CBSE Class 12 Maths Most Repeated Questions with Solutions

Matrix Representation AX = B [½ Mark]

Write the system as AX = B , where: A = \begin{pmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 5 \\ -4 \\ 3 \end{pmatrix}

Evaluate Determinant |A| [1 Mark]

|A| = 2(4 - (-1)) - 3(-2 - 3) + 3(-1 - (-6)) = 2(5) - 3(-5) + 3(5) = 10 + 15 + 15 = 40 \neq 0 . Since |A| \neq 0 , A is non-singular and A^{-1} exists uniquely.

Cofactor Matrix & Adjoint [1½ Marks]

Cofactors of matrix A : C_{11} = +5, \quad C_{12} = -(-5) = 5, \quad C_{13} = +5 C_{21} = -(-3) = 3, \quad C_{22} = -13, \quad C_{23} = -(-11) = 11 C_{31} = +9, \quad C_{32} = -(2-3) = 1, \quad C_{33} = -7 \text{adj } A = \begin{pmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{pmatrix}

Compute X = A⁻¹B [2 Marks]

X = A^{-1}B = \frac{1}{40} \begin{pmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{pmatrix} \begin{pmatrix} 5 \\ -4 \\ 3 \end{pmatrix} x = \frac{1}{40}[5(5) + 3(-4) + 9(3)] = \frac{1}{40}[25 - 12 + 27] = \frac{40}{40} = 1 y = \frac{1}{40}[5(5) - 13(-4) + 1(3)] = \frac{1}{40}[25 + 52 + 3] = \frac{80}{40} = 2 z = \frac{1}{40}[5(5) + 11(-4) - 7(3)] = \frac{1}{40}[25 - 44 - 21] = \frac{-40}{40} = -1 Therefore: x = 1, y = 2, z = -1 .

Define Integral & Apply King's Property [1½ Marks]

Let I = \int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx ... (1) Using the property \int_0^a f(x) dx = \int_0^a f(a-x) dx : I = \int_0^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} dx Since \sin(\pi - x) = \sin x and \cos(\pi - x) = -\cos x \implies \cos^2(\pi - x) = \cos^2 x : I = \int_0^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} dx ... (2)

Add Equations to Eliminate x [1 Mark]

Adding (1) and (2): 2I = \int_0^{\pi} \frac{[x + (\pi - x)] \sin x}{1 + \cos^2 x} dx = \pi \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} dx

Substitution and Limit Conversion [1½ Marks]

Put \cos x = t \implies -\sin x dx = dt \implies \sin x dx = -dt . When x = 0 , t = \cos 0 = 1 . When x = \pi , t = \cos \pi = -1 . 2I = \pi \int_1^{-1} \frac{-dt}{1 + t^2} = \pi \int_{-1}^1 \frac{dt}{1 + t^2}

Integration and Final Value [1 Mark]

2I = \pi [\tan^{-1} t]_{-1}^1 = \pi [\tan^{-1}(1) - \tan^{-1}(-1)] = \pi \left[\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right] = \pi \left(\frac{\pi}{2}\right) = \frac{\pi^2}{2} Therefore: I = \frac{\pi^2}{4} .

Identify Position Vectors and Direction Vectors [1 Mark]

Comparing with \vec{r} = \vec{a}_1 + \lambda \vec{b}_1 and \vec{r} = \vec{a}_2 + \mu \vec{b}_2 : \vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b}_1 = \hat{i} - 3\hat{j} + 2\hat{k} \vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k}, \quad \vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k} \vec{a}_2 - \vec{a}_1 = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}

Cross Product b₁ × b₂ [1½ Marks]

\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 - (-6)) = -9\hat{i} + 3\hat{j} + 9\hat{k} |\vec{b}_1 \times \vec{b}_2| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19}

Scalar Product (a₂ - a₁) · (b₁ × b₂) [1 Mark]

(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-9\hat{i} + 3\hat{j} + 9\hat{k}) = 3(-9) + 3(3) + 3(9) = -27 + 9 + 27 = 9

Shortest Distance Formula and Evaluation [1½ Marks]

Shortest distance formula: d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \frac{|9|}{3\sqrt{19}} = \frac{3}{\sqrt{19}} = \frac{3\sqrt{19}}{19}\text{ units} .

Define Events & Prior Probabilities [1 Mark]

Let E_1 = Event of choosing Bag I, E_2 = Event of choosing Bag II, A = Event of drawing a red ball. Since the bags are chosen at random: P(E_1) = \frac{1}{2}, \quad P(E_2) = \frac{1}{2}

Calculate Conditional Probabilities [1 Mark]

Bag I has 3 red and 4 black (Total = 7 balls): P(A|E_1) = \frac{3}{7} Bag II has 5 red and 6 black (Total = 11 balls): P(A|E_2) = \frac{5}{11}

State Bayes' Theorem Formula Explicitly [1 Mark]

By Bayes' Theorem, the probability that the red ball came from Bag II is: P(E_2|A) = \frac{P(E_2) \cdot P(A|E_2)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}

Substitute and Evaluate [2 Marks]

P(E_2|A) = \frac{\frac{1}{2} \times \frac{5}{11}}{\left(\frac{1}{2} \times \frac{3}{7}\right) + \left(\frac{1}{2} \times \frac{5}{11}\right)} Cancelling \frac{1}{2} from numerator and denominator: P(E_2|A) = \frac{\frac{5}{11}}{\frac{3}{7} + \frac{5}{11}} = \frac{\frac{5}{11}}{\frac{33 + 35}{77}} = \frac{5}{11} \times \frac{77}{68} = \frac{5 \times 7}{68} = \frac{35}{68} .

Geometric Setup & Relation [1 Mark]

Let r be the radius and h be the height of the cylinder inscribed in a sphere of radius R . From the right triangle formed by the center, the base radius, and half-height \frac{h}{2} : r^2 + \left(\frac{h}{2}\right)^2 = R^2 \implies r^2 = R^2 - \frac{h^2}{4}

Express Volume as Single Variable Function [1 Mark]

Volume of cylinder V = \pi r^2 h . Substitute r^2 : V(h) = \pi \left(R^2 - \frac{h^2}{4}\right)h = \pi \left(R^2 h - \frac{h^3}{4}\right)

First Derivative & Critical Point [1½ Marks]

Differentiating with respect to h : \frac{dV}{dh} = \pi \left(R^2 - \frac{3h^2}{4}\right) For maximum volume, \frac{dV}{dh} = 0 \implies R^2 - \frac{3h^2}{4} = 0 \implies 3h^2 = 4R^2 \implies h = \frac{2R}{\sqrt{3}} .

Second Derivative Test & Maximum Volume [1½ Marks]

Second derivative: \frac{d^2V}{dh^2} = \pi \left(0 - \frac{6h}{4}\right) = -\frac{3\pi h}{2} At h = \frac{2R}{\sqrt{3}} , \frac{d^2V}{dh^2} = -\frac{3\pi(2R)}{2\sqrt{3}} = -\sqrt{3}\pi R < 0 . By Second Derivative Test, V is maximum at h = \frac{2R}{\sqrt{3}} . Maximum volume: V_{\max} = \pi \left[R^2 \left(\frac{2R}{\sqrt{3}}\right) - \frac{1}{4}\left(\frac{2R}{\sqrt{3}}\right)^3\right] = \pi \left[\frac{2R^3}{\sqrt{3}} - \frac{2R^3}{3\sqrt{3}}\right] = \frac{4\pi R^3}{3\sqrt{3}} .