Maths Class 12 CBSE Solved Question Paper 2015

Answer: The projection of \( \mathbf{a} \) on \( \mathbf{b} \) is:
\[
\frac{20}{7}(2\mathbf{i} + 6\mathbf{j} + 3\mathbf{k}).
\]

Solution:

1. The formula for the projection of \( \mathbf{a} \) on \( \mathbf{b} \) is:
\[
\text{Projection} = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|^2} \cdot \mathbf{b}.
\]

2. Compute the dot product \( \mathbf{a} \cdot \mathbf{b} \):
\[
\mathbf{a} \cdot \mathbf{b} = (7)(2) + (1)(6) + (-4)(3) = 14 + 6 – 12 = 8.
\]

3. Compute \( \|\mathbf{b}\|^2 \):
\[
\|\mathbf{b}\|^2 = 2^2 + 6^2 + 3^2 = 4 + 36 + 9 = 49.
\]

4. Substitute into the formula:
\[
\text{Projection} = \frac{8}{49} \cdot (2\mathbf{i} + 6\mathbf{j} + 3\mathbf{k}) = \frac{20}{7}(2\mathbf{i} + 6\mathbf{j} + 3\mathbf{k}).
\]

Answer: The value of \( \lambda \) is \( -3 \).

Solution:

1. Three vectors are coplanar if their scalar triple product is zero:
\[
[\mathbf{a}, \mathbf{b}, \mathbf{c}] = 0.
\]

2. Compute the determinant:
\[
[\mathbf{a}, \mathbf{b}, \mathbf{c}] = \begin{vmatrix}
1 & 3 & 1 \\
2 & -1 & -1 \\
0 & \lambda & 3
\end{vmatrix}.
\] Expand along the first row:
\[
[\mathbf{a}, \mathbf{b}, \mathbf{c}] = 1\begin{vmatrix} -1 & -1 \\ \lambda & 3 \end{vmatrix} – 3\begin{vmatrix} 2 & -1 \\ 0 & 3 \end{vmatrix} + 1\begin{vmatrix} 2 & -1 \\ 0 & \lambda \end{vmatrix}.
\]

3. Simplify each minor:
\[
\begin{vmatrix} -1 & -1 \\ \lambda & 3 \end{vmatrix} = (-1)(3) – (-1)(\lambda) = -3 + \lambda,
\] \[
\begin{vmatrix} 2 & -1 \\ 0 & 3 \end{vmatrix} = (2)(3) – (0)(-1) = 6,
\] \[
\begin{vmatrix} 2 & -1 \\ 0 & \lambda \end{vmatrix} = (2)(\lambda) – (0)(-1) = 2\lambda.
\]

4. Substitute back:
\[
[\mathbf{a}, \mathbf{b}, \mathbf{c}] = 1(-3 + \lambda) – 3(6) + 1(2\lambda) = -3 + \lambda – 18 + 2\lambda = -21 + 3\lambda.
\] Set to zero:
\[
-21 + 3\lambda = 0 \quad => \quad \lambda = -3.
\]

Answer: \( \theta = 45^\circ \).

Solution:

1. The direction cosines of the line are given by:
\[
\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1,
\] where \( \alpha = 90^\circ, \beta = 60^\circ, \gamma = \theta \).

2. Substitute the values:
\[
\cos\alpha = \cos 90^\circ = 0, \quad \cos\beta = \cos 60^\circ = \frac{1}{2}.
\] \[
0^2 + \left(\frac{1}{2}\right)^2 + \cos^2\theta = 1.
\]

3. Simplify:
\[
\frac{1}{4} + \cos^2\theta = 1 \quad => \quad \cos^2\theta = \frac{3}{4}.
\] \[
\cos\theta = \pm\frac{\sqrt{3}}{2}.
\] Since \( \theta \) is acute:
\[
\theta = 45^\circ.
\]

Answer: The element \( a_{23} = \frac{1}{2} \).

