NCERT Exemplar Solutions Class 12 Maths Continuity and Differentiability — Short Answer and Long Answer Q65–82

NCERT Exemplar Problems, Class 12 Mathematics — Continuity and Differentiability, Short Answer questions 65 to 78 and Long Answer questions 79 to 82. All 18 questions solved.

Questions 65 to 78 are about two theorems. Rolle’s theorem says that if $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then $f'(c)=0$ for some $c\in(a,b)$. The mean value theorem drops the condition $f(a)=f(b)$ and says $f'(c)=\dfrac{f(b)-f(a)}{b-a}$ for some $c\in(a,b)$. Geometrically, there is a point where the tangent is parallel to the chord. Questions 79 to 82 return to differentiability at a point, implicit differentiation and logarithmic differentiation.

A note on the syllabus. 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 does name “Rolle’s theorem and Lagrange’s mean value theorem”, so Questions 65 to 78 on this page are worth working for Advanced even though they are outside the current textbook. Questions 79 to 82 are unaffected.

Key insight. For both theorems, check the hypotheses first (continuous on the closed interval, differentiable on the open one, and for Rolle equal end values), then solve $f'(c)=\text{slope}$ and keep only the roots that lie strictly inside the interval.

Question 65

Verify the Rolle’s theorem for $f(x)=x(x-1)^2$ in $[0,1]$.

Solution. $f$ is a polynomial, so it is continuous and differentiable everywhere. $f(0)=0$ and $f(1)=0$, so Rolle’s theorem applies. Now $f'(x)=(x-1)^2+2x(x-1)=(x-1)(3x-1)$. This is zero at $x=1$ (an endpoint, so not allowed) and at $x=\tfrac13$, which lies in $(0,1)$.

Question 66

Verify the Rolle’s theorem for $f(x)=\sin^4x+\cos^4x$ in $\left[0,\dfrac{\pi}{2}\right]$.

Solution. $f$ is continuous and differentiable everywhere, and $f(0)=1=f\!\left(\tfrac\pi2\right)$. Differentiating, $f'(x)=4\sin^3x\cos x-4\cos^3x\sin x=-4\sin x\cos x\cos 2x=-\sin 4x$. Setting $\sin 4x=0$ gives $4x=\pi$ inside the interval, so $x=\dfrac\pi4$, which lies in $\left(0,\tfrac\pi2\right)$.

Question 67

Verify the Rolle’s theorem for $f(x)=\log(x^2+2)-\log 3$ in $[-1,1]$.

Solution. $x^2+2>0$ always, so $f$ is continuous and differentiable on $\mathbb R$. $f(-1)=\log3-\log3=0=f(1)$. Then $f'(x)=\dfrac{2x}{x^2+2}$, which is zero only at $x=0\in(-1,1)$.

Question 68

Verify the Rolle’s theorem for $f(x)=x(x+3)e^{-x/2}$ in $[-3,0]$.

Solution. The product of a polynomial and an exponential is continuous and differentiable everywhere, and $f(-3)=0=f(0)$. By the product rule, $f'(x)=e^{-x/2}\left[(2x+3)-\tfrac12(x^2+3x)\right]=-\tfrac12e^{-x/2}(x^2-x-6)=-\tfrac12e^{-x/2}(x-3)(x+2)$. The exponential never vanishes, so $f'(x)=0$ at $x=3$ or $x=-2$. Only $x=-2$ lies in $(-3,0)$.

Question 69

Verify the Rolle’s theorem for $f(x)=\sqrt{4-x^2}$ in $[-2,2]$.

Solution. $f$ is continuous on $[-2,2]$ since $4-x^2\ge0$ there, and $f(-2)=0=f(2)$. For $-2<x<2$, $f'(x)=\dfrac{-x}{\sqrt{4-x^2}}$ exists, so $f$ is differentiable on the open interval (the derivative fails only at the endpoints, which the theorem does not require). $f'(x)=0$ gives $x=0\in(-2,2)$.

Question 70

Discuss the applicability of Rolle’s theorem on the function given by $f(x)=\begin{cases}x^2+1, & 0\le x\le1\\ 3-x, & 1\le x\le2.\end{cases}$

Solution. Both pieces equal $2$ at $x=1$, so $f$ is continuous on $[0,2]$, and $f(0)=1=f(2)$. But differentiability fails at $x=1$. The left derivative is $2x=2$ and the right derivative is $-1$. They differ, so $f$ is not differentiable on $(0,2)$ and one hypothesis of Rolle’s theorem breaks. Indeed no $c$ exists: $f'(x)=2x=0$ only at $x=0$ (an endpoint), and $f'(x)=-1\neq0$ on the other piece.

Question 71

Find the points on the curve $y=(\cos x-1)$ in $[0,2\pi]$, where the tangent is parallel to $x$-axis.

