Class 12 Board MCQ Solutions: 20 Questions

This lesson works through twenty one-mark MCQ questions from Class 12 board papers, writing each answer first and then the supporting steps. The questions cover inverse trigonometry, matrices and determinants, vectors, and differential equations.

Combining inverse trigonometric values

Evaluate $3\sin^{-1}\tfrac{1}{2} + 2\cos^{-1}\tfrac{1}{2}$.

Since $\sin^{-1}\tfrac{1}{2} = \tfrac{\pi}{6}$ and $\cos^{-1}\tfrac{1}{2} = \tfrac{\pi}{3}$,

$3\cdot\tfrac{\pi}{6} + 2\cdot\tfrac{\pi}{3} = \tfrac{\pi}{2} + \tfrac{2\pi}{3} = \tfrac{3\pi + 4\pi}{6} = \tfrac{7\pi}{6}.$

A nested inverse trig expression

Evaluate $\tan^{-1}\!\big(2\cos(2\sin^{-1}\tfrac{1}{2})\big)$.

Here $\sin^{-1}\tfrac{1}{2} = \tfrac{\pi}{6}$, so $2\sin^{-1}\tfrac{1}{2} = \tfrac{\pi}{3}$ and $\cos\tfrac{\pi}{3} = \tfrac{1}{2}$.

$\tan^{-1}\!\big(2\cdot\tfrac{1}{2}\big) = \tan^{-1}(1) = \tfrac{\pi}{4}.$

Identity matrix times a column

If $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$, find $x + y + z$.

The matrix is the identity $I$, so $I\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$. Hence $x = 1$, $y = -1$, $z = 0$, and

$x + y + z = 1 + (-1) + 0 = 0.$

Determinant of the adjoint

If $A$ is a square matrix of order $3$ with $\det(\operatorname{adj} A) = 225$, find $\det A$.

Using $\det(\operatorname{adj} A) = (\det A)^{n-1}$ with $n = 3$,

$225 = (\det A)^{2} \implies \det A = \sqrt{225} = 15.$

Inverse of a rotation matrix

Find the inverse of $A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$.

Determinant.

$\det A = \cos^2\theta - (-\sin^2\theta) = \cos^2\theta + \sin^2\theta = 1.$

Adjoint. For a $2\times 2$ matrix, swap the diagonal entries and change the sign of the off-diagonal entries:

$\operatorname{adj} A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}.$

Inverse. Since $A^{-1} = \dfrac{\operatorname{adj} A}{\det A}$ and $\det A = 1$,

$A^{-1} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}.$

Degree of a differential equation

For $\left(\dfrac{d^2y}{dx^2}\right)^2 - \dfrac{dy}{dx} + 1 = 0$, the degree is the power of the highest order derivative. The highest order term $\dfrac{d^2y}{dx^2}$ appears to the power $2$, so the degree is $2$.

Angle from a cross product

Given $|\vec a| = 3$, $|\vec b| = \tfrac{2}{3}$, and $\vec a \times \vec b$ is a unit vector, find the angle between them.

$\sin\theta = \frac{|\vec a \times \vec b|}{|\vec a|\,|\vec b|} = \frac{1}{3 \cdot \tfrac{2}{3}} = \frac{1}{2} \implies \theta = \frac{\pi}{6}.$

Projection of one vector on another

For $\vec a = 7\hat\imath + \hat\jmath - 4\hat k$ and $\vec b = 2\hat\imath + 6\hat\jmath + 3\hat k$, find the projection of $\vec a$ on $\vec b$.

$\vec a \cdot \vec b = 7(2) + 1(6) + (-4)(3) = 14 + 6 - 12 = 8.$

$|\vec b| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{49} = 7.$

$\text{projection} = \frac{\vec a \cdot \vec b}{|\vec b|} = \frac{8}{7}.$

A principal value of inverse cosine

Find $\cos^{-1}\!\left(-\tfrac{\sqrt 3}{2}\right)$.

The value is negative, so the angle lies in the second quadrant. Writing $-\tfrac{\sqrt 3}{2} = \cos\!\left(\pi - \tfrac{\pi}{6}\right)$,

$\cos^{-1}\!\left(-\tfrac{\sqrt 3}{2}\right) = \pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}.$

Solving a determinant equation

Find $x$ if $\begin{vmatrix} 3 & 4 \\ -5 & 2 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ -5 & 3 \end{vmatrix}$.

$3(2) - 4(-5) = 3(2x) - 4(-5)$

$6 + 20 = 6x + 20 \implies 6 = 6x \implies x = 1.$

Another adjoint determinant

If $A$ is of order $3$ with $\det(\operatorname{adj} A) = 64$, find $\det A'$.

$64 = (\det A)^{2} \implies \det A = \pm 8.$

Since $\det A' = \det A$ for the transpose, $\det A' = \pm 8$.

Inverse when $A^2 = I$

If $A^2 = I$, find $A^{-1}$. Pre-multiplying $A^2 = I$ by $A^{-1}$ gives $A^{-1}A^2 = A^{-1}I$, so $A = A^{-1}$.

Degree of a second differential equation

For $y\,\dfrac{d^2y}{dx^2} + \left(\dfrac{dy}{dx}\right)^2 = x\left(\dfrac{d^2y}{dx^2}\right)^2$, the highest order derivative is $\dfrac{d^2y}{dx^2}$ and its highest power is $2$, so the degree is $2$.

Area of a parallelogram

If $\vec a$ and $\vec b$ are two adjacent sides of a parallelogram, its area is $|\vec a \times \vec b|$.

Angle from a dot product

Given $|\vec a| = 3$, $|\vec b| = 4$, and $\vec a \cdot \vec b = 6$,

$\cos\theta = \frac{\vec a \cdot \vec b}{|\vec a|\,|\vec b|} = \frac{6}{3 \cdot 4} = \frac{1}{2} \implies \theta = \frac{\pi}{3}.$

Direction cosines making equal angles

A line through the origin in the first quadrant makes equal angles with the three axes. With $l^2 + m^2 + n^2 = 1$ and equal angles, $3\cos^2\alpha = 1$, so $\cos\alpha = \tfrac{1}{\sqrt 3}$ (positive, since the line is in the first quadrant). The direction cosines are $\left(\tfrac{1}{\sqrt 3}, \tfrac{1}{\sqrt 3}, \tfrac{1}{\sqrt 3}\right)$.

Perpendicular vectors

Find $\lambda$ so that $\vec a = 2\hat\imath + \lambda\hat\jmath + \hat k$ and $\vec b = \hat\imath - 2\hat\jmath + 3\hat k$ are perpendicular. Then $\vec a \cdot \vec b = 0$:

$2(1) + \lambda(-2) + 1(3) = 0 \implies 5 - 2\lambda = 0 \implies \lambda = \frac{5}{2}.$

Order and degree

For $\left(\dfrac{dy}{dx}\right)^2 + 3y - \dfrac{d^2y}{dx^2} = 0$, the highest order derivative is $\dfrac{d^2y}{dx^2}$ with power $1$, so the order is $2$ and the degree is $1$.

Equating matrices

If $2\begin{bmatrix} 1 & 2 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$, find $x$ and $y$. Comparing entries, $2 + y = 5$ gives $y = 3$, and $2x + 2 = 8$ gives $x = 3$.

Key takeaways

• For inverse trig MCQs, convert to principal-range angles first, then simplify.
• For order $n$, $\det(\operatorname{adj} A) = (\det A)^{n-1}$, and $\det A' = \det A$.
• The order is the highest derivative present; the degree is the power of that highest order derivative once the equation is polynomial in the derivatives.