Mathematics — Class 12 practice questions

Worked examples for Mathematics — Class 12

1. Question 1

   If\ $\quad \left|\bar{a}\right| = 3, \quad \left|\bar{b}\right| = 5$\ and the angle between $\bar{a}$ and $\bar{b}$ is $\dfrac{\pi}{3}$, then ${\left|\bar{a} + \bar{b}\right| =}$

   1. A. $\sqrt{19}$

      This is $\left|\bar{a} - \bar{b}\right| = \sqrt{9 + 25 - 15}$. The cross term was subtracted when it should have been added. Expanding $(\bar{a} + \bar{b}) \cdot (\bar{a} + \bar{b})$ gives $+2\,\bar{a} \cdot \bar{b}$, and the minus sign belongs only to $\left|\bar{a} - \bar{b}\right|^{2}$.

   2. B. $49$

      This is $\left|\bar{a} + \bar{b}\right|^{2}$. The expansion was done correctly, but the square root was never taken at the end.

   3. C. $7$Correct answer

      $\left|\bar{a} + \bar{b}\right|^{2} = \left|\bar{a}\right|^{2} + \left|\bar{b}\right|^{2} + 2\left|\bar{a}\right|\left|\bar{b}\right|\cos\frac{\pi}{3} = 9 + 25 + 2 \times 3 \times 5 \times \frac{1}{2} = 49$, so $\left|\bar{a} + \bar{b}\right| = 7$.

   4. D. $\sqrt{34}$

      The magnitudes were combined as if $\bar{a}$ and $\bar{b}$ were perpendicular, which drops the $2\,\bar{a} \cdot \bar{b}$ term. That shortcut (Pythagoras) only works when the angle is $\frac{\pi}{2}$; here $\bar{a} \cdot \bar{b} = \frac{15}{2}$, which is not zero.

   Explanation

   The magnitude of a sum is found from its square, and the square is a scalar product of the vector with itself: $\left|\bar{a} + \bar{b}\right|^{2} = (\bar{a} + \bar{b}) \cdot (\bar{a} + \bar{b}) = \left|\bar{a}\right|^{2} + 2\,\bar{a} \cdot \bar{b} + \left|\bar{b}\right|^{2}$. The given angle supplies $\bar{a} \cdot \bar{b} = \left|\bar{a}\right|\left|\bar{b}\right|\cos\theta$, and the last step is to take the square root.

2. Question 2

   Consider the lines $\quad \dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{4}$ $\quad \dfrac{x - 2}{1} = \dfrac{y - k}{2} = \dfrac{z - 1}{1}$ If they are coplanar, then ${k =}$

   1. A. $\dfrac{11}{2}$Correct answer

      Here $\bar{a}_2 - \bar{a}_1 = (1, k - 2, -2)$ and $\bar{b}_1 \times \bar{b}_2 = (2, 3, 4) \times (1, 2, 1) = (-5, 2, 1)$. For coplanar lines the scalar triple product is zero: $-5 + 2(k - 2) - 2 = 2k - 11 = 0$, so $k = \frac{11}{2}$.

   2. B. $-\dfrac{3}{2}$

      In $\bar{b}_1 \times \bar{b}_2$ the sign of the $\hat{j}$ cofactor was not changed. The student got $-5\hat{i} - 2\hat{j} + \hat{k}$ instead of $-5\hat{i} + 2\hat{j} + \hat{k}$. The condition then reads $-5 - 2(k - 2) - 2 = 0$, which gives $k = -\frac{3}{2}$. The middle term of a $3 \times 3$ expansion always carries a minus sign.

   3. C. $4$

      The student made the joining vector perpendicular to the first line only: $(1, k - 2, -2) \cdot (2, 3, 4) = 3k - 12 = 0$. For coplanarity, the joining vector must be perpendicular to the common normal $\bar{b}_1 \times \bar{b}_2$, not to one of the direction vectors.

   4. D. $\dfrac{7}{2}$

      The student added the position vectors of the two given points, $(3, k + 2, 4)$, instead of subtracting them. The condition then reads $-15 + 2(k + 2) + 4 = 2k - 7 = 0$. The vector that lies in the common plane is the one joining the points, $\bar{a}_2 - \bar{a}_1$.