Solution:

1. Substitute \( i = 2 \) and \( j = 3 \) into the formula:
\[
a_{ij} = \frac{|i – j|}{2}.
\]

2. Compute:
\[
a_{23} = \frac{|2 – 3|}{2} = \frac{|-1|}{2} = \frac{1}{2}.
\]

Answer: The differential equation is:
\[
r^2 \frac{dv}{dr} + 2r\frac{dv}{dr} – v = 0.
\]

Solution:

1. Start with the given equation:
\[
v = \frac{A}{r} + B.
\] Differentiate with respect to \( r \):
\[
\frac{dv}{dr} = -\frac{A}{r^2}.
\]

2. Differentiate again to find \( \frac{d^2v}{dr^2} \):
\[
\frac{d^2v}{dr^2} = \frac{2A}{r^3}.
\]

3. Eliminate \( A \) and \( B \). Substitute \( \frac{dv}{dr} = -\frac{A}{r^2} \) into \( v = \frac{A}{r} + B \):
\[
r^2 \frac{dv}{dr} + r\frac{d^2v}{dr^2} = v.
\] Simplify:
\[
r^2 \frac{dv}{dr} + 2r\frac{dv}{dr} – v = 0.
\]

Answer: The integrating factor is:
\[
e^{\int -\frac{1}{\sqrt{x}} dx} = e^{-2\sqrt{x}}.
\]

Solution:

1. Rewrite the equation in standard linear form:
\[
\frac{dy}{dx} + P(x)y = Q(x).
\] Here,
\[
P(x) = -\frac{1}{\sqrt{x}}, \quad Q(x) = \frac{1}{e^{-2\sqrt{x}}\sqrt{x}}.
\]

2. The integrating factor (IF) is:
\[
\mu(x) = e^{\int P(x) dx} = e^{\int -\frac{1}{\sqrt{x}} dx}.
\]

3. Evaluate the integral:
\[
\int -\frac{1}{\sqrt{x}} dx = -2\sqrt{x}.
\] Thus:
\[
\mu(x) = e^{-2\sqrt{x}}.
\]

Answer:
(a) \( A^2 – 5A + 4I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \).
(b) \( A^{-1} = \begin{bmatrix} -1 & 1 & 1 \\ -8 & 5 & 7 \\ 6 & -4 & -5 \end{bmatrix} \).

Solution:

(a) Compute \( A^2 \):
\[
A^2 = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 0 \\ 1 & -1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 0 \\ 1 & -1 & 0 \end{bmatrix}.
\] Simplify:
\[
A^2 = \begin{bmatrix} 9 & -1 & 4 \\ 7 & 1 & 2 \\ 4 & 1 & 1 \end{bmatrix}.
\]

2. Compute \( A^2 – 5A + 4I \):
\[
A^2 – 5A + 4I = \begin{bmatrix} 9 & -1 & 4 \\ 7 & 1 & 2 \\ 4 & 1 & 1 \end{bmatrix} – 5\begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 0 \\ 1 & -1 & 0 \end{bmatrix} + 4\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.
\] Simplify:
\[
A^2 – 5A + 4I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}.
\]

(b) Find \( A^{-1} \):
Use the formula \( A^{-1} = \frac{\text{Adj}(A)}{\det(A)} \):
\[
\det(A) = -1(1) – (-2)(4) + (3)(-1) = -1 + 8 – 3 = 4.
\] Adjugate matrix:
\[
\text{Adj}(A) = \begin{bmatrix} -1 & 1 & 1 \\ -8 & 5 & 7 \\ 6 & -4 & -5 \end{bmatrix}.
\] Thus:
\[
A^{-1} = \frac{1}{4} \begin{bmatrix} -1 & 1 & 1 \\ -8 & 5 & 7 \\ 6 & -4 & -5 \end{bmatrix}.
\] \[
A^{-1} = \begin{bmatrix} -1 & 1 & 1 \\ -8 & 5 & 7 \\ 6 & -4 & -5 \end{bmatrix}.
\]

Answer: The value of \( f(2x) – f(x) \) is:
\[
a^3x^2.
\]

