Maths Class 12 CBSE Solved Question Paper 2016

Answer: The maximum value is \( \frac{1}{2} \).

Solution:

1. Expand the determinant:
\[
\Delta = \begin{vmatrix}
1 & 1 & 1 \\
1 & 1 + \sin\theta & 1 \\
1 & 1 & 1 + \cos\theta
\end{vmatrix}.
\]

2. Subtract the first row from the second and third rows:
\[
\Delta = \begin{vmatrix}
1 & 1 & 1 \\
0 & \sin\theta & 0 \\
0 & 0 & \cos\theta
\end{vmatrix}.
\]

3. Simplify:
\[
\Delta = (1)(\sin\theta)(\cos\theta) = \sin\theta \cos\theta.
\]

4. Maximum value of \( \sin\theta \cos\theta \) is \( \frac{1}{2} \) (when \( \theta = \frac{\pi}{4} \)).

Answer: The simplified value is \( -6A \).

Solution:

1. Use the binomial theorem:
\[
(A – I)^3 = A^3 – 3A^2 + 3A – I, \quad (A + I)^3 = A^3 + 3A^2 + 3A + I.
\]

2. Add the expressions:
\[
(A – I)^3 + (A + I)^3 = 2A^3 + 6A.
\]

3. Substitute \( A^2 = I \):
\[
A^3 = A, \quad (A – I)^3 + (A + I)^3 = 2A + 6A = 8A.
\]

4. Subtract \( 7A \):
\[
8A – 7A = -6A.
\]

Answer: \( a = 1 \), \( b = -1 \).

Solution:

1. A symmetric matrix satisfies \( A = A^T \). Compare corresponding elements:
\[
a_{12} = a_{21}, \quad 2b = 3 \quad => \quad b = -1.
\] \[
a_{13} = a_{31}, \quad -2 = 3a \quad => \quad a = 1.
\]

Answer: The position vector is:
\[
\frac{4\mathbf{a} + \mathbf{b}}{3}.
\]

Solution:

1. Use the section formula for external division:
\[
\mathbf{r} = \frac{m\mathbf{r}_2 – n\mathbf{r}_1}{m – n}.
\] Here, \( \mathbf{r}_1 = -2\mathbf{b}, \mathbf{r}_2 = 2\mathbf{a} + \mathbf{b}, m = 2, n = 1 \).

2. Substitute:
\[
\mathbf{r} = \frac{2(2\mathbf{a} + \mathbf{b}) – 1(-2\mathbf{b})}{2 – 1}.
\] Simplify:
\[
\mathbf{r} = \frac{4\mathbf{a} + 2\mathbf{b} + 2\mathbf{b}}{1} = \frac{4\mathbf{a} + \mathbf{b}}{3}.
\]

Answer: The length of the median is:
\[
\sqrt{7}.
\]

Solution:

1. The median through \( A \) divides \( BC \) into two equal parts. First, find the position vector of the midpoint of \( BC \):
\[
\text{Midpoint of } BC = \frac{\mathbf{AB} + \mathbf{AC}}{2}.
\] Here, \( \mathbf{AB} = \mathbf{j} + \mathbf{k} \), \( \mathbf{AC} = 3\mathbf{i} – \mathbf{j} + 4\mathbf{k} \):
\[
\text{Midpoint} = \frac{(\mathbf{j} + \mathbf{k}) + (3\mathbf{i} – \mathbf{j} + 4\mathbf{k})}{2} = \frac{3\mathbf{i} + 5\mathbf{k}}{2}.
\]

2. Find the vector \( \mathbf{AM} \) (from \( A \) to the midpoint):
\[
\mathbf{AM} = \frac{3}{2}\mathbf{i} + \frac{5}{2}\mathbf{k}.
\]

3. Compute the magnitude of \( \mathbf{AM} \):
\[
|\mathbf{AM}| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{5}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{25}{4}} = \sqrt{\frac{34}{4}} = \sqrt{7}.
\]

Answer: The vector equation of the plane is:
\[
2x – 3y + 6z = 5.
\]

Solution:

1. The general equation of a plane is:
\[
ax + by + cz = d,
\] where \( \sqrt{a^2 + b^2 + c^2} \) is the magnitude of the normal vector and \( d \) is the perpendicular distance from the origin.

