite the equation

$$ 5^{\frac{1}{1 - |\sin x|}} = 25 = 5^2 $$

Equating exponents: $$ \frac{1}{1 - |\sin x|} = 2 \Rightarrow 1 - |\sin x| = \frac{1}{2} \Rightarrow |\sin x| = \frac{1}{2} $$

Step 3: Solve for $ x \in (-\pi, \pi) $

We want all $ x \in (-\pi, \pi) $ such that $ |\sin x| = \frac{1}{2} $

So $ \sin x = \pm \frac{1}{2} $. Within $ (-\pi, \pi) $, the values of $ x $ satisfying this are:

$x = \frac{\pi}{6}$
$x = \frac{5\pi}{6}$
$x = -\frac{\pi}{6}$
$x = -\frac{5\pi}{6}$

✅ Final Answer: $\boxed{4}$ solutions

Qus : 116NIMCET PYQ

Let $a$ be the distance between the lines $−2x + y = 2$ and $2x − y = 2$, and $b$ be the distance between the lines $4x − 3y= 5$ and $6y − 8x = 1$, then

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Solution

Qus : 117NIMCET PYQ

The system of equations $x+2y+2z=5$, $x+2y+3z=6$, $x+2y+\lambda z=\mu$ has infinitely many solutions if

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Solution

Given System of Equations:

$x + 2y + 2z = 5$
$x + 2y + 3z = 6$
$x + 2y + \lambda z = \mu$

Goal: Find values of $\lambda$ and $\mu$ such that the system has infinitely many solutions

Step 1: Write Augmented Matrix

$ [A|B] = \begin{bmatrix} 1 & 2 & 2 & 5 \\ 1 & 2 & 3 & 6 \\ 1 & 2 & \lambda & \mu \end{bmatrix} $

Step 2: Row operations: Subtract $R_1$ from $R_2$ and $R_3$

$ \Rightarrow \begin{bmatrix} 1 & 2 & 2 & 5 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & \lambda - 2 & \mu - 5 \end{bmatrix} $

Step 3: For infinitely many solutions, rank of coefficient matrix = rank of augmented matrix < number of variables (3)

This happens when the third row becomes all zeros:

$ \lambda - 2 = 0 \quad \text{and} \quad \mu - 5 = 0 $

$\Rightarrow \lambda = 2,\quad \mu = 5$

✅ Final Answer: $\boxed{\lambda = 2,\ \mu = 5}$

Qus : 118NIMCET PYQ

If $cosec\theta-cot \theta=2$, then the value of $cosec\theta$ is

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Solution

Qus : 119NIMCET PYQ

Which of the following is TRUE?

A. If $f$ is continuous on $[a,b]$, then $\int ^b_axf(x)\mathrm{d}x=x\int ^b_af(x)\mathrm{d}x$

B. $\int ^3_0{e}^{{x}^2}dx=\int ^5_0e^{{x}^2}dx+{\int ^5_3e}^{{x}^2}dx$

C. If $f$ is continuous on $[a,b]$, then $\frac{d}{\mathrm{d}x}\Bigg{(}\int ^b_af(x)dx\Bigg{)}=f(x)$

D. Both (a) and (b)

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Solution

Qus : 120NIMCET PYQ

The solution of the equation ${4\cos }^2x+6{\sin }^2x=5$ are

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Solution

Qus : 121NIMCET PYQ

If F|= 40N (Newtons), |D| = 3m, and $\theta={60^{\circ}}$, then the work done by F acting

from P to Q is

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Solution

Formula for work done:

\[ W = |F| \cdot |D| \cdot \cos\theta \]

Given:

$ |F| = 40 \, \text{N} $
$ |D| = 3 \, \text{m} $
$ \theta = 60^\circ $

Step 1: Plug in the values:

\[ W = 40 \cdot 3 \cdot \cos(60^\circ) \]

Step 2: Use $ \cos(60^\circ) = \frac{1}{2} $

\[ W = 40 \cdot 3 \cdot \frac{1}{2} = 60 \, \text{J} \]

✅ Final Answer: $\boxed{60 \, \text{J}}$

Qus : 122NIMCET PYQ

The function $f(x)=\begin{cases}{{(1+2x)}^{1/x}} & {,x\ne0} \\ {{e}^2} & {,x=0}\end{cases}$, is

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Solution

Qus : 123NIMCET PYQ

A committee of 5 is to be chosen from a group of 9 people. The probability that a certain married couple will either serve together or not at all is

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Solution

Total people: 9

Married couple: 2 specific people among them

Total ways to choose 5 people from 9:

\[ \text{Total} = \binom{9}{5} = 126 \]

✅ Case 1: Both are selected

We fix the married couple (2 people), then choose 3 more from remaining 7:

\[ \binom{7}{3} = 35 \]

✅ Case 2: Both are NOT selected

We remove both from the pool, then choose 5 from remaining 7:

\[ \binom{7}{5} = \binom{7}{2} = 21 \]

✅ Favorable outcomes:

\[ \text{Favorable} = 35 + 21 = 56 \]

✅ Probability:

\[ \text{Required Probability} = \frac{56}{126} = \frac{28}{63} = \frac{4}{9} \]

✅ Final Answer: $\boxed{\dfrac{4}{9}}$

Qus : 124NIMCET PYQ

Let $\vec{a}=2\hat{i}+2\hat{j}+\hat{k}$ and $\vec{b}$ be another vector such that $\vec{a}.\vec{b}=14$ and $\vec{a} \times \vec{b}=3\hat{i}+\hat{j}-8\hat{k}$ the vector $\vec{b}$ =

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Solution

Qus : 125NIMCET PYQ

Find the cardinality of the set C which is defined as $C={\{x|\, \sin 4x=\frac{1}{2}\, forx\in(-9\pi,3\pi)}\}$.

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Solution

We are given:

\[ \sin(4x) = \frac{1}{2}, \quad x \in (-9\pi,\ 3\pi) \]

Step 1: General solutions for $ \sin(θ) = \frac{1}{2} $

\[ θ = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad θ = \frac{5\pi}{6} + 2n\pi \]

Let $ θ = 4x $, so we get:

$ x = \frac{\pi}{24} + \frac{n\pi}{2} $
$ x = \frac{5\pi}{24} + \frac{n\pi}{2} $

✅ Step 2: Count how many such $ x $ fall in the interval $ (-9\pi, 3\pi) $

By checking all possible $ n $ values, we find:

For $ x = \frac{\pi}{24} + \frac{n\pi}{2} $: 24 valid values
For $ x = \frac{5\pi}{24} + \frac{n\pi}{2} $: 24 valid values

? Total distinct values = 24 + 24 = 48

✅ Final Answer: $\boxed{48}$

Qus : 126NIMCET PYQ

Which term of the series $\frac{\sqrt[]{5}}{3},\, \frac{\sqrt[]{5}}{4},\frac{1}{\sqrt[]{5}},\, ...$ is $\frac{\sqrt{5}}{13}$ ?

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Solution

Qus : 127NIMCET PYQ

At how many points the following curves intersect $\frac{{y}^2}{9}-\frac{{x}^2}{16}=1$ and $\frac{{x}^2}{4}+\frac{{(y-4)}^2}{16}=1$

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Solution

Qus : 128NIMCET PYQ

The first three moments of a distribution about 2 are 1, 16, -40 respectively. The mean and variance of the distribution are

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Qus : 129NIMCET PYQ

If for non-zero x, $cf(x)+df\Bigg{(}\frac{1}{x}\Bigg{)}=|\log |x||+3,$ where $c\ne 0$, then $\int ^e_1f(x)dx=$

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Qus : 130NIMCET PYQ

A survey is done among a population of 200 people who like either tea or coffee. It is found that 60% of the pop lation like tea and 72% of the population like coffee. Let $x$ be the number of people who like both tea & coffee. Let $m{\leq x\leq n}$, then choose the correct option.

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Solution

Qus : 131NIMCET PYQ

A critical orthopedic surgery is performed on 3 patients. The probability of recovering a patient is 0.6. Then the probability that after surgery, exactly two of them will recover is

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Solution

Given:

Number of patients = 3
Probability of recovery $ p = 0.6 $
Probability of failure $ q = 1 - p = 0.4 $

We want: Probability that exactly 2 recover out of 3.

? Use Binomial Probability Formula:

\[ P(X = r) = \binom{n}{r} p^r (1 - p)^{n - r} \] where $ n = 3, r = 2, p = 0.6 $

? Calculation:

\[ P(X = 2) = \binom{3}{2} (0.6)^2 (0.4)^1 = 3 \times 0.36 \times 0.4 = 0.432 \]

✅ Final Answer: $\boxed{0.432}$

Qus : 132NIMCET PYQ

The value of $\cot \Bigg{(}{cosec}^{-1}\frac{5}{3}+{\tan }^{-1}\frac{2}{3}\Bigg{)}$ is

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Solution

Qus : 133NIMCET PYQ

The value of $\tan \Bigg{(}\frac{\pi}{4}+\theta\Bigg{)}\tan \Bigg{(}\frac{3\pi}{4}+\theta\Bigg{)}$ is

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Solution

We are given:

\[ \text{Evaluate } \tan\left(\frac{\pi}{4} + \theta\right) \cdot \tan\left(\frac{3\pi}{4} + \theta\right) \]

✳ Step 1: Use identity

\[ \tan\left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] But we don’t need expansion — use known angle values:

\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta} \]

\[ \tan\left(\frac{3\pi}{4} + \theta\right) = \frac{-1 + \tan\theta}{1 + \tan\theta} \]

✳ Step 2: Multiply

\[ \left(\frac{1 + \tan\theta}{1 - \tan\theta}\right) \cdot \left(\frac{-1 + \tan\theta}{1 + \tan\theta}\right) \]

Simplify:

\[ = \frac{(1 + \tan\theta)(-1 + \tan\theta)}{(1 - \tan\theta)(1 + \tan\theta)} = \frac{(\tan^2\theta - 1)}{1 - \tan^2\theta} = \boxed{-1} \]

✅ Final Answer:

\[ \boxed{-1} \]

Qus : 134NIMCET PYQ

If $0{\lt}P(A){\lt}1$ and $0{\lt}P(B){\lt}1$ and $P(A\cap B)=P(A)P(B)$, then

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Solution

Qus : 135NIMCET PYQ

If $\sin x=\sin y$ and $\cos x=\cos y$, then the value of x-y is

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Solution

Given:

\[ \sin x = \sin y \quad \text{and} \quad \cos x = \cos y \]

✳ Step 1: Use the identity for sine

\[ \sin x = \sin y \Rightarrow x = y + 2n\pi \quad \text{or} \quad x = \pi - y + 2n\pi \]

✳ Step 2: Use the identity for cosine

\[ \cos x = \cos y \Rightarrow x = y + 2m\pi \quad \text{or} \quad x = -y + 2m\pi \]

? Combine both conditions

For both $ \sin x = \sin y $ and $ \cos x = \cos y $ to be true, the only consistent solution is:

\[ x = y + 2n\pi \Rightarrow x - y = 2n\pi \]

✅ Final Answer:

\[ \boxed{x - y = 2n\pi \quad \text{for } n \in \mathbb{Z}} \]

Qus : 136NIMCET PYQ

Which of the following is NOT true?

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Solution

Qus : 137NIMCET PYQ

For an invertible matrix A, which of the following is not always true:

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Solution

Qus : 138NIMCET PYQ

$f(x)=x+|x|$ is continuous for

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Qus : 139NIMCET PYQ

For what values of $\lambda$ does the equation $6x^2 - xy + \lambda y^2 = 0$ represents two perpendicular lines and two lines inclined at an angle of $\pi/4$.

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Solution

Qus : 140NIMCET PYQ

If ${{a}}_1,{{a}}_2,\ldots,{{a}}_n$ are any real numbers and $n$ is any positive integer, then

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Solution

Qus : 141NIMCET PYQ

A speaks truth in 40% and B in 50% of the cases. The probability that they contradict each other while narrating some incident is:

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Solution

A speaks the truth in 40% of the cases and B in 50% of the cases.

What is the probability that they contradict each other while narrating an incident?

? Let’s Define:

$ P(A_T) = 0.4 $ → A tells the truth
$ P(A_L) = 0.6 $ → A lies
$ P(B_T) = 0.5 $ → B tells the truth
$ P(B_L) = 0.5 $ → B lies

? Contradiction happens in two cases:

A tells the truth, B lies → $ 0.4 \times 0.5 = 0.2 $
A lies, B tells the truth → $ 0.6 \times 0.5 = 0.3 $

Total probability of contradiction: \[ P(\text{Contradiction}) = 0.2 + 0.3 = \boxed{0.5} \]

✅ Final Answer:

\[ \boxed{\frac{1}{2}} \]

Qus : 142NIMCET PYQ

In a reality show, two judges independently provided marks base do the performance of the participants. If the marks provided by the second judge are given by Y = 10.5 + 2x, where X is the marks provided by the first judge. If the variance of the marks provided by the second judge is 100, then the variance of the marks provided by the first judge is:

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Solution

Qus : 143NIMCET PYQ

If $D={\begin{vmatrix}{1} & 1 & {1} \\ 1 & {2+x} & {1} \\ {1} & {1} & {2+y}\end{vmatrix}}\, for\, x\ne0,\, y\ne0$ then D is

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Solution

Qus : 144NIMCET PYQ

The two parabolas $y^2 = 4a(x + c)$ and $y^2 = 4bx, a > b > 0$ cannot have a common normal unless

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Solution

Qus : 145NIMCET PYQ

Area of the parallelogram formed by the lines y=4x, y=4x+1, x+y=0 and x+y=1

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Solution

Qus : 146NIMCET PYQ

A man starts at the origin O and walks a distance of 3 units in the north- east direction and then walks a distance of 4 units in the north-west direction to reach the point P. then $\vec{OP}$ is equal to

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Solution

A man starts at the origin $ O $, walks 3 units in the north-east direction, then 4 units in the north-west direction to reach point $ P $. Find the displacement vector $ \vec{OP} $.

