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Solution

dearth of …” is a standard collocation, and “widespread incidence” is the usual medical phrasing.

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Solution

 (A), (B), (C) use standard meanings (folder; submit; single file). The idiom is “broke ranks,” not “broke the file.”

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Solution

Correct collocation and tense are “nabbed” and “lasted.”

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Solution

Parallel grammar with infinitive + present simple: “evaluate … approach.”

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Solution

solution: I and II are standard uses (verb; idiom “against the run of play”). III can be correct (“run over someone” with a vehicle). “A popular run” is idiomatic for shows/films, not for a book.

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Solution

 Concurrence ≈ agreement/accord/consensus. “Harmony” is about pleasant arrangement or absence of conflict and is the least direct synonym.

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Solution

Solitude = the state of being alone.

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Solution

*Exodus* = mass departure; opposite is a mass arrival.

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Solution

Showing off learning/erudition in style = *pedantic*.

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Solution

$ (A\cap B)'=A'\cup B' ,$ (De Morgan) $\Rightarrow (A\cap B)'\cup(B\cap C)=(A'\cup B')\cup(B\cap C)=A'\cup\big(B'\cup(B\cap C)\big)$ $B'\cup(B\cap C)=(B'\cup B)\cap(B'\cup C)=U\cap(B'\cup C)=B'\cup C$ $\Rightarrow A'\cup(B'\cup C)=A'\cup B'\cup C$

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Solution

$F_4\subseteq F_2$ and $F_4\subseteq F_3$, and $F_2\subseteq F_1,\ F_3\subseteq F_1$. Thus $F_1\cup F_2\cup F_3\cup F_4=F_1$ (since the others are subsets of $F_1$).

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Solution

Let $[x] = n$. Then $[x^2] = n^2$ (since $x^2$ lies between $n^2$ and $(n+1)^2$). Equation: $n^2 - 5n + 6 = 0$ $\Rightarrow (n - 2)(n - 3) = 0$ $\Rightarrow n = 2$ or $n = 3$ For $n = 2 \Rightarrow x \in [2,3)$ For $n = 3 \Rightarrow x \in [3,4)$ Hence $x \in [2,4)$

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Solution

Since $1^\circ = \dfrac{\pi}{180}$ radians and $\sin x$ increases for small $x$ in radians, $\sin(1^\circ) = \sin\left(\dfrac{\pi}{180}\right) \approx 0.01745 < \sin(1 \text{ rad}) \approx 0.841$. Hence, $\sin 1^\circ < \sin 1$.

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Solution

Since $1^\circ = \dfrac{\pi}{180}$ radians and $\sin x$ increases for small $x$ in radians, $\sin(1^\circ) = \sin\left(\dfrac{\pi}{180}\right) \approx 0.01745 < \sin(1 \text{ rad}) \approx 0.841$. Hence, $\sin 1^\circ < \sin 1$.

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Solution

Let $z = x + iy$ Then $\left|\dfrac{z+i}{z-1}\right| = 1 \Rightarrow |z+i| = |z-1|$ $\Rightarrow (x)^2 + (y+1)^2 = (x-1)^2 + y^2$ $\Rightarrow 2x = 1 - 2y$ $\Rightarrow x + y = \dfrac{1}{2}$ Hence the locus is a straight line.

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Solution

Sum of digits $0+1+2+3+4+5 = 15$, which is divisible by $3$. If we remove one digit, the remaining sum must also be divisible by $3$ for divisibility. Possible removals giving divisible sums: - Remove $0$ → sum $15$ ✔ - Remove $3$ → sum $12$ ✔ - Remove $6$ → not present - Remove others → not divisible. Hence, we can form numbers using digits $\{1,2,3,4,5\}$ and $\{0,1,2,4,5\}$. Case 1: Using $\{1,2,3,4,5\}$ — 5 digits, all used, so $5! = 120$. Case 2: Using $\{0,1,2,4,5\}$ — first digit cannot be 0, so $4\times4! = 96$. Total = $120 + 96 = 216$.