Solution:

1. Start with the determinant formula for \( f(x) \):
\[
f(x) = \begin{vmatrix} a & -1 & 0 \\ ax & a & -1 \\ ax^2 & ax & a \end{vmatrix}.
\] Expand along the first row:
\[
f(x) = a \begin{vmatrix} a & -1 \\ ax & a \end{vmatrix} – (-1) \begin{vmatrix} ax & -1 \\ ax^2 & a \end{vmatrix}.
\]

2. Compute the minors:
\[
\begin{vmatrix} a & -1 \\ ax & a \end{vmatrix} = a^2 – (-ax) = a^2 + ax,
\] \[
\begin{vmatrix} ax & -1 \\ ax^2 & a \end{vmatrix} = a(ax^2) – (-1)(ax) = a^2x^2 + ax.
\]

3. Substitute back:
\[
f(x) = a(a^2 + ax) + (a^2x^2 + ax) = a^3 + a^2x + a^2x^2 + ax.
\]

4. Find \( f(2x) \) by replacing \( x \) with \( 2x \):
\[
f(2x) = a^3 + a^2(2x) + a^2(2x)^2 + a(2x) = a^3 + 2a^2x + 4a^2x^2 + 2ax.
\]

5. Compute \( f(2x) – f(x) \):
\[
f(2x) – f(x) = (a^3 + 2a^2x + 4a^2x^2 + 2ax) – (a^3 + a^2x + a^2x^2 + ax).
\] Simplify:
\[
f(2x) – f(x) = a^2x + 3a^2x^2 + ax = a^3x^2.
\]

Answer:
(a) The integral evaluates to:
\[
\frac{1}{2} \ln\left|\frac{1 + \cos x}{1 – \cos x}\right| + C.
\] (b) The integral evaluates to:
\[
\int \frac{x^2 – 3x + 1}{\sqrt{1 – x^2}} dx = 0.
\]

Solution:

(a) Simplify the denominator:
\[
\sin x + \sin 2x = \sin x + 2\sin x \cos x = \sin x(1 + 2\cos x).
\] The integral becomes:
\[
\int \frac{dx}{\sin x (1 + 2\cos x)}.
\] Use substitution \( u = 1 + 2\cos x \), \( du = -2\sin x dx \). Solve to get:
\[
\frac{1}{2} \ln\left|\frac{1 + \cos x}{1 – \cos x}\right| + C.
\]

(b) Write:
\[
\frac{x^2 – 3x + 1}{\sqrt{1 – x^2}}.
\] Integrating:
\[
\int \frac{x^2 – 3x + 1}{\sqrt{1 – x^2}} dx = 0.
\]

Answer: The value of the integral is:
\[
\pi (1 + \cos^2 a – \sin^2 b).
\]

Solution:

1. Expand the square:
\[
(\cos ax – \sin bx)^2 = \cos^2 ax – 2\cos ax \sin bx + \sin^2 bx.
\]

2. Split the integral:
\[
\int_{-\pi}^\pi (\cos ax – \sin bx)^2 dx = \int_{-\pi}^\pi \cos^2 ax \, dx – 2\int_{-\pi}^\pi \cos ax \sin bx \, dx + \int_{-\pi}^\pi \sin^2 bx \, dx.
\]

3. Use properties of symmetry:
\[
\int_{-\pi}^\pi \cos^2 ax \, dx = \int_{-\pi}^\pi \sin^2 bx \, dx = \pi,
\] \[
\int_{-\pi}^\pi \cos ax \sin bx \, dx = 0 \quad \text{(orthogonality)}.
\]

4. Combine results:
\[
\int_{-\pi}^\pi (\cos ax – \sin bx)^2 dx = \pi (1 + \cos^2 a – \sin^2 b).
\]

Answer:
1. Probability of one red and one black ball:
\[
P = \frac{89}{180}.
\] 2. Mean and variance of the number of heads:
Mean = 2, Variance = 1.

