CBSE Class 12 Maths MCQs, chapter by chapter

987 practice questions for CBSE Class 12 Mathematics, organised by NCERT chapter (13 chapters). Every question includes a written explanation. Pick a chapter below, or try the samples first.

CBSE Class 12 Maths mock tests and papers

Practice by chapter

Relations and Functions

75 questions in timed practice

Inverse Trigonometric Functions

75 questions in timed practice

Matrices

81 questions in timed practice

Determinants

81 questions in timed practice

Continuity and Differentiability

75 questions in timed practice

Applications of Derivatives

75 questions in timed practice

Integrals

75 questions in timed practice

Applications of the Integrals

75 questions in timed practice

Differential Equations

75 questions in timed practice

Vector Algebra

75 questions in timed practice

Three Dimensional Geometry

75 questions in timed practice

Linear Programming

75 questions in timed practice

Probability

75 questions in timed practice

Sample questions

Q1
medium
Take $A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Compute $AB - BA$.
  1. A $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$
  2. B $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$
  3. C $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$
  4. D $\begin{pmatrix} 0 & 3 \\ 3 & 0 \end{pmatrix}$
Show answer and explanation

Correct answer: A - $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$

$AB$ has entries $0, 1, 2, 0$ and $BA$ has entries $0, 2, 1, 0$, so $AB - BA$ has entries $0, -1, 1, 0$. The zero matrix would follow only if the two products were equal, which they are not. The array with entries $0, 1, -1, 0$ is $BA - AB$, and the array with two threes is $AB + BA$.
Q2
easy
Suppose $F$ and $G$ are both antiderivatives of the same function $f$ on an interval. What can be said about $G(x)-F(x)$ on that interval?
  1. A It is a constant multiple of $x$.
  2. B It is a constant.
  3. C It equals $f(x)$.
  4. D It is zero for every such pair.
Show answer and explanation

Correct answer: B - It is a constant.

Since $F'(x)=G'(x)=f(x)$, the derivative of $G-F$ is $0$ throughout the interval, and a function with zero derivative on an interval is constant. It cannot equal $f(x)$, which is the shared derivative rather than the difference. A constant multiple of $x$ has a non-zero derivative, so it is ruled out. The difference need not be zero, because the two antiderivatives may carry different constants.
Q3
easy
If $\vec{a} = \hat{i} + 2\hat{j}$ and $\vec{b} = 3\hat{i} - \hat{j}$, find $\vec{a} + \vec{b}$.
  1. A $3\hat{i} - 2\hat{j}$
  2. B $2\hat{i} - 3\hat{j}$
  3. C $4\hat{i} + \hat{j}$
  4. D $-2\hat{i} + 3\hat{j}$
Show answer and explanation

Correct answer: C - $4\hat{i} + \hat{j}$

Vectors add component by component, giving $(1+3)\hat{i} + (2-1)\hat{j} = 4\hat{i}+\hat{j}$. The vector $-2\hat{i}+3\hat{j}$ is $\vec{a}-\vec{b}$, $2\hat{i}-3\hat{j}$ is $\vec{b}-\vec{a}$, and $3\hat{i}-2\hat{j}$ comes from multiplying matching components instead of adding them.
Q4
medium
For $0<x<\pi$, let $y=(\sin x)^x$. Find $\dfrac{dy}{dx}$.
  1. A $(\sin x)^x\left(\ln \sin x + x\tan x\right)$
  2. B $(\sin x)^x\ln \sin x$
  3. C $x(\sin x)^{x-1}\cos x$
  4. D $(\sin x)^x\left(\ln \sin x + x\cot x\right)$
Show answer and explanation

Correct answer: D - $(\sin x)^x\left(\ln \sin x + x\cot x\right)$

Take logarithms: $\ln y=x\ln(\sin x)$. Differentiating, $\frac{1}{y}\frac{dy}{dx}=\ln\sin x+x\cdot\frac{\cos x}{\sin x}=\ln\sin x+x\cot x$, and multiplying back by $y$ gives the result. Using the power rule as though the exponent were constant gives $x(\sin x)^{x-1}\cos x$, differentiating only the first product term drops $x\cot x$, and $\frac{\cos x}{\sin x}$ is $\cot x$, not $\tan x$.
Q5
medium
A relation $R$ is defined on the set of all straight lines in a plane by taking $L_1\,R\,L_2$ to mean that $L_1$ is parallel to $L_2$, where every line counts as parallel to itself. Statement 1: $R$ is an equivalence relation. Statement 2: The equivalence class of a line $L$ is the set of all lines perpendicular to $L$. Which is correct?
  1. A Statement 1 is false and Statement 2 is true
  2. B Both statements are true
  3. C Both statements are false
  4. D Statement 1 is true and Statement 2 is false
Show answer and explanation

Correct answer: D - Statement 1 is true and Statement 2 is false

With a line counted as parallel to itself, $R$ is reflexive; parallelism works both ways, so $R$ is symmetric; and two lines parallel to a common line are parallel to each other, so $R$ is transitive. Statement 1 is therefore true. The class of $L$ collects the lines related to $L$, which are the lines parallel to $L$, not the ones perpendicular to it, so Statement 2 is false.
Q6
hard
A parallelogram has adjacent sides $\vec{a}$ and $\vec{b}$ with $|\vec{a}| = 4$, $|\vec{b}| = 5$ and $\vec{a} \cdot \vec{b} = 10$. Find its area.
  1. A $20$ square units
  2. B $10$ square units
  3. C $10\sqrt{3}$ square units
  4. D $5\sqrt{3}$ square units
Show answer and explanation

Correct answer: C - $10\sqrt{3}$ square units

From $\cos\theta = \frac{10}{4 \times 5} = \frac{1}{2}$ the angle is $60^\circ$, so $\sin\theta = \frac{\sqrt{3}}{2}$ and the area $|\vec{a}||\vec{b}|\sin\theta = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$. Leaving out $\sin\theta$ gives $20$, using the scalar product itself as the area gives $10$, and $5\sqrt{3}$ is the area of the triangle on the same two sides.

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