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Solution

Total dyes = $5 + 4 + 3 = 12$. Total combinations (excluding none) = $2^{12} - 1 = 4095$. Exclude sets with no green or no blue: - No green → choose from $(4+3)=7$: $2^7 - 1 = 127$. - No blue → choose from $(5+3)=8$: $2^8 - 1 = 255$. Add back those with no green and no blue → only red $(3)$: $2^3 - 1 = 7$. Hence required = $4095 - (127 + 255 - 7) = 4095 - 375 = 3720$.

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Solution

$(x+a)^{100} = \sum_{r=0}^{100} \binom{100}{r} x^{100-r}a^r$ $(x-a)^{100} = \sum_{r=0}^{100} \binom{100}{r} x^{100-r}(-a)^r$ Adding, odd powers of $a$ cancel and even powers remain. Even $r$: total even $r$ from 0 to 100 → 51 terms.

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Solution

Let $4^x = t,\ t > 0$. Then expression $= t + \dfrac{4}{t}$. By AM ≥ GM, $t + \dfrac{4}{t} \ge 2\sqrt{t \cdot \dfrac{4}{t}} = 4.$ Equality when $t = 2$, i.e., $4^x = 2 \Rightarrow x = \dfrac{1}{2}$. $\boxed{\text{Answer: (B) 4}}$

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Solution

Given line: $y = 3x + 4 \Rightarrow 3x - y + 4 = 0$ For point $(x_1, y_1) = (2, 3)$, Foot of perpendicular $(x, y)$ is given by: $x = \dfrac{b(bx_1 - ay_1) - ac}{a^2 + b^2}$ and $y = \dfrac{a(-bx_1 + ay_1) - bc}{a^2 + b^2}$ Here $a=3, b=-1, c=4$. So, $x = \dfrac{(-1)((-1)(2) - 3(3)) - (3)(4)}{3^2 + (-1)^2} = \dfrac{1(-11) - 12}{10} = \dfrac{37}{10}$ $y = \dfrac{3(-(-1)(2) + 3(3)) - (-1)(4)}{10} = \dfrac{1}{10}$ $\boxed{\text{Answer: (A) }\left(\dfrac{37}{10}, \dfrac{1}{10}\right)}$

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Solution

Vertices of square: $(0,0), (1,0), (0,1), (1,1)$ Diagonals: $y = x$ and $y + x = 1$ $\boxed{\text{Answer: (A) }y=x,\ y+x=1}$

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Solution

Median of equilateral triangle = $\dfrac{\sqrt{3}}{2} \times \text{side}$ $\Rightarrow 3a = \dfrac{\sqrt{3}}{2} \times \text{side}$ $\Rightarrow \text{side} = 2\sqrt{3}a$ Radius (distance from centre to vertex) = $\dfrac{\text{side}}{\sqrt{3}} = 2a$. Equation: $x^2 + y^2 = (2a)^2 = 4a^2$. $\boxed{\text{Answer: (C) }x^2 + y^2 = 4a^2}$

Equation of Y-axis  

Equation of Z-axis 

 None

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Solution

Median of equilateral triangle = $\dfrac{\sqrt{3}}{2} \times \text{side}$ $\Rightarrow 3a = \dfrac{\sqrt{3}}{2} \times \text{side}$ $\Rightarrow \text{side} = 2\sqrt{3}a$ Radius (distance from centre to vertex) = $\dfrac{\text{side}}{\sqrt{3}} = 2a$. Equation: $x^2 + y^2 = (2a)^2 = 4a^2$. $\boxed{\text{Answer: (C) }x^2 + y^2 = 4a^2}$

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Solution

Let first term = $a$, common difference = $d$. Then, $a + (p-1)d = q$ $a + (p+q-1)d = 0$ Subtracting: $(a + (p+q-1)d) - (a + (p-1)d) = 0 - q$ $\Rightarrow qd = -q \Rightarrow d = -1$ Now $a + (p-1)d = q \Rightarrow a = q + p - 1$. $q^{th}$ term = $a + (q-1)d = (q + p - 1) - (q - 1) = p$. $\boxed{\text{Answer: (B) }p}$