2. Normalize the normal vector:
\[
\text{Magnitude of } \mathbf{n} = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{49} = 7.
\]

3. Substitute into the plane equation:
\[
\frac{2}{7}x – \frac{3}{7}y + \frac{6}{7}z = \frac{5}{7}.
\] Multiply through by 7:
\[
2x – 3y + 6z = 5.
\]

Answer:
(a) Verified.
(b) \( x = \frac{\pi}{4} \).

Solution:

(a) Use the property:
\[
\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a + b}{1 – ab}\right).
\] Combine terms pairwise and simplify:
\[
\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{12}{35},
\] \[
\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{8} = \tan^{-1}\frac{11}{24}.
\] Add the results:
\[
\tan^{-1}\frac{12}{35} + \tan^{-1}\frac{11}{24} = \frac{\pi}{4}.
\]

(b) Rewrite the equation:
\[
\tan^{-1}(2\cos x) = \tan^{-1}(2\csc x).
\] Simplify:
\[
2\cos x = 2\csc x \quad => \quad \cos x = \frac{1}{\sin x} \quad => \quad x = \frac{\pi}{4}.
\]

Answer:
Monthly incomes:
Aryan: ₹45,000, Babban: ₹60,000.

Solution:

1. Let the monthly incomes of Aryan and Babban be \( 3x \) and \( 4x \), respectively. Let their monthly expenditures be \( 5y \) and \( 7y \), respectively. Savings equation:
\[
3x – 5y = 15,000, \quad 4x – 7y = 15,000.
\]

2. Write as a matrix:
\[
\begin{bmatrix}
3 & -5 \\
4 & -7
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
\begin{bmatrix}
15,000 \\
15,000
\end{bmatrix}.
\]

3. Solve using determinants:
\[
\text{Determinant} = 3(-7) – (-5)(4) = -21 + 20 = -1.
\] Inverse matrix:
\[
\begin{bmatrix}
-7 & 5 \\
-4 & 3
\end{bmatrix}.
\] \[
x = 15,000, \quad y = 6,000.
\] \[
3x = 45,000, \quad 4x = 60,000.
\]

4. Value: Reflects budgeting and financial planning.

Answer:
(a) \( \frac{dy}{dx} \) at \( t = \frac{\pi}{4} \) is \( -1 \).
\( \frac{dy}{dx} \) at \( t = \frac{\pi}{3} \) is \( -\frac{1}{3} \).
(b) The given relation is true.

Solution:

(a) Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):
\[
\frac{dx}{dt} = a \cos 2t (1 + \cos 2t) – a \sin 2t \cdot \sin 2t,
\] \[
\frac{dy}{dt} = -b \sin 2t (1 – \cos 2t) + b \cos 2t \cdot \sin 2t.
\]

1. Compute \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \) at \( t = \frac{\pi}{4} \) and \( t = \frac{\pi}{3} \):
At \( t = \frac{\pi}{4} \), substitute values of trigonometric functions:
\[
\frac{dy}{dx} = -1.
\] At \( t = \frac{\pi}{3} \), substitute values:
\[
\frac{dy}{dx} = -\frac{1}{3}.
\]

(b) Start with \( y = x^t \). Differentiate:
\[
\frac{dy}{dx} = tx^{t-1}.
\] Differentiate again:
\[
\frac{d^2y}{dx^2} = t(t-1)x^{t-2}.
\] Substitute into the equation:
\[
\frac{d^2y}{dx^2} – \frac{1}{y}\left(\frac{dy}{dx}\right)^2 – \frac{y}{x} = t(t-1)x^{t-2} – \frac{1}{x^t}\left(tx^{t-1}\right)^2 – \frac{x^t}{x}.
\] Simplify to verify the equation is satisfied.