? Solution:

North-East (45°): \[ \vec{A} = 3 \cdot \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( \frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}} \right) \]
North-West (135°): \[ \vec{B} = 4 \cdot \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( -\frac{4}{\sqrt{2}}, \frac{4}{\sqrt{2}} \right) \]
Total Displacement: \[ \vec{OP} = \vec{A} + \vec{B} = \left( \frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}} \right) \]

✅ Final Answer:

\[ \boxed{ \vec{OP} = \left( \frac{-1}{\sqrt{2}},\ \frac{7}{\sqrt{2}} \right) } \]

Qus : 147NIMCET PYQ

A four-digit number is formed using the digits 1, 2, 3, 4, 5 without repetition. The probability that is divisible by 3 is

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Solution

Qus : 148NIMCET PYQ

Among the given numbers below, the smallest number which will be divided by 9, 10, 15 and 20, leaves the remainders 4, 5, 10, and 15, respectively

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Solution

Find the smallest number which when divided by 9, 10, 15 and 20 leaves remainders 4, 5, 10 and 15 respectively.

✅ Solution:

Let the number be $ x $.

$ x \equiv 4 \mod 9 \Rightarrow x - 4 $ divisible by 9
$ x \equiv 5 \mod 10 \Rightarrow x - 5 $ divisible by 10
$ x \equiv 10 \mod 15 \Rightarrow x - 10 $ divisible by 15
$ x \equiv 15 \mod 20 \Rightarrow x - 15 $ divisible by 20

So, $ x + 5 $ is divisible by LCM of 9, 10, 15, 20

LCM = $ 2^2 \cdot 3^2 \cdot 5 = 180 $

$ x + 5 = 180 \times 2 = 360 \Rightarrow x = 355 $

? Final Answer: $ \boxed{355} $

Qus : 149NIMCET PYQ

For $a\in R$ (the set of al real numbers), $a \ne 1$, $\lim _{{n}\rightarrow\infty}\frac{({1}^a+{2}^a+{\ldots+{n}^a})}{{(n+1)}^{a-1}\lbrack(na+1)(na+b)\ldots(na+n)\rbrack}=\frac{1}{60}$ . Then one of the value of $a$ is

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Solution

Qus : 150NIMCET PYQ

The value of $\sum ^n_{r=1}\frac{{{{}^nP}}_r}{r!}$ is:

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Solution

Question: Find the value of:

$$ \sum_{r=1}^{n} \frac{nP_r}{r!} $$

Solution:

We know: $ nP_r = \frac{n!}{(n - r)!} \Rightarrow \frac{nP_r}{r!} = \frac{n!}{(n - r)! \cdot r!} = \binom{n}{r} $

Therefore,

$$ \sum_{r=1}^{n} \frac{nP_r}{r!} = \sum_{r=1}^{n} \binom{n}{r} = 2^n - 1 $$

Final Answer: $$ \boxed{2^n - 1} $$

Qus : 151NIMCET PYQ

If $\vec{a}=\lambda \hat{i}+\hat{j}-2\hat{k}$ , $\vec{b}=\hat{i}+\lambda \hat{j}-2\hat{k}$ and $\vec{c}=\hat{i}+\hat{j}+\hat{k}$ and $\begin{bmatrix}{\vec{a}} & {\vec{b}} & {\vec{c}} \end{bmatrix}=7$, then the values of the $\lambda$ are

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Solution

Qus : 152NIMCET PYQ

Let A and B be two events defined on a sample space $\Omega$. Suppose $A^C$ denotes the complement of A relative to the sample space $\Omega$. Then the probability $P\Bigg{(}(A\cap{B}^C)\cup({A}^C\cap B)\Bigg{)}$ equals

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Solution

Given: Two events $ A $ and $ B $ defined on sample space $ \Omega $. We are to find the probability:

$$ P\left((A \cap B^c) \cup (A^c \cap B)\right) $$

Step 1: This is the probability of events that are in exactly one of A or B (but not both), i.e., symmetric difference of A and B:

$$ (A \cap B^c) \cup (A^c \cap B) = A \Delta B $$

Step 2: So, we use:

$$ P(A \Delta B) = P(A) + P(B) - 2P(A \cap B) $$

Final Answer:

$$ \boxed{P(A) + P(B) - 2P(A \cap B)} $$

Qus : 153NIMCET PYQ

The function $f(x)=\log (x+\sqrt[]{{x}^2+1})$ is

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Solution

Qus : 154NIMCET PYQ

Let Z be the set of all integers, and consider the sets $X=\{(x,y)\colon{x}^2+2{y}^2=3,\, x,y\in Z\}$ and $Y=\{(x,y)\colon x{\gt}y,\, x,y\in Z\}$. Then the number of elements in $X\cap Y$ is:

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Solution

Given: $$x^2 + 2y^2 = 3 \text{ and } x > y \text{ with } x, y \in \mathbb{Z}$$

Solutions to the equation are: $$\{(1,1), (1,-1), (-1,1), (-1,-1)\}$$

Among them, only $ (1, -1) $ satisfies $ x > y $.

Answer: $$\boxed{1}$$

Qus : 155NIMCET PYQ

The mean of 25 observations was found to be 38. It was later discovered that 23 and 38 were misread as 25 and 36, then the mean is

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Solution

Qus : 156NIMCET PYQ

The value of $f(1)$ for $f\Bigg{(}\frac{1-x}{1+x}\Bigg{)}=x+2$ is

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Solution

Given:
$$f\left(\frac{1 - x}{1 + x}\right) = x + 2$$

To Find: $ f(1) $

Let $ \frac{1 - x}{1 + x} = 1 \Rightarrow x = 0 $

Then, $ f(1) = f\left(\frac{1 - 0}{1 + 0}\right) = 0 + 2 = 2 $

Answer: $$\boxed{2}$$

Qus : 157NIMCET PYQ

The area enclosed within the curve |x|+|y|=2 is

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Solution

Qus : 158NIMCET PYQ

Given a set A with median $m_1 = 2$ and set B with median $m_2 = 4$

What can we say about the median of the combined set?

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Solution

Given two sets:

Set $ A $ has median $ m_1 = 2 $
Set $ B $ has median $ m_2 = 4 $

What can we say about the median of the combined set $ A \cup B $?

✅ Answer:

The combined median depends on the size and values of both sets.

Without that information, we only know that:

\[ \text{Combined Median} \in [2, 4] \]

So, the exact median cannot be determined with the given data.

Qus : 159NIMCET PYQ

If ${\Bigg{(}\frac{x}{a}\Bigg{)}}^2+{\Bigg{(}\frac{y}{b}\Bigg{)}}^2=1$, $(a{\gt}b)$ and ${x}^2-{y}^2={c}^2$ cut at right angles, then

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Solution

If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ are orthogonal.

Then

$a^2-b^2=c^2-d^2$

Similarly

If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $x^2-y^2=c^2$ are orthogonal.

It means

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{c^2}+\frac{y^2}{-c^2}=1$ are orthogonal

Then

$a^2-b^2=c^2-(-c^2)$

$a^2-b^2=2c^2$

Qus : 160NIMCET PYQ

Let $f(x)=\begin{cases}{{x}^2\sin \frac{1}{x}} & {,\, x\ne0} \\ {0} & {,x=0}\end{cases}$

Then which of the follwoing is true

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Solution

Qus : 161NIMCET PYQ

If $\alpha , \beta$ are the roots of $x^2-x-1=0$ and $A_n=\alpha^n+\beta^n$, the Arithmetic mean of $A_{n-1}$ and $A_n$ is

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Solution

Qus : 162NIMCET PYQ

A coin is thrown 8 number of times. What is the probability of getting a head in an odd number of throw?

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Solution

Total outcomes = $ 2^8 = 256 $

Favorable outcomes (odd heads):

$ \binom{8}{1} = 8 $
$ \binom{8}{3} = 56 $
$ \binom{8}{5} = 56 $
$ \binom{8}{7} = 8 $

Total favorable = $ 8 + 56 + 56 + 8 = 128 $

So, Probability = $ \frac{128}{256} = \boxed{\frac{1}{2}} $

? Final Answer: $ \boxed{\frac{1}{2}} $

Qus : 163NIMCET PYQ

If $a_1, a_2, a_3,...a_n$, are in Arithmetic Progression with common difference d, then the sum $(sind) (cosec a_1 . cosec a_2+cosec a_2.cosec a_2+...+cosec a_{n-1}.cosec a_n)$ is equal to

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Solution

Qus : 164NIMCET PYQ

Consider the function $f(x)={x}^{2/3}{(6-x)}^{1/3}$. Which of the following statement is false?

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Solution

Given Function: $$f(x) = x^{2/3}(6 - x)^{1/3}$$

f is increasing in (0, 4): ✅ True
f has a point of inflection at x = 0: ✅ True
f has a point of inflection at x = 6: ✅ True
f is decreasing in (6, ∞): ❌ False (function not defined there)

Correct Answer (False Statement): $$\boxed{\text{f is decreasing in } (6, \infty)}$$

Qus : 165NIMCET PYQ

Solutions of the equation ${\tan }^{-1}\sqrt[]{{x}^2+x}+{\sin }^{-1}\sqrt[]{{x}^2+x+1}=\frac{\pi}{2}$ are

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Solution

Qus : 166NIMCET PYQ

The value of ${{Lt}}_{x\rightarrow0}\frac{{e}^x-{e}^{-x}-2x}{1-\cos x}$ is equal to

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Solution

Evaluate: $$\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{1 - \cos x}$$

Step 1: Apply L'Hôpital's Rule (since it's 0/0):

First derivative: $$\frac{e^x + e^{-x} - 2}{\sin x}$$

Still 0/0 → Apply L'Hôpital's Rule again: $$\frac{e^x - e^{-x}}{\cos x}$$

Now, $$\lim_{x \to 0} \frac{1 - 1}{1} = 0$$

Final Answer: $$\boxed{0}$$

Qus : 167NIMCET PYQ

In a Harmonic Progression, $p^{th}$ term is $q$ and the $q^{th}$ term is $p$. Then $pq^{th}$ term is

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Solution

Qus : 168NIMCET PYQ

Consider the function $$f(x)=\begin{cases}{-{x}^3+3{x}^2+1,} & {if\, x\leq2} \\ {\cos x,} & {if\, 2{\lt}x\leq4} \\ {{e}^{-x},} & {if\, x{\gt}4}\end{cases}$$ Which of the following statements about f(x) is true:

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Solution

Qus : 169NIMCET PYQ

If the roots of the quadratic equation $x^2+px+q=0$ are tan 30° and tan 15° respectively, then the value of 2 + p - q is

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Solution

Qus : 170NIMCET PYQ

If one AM (Arithmetic mean) 'a' and two GM's (Geometric means) p and q be inserted between any two positive numbers, the value of p^3+q^3 is

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Solution

Problem:

If one Arithmetic Mean (AM) $ a $ and two Geometric Means $ p $ and $ q $ are inserted between any two positive numbers, find the value of: \[ p^3 + q^3 \]

Given:

Let two positive numbers be $ A $ and $ B $.
One AM: $ a = \frac{A + B}{2} $
Two GMs inserted: so the four terms in G.P. are: \[ A, \ p = \sqrt[3]{A^2B}, \ q = \sqrt[3]{AB^2}, \ B \]

Now calculate:

\[ pq = \sqrt[3]{A^2B} \cdot \sqrt[3]{AB^2} = \sqrt[3]{A^3B^3} = AB \]
\[ p^3 = A^2B, \quad q^3 = AB^2 \]
\[ p^3 + q^3 = A^2B + AB^2 = AB(A + B) \]

Also,

\[ 2apq = 2 \cdot \frac{A + B}{2} \cdot AB = AB(A + B) \]

✅ Therefore,

$ \boxed{p^3 + q^3 = 2apq} $

Qus : 171NIMCET PYQ

A straight line through the point (4, 5) is such that its intercept between the axes is bisected at A, then its equation is

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Solution

Qus : 172NIMCET PYQ

The equation $3x^2 + 10xy + 11y^2 + 14x + 12y + 5 = 0$ represents

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Solution

Rule for Classifying Conics Using Discriminant

Given the equation: $ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $

Compute: $ \Delta = B^2 - 4AC $

? Based on value of $ \Delta $:

Ellipse: $ \Delta < 0 $ and $ A \ne C $, $ B \ne 0 $ → tilted ellipse
Circle: $ \Delta < 0 $ and $ A = C $, $ B = 0 $
Parabola: $ \Delta = 0 $
Hyperbola: $ \Delta > 0 $

Example:

For the equation: $ 3x^2 + 10xy + 11y^2 + 14x + 12y + 5 = 0 $

$ A = 3 $, $ B = 10 $, $ C = 11 $ →
$ \Delta = 10^2 - 4(3)(11) = 100 - 132 = -32 $

Since $ \Delta < 0 $, it represents an ellipse.

Qus : 173NIMCET PYQ

The value of $\int \frac{({x}^2-1)}{{x}^3\sqrt[]{2{x}^4-2{x}^2+1}}dx$

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Solution

Qus : 174NIMCET PYQ

The points (1,1/2) and (3,-1/2) are

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Solution

Given:

Points: $ A = (1, \frac{1}{2}) $, $ B = (3, -\frac{1}{2}) $

Line: $ 2x + 3y = k $

Step 1: Evaluate $ 2x + 3y $

For A: $ 2(1) + 3\left(\frac{1}{2}\right) = \frac{7}{2} $
For B: $ 2(3) + 3\left(-\frac{1}{2}\right) = \frac{9}{2} $

✅ Option-wise Check:

In between the lines $ 2x + 3y = -6 $ and $ 2x + 3y = 6 $: ✔️ True since $ \frac{7}{2}, \frac{9}{2} \in (-6, 6) $
On the same side of $ 2x + 3y = 6 $: ✔️ True, both values are less than 6
On the same side of $ 2x + 3y = -6 $: ✔️ True, both values are greater than -6
On the opposite side of $ 2x + 3y = -6 $: ❌ False, both are on the same side

✅ Final Answer:

The correct statements are:

In between the lines $ 2x + 3y = -6 $ and $ 2x + 3y = 6 $
On the same side of the line $ 2x + 3y = 6 $
On the same side of the line $ 2x + 3y = -6 $

Qus : 175NIMCET PYQ

Coordinate of the focus of the parabola $4y^2+12x-20y+67=0$ is

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Solution

Qus : 176NIMCET PYQ

How much work does it take to slide a crate for a distance of 25m along a loading dock by pulling on it with a 180 N force where the dock is at an angle of 45° from the horizontal?

