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: $\boxed{\dfrac{\sqrt{17}}{9}}$

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: $\boxed{60}$

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

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

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: $\boxed{1134}$

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

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: $\boxed{16}$

Step 1: Recognize the series

The exponent is an infinite geometric series: $$ 1 + |\sin x| + |\sin x|^2 + |\sin x|^3 + \cdots $$

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

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

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: $\boxed{60 \, \text{J}}$

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: $\boxed{\dfrac{4}{9}}$

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: $\boxed{48}$

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: $\boxed{0.432}$

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:

\[ \boxed{-1} \]

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:

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

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:

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

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:

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

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: \( \boxed{355} \)

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

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:

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

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: $$\boxed{1}$$

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: $$\boxed{2}$$

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.

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: \( \boxed{\frac{1}{2}} \)

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

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: $$\boxed{0}$$

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,

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

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.

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:

1. In between the lines \( 2x + 3y = -6 \) and \( 2x + 3y = 6 \)
2. On the same side of the line \( 2x + 3y = 6 \)
3. On the same side of the line \( 2x + 3y = -6 \)

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):

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

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.

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:

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

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

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:

\[ \boxed{a < 0} \]

? 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: \(\boxed{14}\)

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 \( x = 0 \)
• \( f'(0) = \boxed{-\frac{1}{2}} \)

? 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:

\[ \boxed{-1} \]

✔️ 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: \(\boxed{\frac{4}{1155}}\)

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: \(\boxed{1 \text{ or } 4}\)

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: \(\boxed{e^{-1}}\)