   Explanation

   Two non-parallel lines are coplanar exactly when the vector joining a point of one line to a point of the other is perpendicular to $\bar{b}_1 \times \bar{b}_2$. In other words, $(\bar{a}_2 - \bar{a}_1) \cdot (\bar{b}_1 \times \bar{b}_2) = 0$. In Cartesian form this is the determinant with rows $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$, $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ set equal to zero. Here that determinant is linear in $k$, so it gives exactly one value of $k$.

3. Question 3

   If $y = \log\left(\sec x + \tan x\right)$, $0 < x < \dfrac{\pi}{2}$, then ${\dfrac{dy}{dx} =}$

   1. A. $\sec x$Correct answer

      $\frac{dy}{dx} = \frac{1}{\sec x + \tan x}\left(\sec x\tan x + \sec^{2} x\right) = \frac{\sec x\left(\tan x + \sec x\right)}{\sec x + \tan x} = \sec x$.

   2. B. $\dfrac{1}{\sec x + \tan x}$

      Only the outer derivative, $\frac{d}{du}\log u = \frac{1}{u}$, was taken. By the chain rule it must be multiplied by the derivative of the inner function $u = \sec x + \tan x$, which is $\sec x\tan x + \sec^{2} x$.

   3. C. $\sec x\tan x + \sec^{2} x$

      This is the derivative of the inner function $\sec x + \tan x$ alone. The factor $\frac{1}{\sec x + \tan x}$ from differentiating the logarithm was never applied.

   4. D. $\dfrac{\sec^{2} x - \sec x\tan x}{\sec x + \tan x}$

      The derivative of $\sec x$ was taken as $-\sec x\tan x$. That minus sign belongs to $\frac{d}{dx}\cos x$ and $\frac{d}{dx}\cosec x$; the derivative of $\sec x$ is $+\sec x\tan x$.

   Explanation

   The function is a composite $\log u$ with $u = \sec x + \tan x$, so by the chain rule $\frac{dy}{dx} = \frac{1}{u}\cdot\frac{du}{dx}$. The inner derivative $\sec x\tan x + \sec^{2} x$ factors as $\sec x\left(\sec x + \tan x\right)$, and the bracket cancels with $u$ to leave $\sec x$. This is also why $\int \sec x\,dx = \log\left(\sec x + \tan x\right) + c$.

4. Question 4

   If $y = y(x)$ is the solution of the differential equation $\quad \dfrac{dy}{dx} + \dfrac{y}{x} = x^{2}, \quad x > 0$, satisfying $y(1) = 1$, then ${y(2) =}$

   1. A. $\dfrac{19}{8}$Correct answer

      Here $P = \frac{1}{x}$, so $\text{I.F.} = e^{\int \frac{1}{x}dx} = e^{\log x} = x$. Then $xy = \int x^{3}\,dx = \frac{x^{4}}{4} + c$. From $y(1) = 1$, $c = \frac{3}{4}$, so $2y(2) = 4 + \frac{3}{4}$, which gives $y(2) = \frac{19}{8}$.

   2. B. $\dfrac{5}{3}$

      The integrating factor $x$ was applied to the left side only, giving $xy = \int x^{2}\,dx = \frac{x^{3}}{3} + c$. The solution is $y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.})\,dx + c$, so the right side must also be multiplied by $x$ before integrating: $\int x^{3}\,dx$.

   3. C. $2$

      The constant of integration was left out, so $xy = \frac{x^{4}}{4}$, and the condition $y(1) = 1$ was never used. That curve does not even pass through $(1, 1)$. The constant is exactly what the initial condition fixes.

   4. D. $\dfrac{19}{2}$

      The constant was found correctly, but at the last step $y$ was taken as $x\left(\frac{x^{4}}{4} + \frac{3}{4}\right)$, which multiplies by the integrating factor instead of dividing by it. From $xy = \frac{x^{4}}{4} + \frac{3}{4}$ you get $y = \frac{1}{x}\left(\frac{x^{4}}{4} + \frac{3}{4}\right)$.