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Solution

Work Done Problem:

A crate is pulled 25 m along a dock with a force of 180 N at an angle of 45°.

✅ Formula Used:

\[ \text{Work} = F \cdot d \cdot \cos(\theta) \]

✅ Substituting Values:

\[ W = 180 \times 25 \times \cos(45^\circ) = 180 \times 25 \times 0.70710678118 = 3181.98052\, \text{J} \]

✅ Final Answer (to 5 decimal places):

\[ \boxed{3.181\times 10^3 \, \text{Joules}} \]

Qus : 177NIMCET PYQ

There are two circles in xy −plane whose equations are $x^2+y^2-2y=0$ and $x^2+y^2-2y-3=0$. A point $(x,y)$ is chosen at random inside the larger circle. Then the probability that the point has been taken from smaller circle is

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Solution

Qus : 178NIMCET PYQ

The vector $\vec{A}=(2x+1)\hat{i}+(x^2-6y)\hat{j}+(xy^2+3z)\hat{k}$ is a

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Solution

Vector Field:

\[ \vec{A} = (2x + 1)\hat{i} + (x^2 - 6y)\hat{j} + (xy^2 + 3z)\hat{k} \]

Divergence:

\[ \nabla \cdot \vec{A} = 2 - 6 + 3 = -1 \neq 0 \]

Not solenoidal ❌

Curl:

\[ \nabla \times \vec{A} = (2xy)\hat{i} - (y^2)\hat{j} + (2x)\hat{k} \neq \vec{0} \]

Not conservative ❌

Final Answer:

$ \vec{A} $ is neither conservative nor solenoidal.

Vector Sink Field Analysis

Given vector field:

\[ \vec{A} = (2x + 1)\hat{i} + (x^2 - 6y)\hat{j} + (xy^2 + 3z)\hat{k} \]

Divergence:

\[ \nabla \cdot \vec{A} = 2 - 6 + 3 = -1 \]

✅ Conclusion:

The divergence is negative at every point, so $ \vec{A} $ is a sink field.

Qus : 179NIMCET PYQ

In a triangle ABC, if the tangent of half the difference of two angles is equal to one third of the tangent of half the sum of the angles, then the ratio of the sides opposite to the angles is

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Solution

Qus : 180NIMCET PYQ

Region R is defined as region in first quadrant satisfying the condition $x^2 + y^2 < 4$. Given that a point P=(r,s) lies in R, what is the probability that r>s?

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Solution

Probability that $ r > s $ in Region $ R $

Given: $ R = \{ (x, y) \in \mathbb{R}^2 \mid x^2 + y^2 < 4 \} $ in the first quadrant

Area of region $ R $ in first quadrant: \[ A = \frac{1}{4} \pi (2)^2 = \pi \]

Region where $ r > s $ (i.e., below line $ x = y $) occupies half of that quarter-circle: \[ A_{\text{favorable}} = \frac{1}{2} \pi \]

Therefore, the required probability is:

\[ \text{Probability} = \frac{\frac{1}{2} \pi}{\pi} = \boxed{\frac{1}{2}} \]

Qus : 181NIMCET PYQ

If $x^my^n$=$(x+y)^{m+n}$, then $\frac{dy}{dx}$ is

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Solution

Qus : 182NIMCET PYQ

Lines $L_1, L_2, .., L_10 $are distinct among which the lines $L_2, L_4, L_6, L_8, L_{10}$ are parallel to each other and the lines $L_1, L_3, L_5, L_7, L_9$ pass through a given point C. The number of point of intersection of pairs of lines from the complete set $L_1, L_2, L_3, ..., L_{10}$ is

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Solution

Total Number of Intersection Points

Given:

10 distinct lines: $ L_1, L_2, \ldots, L_{10} $
$ L_2, L_4, L_6, L_8, L_{10} $: parallel (no intersections among them)
$ L_1, L_3, L_5, L_7, L_9 $: concurrent at point $ C $ (intersect at one point)

? Calculation:

\[ \text{Total line pairs: } \binom{10}{2} = 45 \]

\[ \text{Subtract parallel pairs: } \binom{5}{2} = 10 \Rightarrow 45 - 10 = 35 \]

\[ \text{Concurrent at one point: reduce } 10 \text{ pairs to 1 point} \Rightarrow 35 - 9 = \boxed{26} \]

✅ Final Answer: $\boxed{26}$ unique points of intersection

Qus : 183NIMCET PYQ

If $cos^{-1} \frac{x}{2}+cos^{-1} \frac{y}{3}=\phi$, then $9x^2-12xy cos\phi+4y^2$ is

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Solution

Qus : 184NIMCET PYQ

If the line $a^2 x + ay +1=0$, for some real number $a$, is normal to the curve $xy=1$ then

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Solution

Problem:

The line \[ a^2x + ay + 1 = 0 \] is normal to the curve \[ xy = 1 \]. Find possible values of $ a \in \mathbb{R} $.

Step 1: Slope of Line

Rewrite: \[ y = -a x - \frac{1}{a} \] → slope = $ -a $

Step 2: Curve Derivative

\[ xy = 1 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} \] Slope of normal = $ \frac{x}{y} $

Match Slopes

\[ -a = \frac{x}{y} \Rightarrow x = -a y \]

Plug into Curve

\[ xy = 1 \Rightarrow (-a y)(y) = 1 \Rightarrow y^2 = -\frac{1}{a} \]

For real $ y $, we need $ a < 0 $

✅ Final Answer:

\[ \boxed{a < 0} \]

Qus : 185NIMCET PYQ

The value of ${3}^{3-\log _35}$ is

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Solution

Qus : 186NIMCET PYQ

Out of a group of 50 students taking examinations in Mathematics, Physics, and Chemistry, 37 students passed Mathematics, 24 passed Physics, and 43 passed Chemistry. Additionally, no more than 19 students passed both Mathematics and Physics, no more than 29 passed both Mathematics and Chemistry, and no more than 20 passed both Physics and Chemistry. What is the maximum number of students who could have passed all three examinations?

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Solution

? Maximum Students Passing All Three Exams

Given:

Total students = 50
$ |M| = 37 $, $ |P| = 24 $, $ |C| = 43 $
$ |M \cap P| \leq 19 $, $ |M \cap C| \leq 29 $, $ |P \cap C| \leq 20 $

We use the inclusion-exclusion principle:

\[ |M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |M \cap C| - |P \cap C| + |M \cap P \cap C| \]

Let $ x = |M \cap P \cap C| $. Then:

\[ 50 \geq 37 + 24 + 43 - 19 - 29 - 20 + x \Rightarrow 50 \geq 36 + x \Rightarrow x \leq 14 \]

✅ Final Answer: $\boxed{14}$

Qus : 187NIMCET PYQ

There are two sets A and B with |A| = m and |B| = n. If |P(A)| − |P(B)| = 112 then choose the wrong option (where |A| denotes the cardinality of A, and P(A) denotes the power set of A)

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Solution

Qus : 188NIMCET PYQ

Let $f\colon\mathbb{R}\rightarrow\mathbb{R}$ be a function such that $f(0)=\frac{1}{\pi}$ and $f(x)=\frac{x}{e^{\pi x}-1}$ for $x\ne0$, then

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Solution

Analysis of Continuity and Differentiability

Function:

\[ f(x) = \begin{cases} \dfrac{x}{e^{\pi x} - 1}, & x \neq 0 \\ \dfrac{1}{\pi}, & x = 0 \end{cases} \]

✅ Continuity at $ x = 0 $:

\[ \lim_{x \to 0} f(x) = \frac{1}{\pi} = f(0) \quad \Rightarrow \quad \text{Function is continuous at } x = 0 \]

✏️ Differentiability at $ x = 0 $:

\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = -\frac{1}{2} \]

✅ Final Result:

Function is continuous at $ x = 0 $
Function is differentiable at $ x = 0 $
$ f'(0) = \boxed{-\frac{1}{2}} $

Qus : 189NIMCET PYQ

The eccentricity of an ellipse, with its center at the origin is $\frac{1}{3}$ . If one of the directrices is $x=9$, then the equation of ellipse is:

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Solution

Qus : 190NIMCET PYQ

If f(x)=cos[$\pi$^2]x+cos[-$\pi$^2]x, where [.] stands for greatest integer function, then $f(\pi/2)$=

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Solution

? Function with Greatest Integer and Cosine

Given:

\[ f(x) = \cos\left([\pi^2]x\right) + \cos\left([-\pi^2]x\right) \]

Find: \[ f\left(\frac{\pi}{2}\right) \]

Step 1: Estimate Floor Values

\[ \pi^2 \approx 9.8696 \Rightarrow [\pi^2] = 9,\quad [-\pi^2] = -10 \]

Step 2: Plug into the Function

\[ f\left(\frac{\pi}{2}\right) = \cos\left(9 \cdot \frac{\pi}{2}\right) + \cos\left(-10 \cdot \frac{\pi}{2}\right) = \cos\left(\frac{9\pi}{2}\right) + \cos(-5\pi) \]

Step 3: Simplify

\[ \cos\left(\frac{9\pi}{2}\right) = 0,\quad \cos(-5\pi) = -1 \]

✅ Final Answer:

\[ \boxed{-1} \]

Qus : 191NIMCET PYQ

If the angle of elevation of the top of a hill from each of the vertices A, B and C of a horizontal triangle is $\alpha$, then the height of the hill is

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Solution

Qus : 192NIMCET PYQ

If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is

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Solution

✔️ Verified Probability

Total numbers divisible by 6 from 1 to 100: 16

\[ \binom{100}{3} = 161700, \quad \binom{16}{3} = 560 \]

Probability: \[ \frac{560}{161700} = \frac{4}{1155} \]

✅ Final Answer: $\boxed{\frac{4}{1155}}$

Qus : 193NIMCET PYQ

If $(\vec{a} \times \vec{b}) \times \vec{c}= \vec{a} \times (\vec{b} \times \vec{c})$, then

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Solution

Qus : 194NIMCET PYQ

It is given that the mean, median and mode of a data set is $1, 3^x$ and $9^x$ respectively. The possible values of the mode is

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Solution

Mean, Median, and Mode Relation

Given:

Mean = 1
Median = $ 3^x $
Mode = $ 9^x $

Use empirical formula:

\[ \text{Mode} = 3 \cdot \text{Median} - 2 \cdot \text{Mean} \]

\[ 9^x = 3 \cdot 3^x - 2 \Rightarrow (3^x)^2 = 3 \cdot 3^x - 2 \]

Let $ y = 3^x $, then:

\[ y^2 = 3y - 2 \Rightarrow y^2 - 3y + 2 = 0 \Rightarrow (y - 1)(y - 2) = 0 \]

So, $ y = 1 \text{ or } 2 \Rightarrow 9^x = y^2 = 1 \text{ or } 4 $

✅ Final Answer: $\boxed{1 \text{ or } 4}$

Qus : 195NIMCET PYQ

The value of the series $\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+\cdots$ is

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Solution

Given the infinite series:

\[ S = \frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \cdots = \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!} \]

This is a known convergent series, and its sum is:

\[ \boxed{e^{-1}} \]

✅ Final Answer: $\boxed{e^{-1}}$

Qus : 196NIMCET PYQ

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Solution

Qus : 197NIMCET PYQ

The function

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Solution

Qus : 198NIMCET PYQ

Two person A and B agree to meet 20 april 2018 between 6pm to 7pm with understanding that they will wait no longer than 20 minutes for the other. What is the probability that they meet?

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Solution

Qus : 199NIMCET PYQ

Three numbers a,b and c are chosen at random (without replacement) from among the numbers 1, 2, 3, ..., 99. The probability that

is divisible by 3 is,

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Solution

Qus : 200NIMCET PYQ

A and B play a game where each is asked to select a number from 1 to 25. If the two number match, both of them win a prize. The probability that they will not win a prize in a single trial is :

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NIMCET Previous Year PYQNIMCET NIMCET 2016 PYQ

Solution

The total number of ways in which numbers can be choosed =25 x 25=625 The number of ways in which either players can choose same numbers = 25

Probability that they win a prize = 25/625 = 1/25

Thus, the probability that they will not win a prize in a single trial = 1 - 1/25 = 24/25

Qus : 201NIMCET PYQ

The number of common terms in the two sequences 17, 21, 25, ..........., 817 and 16, 21, 26, ..........., 851 is

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Solution

Qus : 202NIMCET PYQ

The quadratic equation whose roots are

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Solution

Qus : 203NIMCET PYQ

Sum to infinity of a geometric is twice the sum of the first two terms. Then what are the possible values of common ratio?

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Solution

Qus : 204NIMCET PYQ

The area of the region bounded by x-axis and the curves defined by $y=tanx$, $-\frac{\pi}{3}\leq x\leq \frac{\pi}{3}$ and $y=cotx$, $-\frac{\pi}{6}\leq x\leq \frac{3\pi}{2}$ is

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Solution

Qus : 205NIMCET PYQ

Suppose that m and n are fixed numbers such that the mth term am is equal to n and nth term an is equal to m, (m≠n), the the value of (m+n)th term is

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Solution

Qus : 206NIMCET PYQ

If $\log (1-x+x^2)={{a}}_1x+{{a}}_{2{}^{{}^{}}}{x}^2+{{{}{{a}}_3{x}^3+.\ldots.}}^{}$ then ${{a}}_3+{{a}}_6+{{a}}_9+.\ldots.$ is equal to

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3
4
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Solution

Qus : 207NIMCET PYQ

If A is an invertible skew-symmetric matrix, then

is a

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Solution

Qus : 208NIMCET PYQ

$\lim_{x\to \infty} (\frac{x+7}{x+2})^{x+5}$ equal to

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3
4
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Solution

Qus : 209NIMCET PYQ

If the mean of the squares of first n natural numbers be 11, then n is equal to?

