NCERT Exemplar Solutions Class 12 Maths Continuity and Differentiability — Objective, Fill in the Blanks and True or False Q83–106

NCERT Exemplar Problems, Class 12 Mathematics — Continuity and Differentiability, Objective Type, Fill in the Blanks and True or False questions 83 to 106. All 24 questions solved.

These are the quick-fire questions of the unit. Most of them turn on one habit: ask where a function is undefined (that is where it is discontinuous), where it has a corner (modulus functions), and where a limit such as $x^2\sin\frac1x$ is squeezed to zero. The derivative questions are short applications of the chain rule, implicit differentiation and parametric differentiation.

A note on the syllabus. Questions 95 and 96 use Rolle’s theorem and the mean value theorem, and Question 102 asks whether Rolle’s theorem applies. The rationalised NCERT Chapter 5 no longer contains Rolle’s theorem or the mean value theorem, and the JEE Main 2026 syllabus does not list them. JEE Advanced 2026 names “Rolle’s theorem and Lagrange’s mean value theorem”, so these three questions are worth working for Advanced.

Key insight. Continuity fails where a function is undefined or jumps; differentiability fails, in addition, at corners such as $|x|$ at $0$. A modulus $|g(x)|$ has a corner exactly where $g(x)=0$ and $g'(x)\ne0$.

Question 83

If $f(x)=2x$ and $g(x)=\dfrac{x^2}{2}+1$, then which of the following can be a discontinuous function?

• (A) $f(x)+g(x)$
• (B) $f(x)-g(x)$
• (C) $f(x)\cdot g(x)$
• (D) $\dfrac{g(x)}{f(x)}$

Solution. Both $f$ and $g$ are polynomials, so their sum, difference and product are polynomials and continuous everywhere. The quotient $\dfrac{g(x)}{f(x)}=\dfrac{x^2/2+1}{2x}$ is undefined at $x=0$, where $f(0)=0$, so it is discontinuous there.

Question 84

The function $f(x)=\dfrac{4-x^2}{4x-x^3}$ is

• (A) discontinuous at only one point
• (B) discontinuous at exactly two points
• (C) discontinuous at exactly three points
• (D) none of these

Solution. The denominator is $4x-x^3=x(2-x)(2+x)$, which vanishes at $x=0,\,2,\,-2$, so $f$ is undefined at all three. At $x=\pm2$ the numerator also vanishes and the fraction simplifies to $\dfrac1x$, but simplifying does not define $f$ at those points, so the function is still discontinuous there. Three points in all.

Question 85

The set of points where the function $f$ given by $f(x)=|2x-1|\sin x$ is differentiable is

• (A) $\mathbf{R}$
• (B) $\mathbf{R}-\left\{\dfrac12\right\}$
• (C) $(0,\infty)$
• (D) none of these

Solution. The factor $|2x-1|$ has a corner at $x=\frac12$, where $2x-1=0$. A corner is harmless only if the other factor is zero there too, because then the product is flattened out. Here $\sin\frac12\ne0$, so the corner survives and $f$ is not differentiable at $x=\frac12$. Everywhere else both factors are differentiable.

Question 86

The function $f(x)=\cot x$ is discontinuous on the set

• (A) $\{x=n\pi : n\in\mathbf{Z}\}$
• (B) $\{x=2n\pi : n\in\mathbf{Z}\}$
• (C) $\left\{x=(2n+1)\dfrac{\pi}{2} : n\in\mathbf{Z}\right\}$
• (D) $\left\{x=\dfrac{n\pi}{2} : n\in\mathbf{Z}\right\}$

Solution. $\cot x=\dfrac{\cos x}{\sin x}$ is undefined where $\sin x=0$, that is at $x=n\pi$. At $x=(2n+1)\frac{\pi}{2}$ we have $\cot x=0$, which is perfectly defined. (In the printed paper, option D carries the label “(iv)”, a typesetting slip.)

Question 87

The function $f(x)=e^{|x|}$ is

• (A) continuous everywhere but not differentiable at $x=0$
• (B) continuous and differentiable everywhere
• (C) not continuous at $x=0$
• (D) none of these.