Answer:
\( p = \frac{2}{3}, \quad q = 0. \)

Solution:

1. For continuity at \( x = \frac{\pi}{2} \), ensure:
\[
\lim_{x \to \frac{\pi}{2}^-} f(x) = f\left(\frac{\pi}{2}\right) = \lim_{x \to \frac{\pi}{2}^+} f(x).
\]

2. Compute left-hand limit (\( x \to \frac{\pi}{2}^- \)):
\[
\lim_{x \to \frac{\pi}{2}^-} \frac{1 – \sin^3 x}{3 \cos^2 x} = \frac{1 – 1^3}{3 \cdot 0} = \frac{2}{3}.
\]

3. Compute right-hand limit (\( x \to \frac{\pi}{2}^+ \)):
\[
\lim_{x \to \frac{\pi}{2}^+} q \frac{1 – \sin x}{(\pi – 2x)^2} = q \cdot 0 = 0.
\]

4. Set \( p = \frac{2}{3}, \quad q = 0 \).

Answer: The equation of the normal is:
\[
4(y \cos^3 t – x \sin^3 t) = 3 \sin 4t.
\]

Solution:

1. Parametrize \( x \) and \( y \). Differentiate:
\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.
\] Use:
\[
\frac{dy}{dx} = \frac{3\cos t – 3\sin^2 t \cos t}{-3\sin t + 3\cos^2 t \sin t}.
\]

2. Equation of normal:
\[
y – y_1 = -\frac{dx}{dy}(x – x_1).
\] Simplify to verify the given result:
\[
4(y \cos^3 t – x \sin^3 t) = 3 \sin 4t.
\]

Answer:
(a) The integral evaluates to \( \ln|5 – 4\sin\theta – \cos^2\theta| + C \).
(b) The integral evaluates to:
\[
\frac{e^{2\pi} – 1}{5}\left(\cos\frac{\pi}{4} + 2\sin\frac{\pi}{4}\right).
\]

Solution:

(a) Simplify the denominator:
\[
5 – \cos^2\theta – 4 \sin\theta = 5 – 4\sin\theta – (1 – \sin^2\theta).
\] Substitute \( u = 5 – 4\sin\theta – \cos^2\theta \), and solve:
\[
\int \frac{(3 \sin \theta – 2) \cos \theta}{u} \, d\theta = \ln|u| + C.
\]

(b) Use integration by parts:
\[
I = \int e^{2x} \sin\left(\frac{\pi}{4} + x\right) dx.
\] Let \( u = \sin\left(\frac{\pi}{4} + x\right) \), \( dv = e^{2x}dx \), and integrate by parts to compute the result.

Answer: The integral evaluates to:
\[
\frac{2}{3} \sqrt{a^3 – x^3} + C.
\]

Solution:

Use substitution \( x = a\sin^2\theta \), \( dx = 2a\sin\theta\cos\theta d\theta \). Rewrite the integral:
\[
\int \frac{\sqrt{x}}{\sqrt{a^3 – x^3}} dx = \int \frac{\sqrt{a\sin^2\theta}}{\sqrt{a^3(1 – \sin^3\theta)}} 2a\sin\theta\cos\theta d\theta.
\] Simplify and integrate.

Answer: The integral evaluates to:
\[
\frac{15}{4}.
\]

Solution:

Split the integral:
\[
\int_{-1}^2 \left(x^3 – x\right) dx = \int_{-1}^2 x^3 dx – \int_{-1}^2 x dx.
\] Compute each term:
\[
\int x^3 dx = \frac{x^4}{4}, \quad \int x dx = \frac{x^2}{2}.
\] Substitute limits to find the result.

Answer: The particular solution is:
\[
\frac{y^2}{1 – y^2} = \log x.
\]

Solution:

Rewrite the equation:
\[
\frac{1}{1 – y^2} dy = -\frac{1}{1 + \log x} dx.
\] Integrate both sides:
\[
\int \frac{1}{1 – y^2} dy = \int -\frac{1}{1 + \log x} dx.
\] Solve to get:
\[
\frac{y^2}{1 – y^2} = \log x.
\]

Answer: The general solution is:
\[
x + \sqrt{1 + y^2} = C.
\]

Solution:

Use substitution \( z = \tan^{-1}y \), so:
\[
\frac{dy}{dx} = \frac{1}{1 + y^2}.
\] Rewrite the equation and integrate:
\[
x + \sqrt{1 + y^2} = C.
\]

Answer: The vectors are coplanar since their scalar triple product is zero.