Solution. A tangent parallel to the $x$-axis has zero slope. $\dfrac{dy}{dx}=-\sin x=0$ gives $x=0,\pi,2\pi$ in $[0,2\pi]$, and $y=\cos x-1$ there is $0,-2,0$. NCERT’s printed answer is $(\pi,-2)$, the interior point that Rolle’s theorem guarantees (the end values are both $0$). On the closed interval the endpoints $(0,0)$ and $(2\pi,0)$ also have horizontal tangents, so the key is incomplete in that sense.

Question 72

Using Rolle’s theorem, find the point on the curve $y=x(x-4)$, $x\in[0,4]$, where the tangent is parallel to $x$-axis.

Solution. $y=x^2-4x$ is a polynomial with $y(0)=0=y(4)$, so Rolle’s theorem applies. $y’=2x-4=0$ gives $x=2$, and $y(2)=2(-2)=-4$.

Question 73

Verify mean value theorem for $f(x)=\dfrac{1}{4x-1}$ in $[1,4]$.

Solution. $f$ is continuous and differentiable wherever $4x-1\neq0$, that is away from $x=\tfrac14$, so on $[1,4]$ the theorem applies. $f(1)=\tfrac13$ and $f(4)=\tfrac1{15}$, so the chord slope is $\dfrac{\frac1{15}-\frac13}{3}=-\dfrac{4}{45}$. Since $f'(x)=-\dfrac{4}{(4x-1)^2}$, we need $(4c-1)^2=45$, so $4c-1=3\sqrt5$ (the positive root, because $4c-1>0$ on the interval). Hence $c=\dfrac{1+3\sqrt5}{4}\approx1.93$, which lies in $(1,4)$.

Question 74

Verify mean value theorem for $f(x)=x^3-2x^2-x+3$ in $[0,1]$.

Solution. $f$ is a polynomial. $f(0)=3$ and $f(1)=1$, so the chord slope is $-2$. $f'(x)=3x^2-4x-1=-2$ gives $3x^2-4x+1=0$, so $(3x-1)(x-1)=0$. The roots are $x=\tfrac13$ and $x=1$. The root $1$ is an endpoint and is rejected.

Question 75

Verify mean value theorem for $f(x)=\sin x-\sin 2x$ in $[0,\pi]$.

Solution. $f$ is continuous and differentiable everywhere, and $f(0)=0=f(\pi)$, so the chord slope is $0$. $f'(x)=\cos x-2\cos 2x=\cos x-4\cos^2x+2$. Setting this to zero gives $4\cos^2x-\cos x-2=0$, so $\cos x=\dfrac{1\pm\sqrt{33}}{8}$. Both values lie in $(-1,1)$ ($\approx0.843$ and $\approx-0.593$), and $\cos$ is one-to-one on $(0,\pi)$, so each gives exactly one point there: $c\approx0.568$ and $c\approx2.206$.

Question 76

Verify mean value theorem for $f(x)=\sqrt{25-x^2}$ in $[1,5]$.

Solution. $f$ is continuous on $[1,5]$ and differentiable on $(1,5)$, with $f'(x)=\dfrac{-x}{\sqrt{25-x^2}}$. $f(1)=\sqrt{24}=2\sqrt6$ and $f(5)=0$, so the chord slope is $\dfrac{0-2\sqrt6}{4}=-\dfrac{\sqrt6}{2}$. Then $\dfrac{c}{\sqrt{25-c^2}}=\dfrac{\sqrt6}{2}$; squaring, $4c^2=6(25-c^2)$, so $c^2=15$. Taking the positive root (the left side is positive), $c=\sqrt{15}\approx3.87\in(1,5)$.

Question 77

Find a point on the curve $y=(x-3)^2$, where the tangent is parallel to the chord joining the points $(3,0)$ and $(4,1)$.

Solution. The chord has slope $\dfrac{1-0}{4-3}=1$. The tangent slope is $\dfrac{dy}{dx}=2(x-3)$, so $2(x-3)=1$ gives $x=\tfrac72$, which lies between $3$ and $4$. Then $y=\left(\tfrac12\right)^2=\tfrac14$.

Question 78

Using mean value theorem, prove that there is a point on the curve $y=2x^2-5x+3$ between the points A$(1,0)$ and B$(2,1)$, where tangent is parallel to the chord AB. Also, find that point.

Solution. The curve passes through $A$: $2-5+3=0$, and through $B$: $8-10+3=1$. As a polynomial it is continuous and differentiable, so the mean value theorem guarantees a point strictly between them with tangent slope equal to the chord slope $\dfrac{1-0}{2-1}=1$. Solving $y’=4x-5=1$ gives $x=\tfrac32\in(1,2)$, and $y=2\cdot\tfrac94-\tfrac{15}{2}+3=0$.