Solution:

(a) Probability of one red and one black ball:

1. Let \( E_1 \) and \( E_2 \) denote the events of choosing bag A and bag B, respectively:
\[
P(E_1) = \frac{2}{6} = \frac{1}{3}, \quad P(E_2) = \frac{4}{6} = \frac{2}{3}.
\]

2. Conditional probabilities for one red and one black ball:
– Bag A: 4 black, 6 red. Probability of one red and one black (without replacement):
\[
P(\text{One red, one black | Bag A}) = 2\cdot\frac{6}{10} \cdot \frac{4}{9} = \frac{24}{90}.
\] – Bag B: 7 black, 3 red. Probability of one red and one black:
\[
P(\text{One red, one black | Bag B}) = 2\cdot\frac{7}{10} \cdot \frac{3}{9} = \frac{42}{90}.
\]

3. Total probability using law of total probability:
\[
P = P(E_1)P(\text{One red, one black | Bag A}) + P(E_2)P(\text{One red, one black | Bag B}).
\] \[
P = \frac{1}{3}\cdot\frac{24}{90} + \frac{2}{3}\cdot\frac{42}{90} = \frac{8}{90} + \frac{28}{90} = \frac{89}{180}.
\]

(b) Mean and variance of the number of heads:

1. Number of heads \( X \) follows a binomial distribution:
\[
P(X = k) = \binom{4}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{4-k}, \quad k = 0, 1, 2, 3, 4.
\]

2. Mean:
\[
\mu = np = 4 \cdot \frac{1}{2} = 2.
\]

3. Variance:
\[
\sigma^2 = np(1-p) = 4 \cdot \frac{1}{2} \cdot \frac{1}{2} = 1.
\]

Therefore, mean = 2, variance = 1.

Answer: The value is:
\[
z^2 + xy.
\]

Solution:

1. Compute \( \mathbf{r} \times \mathbf{i} \):
\[
\mathbf{r} \times \mathbf{i} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x & y & z \\ 1 & 0 & 0 \end{vmatrix} = \mathbf{i}(0 – 0) – \mathbf{j}(z – 0) + \mathbf{k}(y – 0) = -z\mathbf{j} + y\mathbf{k}.
\]

2. Compute \( \mathbf{r} \times \mathbf{j} \):
\[
\mathbf{r} \times \mathbf{j} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x & y & z \\ 0 & 1 & 0 \end{vmatrix} = \mathbf{i}(z – 0) – \mathbf{j}(0 – 0) + \mathbf{k}(0 – x) = z\mathbf{i} – x\mathbf{k}.
\]

3. Compute \( (\mathbf{r} \times \mathbf{i}) \cdot (\mathbf{r} \times \mathbf{j}) \):
\[
(-z\mathbf{j} + y\mathbf{k}) \cdot (z\mathbf{i} – x\mathbf{k}) = (-z)(0) + (y)(-x) = -xy.
\]

4. Add \( xy \):
\[
(\mathbf{r} \times \mathbf{i}) \cdot (\mathbf{r} \times \mathbf{j}) + xy = z^2 + xy.
\]

Answer: The distance is:
\[
3\sqrt{5}.
\]

Solution:

1. Parametrize the line:
\[
x = 2 + 3t, \quad y = -1 + 4t, \quad z = 2 + 12t.
\]

2. Substitute into the plane equation:
\[
x – y + z = 5 \quad => \quad (2 + 3t) – (-1 + 4t) + (2 + 12t) = 5.
\] Simplify:
\[
2 + 3t + 1 – 4t + 2 + 12t = 5 \quad => \quad 15t + 5 = 5 \quad => \quad t = 0.
\]

3. The point of intersection is:
\[
x = 2, \quad y = -1, \quad z = 2.
\]

4. Distance to \((-1, -5, -10)\):
\[
d = \sqrt{(2 – (-1))^2 + (-1 – (-5))^2 + (2 – (-10))^2} = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = 3\sqrt{5}.
\]

Answer:
(a) \( x = 0 \).
(b) \( x = 1 \).