   Explanation

   The equation is linear, $\frac{dy}{dx} + Py = Q$ with $P = \frac{1}{x}$ and $Q = x^{2}$. Its integrating factor $e^{\int P\,dx} = x$ turns the left side into $\frac{d}{dx}(xy)$. Integrating $Q \cdot (\text{I.F.}) = x^{3}$ gives the general solution, and the initial condition fixes the constant before $x = 2$ is substituted.

5. Question 5

   Evaluate $\displaystyle\int_{1}^{4}\sqrt{x}\,dx$.

   1. A. $\dfrac{16}{3}$

      Only the upper limit was substituted: $\frac{2}{3}\left(4\right)^{3/2} = \frac{16}{3}$. A definite integral is $F(b) - F(a)$, and here $F(1) = \frac{2}{3}$ is not zero, so it must be subtracted.

   2. B. $\dfrac{21}{2}$

      The antiderivative was taken as $\frac{3}{2}x^{3/2}$ — multiplied by the new exponent instead of divided by it. Differentiating $\frac{3}{2}x^{3/2}$ returns $\frac{9}{4}\sqrt{x}$, not $\sqrt{x}$.

   3. C. $\dfrac{14}{3}$Correct answer

      $\displaystyle\int x^{1/2}\,dx = \frac{2}{3}x^{3/2}$, so the value is $\frac{2}{3}\left(4^{3/2} - 1^{3/2}\right) = \frac{2}{3}(8 - 1) = \frac{14}{3}$.

   4. D. $42$

      The substitution $u = \sqrt{x}$ turns the integral into $\displaystyle\int 2u^{2}\,du = \frac{2}{3}u^{3}$, but the original $x$-limits were kept: $\frac{2}{3}\left(4^{3} - 1^{3}\right) = 42$. When the variable changes, the limits change with it — here to $u = 1$ and $u = 2$, giving $\frac{2}{3}(8 - 1)$.

   Explanation

   Write the radical as a power, $\sqrt{x} = x^{1/2}$, so the power rule gives the antiderivative $\frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$: the exponent is raised by one and the NEW exponent divides. The fundamental theorem of calculus then asks for the difference of its values at the two limits, using $4^{3/2} = \left(\sqrt{4}\right)^{3} = 8$ and $1^{3/2} = 1$.

6. Question 6

   If $A = \begin{pmatrix} 2 & -1 & 3 \\ 1 & 4 & -2 \\ 3 & 0 & 5 \end{pmatrix}$, then the cofactor of the element $a_{32}$ is

   1. A. $-7$

      This is the minor $M_{32}$, reported as though it were the cofactor. The cofactor carries the factor $(-1)^{i+j}$, and here $(-1)^{3+2} = -1$, so the minor must still be negated.

   2. B. $-3$

      This is the cofactor $A_{23}$: the row and column indices have been interchanged, so the second row and third column were deleted instead of the third row and second column. In $a_{ij}$ the first index is always the row.

   3. C. $7$Correct answer

      Deleting the third row and the second column leaves $\begin{pmatrix} 2 & 3 \\ 1 & -2 \end{pmatrix}$, so $M_{32} = 2 \times (-2) - 3 \times 1 = -7$. Hence $A_{32} = (-1)^{3+2}M_{32} = -(-7) = 7$.

   4. D. $1$

      The $2 \times 2$ minor was evaluated as $2 \times (-2) + 3 \times 1 = -1$, adding the two products instead of subtracting. A second-order determinant is $ad - bc$; the anti-diagonal product is subtracted.

   Explanation

   The minor of an element is the determinant left after striking out the row and the column through that element, and the cofactor is that minor multiplied by the chequerboard sign $(-1)^{i+j}$. For an element in the third row and second column the indices sum to an odd number, so the sign factor is $-1$ and the cofactor is the negative of the minor. The value of the element itself, $0$ here, plays no part in forming its own cofactor.