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Solution

Qus : 210NIMCET PYQ

A and B are independent witness in a case. The chance that A speaks truth is x and B speaks truth is y. If A and B agree on certain statement, the probability that the statement is true is

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Solution

P(A speaks truth) = x

P(B speaks truth) = y

Since, both A and B agree on certain statement.

Hence, Total Probability =P(A)P(B)+P(A')P(B')

=xy + (1-x)(1-y)

If statement is true then it means both A and B speaks truth.

∴ Required Probability =

Qus : 211NIMCET PYQ

The number of common tangents to the circle $x^2+y^2=4$ and $x^2+y^2-6x-8y=24$ is

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Solution

Qus : 212NIMCET PYQ

The set of points, where

is differential in

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Solution

Qus : 213NIMCET PYQ

The harmonic mean of two numbers is 4. Their arithmetic mean A and the geometric mean G satisfy the relation 2A+G² = 27, then the two numbers are

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Solution

Qus : 214NIMCET PYQ

In a ΔABC, if $\tan ^2\frac{A}{2}+\tan ^2\frac{B}{2}+\tan ^2\frac{C}{2}=k$ , then k is always

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Solution

Qus : 215NIMCET PYQ

is equal to

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Solution

Qus : 216NIMCET PYQ

In an entrance test there are multiple choice questions, with four possible answer to each question of which one is correct. The probability that a student knows the answer to a question is 90%. If the student gets the correct answer to a question, then the probability that he as guessing is

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Solution

Qus : 217NIMCET PYQ

Let $\vec{a}=2\widehat{i}\, +\widehat{j}\, +2\widehat{k}$ , $\vec{b}=\widehat{i}-\widehat{j}+2\widehat{k}$ and $\vec{c}=\widehat{i}+\widehat{j}-2\widehat{k}$ are are three vectors. Then, a vector in the plane of $\vec{a}$ and $\vec{c}$ whose projection on $\vec{b}$ is of magnitude $\frac{1}{\sqrt{6}}$ is

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Solution

Qus : 218NIMCET PYQ

Let

be defined by

. Find

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Solution

Qus : 219NIMCET PYQ

A man is known to speak the truth 2 out of 3 times. He threw a dice cube with 1 to 6 on its faces and reports that it is 1. Then the probability that it is actually 1 is

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Solution

Qus : 220NIMCET PYQ

For what value of p, the polynomial $x^4-3x^3+2px^2-6$ is exactly divisible by $(x-1)$

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Solution

Qus : 221NIMCET PYQ

The slope of two-lines

differ by

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Solution

Qus : 222NIMCET PYQ

Let A and B be two events such that

and

where

stands for the complement of event A. Then the events A and B are

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Solution

Qus : 223NIMCET PYQ

If $F(\theta)=\begin{bmatrix}{\cos \theta} & {-\sin \theta} & {0} \\ {\sin \theta} & {\cos \theta} & {0} \\ {0} & {0} & {1}\end{bmatrix}$ , then $F(\theta)F(\alpha)$ is equal to

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Solution

Qus : 224NIMCET PYQ

If the radius of the circle changes at the rate of

, at what rate does the circle's area change when the radius is 10m?

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Solution

Qus : 225NIMCET PYQ

The mean and variance of a random variable X having binomial distribution are 4 and 2 respectively. The P(X = 1) is

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4
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Solution

np = 4

npq = 2

q = 1/2, p = 1/2, n = 8

p(X = 1) = 8C1 (1/2)(1/2)7

Qus : 226NIMCET PYQ

If $\frac{n!}{2!(n-2)!}$ and $\frac{n!}{4!(n-4)!}$ are in the ratio 2:1, then the value of n is

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Solution

Qus : 227NIMCET PYQ

The point of intersection os circle

and the line

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Solution

Qus : 228NIMCET PYQ

is the mean of distribution of x, then usual notation

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Solution

Mean Deviation from mean is Zero.

Qus : 229NIMCET PYQ

The locus of the point of intersection of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which meet right angles is

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Solution

Qus : 230NIMCET PYQ

The number of solutions of the equation sinx + sin5x = sin3x lying in the interval

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Solution

Qus : 231NIMCET PYQ

If E₁ and E₂ are two events associated with a random experiment such that P (E₂) = 0.35, P (E₁ or E₂) = 0.85 and P (E₁ & E₂) = 0.15 then P(E₁) is

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3
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Solution

Qus : 232NIMCET PYQ

If the position vector of A and B relative to O be $\widehat{i}\, -4\widehat{j}+3\widehat{k}$ and $-\widehat{i}\, +2\widehat{j}-\widehat{k}$ respectively, then the median through O of ΔABC is:

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Solution

Qus : 233NIMCET PYQ

In an acute-angled ΔABC the least value of secA+secB+secC is

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Solution

Qus : 234NIMCET PYQ

Find a matrix X such that 2A + B + X = 0, whose A =

and B =

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Solution

Qus : 235NIMCET PYQ

The general value of $\theta$, satisfying the equation $\sin \theta=\frac{-1}{2},\, \tan \theta=\frac{1}{\sqrt[]{3}}$

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Solution

Qus : 236NIMCET PYQ

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Solution

Qus : 237NIMCET PYQ

If in a triangle ABC, the altitudes from the vertices A, B, C on opposite sides are in HP, then sin A, sin B, sin C are in

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Solution

If altitudes of a triangle are in HP then its side will be in AP because sides are inverse proportion to height as area is constant. a, b, c are sides of triangle.

a, b, c are in A.P.

sin A, sin B, sin C are in A.P.

Qus : 238NIMCET PYQ

The area of the triangle formed by the vertices whose position vectors are $3\widehat{i}+\widehat{j}$ , $5\widehat{i}+2\widehat{j}+\widehat{k}$ , $\widehat{i}-2\widehat{j}+3\widehat{k}$ is

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3
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Solution

Qus : 239NIMCET PYQ

then the value of

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Solution

Qus : 240NIMCET PYQ

α, β are the roots of the an equation $x^2- 2x cosθ + 1 = 0$, then the equation having roots αn and βn is

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Solution

Qus : 241NIMCET PYQ

The standard deviation of 20 numbers is 30. If each of the numbers is increased by 4, then the new standard deviation will be

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3
4
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Solution

Qus : 242NIMCET PYQ

The circles whose equations are

and

will touch one another externally, if

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Solution

Qus : 243NIMCET PYQ

The equation (x-a)³+(x-b)³+(x-c)³ = 0 has

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4
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Solution

Let f(x) = (x – a)³ + (x – b)³ + (x – c)³.

Then f'(x) = 3{(x – a)² + (x – b)² + (x –c)²}

clearly , f'(x) > 0 for all x.

so, f'(x) = 0 has no real roots.

Hence, f(x) = 0 has two imaginary and one real root

Qus : 244NIMCET PYQ

The function $f(x)=\frac{x}{1+x\tan x}$ , $0\leq x\leq\frac{\pi}{2}$ is maximum when

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Solution

Qus : 245NIMCET PYQ

The locus of the orthocentre of the triangle formed by the lines (1+p)x-py+p(1+p)=0, (1+p)(x-q)+q(1+ q)=0 and y=0 where p≠q is

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Solution

Straight Line

Qus : 246NIMCET PYQ

Three positive number whose sum is 21 are in arithmetic progression. If 2, 2, 14 are added to them respectively then resulting numbers are in geometric progression. Then which of the following is not among the three numbers?

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Solution

Let the three terms in A.P. be a – d, a, a + d.

given that a – d + a + a + d = 21

a = 7

then the three term in A.P. are 7 – d, 7, 7 + d

According to given condition 9 – d, 9, 21 + d are in G.P.

(9)² = (9 – d) (21 + d)

81 = 189 + 9d – 21d – d²

81 = 189 – 12d – d²

d² + 12d – 108 = 0

d(d + 18) – 6 (d + 18) = 0

(d – 6) (d + 18) = 0

We get, d = 6, –18

Putting d = 6 in the term 7 – d, 7, 7 + d we get 1, 7, 13.

Qus : 247NIMCET PYQ

If $f\colon R\rightarrow R$ is defined by $f(x)=\begin{cases}{\frac{x+2}{{x}^2+3x+2}} & {,\, if\, x\, \in R-\{-1,-2\}} \\ {-1} & {,if\, x=-2} \\ {0} & {,if\, x=-1}\end{cases}$ , then f(x) is continuous on the set

1
2
3
4
Go to Discussion

Solution

Qus : 248NIMCET PYQ

Equation of the common tangents with a positive slope to the circle

and

1
2
3
4
Go to Discussion

Solution

Qus : 249NIMCET PYQ

, then

1
2
3
4
Go to Discussion

Solution

Qus : 250NIMCET PYQ

A polygon has 44 diagonals, the number of sides are

1
2
3
4
Go to Discussion

Solution

Qus : 251NIMCET PYQ

The area enclosed between the curves

and

1
2
3
4
Go to Discussion

Solution

Qus : 252NIMCET PYQ

The value of A that satisfies the equation asinA + bcosA = c is equal to

1
2
3
4
Go to Discussion

Solution

Qus : 253NIMCET PYQ

The probability of occurrence of two events E and F are 0.25 and 0.50, respectively. the probability of their simultaneous occurrence is 0.14. the probability that neither E nor F occur is

1
2
3
4
Go to Discussion

Solution

Qus : 254NIMCET PYQ

Equation of the line perpendicular to x-2y=1 and passing through (1,1) is

1
2
3
4
Go to Discussion

Solution

Qus : 255NIMCET PYQ

If tan x = - 3/4 and 3π/2 < x < 2π, then the value of sin2x is

1
2
3
4
Go to Discussion

Solution

Qus : 256NIMCET PYQ

If $y={\tan }^{-1}\lgroup{\frac{3x-{x}^3}{1-3{x}^2}}\rgroup\, ,\, \frac{-1}{\sqrt[]{3}}{\lt}x{\lt}\frac{1}{\sqrt[]{3}}$ then $\frac{dy}{dx}$ is

1
2
3
4
Go to Discussion

Solution

Incorrect question

Qus : 257NIMCET PYQ

and

, then f(A)=

1
2
3
4
Go to Discussion

Solution

Qus : 258NIMCET PYQ

Find the principal value of

1
2
3
4
Go to Discussion

Solution

Qus : 259NIMCET PYQ

The value of $\tan 9{^{\circ}}-\tan 27{^{\circ}}-\tan 63{^{\circ}}+\tan 81{^{\circ}}$ is equal to

1
2
3
4
Go to Discussion

Solution

Qus : 260NIMCET PYQ

9 balls are to be placed in 9 boxes and 5 of the balls cannot fit into 3 small boxes. The number of ways of arranging one ball in each of the boxes is

1
2
3
4
Go to Discussion

Solution

First off all select 5 boxes out 6 boxes in which 5 big ball can fit then arrange these ball in these 5 boxes and then put remaining 4 ball in any remaining box.

So Ans is [(6C5)5!](4!) = 6!4! = 17280

Qus : 261NIMCET PYQ

If cosθ = 4/5 and cosϕ = 12/13, θ and ϕ both in the fourth quadrant, the value of cos( θ + ϕ )is

1
2
3
4
Go to Discussion

Solution

Qus : 262NIMCET PYQ

In a triangle, if the sum of two sides is x and their product is y such that (x+z)(x-z)=y, where z is the third side of the triangle , then triangle is

1
2
3
4
Go to Discussion

Solution

Qus : 263NIMCET PYQ

Which of the following function is the inverse of itself?

1
2
3
4
Go to Discussion

Solution

Qus : 264NIMCET PYQ

The value of sin36^o is

1
2
3
4
Go to Discussion

Solution

Qus : 265NIMCET PYQ

If $H_1,H_2,\ldots,H_n$ are n harmonic means between a and b $(b\ne a)$;,then $\frac{{{H}}_n+a}{{{H}}_n-a}+\frac{{{H}}_n+b}{{{H}}_n-b}$

1
2
3
4
Go to Discussion

Solution

Qus : 266NIMCET PYQ

A student council has 10 members. From this one President, one Vice-President, one Secretary, one Joint-Secretary and two Executive Committee members have to be elected. In how many ways this can be done?

1
2
3
4
Go to Discussion

Solution

Qus : 267NIMCET PYQ

Express (cos 5x – cos7x) as a product of sines or cosines or sines and cosines,

1
2
3
4
Go to Discussion

Solution

Qus : 268NIMCET PYQ

If $y=\sin ^{-1}(\frac{{x}^2+1}{\sqrt[]{1+3{x}^2+{x}^4}}),\, (x>0),$ then $\frac{dy}{dx}$=

1
2
3
4
Go to Discussion

Solution

Qus : 269NIMCET PYQ

In a survey where 100 students reported which subject they like, 32 students in total liked Mathematics, 38 students liked Business and 30 students liked Literature. Moreover, 7 students liked both Mathematics and Literature, 10 students liked both Mathematics and Business. 8 students like both Business and Literature, 5 students liked all three subjects. Then the number of people who liked exactly one subject is

1
2
3
4
Go to Discussion

Solution

Qus : 270NIMCET PYQ

If non-zero numbers a, b, c are in A.P., then the straight line ax + by + c = 0 always passes through a fixed point, then the point is

1
2
3
4
Go to Discussion

Solution

Since a, b and c are in A. P. 2b = a + c

a –2b + c =0

The line passes through (1, –2).