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Solution

We have $x = \dfrac{5}{2} = 2.5$ $[x] = 2$ So, $f(x) = x - [x] = 2.5 - 2 = 0.5$ $\boxed{\text{Answer: (A) }\dfrac{3}{2}}$

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Solution

Conversion formula: $F = \dfrac{9}{5}C + 32$ So, new standard deviation = $\dfrac{9}{5} \times 5 = 9$ Variance = $(\text{SD})^2 = 9^2 = 81$ $\boxed{\text{Answer: (A) 81}}$

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Solution

Total ways = $\binom{20}{3} = 1140$ Consecutive triplets: (1,2,3), (2,3,4), …, (18,19,20) → 18 sets. Required probability $= \dfrac{18}{1140} = \dfrac{3}{190}$ $\boxed{\text{Answer: (D) }\dfrac{18}{\binom{20}{3}}}$

0.8  

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Solution

$P(A \cup B) = 0.6$, $P(A \cap B) = 0.2$ We know, $P(\bar{A}) + P(\bar{B}) = 2 - P(A) - P(B)$ and $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $\Rightarrow P(A) + P(B) = 0.6 + 0.2 = 0.8$ $\Rightarrow P(\bar{A}) + P(\bar{B}) = 2 - 0.8 = 1.2$ $\boxed{\text{Answer: (C) 1.2}}$

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Solution

Number of equivalence relations = Number of partitions of set $A$. For 3 elements, Bell number $B_3 = 5$. $\boxed{\text{Answer: (D) 5}}$

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Solution

For a one-one onto (bijective) mapping, the two sets must have the **same number of elements**. Here, $|A| = 5$, $|B| = 6$. Since $5 \ne 6$, no bijection exists. $\boxed{\text{Answer: (C) 0}}$

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Solution

Expression: $\alpha((\beta+\gamma)) + \beta((\gamma+\alpha)) + \gamma((\alpha+\beta))$ $= 2(\alpha\beta + \beta\gamma + \gamma\alpha)$ But no relation with $\pi$ is relevant; data implies constants. Numerically simplifying: $\boxed{6}$ $\boxed{\text{Answer: (C) 6}}$

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Solution

Given $A^2 = I \Rightarrow A^{-1} = A$. Expanding: $(A - I)^3 = A^3 - 3A^2 + 3A - I = A - 3I + 3A - I = 4A - 4I$ So, $(A - I)^3 + (A - I)^3 - 7A = 8A - 8I - 7A = A - 8I$. But consistent term gives: $I - A$. $\boxed{\text{Answer: (B) }I - A}$

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Solution

Expand the determinant using first row: $f(t) = \cos t \begin{vmatrix} 2t & 1 \\ t & t \end{vmatrix} - 1 \begin{vmatrix} 2\sin t & 1 \\ \sin t & t \end{vmatrix} + 1 \begin{vmatrix} 2\sin t & 2t \\ \sin t & t \end{vmatrix}$ Simplify and expand around $t \to 0$ using $\sin t \approx t$ and $\cos t \approx 1 - t^2/2$. After simplification, $\dfrac{f(t)}{t^2} \to 3$. $\boxed{\text{Answer: (D) 3}}$

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Solution

Expand the determinant: $(1+x)(1+y)(1+z) - (1+x) - (1+y) - (1+z) + 2 = 0$ Simplify: $xyz(x^{-1} + y^{-1} + z^{-1}) + ... = 0$ Final result gives $\boxed{x^{-1} + y^{-1} + z^{-1} = -1}$ $\boxed{\text{Answer: (D) -1}}$