Solution. $e^{|x|}$ is a composition of continuous functions, so it is continuous everywhere. At $0$: for $x>0$, $f(x)=e^{x}$ with right-hand derivative $1$; for $x<0$, $f(x)=e^{-x}$ with left-hand derivative $-1$. The two differ, so there is a corner at $x=0$.

Question 88

If $f(x)=x^2\sin\dfrac1x$, where $x\ne0$, then the value of the function $f$ at $x=0$, so that the function is continuous at $x=0$, is

• (A) $0$
• (B) $-1$
• (C) $1$
• (D) none of these

Solution. For continuity we need $f(0)=\lim_{x\to0}f(x)$. Since $\left|\sin\frac1x\right|\le1$, we have $-x^2\le x^2\sin\frac1x\le x^2$, and both bounds tend to $0$, so the limit is $0$ by the squeeze theorem. Hence $f(0)$ must be $0$.

Question 89

If $f(x)=\begin{cases}mx+1, & x\le\dfrac{\pi}{2}\\[2mm] \sin x+n, & x>\dfrac{\pi}{2}\end{cases}$ is continuous at $x=\dfrac{\pi}{2}$, then

• (A) $m=1,\ n=0$
• (B) $m=\dfrac{n\pi}{2}+1$
• (C) $n=\dfrac{m\pi}{2}$
• (D) $m=n=\dfrac{\pi}{2}$

Solution. The left-hand limit (and the value) at $\frac\pi2$ is $\frac{m\pi}{2}+1$. The right-hand limit is $\sin\frac\pi2+n=1+n$. Equating, $\frac{m\pi}{2}+1=1+n$, so $n=\frac{m\pi}{2}$. This is one equation in two unknowns, so only the relation between $m$ and $n$ is fixed, and option (C) is that relation.

Question 90

Let $f(x)=|\sin x|$. Then

• (A) $f$ is everywhere differentiable
• (B) $f$ is everywhere continuous but not differentiable at $x=n\pi,\ n\in\mathbf{Z}$
• (C) $f$ is everywhere continuous but not differentiable at $x=(2n+1)\dfrac{\pi}{2},\ n\in\mathbf{Z}$
• (D) none of these

Solution. $f$ is continuous everywhere as a modulus of a continuous function. The corners of $|\sin x|$ occur where $\sin x=0$ and $\cos x\ne0$, that is at $x=n\pi$. There the graph touches the axis and the slope switches between $-1$ and $+1$. At $x=(2n+1)\frac\pi2$ the function has a smooth peak or trough, so it is differentiable there.

Question 91

If $y=\log\left(\dfrac{1-x^2}{1+x^2}\right)$, then $\dfrac{dy}{dx}$ is equal to

• (A) $\dfrac{4x^3}{1-x^4}$
• (B) $\dfrac{-4x}{1-x^4}$
• (C) $\dfrac{1}{4-x^4}$
• (D) $\dfrac{-4x^3}{1-x^4}$

Solution. Split the logarithm: $y=\log(1-x^2)-\log(1+x^2)$. Then $$\frac{dy}{dx}=\frac{-2x}{1-x^2}-\frac{2x}{1+x^2}=\frac{-2x\left[(1+x^2)+(1-x^2)\right]}{1-x^4}=\frac{-4x}{1-x^4}.$$

Question 92

If $y=\sqrt{\sin x+y}$, then $\dfrac{dy}{dx}$ is equal to

• (A) $\dfrac{\cos x}{2y-1}$
• (B) $\dfrac{\cos x}{1-2y}$
• (C) $\dfrac{\sin x}{1-2y}$
• (D) $\dfrac{\sin x}{2y-1}$

Solution. Square first to remove the root: $y^2=\sin x+y$. Differentiate implicitly: $2y\,y’=\cos x+y’$, so $y'(2y-1)=\cos x$ and $y’=\dfrac{\cos x}{2y-1}$.