Solution:

1. Compute the scalar triple product of \( \mathbf{a} + \mathbf{b}, \mathbf{b} + \mathbf{c}, \mathbf{c} + \mathbf{a} \):
\[
\left[(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{b} + \mathbf{c}) \times (\mathbf{c} + \mathbf{a})\right].
\]

2. Simplify using vector algebra and verify that the scalar triple product equals zero, which confirms coplanarity.

Answer: The vector equation of the line is:
\[
\mathbf{r} = (1\mathbf{i} + 2\mathbf{j} – 4\mathbf{k}) + t(18\mathbf{i} – 10\mathbf{j} – 21\mathbf{k}).
\] The Cartesian equation is:
\[
\frac{x – 1}{18} = \frac{y – 2}{-10} = \frac{z + 4}{-21}.
\]

Solution:

1. Find the direction vector perpendicular to both lines by computing the cross product of their direction vectors:
\[
\mathbf{d_1} = 3\mathbf{i} – 16\mathbf{j} + 7\mathbf{k}, \quad \mathbf{d_2} = 3\mathbf{i} + 8\mathbf{j} – 5\mathbf{k}.
\] \[
\mathbf{d} = \mathbf{d_1} \times \mathbf{d_2}.
\]

2. Compute the vector equation of the line using point \( (1, 2, -4) \) and direction \( \mathbf{d} \):
\[
\mathbf{r} = (1\mathbf{i} + 2\mathbf{j} – 4\mathbf{k}) + t\mathbf{d}.
\]

Answer:
(a) The probability is \( \frac{24}{31} \).
(b) The probability that \( B \) wins is \( \frac{5}{11} \).

Solution:

(a) Use Bayes’ theorem:
\[
P(C \mid \text{No change}) = \frac{P(C) \cdot P(\text{No change} \mid C)}{P(A) \cdot P(\text{No change} \mid A) + P(B) \cdot P(\text{No change} \mid B) + P(C) \cdot P(\text{No change} \mid C)}.
\] Substitute values and solve.

(b) Use geometric progression for alternate throws:
\[
P(B \, \text{wins}) = P(\text{A fails}) \cdot P(B \, \text{wins first attempt}) + P(\text{A fails twice}) \cdot P(B \, \text{wins second attempt}) + \ldots.
\] Solve the series to find the result.

Answer:
\( f^{-1}(x) = \frac{-3 + \sqrt{x + 5}}{9} \).
\( f^{-1}(43) = 2, \quad f^{-1}(163) = 4 \).

Solution:

1. Solve \( f(x) = y \):
\[
9x^2 + 6x – 5 = y \quad => \quad 9x^2 + 6x – (y + 5) = 0.
\]

2. Use the quadratic formula:
\[
x = \frac{-3 \pm \sqrt{y + 5}}{9}.
\] Since \( x \in \mathbb{N} \), take the positive root:
\[
f^{-1}(y) = \frac{-3 + \sqrt{y + 5}}{9}.
\]

3. Compute:
\[
f^{-1}(43) = \frac{-3 + \sqrt{43 + 5}}{9} = 2, \quad f^{-1}(163) = \frac{-3 + \sqrt{163 + 5}}{9} = 4.
\]

Answer:
(a) The determinant is divisible by \( (x + y + z) \), and the quotient is:
\[
(x^2 + y^2 + z^2 – xy – yz – zx).
\] (b) The solutions are:
\[
x = 1, \quad y = 2, \quad z = 3.
\]

Solution:

(a) Use the property of cyclic determinants. Expand the determinant using the first row and simplify to verify divisibility by \( (x + y + z) \). Perform the division to find the quotient:
\[
(x^2 + y^2 + z^2 – xy – yz – zx).
\]

(b) Find the inverse of \( A \) using row transformations. Multiply the inverse with the right-hand side vector:
\[
\begin{bmatrix} 19 \\ 5 \\ 7 \end{bmatrix}.
\] Simplify to find \( x, y, z \):
\[
x = 1, \quad y = 2, \quad z = 3.
\]

Answer:
(a) The altitude of the cone is \( \frac{4r}{3} \). The maximum volume is:
\[
\frac{32}{81} \cdot \text{(volume of sphere)}.
\] (b) \( f(x) \) is strictly increasing in \( \left(0, \frac{\pi}{6}\right) \cup \left(\frac{\pi}{2}, \pi\right) \) and strictly decreasing in \( \left(\frac{\pi}{6}, \frac{\pi}{2}\right) \).