7. Question 7

   ${\displaystyle\int_{-2}^{2}\left|x^{2} - 1\right|dx =}$

   1. A. $\dfrac{4}{3}$

      The student integrated $x^{2} - 1$ in one piece over $[-2, 2]$: $\left[\frac{x^{3}}{3} - x\right]_{-2}^{2} = \frac{4}{3}$. But $x^{2} - 1 < 0$ on $(-1, 1)$. On that interval $\left|x^{2} - 1\right| = 1 - x^{2}$, so the integral has to be split at $x = \pm 1$ (Property IV).

   2. B. $4$Correct answer

      $\left|x^{2} - 1\right|$ is even, so the integral is $2\int_{0}^{2}\left|x^{2} - 1\right|dx$. Splitting at $x = 1$ gives $2\left[\int_{0}^{1}(1 - x^{2})\,dx + \int_{1}^{2}(x^{2} - 1)\,dx\right] = 2\left(\frac{2}{3} + \frac{4}{3}\right) = 4$.

   3. C. $2$

      The student noticed that the integrand is even and correctly found $\int_{0}^{2}\left|x^{2} - 1\right|dx = \frac{2}{3} + \frac{4}{3} = 2$. Then they forgot the factor $2$. For an even function, Property VIII gives $\int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx$.

   4. D. $\dfrac{8}{3}$

      The student took the unsplit value $\frac{4}{3}$ and added the size of the negative piece, $\left|\int_{-1}^{1}(x^{2} - 1)\,dx\right| = \frac{4}{3}$, only once. Flipping that piece changes its contribution from $-\frac{4}{3}$ to $+\frac{4}{3}$, so the correction is $\frac{8}{3}$, not $\frac{4}{3}$.

   Explanation

   A modulus has to be removed before integrating. Find where $x^{2} - 1$ changes sign ($x = \pm 1$), and on each sub-interval replace $\left|x^{2} - 1\right|$ by $x^{2} - 1$ or $1 - x^{2}$ as the sign requires (Property IV). Since $f(-x) = f(x)$, Property VIII lets you work on $[0, 2]$ only and double the result.

8. Question 8

   The value of $\tan\left(2\tan^{-1}\dfrac{1}{3}\right)$ is

   1. A. $\dfrac{2}{3}$

      Here $\tan 2\theta$ was taken to be $2\tan\theta = 2 \times \frac{1}{3}$. The tangent of a double angle is $\frac{2\tan\theta}{1 - \tan^{2}\theta}$, not twice the tangent, so the denominator $1 - \frac{1}{9}$ was left out.

   2. B. $\dfrac{3}{4}$Correct answer

      Let $\theta = \tan^{-1}\frac{1}{3}$, so $\tan\theta = \frac{1}{3}$. Then $\tan 2\theta = \frac{2 \cdot \frac{1}{3}}{1 - \frac{1}{9}} = \frac{2/3}{8/9} = \frac{3}{4}$. Equivalently, $2\tan^{-1}\frac{1}{3} = \tan^{-1}\frac{3}{4}$, which is valid because $\left|\frac{1}{3}\right| < 1$.

   3. C. $\dfrac{3}{5}$

      This is $\frac{2x}{1 + x^{2}}$ at $x = \frac{1}{3}$. That is the sine form, $2\tan^{-1}x = \sin^{-1}\frac{2x}{1 + x^{2}}$, so it gives $\sin\left(2\tan^{-1}\frac{1}{3}\right)$. The question asks for the tangent, which needs $1 - x^{2}$ in the denominator.

   4. D. $\dfrac{4}{5}$

      This is $\frac{1 - x^{2}}{1 + x^{2}}$ at $x = \frac{1}{3}$. That is the cosine form, $2\tan^{-1}x = \cos^{-1}\frac{1 - x^{2}}{1 + x^{2}}$, so it gives $\cos\left(2\tan^{-1}\frac{1}{3}\right)$, not the tangent.

   Explanation

   Put $\theta = \tan^{-1}\frac{1}{3}$ and use the double-angle formula $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^{2}\theta}$. This is the same as the identity $2\tan^{-1}x = \tan^{-1}\frac{2x}{1 - x^{2}}$ for $|x| < 1$. The three forms of $2\tan^{-1}x$ (sine, cosine and tangent) have different fractions, and only the tangent form answers this question.