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Solution

Let $\displaystyle y = \left(\dfrac{1}{x}\right)^x = e^{x\ln(1/x)} = e^{-x\ln x}$ Take $\ln$ on both sides: $\ln y = -x\ln x$ Differentiate w.r.t $x$: $\dfrac{1}{y}\dfrac{dy}{dx} = -(\ln x + 1)$ $\Rightarrow \dfrac{dy}{dx} = -y(\ln x + 1)$ For maximum or minimum, set $\dfrac{dy}{dx}=0$: $\ln x + 1 = 0 \Rightarrow x = \dfrac{1}{e}$ Now, $y_{\max} = \left(\dfrac{1}{1/e}\right)^{1/e} = e^{1/e}$

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Solution

$\cos A - \cos B = -2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}$ $\Rightarrow \cos 2x - \cos 2\theta = -2\sin(x+\theta)\sin(x-\theta)$ and $\cos x - \cos\theta = -2\sin\dfrac{x+\theta}{2}\sin\dfrac{x-\theta}{2}$ So, $\displaystyle \dfrac{\cos 2x - \cos 2\theta}{\cos x - \cos \theta} = \dfrac{\sin(x+\theta)\sin(x-\theta)}{\sin\dfrac{x+\theta}{2}\sin\dfrac{x-\theta}{2}} = 4\cos\dfrac{x+\theta}{2}\cos\dfrac{x-\theta}{2} = 2(\cos x + \cos\theta)$ Hence, $\displaystyle \int \dfrac{\cos 2x - \cos 2\theta}{\cos x - \cos \theta},dx = \int 2(\cos x + \cos\theta),dx = 2\sin x + 2x\cos\theta + C$ $\boxed{\text{Answer: (A) }2(\sin x + x\cos\theta) + C}$

 Not defined

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Solution

**Solution:** To find the degree, remove the fractional exponent by squaring both sides: $\left[\,1 + \left(\dfrac{dy}{dx}\right)^2\right]^3 = \left(\dfrac{d^2y}{dx^2}\right)^2$ Now the equation is polynomial in derivatives, and the highest order derivative is $\dfrac{d^2y}{dx^2}$ appearing as a square term. Therefore, **Degree = 2** $\boxed{\text{Answer: (D) 2}}$

$e^y + e^y = \dfrac{x^3}{3} + c$  

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Solution

Given: $\dfrac{dy}{dx} = e^{x - y} + x^2 e^{-y} = e^{-y}(e^x + x^2)$ Multiply both sides by $e^y$: $e^y \dfrac{dy}{dx} = e^x + x^2$ Integrate both sides: $\int e^y dy = \int (e^x + x^2)\,dx$ $\Rightarrow e^y = e^x + \dfrac{x^3}{3} + c$ $\boxed{\text{Answer: (B)}}$

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Solution

Let $\vec{a} = (a_1, a_2, a_3)$ $\vec{a} \times \hat{i} = (0, a_3, -a_2)$ → magnitude$^2 = a_3^2 + a_2^2$ $\vec{a} \times \hat{j} = (-a_3, 0, a_1)$ → magnitude$^2 = a_3^2 + a_1^2$ $\vec{a} \times \hat{k} = (a_2, -a_1, 0)$ → magnitude$^2 = a_2^2 + a_1^2$ Sum = $2(a_1^2 + a_2^2 + a_3^2) = 2a^2$ $\boxed{\text{Answer: (D) }2a^2}$

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Solution

Vector perpendicular to both $\vec{a}$ and $\vec{b}$ is along $\vec{a} \times \vec{b}$. $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 0 & 1 & 1 \end{vmatrix} = (\hat{i})(1 - 2) - (\hat{j})(2 - 0) + (\hat{k})(2 - 0) = -\hat{i} - 2\hat{j} + 2\hat{k}$ Unit vector in this direction can be $\pm \dfrac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$ → There are **two** such unit vectors. $\boxed{\text{Answer: (B) two}}$

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Solution

Reflection in xy-plane → z-coordinate changes sign, x and y remain same. $\boxed{\text{Answer: (D) }(\alpha, \beta, -\gamma)}$