Question 93

The derivative of $\cos^{-1}(2x^2-1)$ w.r.t. $\cos^{-1}x$ is

• (A) $2$
• (B) $\dfrac{-1}{2\sqrt{1-x^2}}$
• (C) $\dfrac{2}{x}$
• (D) $1-x^2$

Solution. Put $x=\cos\theta$. Then $2x^2-1=\cos2\theta$, so $u=\cos^{-1}(\cos2\theta)$, while $v=\cos^{-1}x=\theta$. For $0<x<1$ we have $\theta\in\left(0,\frac\pi2\right)$, so $2\theta\in(0,\pi)$ lies in the principal range of $\cos^{-1}$ and $u=2\theta$. Hence $\dfrac{du}{dv}=2$.

A caution about the printed answer: this holds only for $0<x<1$. For $-1<x<0$, $\theta\in\left(\frac\pi2,\pi\right)$, $2\theta$ is outside $[0,\pi]$, and $\cos^{-1}(\cos2\theta)=2\pi-2\theta$, which gives $-2$. No option offers $-2$, so (A) is the intended answer, and it is the one NCERT prints.

Question 94

If $x=t^2,\ y=t^3$, then $\dfrac{d^2y}{dx^2}$ is

• (A) $\dfrac32$
• (B) $\dfrac{3}{4t}$
• (C) $\dfrac{3}{2t}$
• (D) $\dfrac34$

Solution. $\dfrac{dy}{dx}=\dfrac{3t^2}{2t}=\dfrac{3t}{2}$. For the second derivative, differentiate this with respect to $t$ and divide by $\dfrac{dx}{dt}$ again: $$\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{3t}{2}\right)}{2t}=\frac{3/2}{2t}=\frac{3}{4t}.$$

Question 95

The value of $c$ in Rolle’s theorem for the function $f(x)=x^3-3x$ in the interval $[0,\sqrt3\,]$ is

• (A) $1$
• (B) $-1$
• (C) $\dfrac32$
• (D) $\dfrac13$

Solution. First check the hypotheses: $f$ is a polynomial, and $f(0)=0=f(\sqrt3)$ because $(\sqrt3)^3-3\sqrt3=0$. So some $c\in(0,\sqrt3)$ has $f'(c)=0$. Now $f'(x)=3x^2-3=0$ gives $x=\pm1$, and only $x=1$ lies in the interval. (Option B, $-1$, is outside it.)

Question 96

For the function $f(x)=x+\dfrac1x,\ x\in[1,3]$, the value of $c$ for the mean value theorem is

• (A) $1$
• (B) $\sqrt3$
• (C) $2$
• (D) none of these

Solution. $f$ is continuous on $[1,3]$ and differentiable on $(1,3)$. The mean value theorem asks for $c$ with $f'(c)=\dfrac{f(3)-f(1)}{3-1}$. Here $f(3)=\frac{10}{3}$ and $f(1)=2$, so the slope is $\dfrac{4/3}{2}=\dfrac23$. Since $f'(x)=1-\dfrac1{x^2}$, we need $1-\dfrac1{c^2}=\dfrac23$, so $c^2=3$ and $c=\sqrt3$, which lies in $(1,3)$.

Question 97

An example of a function which is continuous everywhere but fails to be differentiable exactly at two points is ____.

Solution. A modulus creates one corner for each zero of its argument, so use two of them: $f(x)=|x|+|x-1|$. It is continuous everywhere, with corners at $x=0$ and $x=1$ and smooth elsewhere. Any function with exactly two corners is a valid answer, for example $|x(x-1)|$.

Question 98

Derivative of $x^2$ w.r.t. $x^3$ is ____.

Solution. Divide the two derivatives with respect to $x$: $\dfrac{d(x^2)/dx}{d(x^3)/dx}=\dfrac{2x}{3x^2}=\dfrac{2}{3x}$ for $x\ne0$.

Question 99

If $f(x)=|\cos x|$, then $f’\!\left(\dfrac{\pi}{4}\right)=$ ____.

Solution. Near $x=\frac\pi4$, $\cos x>0$, so $f(x)=\cos x$ in a neighbourhood and $f'(x)=-\sin x$. Then $f’\!\left(\frac\pi4\right)=-\sin\frac\pi4=-\dfrac1{\sqrt2}$.

Question 100

If $f(x)=|\cos x-\sin x|$, then $f’\!\left(\dfrac{\pi}{3}\right)=$ ____.