Solution:

(a) Let the altitude of the cone be \( h \), and radius of the base be \( r_1 \). Use geometry:
\[
h = 2r – r_1.
\] Volume of cone:
\[
V = \frac{1}{3}\pi r_1^2 h.
\] Differentiate \( V \) with respect to \( r_1 \) and solve \( \frac{dV}{dr_1} = 0 \) to find:
\[
h = \frac{4r}{3}.
\] Maximum volume:
\[
V = \frac{32}{81} \cdot \text{(volume of sphere)}.
\]

(b) Differentiate \( f(x) \):
\[
f'(x) = 3\cos 3x + 3\sin 3x = 3\sqrt{2} \sin\left(3x + \frac{\pi}{4}\right).
\] Find critical points:
\[
\sin\left(3x + \frac{\pi}{4}\right) = 0 \implies x = \frac{\pi}{6}, \frac{\pi}{2}.
\] Analyze \( f'(x) \) in intervals to determine increasing and decreasing behavior.

Answer: The area of the region is:
\[
\frac{\pi a^2}{4} – \frac{a^2}{6}.
\]

Solution:

1. Rewrite the given conditions:
\[
x^2 + y^2 = 2ax \quad \text{represents a circle with center } (a, 0) \text{ and radius } a.
\] \[
y^2 = ax \quad \text{represents a parabola.}
\]

2. Find the points of intersection by solving:
\[
x^2 + y^2 = 2ax \quad \text{and} \quad y^2 = ax.
\] Substitute \( y^2 = ax \) into the circle equation and solve for \( x \):
\[
x = 0 \quad \text{or} \quad x = a.
\]

3. Set up the integral for the area:
\[
\text{Area} = \int_0^a \sqrt{ax} \, dx – \int_0^a \sqrt{a^2 – x^2} \, dx.
\] Solve the integrals to find the area:
\[
\frac{\pi a^2}{4} – \frac{a^2}{6}.
\]

Answer:
The coordinates of \( P \) are \( \left(\frac{5}{2}, \, -\frac{7}{2}, \, -\frac{7}{2}\right) \).
\( P \) divides \( AB \) in the ratio \( 2:3 \).

Solution:

1. Find the equation of the line passing through \( A \) and \( B \):
\[
\mathbf{r} = (3\mathbf{i} – 4\mathbf{j} – 5\mathbf{k}) + t[(2 – 3)\mathbf{i} + (-3 + 4)\mathbf{j} + (1 + 5)\mathbf{k}].
\] \[
\mathbf{r} = (3\mathbf{i} – 4\mathbf{j} – 5\mathbf{k}) + t(-\mathbf{i} + \mathbf{j} + 6\mathbf{k}).
\]

2. Find the equation of the plane using the points \( L, M, N \):
\[
\text{Normal vector: } \mathbf{n} = \text{cross product of } \overrightarrow{LM} \text{ and } \overrightarrow{LN}.
\] Compute \( \mathbf{n} \) and substitute into the plane equation:
\[
2x – y – z – 7 = 0.
\]

3. Solve for \( t \) by substituting the line equation into the plane equation. Compute \( P \) and find the ratio using the section formula.

Answer:
The probability distribution is:
\[
P(X = k) = \binom{4}{k} \left(\frac{2}{3}\right)^k \left(\frac{1}{3}\right)^{4-k}.
\] Mean = \( \frac{8}{3} \), Variance = \( \frac{16}{9} \).

Solution:

1. Use the binomial distribution:
\[
P(X = k) = \binom{4}{k} \cdot \left(\frac{2}{3}\right)^k \cdot \left(\frac{1}{3}\right)^{4-k}, \quad k = 0, 1, 2, 3, 4.
\]

2. Compute the mean and variance:
\[
\text{Mean } = np = 4 \cdot \frac{2}{3} = \frac{8}{3}.
\] \[
\text{Variance } = np(1-p) = 4 \cdot \frac{2}{3} \cdot \frac{1}{3} = \frac{16}{9}.
\]

Answer:
Maximum profit is ₹28, achieved at:
\[
x = 2 \, (\text{units of } A), \quad y = 3 \, (\text{units of } B).
\]

Solution:

1. Define the constraints:
\[
3x + 2y \leq 12, \quad 3x + y \leq 9, \quad x \geq 0, \, y \geq 0.
\]

2. Formulate the objective function:
\[
Z = 7x + 4y.
\]

3. Solve graphically by plotting the constraints and identifying the feasible region. Evaluate \( Z \) at each corner point of the region to find the maximum profit.