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.

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

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:

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

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:

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

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

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:

$$\frac{3k}{2} = 4 \Rightarrow k = \frac{8}{3}$$

$$\boxed{k = \frac{8}{3}}$$

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

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

$$\boxed{\text{Probability} = \frac{14}{55}}$$

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, $$e^x \cos x = -1$$

$$\boxed{\text{At least one root}}$$

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:   $\boxed{f(5) = 30}$

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:   $\boxed{(p, q) = (7,\ 3)}$

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:   $\boxed{\binom{1002}{50}}$

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:   $\boxed{0}$

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:   $\boxed{2 \text{ points}}$

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: $\boxed{(4a,\ 4a)}$

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

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: $\boxed{5\sqrt{3}}$

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: $\boxed{\frac{5}{18}}$

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: $\boxed{e^{-1}}$