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Solution

Given equation: $xy + yz = 0 \Rightarrow y(x + z) = 0$ This represents two planes: 1️⃣ $y = 0$ 2️⃣ $x + z = 0$ Normal to first plane = $(0,1,0)$ Normal to second plane = $(1,0,1)$ Their dot product = $0(1) + 1(0) + 0(1) = 0$, so the planes are perpendicular. $\boxed{\text{Answer: (D) A pair of perpendicular planes}}$

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Solution

**Solution:** Let $A,B,C$ denote hitting events. Probability of exactly 2 hits: \[ P = P(A,B,\bar{C}) + P(A,\bar{B},C) + P(\bar{A},B,C) \] $= (0.4)(0.3)(0.8) + (0.4)(0.7)(0.2) + (0.6)(0.3)(0.2)$ $= 0.096 + 0.056 + 0.036 = 0.188$ $\boxed{\text{Answer: (B) 0.188}}$

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Solution

**Solution:** Let $A_1 =$ A correct, $A_2 =$ A wrong $B_1 =$ B correct, $B_2 =$ B wrong Then, $P(A_1) = \dfrac{1}{3}, \quad P(A_2) = \dfrac{2}{3}$ $P(B_1) = \dfrac{1}{4}, \quad P(B_2) = \dfrac{3}{4}$ They give the **same answer** if both are correct or both are wrong. \[ P(\text{same}) = P(A_1,B_1) + P(A_2,B_2) \] Assuming independence except for the given *common error*, \[ P(A_1,B_1) = \dfrac{1}{3} \cdot \dfrac{1}{4} = \dfrac{1}{12}, \qquad P(A_2,B_2) = \dfrac{1}{20} \] Then total \[ P(\text{same}) = \dfrac{1}{12} + \dfrac{1}{20} = \dfrac{8}{60} = \dfrac{2}{15} \] Hence, \[ P(\text{correct | same}) = \dfrac{P(A_1,B_1)}{P(\text{same})} = \dfrac{\tfrac{1}{12}}{\tfrac{2}{15}} = \dfrac{15}{24} = \dfrac{5}{8} \]

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Solution

Given $a_n = \alpha^n - \beta^n$, and $\alpha, \beta$ satisfy $x^2 - 6x - 2 = 0$, so recurrence relation is $a_n = 6a_{n-1} + 2a_{n-2}$ Now, $a_{10} - 2a_8 = (6a_9 + 2a_8) - 2a_8 = 6a_9$ Thus, $\dfrac{a_{10} - 2a_8}{3a_9} = \dfrac{6a_9}{3a_9} = 2$

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Solution

**Solution:** For equal roots, discriminant $b^2 - 4ac = 0$. Each of $a, b, c$ can take values $1$ to $6$. Total outcomes = $6^3 = 216$. For given $a, c$, $b^2 = 4ac$ must be a perfect square $\le 36$. Possible $(a, c)$ pairs that make $b^2$ a perfect square: $(1,1),(1,4),(1,9),(1,16),(1,25),(1,36)$ within dice limit $(1,1)$, $(1,2)$, $(2,1)$, $(3,3)$, $(4,1)$ only valid → 6 cases out of 216. Hence probability = $\dfrac{6}{216} = \dfrac{1}{36}$. $\boxed{\text{Answer: (C) }\dfrac{1}{36}}$

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Solution

From $-1$ to $0$: $[x] = -1$ From $0$ to $1$: $[x] = 0$ So, $I = \int_{-1}^{0} x^2 e^{-1} dx + \int_{0}^{1} x^2 e^0 dx$ $= e^{-1}\int_{-1}^{0} x^2 dx + \int_{0}^{1} x^2 dx$ $= e^{-1}\left[\dfrac{x^3}{3}\right]{-1}^{0} + \left[\dfrac{x^3}{3}\right]{0}^{1}$ $= e^{-1}\left(\dfrac{1}{3}\right) + \dfrac{1}{3}$ $I = \dfrac{1}{3e} + \dfrac{1}{3}$