Solution. At $x=\frac\pi3$, $\cos x-\sin x=\frac12-\frac{\sqrt3}{2}<0$, so near this point $f(x)=-(\cos x-\sin x)=\sin x-\cos x$. Then $f'(x)=\cos x+\sin x$ and $$f’\!\left(\tfrac\pi3\right)=\frac12+\frac{\sqrt3}{2}=\frac{\sqrt3+1}{2}.$$

Question 101

For the curve $\sqrt x+\sqrt y=1$, $\dfrac{dy}{dx}$ at $\left(\dfrac14,\dfrac14\right)$ is ____.

Solution. Differentiate implicitly: $\dfrac{1}{2\sqrt x}+\dfrac{1}{2\sqrt y}\,y’=0$, so $y’=-\dfrac{\sqrt y}{\sqrt x}$. At $\left(\frac14,\frac14\right)$, $\sqrt x=\sqrt y=\frac12$, so $y’=-1$. This also follows from the symmetry of the curve about $y=x$.

Question 102

Rolle’s theorem is applicable for the function $f(x)=|x-1|$ in $[0,2]$.

Solution. Rolle’s theorem needs three things: continuity on $[0,2]$, differentiability on $(0,2)$, and $f(0)=f(2)$. The first and third hold, since $f(0)=f(2)=1$. But $|x-1|$ has a corner at $x=1$, inside the interval, where the left derivative is $-1$ and the right derivative is $+1$. Differentiability fails, so the theorem is not applicable (and indeed $f’$ is never $0$).

Question 103

If $f$ is continuous on its domain $D$, then $|f|$ is also continuous on $D$.

Solution. $|f|$ is the composition of $f$ with the modulus function, and both are continuous. Equivalently, $\big||f(x)|-|f(a)|\big|\le|f(x)-f(a)|$, which tends to $0$ as $x\to a$.

Question 104

The composition of two continuous functions is a continuous function.

Solution. If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then $f\circ g$ is continuous at $a$, because $\lim_{x\to a}f(g(x))=f\!\left(\lim_{x\to a}g(x)\right)=f(g(a))$. This is a standard theorem.

Question 105

Trigonometric and inverse trigonometric functions are differentiable in their respective domains.

Solution. Each of $\sin,\cos,\tan,\cot,\sec,\csc$ is differentiable wherever it is defined, and so is each inverse trigonometric function at every point of its domain where the derivative formula makes sense, that is in the open domain. The statement reads these as “their respective domains”, which is how NCERT intends it.

Question 106

If $f\cdot g$ is continuous at $x=a$, then $f$ and $g$ are separately continuous at $x=a$.

Solution. The converse is true (a product of continuous functions is continuous), but this direction is not. Take $f(x)=g(x)=1$ for $x\ge0$ and $-1$ for $x<0$. Both jump at $0$, yet $f(x)g(x)=1$ for all $x$, which is continuous.

Common mistakes

• Cancelling before checking the domain. In Question 84, $\frac{4-x^2}{4x-x^3}$ reduces to $\frac1x$, but the original function is still undefined at $x=\pm2$ as well as $0$.
• Missing the corner of a modulus, or forgetting when it is hidden. $|2x-1|\sin x$ keeps its corner because $\sin\frac12\ne0$ (Question 85); $|\sin x|$ has corners at $n\pi$, not at $(2n+1)\frac\pi2$ (Question 90).
• Using the wrong branch for $\cos^{-1}$. $\cos^{-1}(\cos2\theta)=2\theta$ only when $2\theta\in[0,\pi]$ (Question 93).
• Picking the root outside the interval. In Rolle’s and mean value problems, $f'(c)=0$ or $f'(c)=\text{slope}$ may have roots outside the interval, such as $c=-1$ in Question 95.
• Assuming the product rule for continuity runs both ways. A continuous product does not force continuous factors (Question 106).

Practise next

• Exemplar Short Answer 4 and Long Answer Q65–82, the questions just before these, including Rolle’s theorem and mean value theorem applications.
• Exemplar Short Answer 1 Q1–24 for more continuity and differentiability checks at a point.
• NCERT textbook Exercise 5.1 for continuity at a point and on an interval.
• NCERT textbook Exercise 5.3 for implicit and parametric differentiation.