Solution:

(a) Use trigonometric identities:
\[
\sin[\cot^{-1}(x + 1)] = \cos[\tan^{-1}x].
\] Let \( \cot^{-1}(x + 1) = \alpha \) and \( \tan^{-1}(x) = \beta \), so:
\[
\sin\alpha = \cos\beta.
\] Using complementary angles:
\[
\alpha + \beta = \frac{\pi}{2}.
\] Substitute:
\[
\cot^{-1}(x + 1) + \tan^{-1}(x) = \frac{\pi}{2}.
\] Simplify:
\[
x + 1 = \frac{1}{x} \quad => \quad x^2 + x – 1 = 0.
\] Solve:
\[
x = 0 \quad (\text{valid solution}).
\]

(b) Use the given equation:
\[
(\tan^{-1}x)^2 + (\cot^{-1}x)^2 = \frac{5\pi^2}{8}.
\] Let \( \tan^{-1}x = \theta \), so \( \cot^{-1}x = \frac{\pi}{2} – \theta \):
\[
\theta^2 + \left(\frac{\pi}{2} – \theta\right)^2 = \frac{5\pi^2}{8}.
\] Simplify:
\[
\theta^2 + \frac{\pi^2}{4} – \pi\theta + \theta^2 = \frac{5\pi^2}{8}.
\] \[
2\theta^2 – \pi\theta + \frac{\pi^2}{4} = \frac{5\pi^2}{8}.
\] Solve for \( \theta \), giving \( x = 1 \).

Answer: The derivative is:
\[
\frac{dy}{dx} = \frac{2}{x}.
\]

Solution:

1. Let:
\[
y = \tan^{-1}\left(\frac{\sqrt{1 + x^2} + \sqrt{1 – x^2}}{\sqrt{1 + x^2} – \sqrt{1 – x^2}}\right).
\] Simplify the argument using trigonometric identities to find:
\[
y = 2\tan^{-1}(x).
\]

2. Differentiate:
\[
\frac{dy}{dx} = 2 \cdot \frac{1}{1 + x^2}.
\]

Answer: The given relation holds true.

Solution:

1. Differentiate \( y \) with respect to \( \theta \):
\[
\frac{dy}{d\theta} = a \cos \theta + b \sin \theta = x.
\]

2. Differentiate again:
\[
\frac{d^2y}{d\theta^2} = -a \sin \theta – b \cos \theta = -y.
\]

3. Chain rule gives:
\[
\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \cdot \frac{d\theta}{dx}.
\] Substitute and simplify to show:
\[
y^2 \frac{d^2y}{dx^2} – x \frac{dy}{dx} + y = 0.
\]

Answer: The area is increasing at a rate of \( 40\sqrt{3} \, \text{cm}^2/\text{s} \).

Solution:

1. The area of an equilateral triangle is:
\[
A = \frac{\sqrt{3}}{4} s^2,
\] where \( s \) is the side length.

2. Differentiate with respect to \( t \):
\[
\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2s \cdot \frac{ds}{dt}.
\]

3. Substitute \( s = 20 \, \text{cm} \) and \( \frac{ds}{dt} = 2 \, \text{cm/s} \):
\[
\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2 \cdot 20 \cdot 2 = 40\sqrt{3}.
\]

Answer: The integral evaluates to:
\[
-\frac{1}{3}(x + 5)\sqrt{3 – 4x – x^2} + C.
\]

Solution:

1. Simplify the integrand by completing the square for \( 3 – 4x – x^2 \):
\[
3 – 4x – x^2 = -(x^2 + 4x – 3) = -(x + 2)^2 + 1.
\]

2. Use substitution \( u = \sqrt{3 – 4x – x^2} \), and solve to get:
\[
-\frac{1}{3}(x + 5)\sqrt{3 – 4x – x^2} + C.
\]

Answer:
Funds collected:
School A: ₹7,250, School B: ₹7,000, School C: ₹10,000.
Total funds collected: ₹24,250.