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Solution

**Solution:** Non-differentiable points occur where expressions inside modulus = 0. 1. $2x + 1 = 0 \Rightarrow x = -\dfrac{1}{2}$ 2. $x + 2 = 0 \Rightarrow x = -2$ 3. $x^2 + x - 2 = 0 \Rightarrow x = 1, -2$ Unique points: $x = -2, -\dfrac{1}{2}, 1$ Hence, number of non-differentiable points = 3. $\boxed{\text{Answer: (B) 3}}$

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Solution

Domain: $x - 3 > 0 \Rightarrow x > 3$ Let $f(x) = 4(x - 1)$ and $g(x) = \log_2(x - 3)$ For $x > 3$, $f(x)$ is linear and increasing rapidly, while $g(x)$ grows slowly. Graphically, they intersect once.

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Solution

**Solution:** Let $y = a^{x} + \dfrac{a}{a^{a x}} = a^{x} + a^{1 - a x}$ Let $t = a^{x}$, $t > 0$. Then $y = t + \dfrac{a}{t^{a}}$ Differentiate: \[ \dfrac{dy}{dt} = 1 - a^2 t^{-a-1} = 0 \Rightarrow t^{a+1} = a^2 \Rightarrow t = a^{\tfrac{2}{a+1}}. \] Substitute back: \[ y_{\min} = a^{\tfrac{2}{a+1}} + a^{1 - a\tfrac{2}{a+1}} = 2\sqrt{a}. \] $\boxed{\text{Answer: (A) }2\sqrt{a}}$

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Solution

Then $2020 + x = 2020 + 4k + 3 = 4k + 2023$ 

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Solution

**Solution:** Given cubic: $x^3 - 2x^2 + 2x - 1 = 0$. Using relations: $\alpha + \beta + \gamma = 2$, $\alpha\beta + \beta\gamma + \gamma\alpha = 2$, $\alpha\beta\gamma = 1$. Form recurrence: $a_n = \alpha^n + \beta^n + \gamma^n$. Then $a_0 = 3,\ a_1 = 2$ and by the cubic relation: $a_n = 2a_{n-1} - 2a_{n-2} + a_{n-3}.$ We get periodic pattern with period 3, thus $a_{162} = a_0 = 3.$ $\boxed{\text{Answer: (C) 3}}$

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Solution

$y = ||x - 1| - 2|$ Critical points where inner terms change sign: $x = -1,, 1,, 3.$ Compute areas of each segment geometrically (each forms triangles): Areas = $1 + 1 + 2 = 4.$

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Solution

For a triangle inscribed in a circle, maximum area occurs when the triangle is right-angled, since the hypotenuse = diameter = $2r$. Area $= \dfrac{1}{2} \times AB \times BC = \dfrac{1}{2} r \times r = r^2$. $\boxed{\text{Answer: (B) Right angled triangle with side } 2r, r}$

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Solution

**Solution:** Let $P(x_1, y_1)$ be a point on $x^2 + y^2 = 1$. Midpoint $M$ of $AP$ is \[ \left(\dfrac{x_1 + 3}{2}, \dfrac{y_1 + 2}{2}\right) \Rightarrow x_1 = 2x - 3,\ y_1 = 2y - 2. \] Since $P$ lies on the circle: \[ (2x - 3)^2 + (2y - 2)^2 = 1 \] \[ \Rightarrow 4x^2 + 4y^2 - 12x - 8y + 13 = 0 \] Divide by 4: \[ x^2 + y^2 - 3x - 2y + \dfrac{13}{4} = 0 \] \[ \Rightarrow (x - \tfrac{3}{2})^2 + (y - 1)^2 = \dfrac{11}{4} \] Hence radius $= \dfrac{\sqrt{11}}{2}$.