Qus : 271NIMCET PYQ

If $32\, \tan ^8\theta=2\cos ^2\alpha-3\cos \alpha$ and $3\, \cos \, 2\theta=1$, then the general value of $\alpha$ =

1
2
3
4
Go to Discussion

Solution

Qus : 272NIMCET PYQ

If A and B are two events and

, the A and B are two events which are

1
2
3
4
Go to Discussion

Solution

Qus : 273NIMCET PYQ

If the lines x + (a – 1)y + 1 = 0 and 2x + a²y – 1 = 0 are perpendicular, then the condition satisfies by a is

1
2
3
4
Go to Discussion

Solution

Qus : 274NIMCET PYQ

In a triangle ABC, $a\cos ^2\frac{C}{2}+\, c\, \, {\cos }^2\frac{A}{2}=\frac{3b}{2}$ then the sides of the triangle are in

1
2
3
4
Go to Discussion

Solution

Qus : 275NIMCET PYQ

are positive real numbers whose product is a fixed number c, then the minimum of

1
2
3
4
Go to Discussion

Solution

Qus : 276NIMCET PYQ

In a triangle ABC, let angle C = π/2. If R is the inradius and R is circumradius of the triangle ABC, then 2(r + R) equals

1
2
3
4
Go to Discussion

Solution

Qus : 277NIMCET PYQ

If |k|=5 and 0° ≤ θ ≤ 360°, then the number of distinct solutions of 3cos⁡θ + 4sin⁡θ = k is

NIMCET 2021

1
2
3
4
Go to Discussion

Solution

Qus : 278NIMCET PYQ

If θ is acute angle between the pair of lines $x^2-7xy+12y^2=0$, then $\frac{2\cos \theta+3\sin \theta}{4\sin \theta+5\cos \theta}=$

1
2
3
4
Go to Discussion

Solution

Qus : 279NIMCET PYQ

If a, b, c are the roots of the equation

, then the value of

1
2
3
4
Go to Discussion

Solution

Qus : 280NIMCET PYQ

If x² + 3xy + 2y² – x – 4y – 6 = 0 represents a pair of straight lines, their point of intersection is

1
2
3
4
Go to Discussion

Solution

Qus : 281NIMCET PYQ

The four geometric means between 2 and 64 are

1
2
3
4
Go to Discussion

Solution

Qus : 282NIMCET PYQ

The coefficient of

in the expansion of

1
2
3
4
Go to Discussion

Solution

Qus : 283NIMCET PYQ

The equation of the tangent line to the curve y = 2x sin x at the point (π/2, π), is

1
2
3
4
Go to Discussion

Solution

Qus : 284NIMCET PYQ

The lines $px+qy=1$ and $qx+py=1$ are respectively the sides AB, AC of the triangle ABC and the base BC is bisected at $(p,q)$. Equation of the median of the triangle through the vertex A is

1
2
3
4
Go to Discussion

Solution

Qus : 285NIMCET PYQ

Let

and

be the roots f the equation

and

are the roots of the equation

, then the value of r,

1
2
3
4
Go to Discussion

Solution

Qus : 286NIMCET PYQ

If the graph of y = (x – 2)² – 3 is shifted by 5 units up along y-axis and 2 units to the right along the x-axis, then the equation of the resultant graph is

1
2
3
4
Go to Discussion

Solution

When y= f (x) is shifted by k units to the right along x

– axis, it become y= f (x - k )

Hence, new equation of

graph is y = (x - 4)² + 2

Qus : 287NIMCET PYQ

If the vectors $a\hat{i}+\hat{j}+\hat{k},\hat{i}+b\hat{j}+\hat{k},\hat{i}+\hat{j}+c\hat{k}$ , $(a,b,c\ne1)$ are coplanar, then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$

1
2
3
4
Go to Discussion

Solution

Qus : 288NIMCET PYQ

How many natural numbers smaller than

can be formed using the digits 1 and 2 only?

1
2
3
4
Go to Discussion

Solution

Qus : 289NIMCET PYQ

The direction cosines of the vector a = (- 2i + j – 5k) are

1
2
3
4
Go to Discussion

Solution

Qus : 290NIMCET PYQ

Let $\vec{a}=\hat{i}+\hat{j}$ and $\vec{b}=2\hat{i}-\hat{k}$, the point of intersection of the lines $\vec{r}\times\vec{a}=\vec{b}\times\vec{a}$ and $\vec{r}\times\vec{b}=\vec{a}\times\vec{b}$ is

1
2
3
4
Go to Discussion

Solution

Qus : 291NIMCET PYQ

The equation of the hyperbola with centre at the region, length of the transverse axis is 6 and one focus (0, 4) is

1
2
3
4
Go to Discussion

Solution

Qus : 292NIMCET PYQ

A password consists of two alphabets from English followed by three numbers chosen from 0 to 3. If repetitions are allowed, the number of different passwords is

1
2
3
4
Go to Discussion

Solution

Qus : 293NIMCET PYQ

If the system of equations $3x-y+4z=3$ , $x+2y-3z=-2$ , $6x+5y+λz=-3 $ has atleast one solution, then $λ=$

1
2
3
4
Go to Discussion

Solution

Qus : 294NIMCET PYQ

and

, then the value of

1
2
3
4
Go to Discussion

Solution

Qus : 295NIMCET PYQ

If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are vectors such that $\vec{a}$+$\vec{b}$+$\vec{c}$ = 0 and |$\vec{a}$| =7, $\vec{b}$=5, |$\vec{c}$| = 3, then the angle between the vectors $\vec{b}$ and $\vec{c}$

1
2
3
4
Go to Discussion

Solution

Qus : 296NIMCET PYQ

An equilateral triangle is inscribed in the parabola $y^{2} = 4ax$, such that one of the vertices of the triangle coincides with the vertex of the parabola. The length of the side of the triangle is:

1
2
3
4
Go to Discussion

Solution

Qus : 297NIMCET PYQ

If a variable takes values 0, 1, 2,…, 50 with frequencies $1,\, {{50}}_{{{C}}_1},{{50}}_{{{C}}_2},\ldots..,{{50}}_{{{C}}_{50}}$, then the AM is

1
2
3
4
Go to Discussion

Solution

Qus : 298NIMCET PYQ

and

, (a ≠ b ≠ c ≠ 1) are co-planar, then the value of

1
2
3
4
Go to Discussion

Solution

Qus : 299NIMCET PYQ

A chain of video stores sells three different brands of DVD players. Of its DVD player sales, 50% are brand 1, 30% are brand 2 and 20% are brand 3. Each manufacturer offers one year warranty on parts and labor. It is known that 25% of brand 1 DVD players require warranty repair work whereas the corresponding percentage for brands 2 and 3 are 20% and 10% respectively. The probability that a randomly selected purchaser has a DVD player that will need repair while under warranty, is:

1
2
3
4
Go to Discussion

Solution

Qus : 300NIMCET PYQ

If A={1,2,3,4} and B={3,4,5}, then the number of elements in (A∪B)×(A∩B)×(AΔB)

1
2
3
4
Go to Discussion

Solution

Qus : 301NIMCET PYQ

Differential coefficient of

with respect to

1
2
3
4
Go to Discussion

Solution

Qus : 302NIMCET PYQ

Let a, b and c be three vectors having magnitudes 1, 1 and 2 respectively. If a x (a x c) - b = 0, then the acute angle between a and c is

1
2
3
4
Go to Discussion

Solution

Qus : 303NIMCET PYQ

The locus of the intersection of the two lines $\sqrt{3} x-y=4k\sqrt{3}$ and $k(\sqrt{3}x+y)=4\sqrt{3}$, for different values of k, is a hyperbola. The eccentricity of the hyperbola is:

1
2
3
4
Go to Discussion

Solution

Qus : 304NIMCET PYQ

If n is an integer between 0 to 21, then find a value of n for which the value of $n!(21-n)!$ is minimum

1
2
3
4
Go to Discussion

Solution

Qus : 305NIMCET PYQ

is continuous for

1
2
3
4
Go to Discussion

Solution

Qus : 306NIMCET PYQ

Let

and

be three vector such that |

| = 2, |

| = 3, |

| = 5 and

= 0. The value of

1
2
3
4
Go to Discussion

Solution

Qus : 307NIMCET PYQ

Constant forces $\vec{P}= 2\hat{i} - 5\hat{j} + 6\hat{k} $ and $\vec{Q}= -\hat{i} + 2\hat{j}- \hat{k}$ act on a particle. The work done when the particle is displaced from A whose position vector is $4\hat{i} - 3\hat{j} - 2\hat{k} $, to B whose position vector is $6\hat{i} + \hat{j} - 3k\hat{k}$ , is:

1
2
3
4
Go to Discussion

Solution

Qus : 308NIMCET PYQ

Suppose $A_1,A_2,\ldots,A_{30}$ are 30 sets each with five elements and $B_1,B_2,B_3,\ldots,B_n$ are n sets (each with three elements) such that $\bigcup ^{30}_{i=1}{{A}}_i={{\bigcup }}^n_{j=1}{{B}}_i=S\, $ and each element of S belongs to exactly ten of the $A_i$'s and exactly 9 of the $B^{\prime}_j$'s. Then $n=$

1
2
3
4
Go to Discussion

Solution

Qus : 309NIMCET PYQ

and

are unit vectors, then

does not exceeds

1
2
3
4
Go to Discussion

Solution

Qus : 310NIMCET PYQ

= (i + 2j - 3k) and

=(3i -j + 2k), then the angle between (

) and (

)

1
2
3
4
Go to Discussion

Solution

Qus : 311NIMCET PYQ

The value of $\int \sqrt{x} e^{\sqrt{x}} dx$ is equal to:

1
2
3
4
Go to Discussion

Solution

Qus : 312NIMCET PYQ

The eccentric angle of the extremities of latus-rectum of the ellipse $\frac{{x}^2}{{a}^2}^{}+\frac{{y}^2}{{b}^2}^{}=1$ are given by

1
2
3
4
Go to Discussion

Solution

Qus : 313NIMCET PYQ

The vector

lies in the plane of the vector

and

and bisects the angle between

and

. Then which of the following gives possible values of

and

1
2
3
4
Go to Discussion

Solution

Qus : 314NIMCET PYQ

The number of elements in the power set P(S) of the set S = [2, (1, 4)] is

1
2
3
4
Go to Discussion

Solution

Qus : 315NIMCET PYQ

For the vectors $\vec{a}=-4\hat{i}+2\hat{j}, \vec{b}=2\hat{i}+\hat{j}$ and $\vec{c}=2\hat{i}+3\hat{j}$, if $\vec{c}=m\vec{a}+n\vec{b}$ then the value of m + n is

1
2
3
4
Go to Discussion

Solution

Qus : 316NIMCET PYQ

If α≠β and $\alpha^2=5\alpha-3,\beta^2=5\beta-3$, then the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is

1
2
3
4
Go to Discussion

Solution

Qus : 317NIMCET PYQ

If (1 - x + x²)ⁿ = a + a₁x + a₂x² + ... + a_2nx²ⁿ , then a₀ + a₂ + a₄ + ... + a_2n is

1
2
3
4
Go to Discussion

Solution

Qus : 318NIMCET PYQ

The value of $\int_{0}^{\pi/4} log(1+tanx)dx$ is equal to:

1
2
3
4
Go to Discussion

Solution

Qus : 319NIMCET PYQ

The probability that a man who is x years old will die in a year is p. Then, amongst n persons $A_1,A_2,\ldots A_n$ each x year old now, the probability that ${{A}}_1$ will die in one year and (be the first to die ) is

1
2
3
4
Go to Discussion

Solution

Qus : 320NIMCET PYQ

A bird is flying in a straight line with velocity vector 10i+6j+k, measured in km/hr. If the starting point is (1,2,3), how much time does it to take to reach a point in space that is 13m high from the ground?

1
2
3
4
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Solution

Qus : 321NIMCET PYQ

m distinct animals of a circus have to be placed in m cages, one is each cage. There are n small cages and p large animal (n < p < m). The large animals are so large that they do not fit in small cage. However, small animals can be put in any cage. The number of putting the animals into cage is

1
2
3
4
Go to Discussion

Solution

Qus : 322NIMCET PYQ

The number of ways in which 5 days can be chosen in each of the 12 months of a non-leap year, is:

1
2
3
4
Go to Discussion

Solution

Qus : 323NIMCET PYQ

Angle between $\vec{a}$ and $\vec{b}$ is $120{^{\circ}}$. If $|\vec{b}|=2|\vec{a}|$ and the vectors , $\vec{a}+x\vec{b}$ , $\vec{a}-\vec{b}$ are at right angle, then $x=$

1
2
3
4
Go to Discussion

Solution

Qus : 324NIMCET PYQ

The value of

1
2
3
4
Go to Discussion

Solution

Qus : 325NIMCET PYQ

Let A and B two sets containing four and two elements respectively. The number of subsets of the A × B, each having at least three elements is

1
2
3
4
Go to Discussion

Solution

n(A) = 4

n(B) = 2

Then the number of subsets in A*B is 2⁸= 256

Qus : 326NIMCET PYQ

If [x] represents the greatest integer not exceeding x, then $\int_{0}^{9} [x] dx $ is

1
2
3
4
Go to Discussion

Solution

Qus : 327NIMCET PYQ

If a number x is selected at random from natural numbers 1,2,…,100, then the probability for $x+\frac{100}{x}{\gt}29$ is

1
2
3
4
Go to Discussion

Solution

Qus : 328NIMCET PYQ

, then value of

1
2
3
4
Go to Discussion

Solution

Qus : 329NIMCET PYQ

The slope of the function

1
2
3
4
Go to Discussion

Solution

Qus : 330NIMCET PYQ

Let $\vec{a}$ and $\vec{b}$ be two vectors, which of the following vectors are not perpendicular to each other?

1
2
3
4
Go to Discussion

Solution

Qus : 331NIMCET PYQ

If X and Y are two sets, then X∩Y ' ∩ (X∪Y) ' is

1
2
3
4
Go to Discussion

Solution

Qus : 332NIMCET PYQ

In a triangle ABC, angle A=90° and D is the midpoint of AC. What is the value of

equal to?

1
2
3
4
Go to Discussion

Solution

Qus : 333NIMCET PYQ

What is the largest area of an isosceles triangle with two edges of length 3?