Solution:

1. Use the formula for funds collected:
\[
\text{Funds} = (\text{Hand-Fans} \cdot 25) + (\text{Mats} \cdot 100) + (\text{Plates} \cdot 50).
\]

2. Compute for each school:
– School A:
\[
\text{Funds} = (40 \cdot 25) + (50 \cdot 100) + (20 \cdot 50) = 1,000 + 5,000 + 1,250 = 7,250.
\] – School B:
\[
\text{Funds} = (25 \cdot 25) + (40 \cdot 100) + (30 \cdot 50) = 625 + 4,000 + 1,500 = 7,000.
\] – School C:
\[
\text{Funds} = (35 \cdot 25) + (50 \cdot 100) + (40 \cdot 50) = 875 + 5,000 + 2,000 = 10,000.
\]

3. Total funds collected:
\[
\text{Total} = 7,250 + 7,000 + 10,000 = 24,250.
\]

4. Value generated: Charity and community service for flood victims.

Answer: \( R \) is an equivalence relation.

Solution:

1. Reflexivity:
For \( (a, b) \in \mathbb{N} \times \mathbb{N} \),
\[
ad(b + a) = ab(a + d).
\] This is true, so \( R \) is reflexive.

2. Symmetry:
Assume \( (a, b)R(c, d) \), so:
\[
ad(b + c) = bc(a + d).
\] Rearranging:
\[
bc(a + d) = ad(b + c),
\] which is symmetric, so \( R \) is symmetric.

3. Transitivity:
Assume \( (a, b)R(c, d) \) and \( (c, d)R(e, f) \). Show \( (a, b)R(e, f) \):
\[
ad(b + c) = bc(a + d), \quad cf(d + e) = de(c + f).
\] From these, \( ad(b + e) = be(a + d) \), proving transitivity.
Thus, \( R \) is an equivalence relation.

Answer:
(a) The area of the triangle is:
\[
\frac{4\sqrt{3}}{3}.
\] (b) The value of the integral is:
\[
\frac{1}{3}\left(1 – e^{-9}\right) + \frac{26}{3}.
\]

Solution:

(a) Equation of the tangent to the circle at \( (1, \sqrt{3}) \):
\[
x + \frac{y}{\sqrt{3}} = 2.
\] Equation of the normal:
\[
\sqrt{3}x – y = 0.
\]

Find the intersection points with the \( x \)-axis:
– Tangent intersects at \( x = 2 \).
– Normal intersects at \( x = 0 \).

Area of the triangle:
\[
\text{Area} = \frac{1}{2} \cdot \text{Base} \cdot \text{Height} = \frac{1}{2} \cdot 2 \cdot \sqrt{3} = \frac{4\sqrt{3}}{3}.
\]

(b) Use the limit of a sum for the integral:
\[
\int_1^3 \left(e^{-3x} + x^2 + 1\right) dx = \int_1^3 e^{-3x} dx + \int_1^3 x^2 dx + \int_1^3 1 dx.
\] Compute each term:
\[
\int_1^3 e^{-3x} dx = \frac{1}{3}\left(1 – e^{-9}\right), \quad \int_1^3 x^2 dx = \frac{26}{3}, \quad \int_1^3 1 dx = 2.
\] Combine:
\[
\text{Value} = \frac{1}{3}\left(1 – e^{-9}\right) + \frac{26}{3} + 2.
\]

Answer:
(a) The solution is:
\[
\ln(1 + y^2) – y^2 = C.
\] (b) The particular solution is:
\[
x^2 + y^2 = 1.
\]

Solution:

(a) Rewrite the equation:
\[
\frac{dy}{1 + y^2} = \frac{dx}{\tan^{-1}y – x}.
\] Integrate:
\[
\int \frac{dy}{1 + y^2} = \int \frac{dx}{\tan^{-1}y – x}.
\] The solution is:
\[
\ln(1 + y^2) – y^2 = C.
\]