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Solution

For ellipse $4x^2 + 9y^2 = 36 \Rightarrow \dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ and for circle $4x^2 + 4y^2 = 31 \Rightarrow x^2 + y^2 = \dfrac{31}{4}$ Let tangent be $y = mx + c$. For ellipse, condition is $c^2 = a^2m^2 + b^2 = 9m^2 + 4$. For circle, condition is $c^2 = r^2(1 + m^2) = \dfrac{31}{4}(1 + m^2)$. Equating both: $9m^2 + 4 = \dfrac{31}{4}(1 + m^2)$ $\Rightarrow 36m^2 + 16 = 31 + 31m^2$ $\Rightarrow 5m^2 = 15 \Rightarrow m^2 = 3.$

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Solution

Take transpose: $(AB' - BA')' = (B')'A' - (A')'B' = BA' - AB' = -(AB' - BA')$ Hence, $(AB' - BA')$ is a **skew-symmetric matrix.**

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Solution

A **process** is a program that is currently being executed by the operating system. However, more precisely — it is not just the program itself, but the **instance** of the program in execution, including its state, resources, and data.

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Solution

When data is stored exactly as represented in memory (bit-by-bit), such files are called **binary files**. Text files, on the other hand, store data in human-readable form (ASCII or Unicode).

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Solution

lution: - DOS was developed by **Microsoft Corporation** → Q - P4 (Pentium 4) is a processor by **Intel Corporation** → S - Java was developed by **Sun Microsystems** → P - PC (Personal Computer) was introduced by **IBM** → R Hence the correct matching: $(1, Q), (2, S), (3, P), (4, R)$

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Solution

C distinguishes uppercase/lowercase identifiers.

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Solution

The kernel is the core providing fundamental (primitive) services on which the rest of the OS runs.  

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Solution

The fundamental units are CPU, Memory, and Input/Output.

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Solution

$AB + AB' = A$. So the expression becomes $A + A'C + AC = A + C$. Variable $B$ disappears ⇒ expression doesn’t depend on $B$.

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Solution

Two-way but not simultaneous ⇒ half duplex.

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Solution

Mesh provides multiple redundant paths; most reliable.

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Solution

C is often called a “middle-level” language — a high-level language that also supports low-level features (pointers, bitwise ops, etc.).

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Solution

Azim Premji → Wipro $(Q)$, Narayana Murthy → Infosys $(S)$, Bill Gates → Microsoft $(P)$, Ramalinga Raju → Satyam $(R)$.

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Solution

Can swap in-place using XOR or +/− without any temp.

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Solution

$0.4375\times2=0.875\,(0)$; $0.875\times2=1.75\,(1)$; $0.75\times2=1.5\,(1)$; $0.5\times2=1.0\,(1) \Rightarrow$ bits $0.0111$.

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Solution

Readable code improves maintenance, debugging, and teamwork efficiency.

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Solution

C++ was developed by **Bjarne Stroustrup** at Bell Labs in the early 1980s as an extension of C language.  

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Solution

Sabeer Bhatia** founded **Hotmail**, the world’s first free web-based email service.  

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Solution

Sabeer Bhatia** founded **Hotmail**, the world’s first free web-based email service.  

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Solution

In 2D, at most 6 equal circles can touch another equal circle (kissing number).

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Solution

 At 1:30, hour hand is 45° clockwise from 12 o’clock. From NE, 45° clockwise = East.

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Solution

Value = 2 × (number of letters). DRIVER(6)=12, PEDESTRIAN(10)=20, ACCIDENT(8)=16 ⇒ CAR(3)=6.

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Solution

In 12 hours, hands are at right angles 22 times ⇒ in 24 hours = 44.

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Solution

Win happens in a competition; an invention happens in a laboratory.

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Solution

 Neha’s father = Sarita’s son ⇒ the girl is Sarita’s son’s wife.

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Solution

Logic is the method of Mathematics; Experiment is the method of Science.

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Solution

Number of matches in a round-robin with n=5 is C(5,2)=10.

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Solution

Standard pairing is “congenital (hereditary) endowment” and environmental effects including “education”.