1
2
3
4
Go to Discussion

Solution

Qus : 334NIMCET PYQ

In a group of 200 students, the mean and the standard deviation of scores were found to be 40 and 15, respectively. Later on it was found that the two scores 43 and 35 were misread as 34 and 53, respectively. The corrected mean of scores is:

1
2
3
4
Go to Discussion

Solution

Qus : 335NIMCET PYQ

$\int {e}^x(\sinh x+\cosh x)dx$

1
2
3
4
Go to Discussion

Solution

Qus : 336NIMCET PYQ

Through any point (x, y) of a curve which passes through the origin, lines are drawn parallel to the coordinate axes. The curve, given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of

1
2
3
4
Go to Discussion

Solution

Qus : 337NIMCET PYQ

The value of

1
2
3
4
Go to Discussion

Solution

Qus : 338NIMCET PYQ

If $A=\begin{bmatrix} a &b &c \\ b & c & a\\ c& a &b \end{bmatrix}$ , where $a, b, c$ are real positive numbers such that $abc = 1$ and $A^{T}A=I$ then the equation that not holds true among the following is

1
2
3
4
Go to Discussion

Solution

Qus : 339NIMCET PYQ

If the matrix $\begin{bmatrix} -1 &3 &2 \\ 1& K &-3 \\ 1&4 &5 \end{bmatrix}$ has an inverse matrix, then the value of K is:

1
2
3
4
Go to Discussion

Solution

Qus : 340NIMCET PYQ

If $\vec{e_1}=(1,1,1)$ and $\vec{e_2}=(1,1,-1)$ and $\vec{a}$ and $\vec{b}$ and two vectors such that $\vec{e_2}=\vec{a}+2\vec{b}$ , then angle between $\vec{a}$ and $\vec{b}$

1
2
3
4
Go to Discussion

Solution

Qus : 341NIMCET PYQ

A line passing through P(4, 2) meets the x and y-axis at P and Q respectively. If O is the origin, then the locus of the centre of the circumcircle of ΔOPQ is -

1
2
3
4
Go to Discussion

Solution

Qus : 342NIMCET PYQ

Let f(x) be a polynomial of degree four, having extreme value at x = 1 and x = 2. If $\lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3$, then f(2) is

1
2
3
4
Go to Discussion

Solution

Given it has extremum values at x=1 and x=2
⇒f′(1)=0 and f′(2)=0
Given f(x) is a fourth degree polynomial
Let $f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e$

Given

$\lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3$

$\lim _{{x}\rightarrow0}\lbrack1+\frac{a{x}^4+b{x}^3+c{x}^2+\mathrm{d}x+e}{{x}^2}\rbrack=3$

$\lim _{{x}\rightarrow0}\lbrack1+a{x}^2+bx+c+\frac{d}{x}+\frac{e}{{x}^2}\rbrack=3$

For limit to have finite value, value of 'd' and 'e' must be 0

⇒d=0 & e=0

Substituting x=0 in limit

⇒ c+1=3

⇒ c=2

$f^{\prime}(x)=4a{x}^3+3b{x}^2+2cx+d$

$x=1$ and $x=2$ are extreme values,

⇒$f^{\prime}(1)=0$ and $f^{\prime}(2)=0

⇒ $4a+3b+4=0$ and $32a+12b+8=0$

By solving these equations

we get, $a=\frac{1}{2}$ and $b=-2$

So,

$f(x)=\frac{x^{4}}{2}-2x^{3}+2x^{2}$

⇒$f(x)=x^{2}(\frac{x^{2}}{2}-2x+2)$

⇒$f(2)=2^{2}(2-4+2)$

⇒$f(2)=0$

Qus : 343NIMCET PYQ

The equation of the tangent at any point of the curve $x=acos2t$, $y=2\sqrt{2}a sint$ with $m$ as its slope is

1
2
3
4
Go to Discussion

Solution

Qus : 344NIMCET PYQ

The mean deviation from the mean of the AP a, a + d, a + 2d, ..., a + 2nd, is:

1
2
3
4
Go to Discussion

Solution

Qus : 345NIMCET PYQ

If P(1,2), Q(4,6), R(5,7) and S(a,b) are the vertices of a parallelogram PQRS, then

1
2
3
4
Go to Discussion

Solution

Qus : 346NIMCET PYQ

The maximum value of 4 sin²x + 3 cos²x + sin(x/2) + cos(x/2) is

1
2
3
4
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Solution

Qus : 347NIMCET PYQ

The locus of the mid points of all chords of the parabola $y^{2}=4x$ which are drawn through its vertex, is

1
2
3
4
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Solution

Qus : 348NIMCET PYQ

If (x₀, y₀) is the solution of the equations (2x)^ln2 = (3y)^ln3 and 3^lnx = 2^lny, then x₀ is:

1
2
3
4
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Solution

Qus : 349NIMCET PYQ

If $a\, \cos \theta+b\, \sin \, \theta=2$ and $a\, \sin \, \theta-b\, \cos \, \theta=3$ , then ${a}^{2^{}}+{b}^2=$

1
2
3
4
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Solution

Qus : 350NIMCET PYQ

The solution of (e^x + 1) y dy = (y + 1) e^xdx is

1
2
3
4
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Solution

Qus : 351NIMCET PYQ

The value of $\lim_{x\to a} \frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}$

1
2
3
4
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Solution

Qus : 352NIMCET PYQ

The value of tan 1° tan 2° tan 3° ... tan 89° is:

1
2
3
4
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Solution

Qus : 353NIMCET PYQ

If three thrown of three dice, the probability of throwing triplets not more than twice is

1
2
3
4
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Solution

Qus : 354NIMCET PYQ

Evaluate

1
2
3
4
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Solution

Using Property

Qus : 355NIMCET PYQ

, then the values of n and r are:

1
2
3
4
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Solution

Qus : 356NIMCET PYQ

The value of $\int_{-\pi/3}^{\pi/3} \frac{x sinx}{cos^{2}x}dx$

1
2
3
4
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Solution

Qus : 357NIMCET PYQ

If $\alpha$ and $\beta$ are the roots of the equation $2x^{2}+ 2px + p^{2} = 0$, where $p$ is a non-zero real number, and $\alpha^{4}$ and $\beta^{4}$ are the roots of $x^{2} - rx + s = 0$, then the roots of $2x^{2} - 4p^{2}x + 4p^{4} - 2r = 0$ are:

1
2
3
4
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Solution

Qus : 358NIMCET PYQ

$\int {3}^{{3}^{{3}^x}}.{3}^{{3}^x}.{3}^xdx$ is equal to

1
2
3
4
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Solution

Qus : 359NIMCET PYQ

The critical point and nature for the function f(x, y) = x² –2x + 2y² + 4y – 2 is

1
2
3
4
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Solution

Qus : 360NIMCET PYQ

In a class of 50 students, it was found that 30 students read "Hitava", 35 students read "Hindustan" and 10 read neither. How many students read both: "Hitavad" and "Hindustan" newspapers?

1
2
3
4
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Solution

P(x)=50,

P(A ∩ B)' = 10

So P(A ∪ B) = 50 - 10 = 40.

So P(A ∩ B) = P(A) + P(B) - P(A ∪ B)

= 30 + 35 - 40 = 25

Solution Contribution by Priyanka Soni

Qus : 361NIMCET PYQ

The foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ and the hyperbola $\frac{x^{2}}{144}-\frac{y^{2}}{{81}}=\frac{1}{25}$ coincide, then the value of $b^{2}$ is

1
2
3
4
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Solution

Qus : 362NIMCET PYQ

The number of ways to arrange the letters of the English alphabet, so that there are exactly 5 letters between a and b, is:

1
2
3
4
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Solution

Qus : 363NIMCET PYQ

There are 50 questions in a paper. Find the number of ways in which a student can attempt one or more questions :

1
2
3
4
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Solution

Qus : 364NIMCET PYQ

If y = cos²x², find dx/dy

1
2
3
4
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Solution

Let y = (cos x²)²

= 2 (cos x2) sin x2 .2x

= -4 (cos x2) sin x2

Qus : 365NIMCET PYQ

If A = {4^x - 3x - 1 : x ∈ N} and B = {9(x - 1) : x ∈ N}, where N is the set of natural numbers, then

1
2
3
4
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Solution

A = {0,9,54...}

B = {0,9,18,27...}

So, A ⊂ B

Qus : 366NIMCET PYQ

If $a+b+c=\pi$ , then the value of $\begin{vmatrix} sin(A+B+C) &sinB &cosC \\ -sinB & 0 &tanA \\ cos(A+B)&-tanA &0 \end{vmatrix}$ is

1
2
3
4
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Solution

Qus : 367NIMCET PYQ

Suppose, the system of linear equations

-2x + y + z = l

x - 2y + z = m

x + y - 2z = n

is such that l + m + n = 0, then the system has:

1
2
3
4
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Solution

Qus : 368NIMCET PYQ

Consider the following frequency distribution table.

Class Interval	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	180	$f_1$	34	180	136	$f_2$	50

If the total frequency is 686 and the median is 42.6, then the value of $f_1$;and $f_2$ are

1
2
3
4
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Solution

Qus : 369NIMCET PYQ

The derivative of (x³ + e^x + 3^x + cotx) with respect to x is

1
2
3
4
Go to Discussion

Solution

Let y=x³+ e^x+ 3^x+ cotx

Now,

= 3x²+ e^x+ 3^x log3 - cosec² x

Qus : 370NIMCET PYQ

If A = { x, y, z }, then the number of subsets in powerset of A is

1
2
3
4
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Solution

Qus : 371NIMCET PYQ

If the mean deviation of the numbers 1, 1 + d, 1 + 2d, ....., 1 + 100d from their mean is 255, then the value of d is

1
2
3
4
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Solution

Qus : 372NIMCET PYQ

If $\vec{A}=4\hat{i}+3\hat{j}+\hat{k}$ and $\vec{B}=2\hat{i}-\hat{j}+2\hat{k}$ , then the unit vector $\hat{N}$ perpendicular to the vectors $\vec{A}$ and $\vec{B}$ ,such that $\vec{A}, \vec{B}$ , and $\hat{N}$ form a right handed system, is:

1
2
3
4
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Solution

Qus : 373NIMCET PYQ

The solution of the differential equation

1
2
3
4
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Solution

Variable separable

Integrating

Qus : 374NIMCET PYQ

How many words can be formed starting with letter D taking all letters from the word DELHI so that the letters are not repeated:

1
2
3
4
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Solution

Qus : 375NIMCET PYQ

If $P=sin^{20} \theta + cos^{48} \theta $ then the inequality that holds for all values of is

1
2
3
4
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Solution

Qus : 376NIMCET PYQ

The value of $\int \frac{(x+1)}{x(xe^{x}+1)} dx$ is equal to

1
2
3
4
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Solution

Qus : 377NIMCET PYQ

Differentiate [- log(log x), x > 1] with respect to x

1
2
3
4
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Solution

Let y = - log(log x), x>1

On differentiating both sides,

= -

(log (log x))

(log x)

Qus : 378NIMCET PYQ

Naresh has 10 friends, and he wants to invite 6 of them to a party. How many times will 3 particular friends never attend the party?

1
2
3
4
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Solution

^(10-3)C₆= ⁷C₆ = 7

Qus : 379NIMCET PYQ

If a, b, c are in geometric progression, then $log_{ax}^{a}, log_{bx}^{a}$ and $log_{cx}^{a}$ are in

1
2
3
4
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Solution

Qus : 380NIMCET PYQ

The sum of two vectors $\vec{a}$ and $\vec{b}$ is a vector $\vec{c}$ such that $|\vec{a}|=|\vec{b}|=|\vec{c}|=2$. Then, the magnitude of $\vec{a}-\vec{b}$ is equal to:

1
2
3
4
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Solution

Qus : 381NIMCET PYQ

If the points

lie in the region corresponding to the acute angle between the lines and then

1
2
3
4
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Solution

Qus : 382NIMCET PYQ

1
2
3
4
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Solution

Function is the form of

therefore using by L'Hospital rule

Again apply L'Hospital Rule,

Putting x = 0, we get

Qus : 383NIMCET PYQ

There is a young boy’s birthday party in which 3 friends have attended. The mother has arranged 10 games where a prize is awarded for a winning game. The prizes are identical. If each of the 4 children receives at least one prize, then how many distributions of prizes are possible?

1
2
3
4
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Solution

Qus : 384NIMCET PYQ

The value of the sum $\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}$ is

1
2
3
4
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Solution

Qus : 385NIMCET PYQ

If x and y are positive real numbers satisfying the system of equations $x^{2}+y\sqrt{xy}=336$ and $y^{2}+x\sqrt{xy}=112$, then x + y is:

1
2
3
4
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Solution

Qus : 386NIMCET PYQ

If $\vec{a}=\hat{i}-\hat{k},\, \vec{b}=x\hat{i}+\hat{j}+(1-x)\hat{k}$ and $\vec{c}=y\hat{i}+x\hat{j}+(1+x-y)\hat{k}$ , then $[\vec{a} , \vec{b}, \vec{c}]$ depends on

1
2
3
4
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Solution

Qus : 387NIMCET PYQ

From three collinear points A, B and C on a level ground, which are on the same side of a tower, the angles of elevation of the top of the tower are 30°, 45° and 60° respectively. If BC = 60 m, then AB is:

1
2
3
4
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Solution

Qus : 388NIMCET PYQ

A problem in Mathematics is given to 3 students A, B, and C. If the probability of A solving the problem is 1/2 and B not solving it is 1/4 . The whole probability of the problem being solved is 63/64 , then what is the probability of solving it by C?