Question 79

Find the values of $p$ and $q$ so that $f(x)=\begin{cases}x^2+3x+p, & x\le1\\ qx+2, & x>1\end{cases}$ is differentiable at $x=1$.

Solution. Differentiability needs continuity first. At $x=1$ the left piece gives $4+p$ and the right piece tends to $q+2$, so $4+p=q+2$. Then the derivatives must agree: the left derivative is $2x+3=5$ at $x=1$ and the right derivative is $q$, so $q=5$. Substituting, $p=q-2=3$.

Question 80

If $x^m\cdot y^n=(x+y)^{m+n}$, prove that (i) $\dfrac{dy}{dx}=\dfrac yx$ and (ii) $\dfrac{d^2y}{dx^2}=0$.

Solution. (i) Take logarithms: $m\log x+n\log y=(m+n)\log(x+y)$. Differentiate with respect to $x$: $$\frac mx+\frac ny\,y’=\frac{(m+n)(1+y’)}{x+y}.$$ Collect the $y’$ terms: $y’\left[\dfrac ny-\dfrac{m+n}{x+y}\right]=\dfrac{m+n}{x+y}-\dfrac mx$. The bracket simplifies to $\dfrac{nx-my}{y(x+y)}$ and the right side to $\dfrac{nx-my}{x(x+y)}$. Cancelling the common factor $\dfrac{nx-my}{x+y}$ (non-zero for a genuine solution) leaves $\dfrac{y’}{y}=\dfrac1x$, so $y’=\dfrac yx$.

(ii) $y’=\dfrac yx$ means $\dfrac{dy}{y}=\dfrac{dx}{x}$, so $y=kx$ for a constant $k$. Hence $y’=k$ and $y”=0$. Equivalently, differentiating $y’=y/x$ gives $y”=\dfrac{xy’-y}{x^2}=\dfrac{y-y}{x^2}=0$.

Question 81

If $x=\sin t$ and $y=\sin pt$, prove that $(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}+p^2y=0$.

Solution. By the chain rule, $\dfrac{dy}{dx}=\dfrac{p\cos pt}{\cos t}$, and $\cos t=\sqrt{1-x^2}$ near the principal branch. So $$\sqrt{1-x^2}\,\frac{dy}{dx}=p\cos pt.$$ Differentiate both sides with respect to $x$, using $\dfrac{dt}{dx}=\dfrac1{\sqrt{1-x^2}}$ on the right: $$\sqrt{1-x^2}\,y”-\frac{x}{\sqrt{1-x^2}}\,y’=-\frac{p^2\sin pt}{\sqrt{1-x^2}}.$$ Multiplying by $\sqrt{1-x^2}$ gives $(1-x^2)y”-xy’=-p^2y$, which is the required relation.

Question 82

Find $\dfrac{dy}{dx}$, if $y=x^{\tan x}+\sqrt{\dfrac{x^2+1}{2}}$.

Solution. Differentiate the two terms separately. For $u=x^{\tan x}$, take logarithms: $\log u=\tan x\log x$, so $\dfrac{u’}{u}=\sec^2x\log x+\dfrac{\tan x}{x}$. For $v=\sqrt{\tfrac{x^2+1}2}$, $v’=\dfrac{1}{2v}\cdot x=\dfrac{x}{2}\cdot\dfrac{\sqrt2}{\sqrt{x^2+1}}=\dfrac{x}{\sqrt2\sqrt{x^2+1}}$.

Common mistakes

• Skipping the hypotheses. A function can have $f(a)=f(b)$ and still fail Rolle’s theorem because it has a corner inside, as in Question 70.
• Keeping roots outside the open interval. In Questions 65, 68 and 74 the equation $f'(c)=\text{slope}$ has a second root that is an endpoint or lies outside $(a,b)$, and it must be discarded.
• Forgetting the sign when squaring. In Questions 73 and 76 squaring introduces a spurious negative root; check the sign of $4c-1$ or $c$ against the original equation.
• Ignoring the closed interval. Question 71 asks about all of $[0,2\pi]$, so the endpoints count as points of the curve even though Rolle’s theorem only promises an interior one.
• Not matching the derivative of the pieces in Question 79. Continuity alone fixes one relation between $p$ and $q$; the equal one-sided derivatives fix the other.

Practise next

• Exemplar Short Answer Q44–64, the implicit, parametric and proof questions just before these.
• Exemplar Objective, Fill in the Blanks and True or False Q83–106, the next questions, which include Rolle’s theorem in Questions 95 and 96.
• NCERT textbook Exercise 5.3 for implicit differentiation, as in Question 80.
• NCERT textbook Exercise 5.5 for logarithmic differentiation, as in Question 82.