Class 12 Board MCQ Solutions: Ten Previous-Paper Questions

This lesson works through ten objective questions from previous Class 12 board papers, covering inverse trigonometry, matrices, determinants, integration, vectors, linear programming, and probability.

Evaluating $\sin^{-1}\!\left(\cos\tfrac{3\pi}{5}\right)$

Write the cosine as a sine using $\cos\theta = \sin\!\left(\tfrac{\pi}{2} - \theta\right)$:

$\cos\tfrac{3\pi}{5} = \sin\!\left(\tfrac{\pi}{2} - \tfrac{3\pi}{5}\right) = \sin\!\left(\tfrac{5\pi - 6\pi}{10}\right) = \sin\!\left(-\tfrac{\pi}{10}\right).$

Since $-\tfrac{\pi}{10}$ lies in the principal range, $\sin^{-1}\!\left(\sin x\right) = x$ gives:

$\sin^{-1}\!\left(\cos\tfrac{3\pi}{5}\right) = -\tfrac{\pi}{10}.$

Combining matrix products $AB$ and $XY$

Let $A = \begin{bmatrix} 2 & -3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 3 \\ 2 \\ 2 \end{bmatrix}$, with $X = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$ and $Y = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}$.

Product $AB$. Multiplying the row by the column:

$AB = 2\cdot 3 + (-3)\cdot 2 + 4\cdot 2 = 6 - 6 + 8 = 8.$

Product $XY$. Similarly:

$XY = 1\cdot 2 + 2\cdot 3 + 3\cdot 4 = 2 + 6 + 12 = 20.$

Adding the two single-entry results:

$AB + XY = 8 + 20 = 28.$

Solving a determinant equation for $x$

We are given a determinant equation whose solution is $x = -1$. Expanding the determinant along the first row, collecting the terms in $x$, and setting the result to zero reduces to:

$3x = -3 \quad\Longrightarrow\quad x = -1.$

Definite integral $\displaystyle\int_0^{\pi/8} \tan^2 2x\,dx$

Use the identity $\tan^2\theta = \sec^2\theta - 1$:

$\int_0^{\pi/8} \tan^2 2x\,dx = \int_0^{\pi/8} \left(\sec^2 2x - 1\right) dx.$

Integrating each term, with $\displaystyle\int \sec^2 2x\,dx = \tfrac{1}{2}\tan 2x$:

$= \left[\tfrac{1}{2}\tan 2x - x\right]_0^{\pi/8}.$

At the upper limit $2x = \tfrac{\pi}{4}$, so $\tan\tfrac{\pi}{4} = 1$:

$= \left(\tfrac{1}{2}\cdot 1 - \tfrac{\pi}{8}\right) - (0 - 0) = \tfrac{1}{2} - \tfrac{\pi}{8}.$

Angle between two vectors

Given $\vec{a}\cdot\vec{b} = \tfrac{1}{2}\,|\vec{a}|\,|\vec{b}|$, use the dot-product formula:

$\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|} = \frac{\tfrac{1}{2}\,|\vec{a}|\,|\vec{b}|}{|\vec{a}|\,|\vec{b}|} = \tfrac{1}{2}.$

Therefore:

$\theta = \tfrac{\pi}{3} = 60^\circ.$

Linear programming: when the maximum repeats

If the objective function $z$ takes the same maximum value at two corner points of the feasible region, then the maximum is not attained only at those two points. Every point on the line segment joining them gives the same value of $z$, so the number of points where the maximum occurs is infinite.

Probability that $\tfrac{a}{b}$ is an integer

Two distinct numbers $a$ and $b$ ($a \ne b$) are chosen from the set $\{1, 2, 3, 4, 5\}$. The number of ordered pairs is:

$n(S) = 5 \times 4 = 20.$

The favourable pairs, where $\tfrac{a}{b}$ is a whole number, are $(2,1), (3,1), (4,1), (5,1), (4,2)$, giving $5$ outcomes. So:

$P\!\left(\tfrac{a}{b}\text{ is an integer}\right) = \frac{5}{20} = \tfrac{1}{4}.$

Probability of drawing two white balls

A bag contains $3$ white, $4$ black, and $2$ red balls, so $9$ in total. Two balls are drawn without replacement. The probability both are white is:

$\frac{{}^{3}C_2}{{}^{9}C_2} = \frac{3}{36} = \tfrac{1}{12}.$

Matrix equality and the product $xy$

Comparing corresponding entries gives the system:

$x + y = 2, \qquad x - y = 4.$

Adding the equations: $2x = 6$, so $x = 3$. Substituting back: $y = 2 - 3 = -1$. Therefore:

$x \cdot y = 3 \times (-1) = -3.$

Rate of change of a circle's area

The area of a circle is $A = \pi r^2$. Differentiating with respect to the radius:

$\frac{dA}{dr} = 2\pi r.$

At $r = 3$:

$\frac{dA}{dr} = 2\pi \cdot 3 = 6\pi \text{ cm}.$

Key takeaways

• Convert $\cos\theta$ to $\sin\!\left(\tfrac{\pi}{2} - \theta\right)$ to evaluate an inverse-sine cleanly.
• A row matrix times a column matrix gives a single number; here $AB + XY = 28$.
• For $\int \tan^2 2x\,dx$, replace $\tan^2$ with $\sec^2 - 1$ before integrating.
• In an LPP, equal maxima at two corners means infinitely many optimal points along the joining segment.