1
2
3
4
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Solution

Qus : 389NIMCET PYQ

If $42 (^nP_2)=(^nP_4)$ then the value of n is

1
2
3
4
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Solution

Qus : 390NIMCET PYQ

If $x = 1$ is the directrix of the parabola $y^{2} = kx - 8$, then k is:

1
2
3
4
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Solution

Qus : 391NIMCET PYQ

A and B play a game where each is asked to select a number from 1 to 25. If the two numbers match, both win a prize. The probability that they will not win a prize in a single trial is

1
2
3
4
Go to Discussion

Solution

Probability of winning a prize = $\frac{1}{25}$
Thus, probability of not winning = $1-\frac{1}{25}=\frac{24}{25}$

Qus : 392NIMCET PYQ

The foot of the perpendicular from the point (2, 4) upon $x + y = 1$ is

1
2
3
4
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Solution

Qus : 393NIMCET PYQ

If $sin x + a cos x = b$, then $|a sin x - cos x|$ is:

1
2
3
4
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Solution

Qus : 394NIMCET PYQ

A, B, C are three sets of values of x:

A: 2,3,7,1,3,2,3

B: 7,5,9,12,5,3,8

C: 4,4,11,7,2,3,4

Select the correct statement among the following:

1
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3
4
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Solution

A: 2, 3, 7, 1, 3, 2, 3

Increasing Order : A: 1, 2, 2, 3, 3, 3, 7

Mode = 3 (occurs maximum number of times)

Median = 3 (the middle term)

Mean =$\frac{(1+2+2+3+3+3+7)}{7}$

$=\frac{21}{7} = 3$

Hence. Mean=Median=Mode

Qus : 395NIMCET PYQ

The value of k for which the equation $(k-2)x^{2}+8x+k+4=0$ has both real, distinct and negative roots is

1
2
3
4
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Solution

Qus : 396NIMCET PYQ

A condition that $x^{3} + ax^{2} + bx + c$ may have no extremum is

1
2
3
4
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Solution

Qus : 397NIMCET PYQ

Standard deviation for the following distribution is

Size of item	6	7	8	9	10	11	12
Frequency	3	6	9	13	8	5	4

1
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4
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Solution

Total number of items in the distribution = Σ f_i = 3 + 6 + 9 + 13 + 8 + 5 + 4 = 48.

The Mean (x̅) of the given set = $\rm \dfrac{\sum f_i x_i}{\sum f_i}$.

⇒ x̅ = $\rm \dfrac{6\times3+7\times6+8\times9+9\times13+10\times8+11\times5+12\times4}{48}=\dfrac{432}{48}$ = 9.

Let's calculate the variance using the formula: $\rm \sigma^2 =\dfrac{\sum x_i^2}{n}-\bar x^2$.

$\rm \dfrac{\sum {x_i}^2}{n}=\dfrac{6^2\times3+7^2\times6+8^2\times9+9^2\times13+10^2\times8+11^2\times5+12^2\times4}{48}=\dfrac{4012}{48}$ = 83.58.

∴ σ² = 83.58 - 9² = 83.58 - 81 = 2.58.

And, Standard Deviation (σ) = $\rm \sqrt{\sigma^2}=\sqrt{Variance}=\sqrt{2.58}$ ≈ 1.607.

Qus : 398NIMCET PYQ

If (2, 1), (–1, –2), (3, 3) are the midpoints of the sides BC, CA, AB of a triangle ABC, then equation of the line BC is

1
2
3
4
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Solution

Qus : 399NIMCET PYQ

If n and r are integers such that 1 ≤ r ≤ n, then the value of n ^n-1C_r-1is

1
2
3
4
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Solution

Qus : 400NIMCET PYQ

1
2
3
4
Go to Discussion

Solution

Qus : 401NIMCET PYQ

If a fair dice is rolled successively, then the probability that 1 appears in an even numbered throw is

1
2
3
4
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Solution

Qus : 402NIMCET PYQ

If the foci of the ellipse $b^{2}x^{2}+16y^{2}=16b^{2}$ and the hyperbola $81x^{2}-144y^{2}=\frac{81 \times 144}{25}$ coincide, then the value of $b$, is

1
2
3
4
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Solution

Qus : 403NIMCET PYQ

Roots of equation are $ax^2-2bx+c=0$ are n and m , then the value of $\frac{b}{an^2+c}+\frac{b}{am^2+c}$ is

1
2
3
4
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Solution

Qus : 404NIMCET PYQ

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. A vector $\vec{v}$ in the plane of $\vec{a}$ and $\vec{b}$ whose projection on $\frac{\vec{c}}{|\vec{c}|}$ is $\frac{1}{\sqrt{3}}$, is

1
2
3
4
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Solution

Qus : 405NIMCET PYQ

There are 8 students appearing in an examination of which 3 have to appear in Mathematics paper and the remaining 5 in different subjects. Then, the number of ways they can be made to sit in a row, if the candidates in Mathematics cannot sit next to each other is

1
2
3
4
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Solution

Qus : 406NIMCET PYQ

The number of values of k for which the linear equations

4x + ky + z = 0

kx + 4y + z = 0

2x + 2y + z = 0

posses a non-zero solution is

1
2
3
4
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Solution

Since, equation has non-zero solution.
Δ = 0

Qus : 407NIMCET PYQ

The number of bit strings of length 10 that contain either five consecutive 0’s or five consecutive 1’s is

1
2
3
4
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Solution

Qus : 408NIMCET PYQ

If $x$ is so small that $x^{2}$ and higher powers of $x$ can be neglected, then $\frac{(9+2x)^{1/2}(3+4x)}{(1-x)^{1/5}}$ is approximately equal to

1
2
3
4
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Solution

Qus : 409NIMCET PYQ

Let A = (a_ij) and B = (b_ij) be two square matricesof order n and det(A) denotes the determinant of A. Then, which of the following is not correct.

1
2
3
4
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Solution

Qus : 410NIMCET PYQ

If $0 < x < \pi $ and $cos x + sin x = \frac{1}{2}$ , then the value of tan x is

1
2
3
4
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Solution

Qus : 411NIMCET PYQ

If the sets A and B are defined as A = {(x, y) | y = 1 / x, 0 ≠ x ∈ R}, B = {(x, y)|y = -x ∈ R} then

1
2
3
4
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Solution

Qus : 412NIMCET PYQ

The tangent to an ellipse x² + 16y² = 16 and making angel 60° with X-axis is:

1
2
3
4
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Solution

Qus : 413NIMCET PYQ

If $\vec{a}, \vec{b}$ and $\vec{c}$ are the position vectors of the vertices A, B, C of a triangle ABC, then the area of the triangle ABC is

1
2
3
4
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Solution

Qus : 414NIMCET PYQ

If A, B and C is three angles of a ΔABC, whose area is Δ. Let a, b and c be the sides opposite to the angles A, B and C respectively. Is $s=\frac{a+b+c}{2}=6$, then the product $\frac{1}{3} s^{2} (s-a)(s-b)(s-c)$ is equal to

1
2
3
4
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Solution

Qus : 415NIMCET PYQ

Find the number of point(s) of intersection of the ellipse

and the circle x² + y² = 4

1
2
3
4
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Solution

Qus : 416NIMCET PYQ

If $\int e^{x}(f(x)-f'(x))dx=\phi(x)$ , then the value of $\int e^x f(x) dx$ is

1
2
3
4
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Solution

Qus : 417NIMCET PYQ

A normal to the curve $x^{2} = 4y$ passes through the point (1, 2). The distance of the origin from the normal is

1
2
3
4
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Solution

Qus : 418NIMCET PYQ

An arithmetic progression has 3 as its first term. Also, the sum of the first 8 terms is twice the sum of the first 5 terms. Then what is the common difference?

1
2
3
4
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Solution

Qus : 419NIMCET PYQ

If $3x + 4y + k = 0$ is a tangent to the hyperbola ,$9x^{2}-16y^{2}=144$ then the value of $K$ is

1
2
3
4
Go to Discussion

Solution

Qus : 420NIMCET PYQ

Suppose r integers, 0 < r < 10, are chosen from (0, 1, 2, ...,9) at random and with replacement. The probability that no two are equal, is

1
2
3
4
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Solution

Qus : 421NIMCET PYQ

If a + b + c = 0, then the value of

1
2
3
4
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Solution

Qus : 422NIMCET PYQ

$a, b, c$ are positive integers such that $a^{2}+2b^{2}-2bc=100$ and $2ab-c^{2}=100$. Then the value of $\frac{a+b}{c}$ is

1
2
3
4
Go to Discussion

Solution

Qus : 423NIMCET PYQ

If $x^{2} + 2ax + 10 - 3a > 0$ for all x ∈ R, then

1
2
3
4
Go to Discussion

Solution

Qus : 424NIMCET PYQ

Find

1
2
3
4
Go to Discussion

Solution

Qus : 425NIMCET PYQ

If $(– 4, 5)$ is one vertex and $7x – y + 8 = 0$ is one diagonal of a square, then the equation of the other diagonal is

1
2
3
4
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Solution

Qus : 426NIMCET PYQ

A box contains 3 coins, one coin is fair, one coin is two headed and one coin is weighted, so that the probability of heads appearing is $\frac{1}{3}$ . A coin is selected at random and tossed, then the probability that head appears is

1
2
3
4
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Solution

Qus : 427NIMCET PYQ

, then x = 0 is

1
2
3
4
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Solution

Qus : 428NIMCET PYQ

Out of $2n + 1$ tickets, which are consecutively numbered, three are drawn at random. Then the probability that the numbers on them are in arithmetic progression is

1
2
3
4
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Solution

Qus : 429NIMCET PYQ

If a vector $\vec{a}$ makes an equal angle with the coordinate axes and has magnitude 3, then the angle between $\vec{a}$ and each of the three coordinate axes is

1
2
3
4
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Solution

Qus : 430NIMCET PYQ

is a continuous function at x = 0, then the value of k is

1
2
3
4
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Solution

Qus : 431NIMCET PYQ

A circle touches the X-axis and also touches another circle with centre at (0, 3) and radius 2. Then the locus of the centre of the first circle is

1
2
3
4
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Solution

Qus : 432NIMCET PYQ

If $f(x)=\left\{\begin{matrix} \frac{sin[x]}{[x]} &, [x]\ne0 \\ 0 &, [x]=0 \end{matrix}\right.$ , where [x] is the largest integer but not larger than x, then $\lim_{x\to0}f(x)$ is

1
2
3
4
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Solution

Qus : 433NIMCET PYQ

Find the interval(s) on which the graph y=2x³e^x isincreasing

1
2
3
4
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Solution

Qus : 434NIMCET PYQ

Let $\bar{P}$ and $\bar{Q}$ denote the complements of two sets P and Q. Then the set $(P-Q)\cup (Q-P) \cup (P \cap Q)$ is

1
2
3
4
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Solution

Qus : 435NIMCET PYQ

If tan A - tan B = x and cot B - cot A = y, then cot (A - B) is equal to

1
2
3
4
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Solution

Qus : 436NIMCET PYQ

then value of k is

1
2
3
4
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Solution

Qus : 437NIMCET PYQ

With the usual notation $\frac{d^{2}x}{dy^{2}}$

1
2
3
4
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Solution

Qus : 438NIMCET PYQ

If $a = log_{12}^{18}$, $b = log_{24}^{54}$, then $ab + 5(a - b)$ is

1
2
3
4
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Solution

Qus : 439NIMCET PYQ

A cube is made up of 125 one cm. square cubes placed on a table. How many squares are visible only on three sides?

1
2
3
4
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Solution

Qus : 440NIMCET PYQ

Evaluate

1
2
3
4
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Solution

Qus : 441NIMCET PYQ

The radius of the circle passing through the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}$and having it centre at (0, 3) is

1
2
3
4
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Solution

Qus : 442NIMCET PYQ

A student takes a quiz consisting of 5 multiple choice questions. Each question has 4 possible answers. If a student is guessing the answer at random and answer to different are independent, then the probability of atleast one correct answer is

1
2
3
4
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Solution

Qus : 443NIMCET PYQ

A computer producing factory has only two plants $T_1$ and $T_2$. Plant $T_1$ produces 20% and plant $T_2$ produces 80% of total computers produced. 7% of computers produced in the factory turn out to be defective. It is known that P (computer turns out to be defective given that it is produced in plant $T_1$) = 10P (computer turns out to be defective given that it is produced in plant $T_2$). where P(E) denotes the probability of an event E. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $T_2$ is

1
2
3
4
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Solution

Qus : 444NIMCET PYQ

where n is a positive integer, then the relation between I_n and I_n-1 is

1
2
3
4
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Solution

Qus : 445NIMCET PYQ

A function $f : (0,\pi) \to R$ defined by $f(x) = 2 sin x + cos 2x$ has

1
2
3
4
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Solution

Qus : 446NIMCET PYQ

The condition that the line lx + my + n = 0 becomes a tangent to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , is

1
2
3
4
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Solution

Qus : 447NIMCET PYQ

The mean of 5 observation is 5 and their variance is 124. If three of the observations are 1,2 and 6; then the mean deviation from the mean of the data is:

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Solution

Qus : 448NIMCET PYQ

The value of

depends on the

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Solution

Qus : 449NIMCET PYQ

A matrix $M_r$ is defined as $M_r=\begin{bmatrix} r &r-1 \\ r-1&r \end{bmatrix} , r \in N$ then the value of $det(M_1) + det(M_2) +...+ det(M_{2015})$ is

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Solution

Qus : 450NIMCET PYQ

The value of sin 20° sin 40° sin 80° is

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Solution

Qus : 451NIMCET PYQ

The perimeter of a $\Delta ABC$ is 6 times the arithmetic mean of the sines of its angles. If the side a is 1, then the angle A is

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Solution

Quick Solution

Given: Perimeter = 6 × Arithmetic Mean of sin A, sin B, sin C

Using Law of Sines: $ \frac{a}{\sin A} = 2R $, and a = 1 ⇒ $ \sin A = \frac{1}{2R} $