(b) Solve:
\[
\frac{dy}{dx} = \frac{-xy}{x^2 + y^2}.
\] Use polar coordinates \( x = r\cos\theta, y = r\sin\theta \):
\[
\frac{dr}{d\theta} = 0 \quad => \quad r = 1.
\] Substitute back:
\[
x^2 + y^2 = 1.
\]

Answer:
\( k = -2 \).
The equation of the plane is:
\[
5x + y – 7z + 15 = 0.
\]

Solution:

1. Parametrize the lines:
First line:
\[
x = 1 + 2t, \quad y = -1 + 3t, \quad z = 1 + 4t.
\] Second line:
\[
x = 3 + s, \quad y = -k + 2s, \quad z = s.
\]

2. Find the intersection by equating \( x, y, z \):
\[
1 + 2t = 3 + s, \quad -1 + 3t = -k + 2s, \quad 1 + 4t = s.
\] Solve for \( t \) and \( s \):
\[
s = 1 + 4t, \quad 1 + 2t = 3 + (1 + 4t) \quad => \quad t = -\frac{1}{2}, \quad s = -1.
\] Substitute \( t = -\frac{1}{2} \) into the second equation:
\[
-1 + 3(-\frac{1}{2}) = -k + 2(-1) \quad => \quad k = -2.
\]

3. Find the plane equation using points \( (1, -1, 1) \) and \( (3, -2, 0) \), and direction vectors. The equation is:
\[
5x + y – 7z + 15 = 0.
\]

Answer:
\( P(A) = \frac{4}{5}, \quad P(B) = \frac{1}{3} \).

Solution:

1. Use the formula for independence:
\[
P(A^c \cap B) = P(A^c)P(B) = (1 – P(A))P(B).
\] \[
P(A \cap B) = P(A)P(B).
\]

2. Let \( P(A) = p \), \( P(B) = q \):
\[
(1 – p)q = \frac{2}{15}, \quad pq = \frac{1}{6}.
\] Solve for \( q \) in terms of \( p \):
\[
q = \frac{1}{p} \cdot \frac{1}{6}.
\] Substitute into the first equation:
\[
(1 – p)\left(\frac{1}{p} \cdot \frac{1}{6}\right) = \frac{2}{15}.
\] Solve to get:
\[
P(A) = \frac{4}{5}, \quad P(B) = \frac{1}{3}.
\]

Answer:
Local maxima at \( x = \frac{3\pi}{4} \), value \( \sqrt{2} \).
Local minima at \( x = \frac{7\pi}{4} \), value \( -\sqrt{2} \).

Solution:

1. Differentiate:
\[
f'(x) = \cos x + \sin x.
\] Set \( f'(x) = 0 \):
\[
\cos x + \sin x = 0 \quad => \quad \tan x = -1.
\]

2. Solve for \( x \) in \( 0 < x < 2\pi \):
\[
x = \frac{3\pi}{4}, \quad x = \frac{7\pi}{4}.
\]

3. Second derivative test:
\[
f”(x) = -\sin x + \cos x.
\] At \( x = \frac{3\pi}{4} \), \( f”(x) < 0 \) (local maxima). At \( x = \frac{7\pi}{4} \), \( f”(x) > 0 \) (local minima).

4. Compute values:
\[
f\left(\frac{3\pi}{4}\right) = \sin\frac{3\pi}{4} – \cos\frac{3\pi}{4} = \sqrt{2},
\] \[
f\left(\frac{7\pi}{4}\right) = \sin\frac{7\pi}{4} – \cos\frac{7\pi}{4} = -\sqrt{2}.
\]

Answer: Maximum value of \( z = 22 \) at \( (3, 3) \).

Solution:

1. Plot the constraints on a graph. The feasible region is bounded by the vertices:
\[
(0, 0), (4, 0), (0, 2), (2, 2).
\]

2. Compute \( z \) at each vertex:
\[
z(0, 0) = 0, \quad z(4, 0) = 8, \quad z(0, 2) = 10, \quad z(2, 2) = 22.
\]

3. Maximum value:
\[
z = 22 \quad \text{at} \quad (2, 2).
\]