Assume $ A = 30^\circ \Rightarrow \sin A = \frac{1}{2} \Rightarrow 2R = 2 $

⇒ b = 2 sin B, c = 2 sin C

Perimeter = $ 1 + b + c = 1 + 2 \sin B + 2 \sin C $

Mean = $ \frac{\sin A + \sin B + \sin C}{3} $

Check: $ 1 + 2\sin B + 2\sin C = 6 \cdot \frac{1/2 + \sin B + \sin C}{3} $ ✅

✅ Final Answer: 30°

Qus : 452NIMCET PYQ

Find the area bounded by the line y = 3 - x, the parabola y = x² - 9 and

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Solution

Qus : 453NIMCET PYQ

If $\vec{AC}=2\hat{i}+\hat{j}+\hat{k}$ and $\vec{BD}=-\hat{i}+3\hat{j}+2\hat{k}$ then the area of the quadrilateral ABCD is

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Solution

Qus : 454NIMCET PYQ

Two non-negative numbers whose sum is 9 and the product of the one number and square of the other number is maximum, are

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Solution

Qus : 455NIMCET PYQ

In an examination of nine papers, a candidate has to pass in more papers than the number of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful is

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Solution

Candidate Passing Criteria

Total Papers: 9

Condition for Success: Passes > Fails

So, candidate is unsuccessful when: Passes ≤ 4

Calculate ways:

\[ \text{Ways} = \sum_{x=0}^{4} \binom{9}{x} = \binom{9}{0} + \binom{9}{1} + \binom{9}{2} + \binom{9}{3} + \binom{9}{4} \]\[= 1 + 9 + 36 + 84 + 126 = \boxed{256}\]

✅ Final Answer: 256 ways

Qus : 456NIMCET PYQ

are three non-coplanar vectors, then

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Solution

Qus : 457NIMCET PYQ

The value of $sin^{-1}\frac{1}{\sqrt{2}}+sin^{-1}\frac{\sqrt{2}-\sqrt{1}}{\sqrt{6}}+sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+...$ to infinity , is equal to

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Solution

Qus : 458NIMCET PYQ

The median AD of ΔABC is bisected at E and BE is produced to meet the side AC at F. Then, AF ∶ FC is

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Solution

Qus : 459NIMCET PYQ

For a group of 100 candidates, the mean and standard deviation of scores were found to be 40 and 15 respectively. Later on, it was found that the scores 25 and 35 were misread as 52 and 53 respectively. Then the corrected mean and standard deviation corresponding to the corrected figures are

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Solution

Corrected Mean and Standard Deviation

Original Mean: 40, Standard Deviation: 15

Two scores were misread: 25 → 52 and 35 → 53

Corrected Mean:

$ \mu' = \frac{3955}{100} = \boxed{39.55} $

Corrected Standard Deviation:

$ \sigma' = \sqrt{\frac{178837}{100} - (39.55)^2} \approx \boxed{14.96} $

✅ Final Answer: Mean = 39.55, Standard Deviation ≈ 14.96

Qus : 460NIMCET PYQ

Two forces F₁ and F₂ are used to pull a car, which met an accident. The angle between the two forces is θ . Find the values of θ for which the resultant force is equal to

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Solution

Qus : 461NIMCET PYQ

If two circles $x^{2}+y^{2}+2gx+2fy=0$ and $x^{2}+y^{2}+2g'x+2f'y=0$ touch each other then whichof the following is true?

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Solution

Qus : 462NIMCET PYQ

If PQ is a double ordinate of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ such that OPQ is an equilateral triangle, where O is the centre of the hyperbola, then which of the following is true?

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Solution

Qus : 463NIMCET PYQ

Consider the following frequency distribution table.

Class interval	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	180	$f_1$	34	180	136	$f_2$	50

If the total frequency is 685 & median is 42.6 then the values of $f_1$ and $f_2$ are

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Solution

Median & Frequency Table

Given: Median = 42.6, Total Frequency = 685

Using Median Formula:

$ \text{Median} = L + \left( \frac{N/2 - F}{f} \right) \cdot h $

Median Class: 40–50
Lower boundary $ L = 40 $
Frequency $ f = 180 $
Class width $ h = 10 $
Cumulative freq before median class $ F = 214 + f_1 $

Substituting values:

$ 42.6 = 40 + \left( \frac{128.5 - f_1}{180} \right) \cdot 10 \Rightarrow f_1 = \boxed{82} $

Using total frequency:

$ 662 + f_2 = 685 \Rightarrow f_2 = \boxed{23} $

✅ Final Answer: $ f_1 = 82,\quad f_2 = 23 $

Qus : 464NIMCET PYQ

are four vectors such that

is collinear with

and

is collinear with

then

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Solution

Qus : 465NIMCET PYQ

$\int_0^\pi [cotx]dx$ where [.] denotes the greatest integer function, is equal to

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Solution

Qus : 466NIMCET PYQ

In ΔABC, if a = 2, b = 4 and ∠C = 60°, then A and B are respectively equal to

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Solution

Qus : 467NIMCET PYQ

The sum of infinite terms of decreasing GP is equal to the greatest value of the function $f(x) = x^3 + 3x – 9$ in the interval [–2, 3] and difference between the first two terms is f '(0). Then the common ratio of the GP is

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Solution

? GP and Function Relation

Given: $ f(x) = x^3 + 3x - 9 $

The sum of infinite GP = max value of $ f(x) $ on [−2, 3]

The difference between first two terms = $ f'(0) $

Step 1: $ f(x) $ is increasing ⇒ Max at $ x = 3 $

$ f(3) = 27 \Rightarrow \frac{a}{1 - r} = 27 $

Step 2: $ f'(x) = 3x^2 + 3 \Rightarrow f'(0) = 3 $

⇒ $ a(1 - r) = 3 $

Step 3: Solve:
$ a = 27(1 - r) $
$ \Rightarrow 27(1 - r)^2 = 3 \Rightarrow (1 - r)^2 = \frac{1}{9} \Rightarrow r = \frac{2}{3} $

✅ Final Answer: $ r = \frac{2}{3} $

Qus : 468NIMCET PYQ

Forces of magnitude 5, 3, 1 units act in the directions 6i + 2j + 3k, 3i - 2j + 6k, 2i - 3j - 6k respectively on a particle which is displaced from the point (2, −1, −3) to (5, −1, 1). The total work done by the force is

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Solution

Qus : 469NIMCET PYQ

In a right angled triangle, the hypotenuse is four times the perpendicular drawn to it from the opposite vertex. The value of one of the acute angles is

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Solution

Qus : 470NIMCET PYQ

If $ \int \frac{xe^{x}}{\sqrt{1+e^{x}}}=f(x)\sqrt{1+e^{x}}-2log \frac{\sqrt{1+e^{x}}-1}{\sqrt{1+e^{x}}+1}+C$ then $f(x)$ is

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Solution

Qus : 471NIMCET PYQ

If $f(x)=\lim _{{x}\rightarrow0}\, \frac{{6}^x-{3}^x-{2}^x+1}{\log _e9(1-\cos x)}$ is a real number then $\lim _{{x}\rightarrow0}\, f(x)$

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Solution

Qus : 472NIMCET PYQ

The position vectors of points A and B are

and

. Then the position vector of point p dividing AB in the ratio m : n is

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Solution

Qus : 473NIMCET PYQ

A is targeting B, B and C are targeting A. Probability of hitting the target by A, B and C are $\frac{2}{3}, \frac{1}{2}$ and $\frac{1}{3}$ respectively. If A is hit then the probability that B hits the target and C does not, is

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Solution

Qus : 474NIMCET PYQ

The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is

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Solution

Qus : 475NIMCET PYQ

The value of $\int ^{\pi/3}_{-\pi/3}\frac{x\sin x}{{\cos }^2x}dx$ is

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Solution

Qus : 476NIMCET PYQ

If a, b, c are three non-zero vectors with no two of which are collinear, a + 2b is collinear with c and b + 3c is collinear with a , then | a + 2b + 6c | will be equal to

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Solution

Qus : 477NIMCET PYQ

If the angles of a triangle are in the ratio 2 : 3 : 7, then the ratio of the sides opposite to these angles is

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Solution

Qus : 478NIMCET PYQ

How many even integers between 4000 and 7000 have four different digits?

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Solution

Qus : 479NIMCET PYQ

The equation of the tangent at any point of curve $x=a cos2t, y=2\sqrt{2} a sint$ with $m$ as its slope is

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Solution

Qus : 480NIMCET PYQ

Vertices of the vectors i - 2j + 2k , 2i + j - k and 3i - j + 2k form a triangle. This triangle is

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Solution

Qus : 481NIMCET PYQ

Suppose that A and B are two events with probabilities $P(A) =\frac{1}{2} \, P(B)=\frac{1}{3}$ Then which of the following is true?

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Solution

Qus : 482NIMCET PYQ

If $\prod ^n_{i=1}\tan ({{\alpha}}_i)=1\, \forall{{\alpha}}_i\, \in\Bigg{[}0,\, \frac{\pi}{2}\Bigg{]}$ where i=1,2,3,...,n. Then maximum value of $\prod ^n_{i=1}\sin ({{\alpha}}_i)$.

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Solution

Qus : 483NIMCET PYQ

If the volume of a parallelepiped whose adjacent edges are

a = 2i + 3j + 4k,

b = i + αj + 2k

c = i + 2k + αk

is 15, then α =

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Solution

Qus : 484NIMCET PYQ

The number of one-to-one functions from {1, 2, 3} to {1, 2, 3, 4, 5} is

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Solution

Qus : 485NIMCET PYQ

A speaks truth in 60% and B speaks the truth in 50% cases. In what percentage of cases they are likely incontradict each other while narrating some incident is

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Solution

Qus : 486NIMCET PYQ

Solve the equation sin²x - sinx - 2 = 0 for for x on the interval 0 ≤ x < 2π

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Solution

Qus : 487NIMCET PYQ

A harbour lies in a direction 60° South of West from a fort and at a distance 30 km from it, a ship sets out from the harbour at noon and sails due East at 10 km an hour. The time at which the ship will be 70 km from the fort is

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Solution

Qus : 488NIMCET PYQ

If $\overrightarrow{{a}}$ and $\overrightarrow{{b}}$ are vectors in space, given by $\overrightarrow{{a}}=\frac{\hat{i}-2\hat{j}}{\sqrt[]{5}}$ and $\overrightarrow{{b}}=\frac{2\hat{i}+\hat{j}+3\hat{k}}{\sqrt[]{14}}$, then the value of$(2\vec{a} + \vec{b}).[(\vec{a} × \vec{b}) × (\vec{a} – 2\vec{b})]$ is

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Solution

Qus : 489NIMCET PYQ

If $\frac{tanx}{2}=\frac{tanx}{3}=\frac{tanx}{5}$ and x + y + z = π, then the value of tan²x + tan²y + tan²z is

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Solution

Qus : 490NIMCET PYQ

If $x, y, z$ are three consecutive positive integers, then $log (1 + xz)$ is

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Solution

Qus : 491NIMCET PYQ

Let $\vec{A} = 2\hat{i} + \hat{j} – 2\hat{k}$ and $\vec{B} = \hat{i} + \hat{j}$, If $\vec{C}$ is a vector such that $|\vec{C} – \vec{A}| = 3$ and the angle between A × B and C is ${30^{\circ}}$, then $|(\vec{A} × \vec{B}) × \vec{C}|$ = 3 then the value of $\vec{A}.\vec{C}$ is equal to

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Solution

Qus : 492NIMCET PYQ

Find the value of sin 12°sin 48°sin 54°

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Solution

Qus : 493NIMCET PYQ

A professor has 24 text books on computer science and is concerned about their coverage of the topics (P) compilers, (Q) data structures and (R) Operating systems. The following data gives the number of books that contain material on these topics: $n(P) = 8, n(Q) = 13, n(R) = 13, n(P \cap R) = 3, n(P \cap R) = 3, n(Q \cap R) = 3, n(Q \cap R) = 6, n(P \cap Q \cap R) = 2 $ where $n(x)$ is the cardinality of the set $x$. Then the number of text books that have no material on compilers is

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Solution

Qus : 494NIMCET PYQ

Let A and B be sets. $A\cap X=B\cap X=\phi$ and $A\cup X=B\cup X$ for some set X, relation between A & B

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Solution

Qus : 495NIMCET PYQ

If cos x = tan y , cot y = tan z and cot z = tan x, then sinx =

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Solution

Qus : 496NIMCET PYQ

The value of $tan(\frac{7\pi}{8})$ is

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Solution

Qus : 497NIMCET PYQ

If a, b, c, d are in HP and arithmetic mean of ab, bc, cd is 9 then which of the following number is the value of ad?

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Solution

Qus : 498NIMCET PYQ

The value of

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Solution

Qus : 499NIMCET PYQ

If $\vec{a}$ and $\vec{b}$ are vectors such that $|\vec{a}|=13$, $|\vec{b}|=5$ and $\vec{a} . \vec{b} =60$then the value of $|\vec{a} \times \vec{b}|$ is

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Solution

Qus : 500NIMCET PYQ

Find foci of the equation $x^2 + 2x – 4y^2 + 8y – 7 = 0$

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Solution

Finding Foci of a Conic

Given Equation: $ x^2 + 2x - 4y^2 + 8y - 7 = 0 $

Step 1: Complete the square

⇒ $ (x + 1)^2 - 4(y - 1)^2 = 4 $

Rewriting: $ \frac{(x + 1)^2}{4} - \frac{(y - 1)^2}{1} = 1 $

This is a horizontal hyperbola with:

Center: $ (-1, 1) $
$ a^2 = 4 $, $ b^2 = 1 $
$ c = \sqrt{a^2 + b^2} = \sqrt{5} $

✅ Foci: $ (-1 \pm \sqrt{5},\ 1) $

Qus : 501NIMCET PYQ

The value of sin 10°sin 50°sin 70° is

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Solution

sin10° sin50° sin70°

= sin10° sin(60°−10°) sin(60°+10°)

= 1/4 sin3x10°

=1/4x1/2=1/8

Qus : 502NIMCET PYQ

Two towers face each other separated by a distance of 25 meters. As seen from the top of the first tower, the angle of depression of the second tower’s base is 60° and that of the top is 30°. The height (in meters) of the second tower is

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Solution

Qus : 503NIMCET PYQ

The locus of the mid-point of all chords of the parabola $y^2 = 4x$ which are drawn through its vertex is

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