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NCERT Solutions for Class 12 Maths Chapter 5: Continuity and Differentiability

Mathematics develops analytical thinking and problem-solving ability, and Chapter 5: Continuity and Differentiability is one of the most concept-driven chapters in Class 12 Maths. This chapter introduces students to the fundamental ideas of continuity of functions and differentiability, which form the base of calculus and advanced mathematical applications.

NCERT Solutions for Class 12 Maths Chapter 5: Continuity and Differentiability

Continuity and Differentiability

Easy

Q.

\begin{array}{l} Prove that the function f (x) = 5 x - 3 is continuous at x = 0, \\ at x = - 3 and at x = 5 . \end{array}

Continuity and Differentiability

Medium

Q.

\begin{array}{l}\mathrm{Find}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{second}\mathrm{\hspace{0.33em}}\mathrm{order}\mathrm{\hspace{0.33em}}\mathrm{derivative}\mathrm{\hspace{0.33em}}\mathrm{of}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{function}:\\ {\mathrm{tan}}^{-1}\mathrm{x}\end{array}

Continuity and Differentiability

Medium

Q.

{If}\phantom{\rule{0ex}{0ex}}\quad {\mathrm{e}}^{\mathrm{y}}\left(\mathrm{x}+1\right)=1,\quad \mathrm{show}\phantom{\rule{0ex}{0ex}}\quad \mathrm{that}\phantom{\rule{0ex}{0ex}}\quad \frac{{\mathrm{d}}^{2}\mathrm{y}}{\mathrm{d}{\mathrm{x}}^{2}}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{\mathbf{2}}

Continuity and Differentiability

Medium

Q.

{If}\phantom{\rule{0ex}{0ex}}\mathrm{y}=500{\mathrm{e}}^{7\mathrm{x}}+600\phantom{\rule{0ex}{0ex}}{\mathrm{e}}^{-7\mathrm{x}},\quad \mathrm{show}\phantom{\rule{0ex}{0ex}}\quad \mathrm{that}\quad \phantom{\rule{0ex}{0ex}}\frac{{\mathrm{d}}^{2}\mathrm{y}}{\mathrm{d}{\mathrm{x}}^{2}}=\quad 49\mathrm{y}

Continuity and Differentiability

Medium

Q.

{If}\phantom{\rule{0ex}{0ex}}\mathrm{y}=\mathrm{A}{\mathrm{e}}^{\mathrm{mx}}+\mathrm{B}{\mathrm{e}}^{\mathrm{nx}},\quad \mathrm{show}\phantom{\rule{0ex}{0ex}}\quad \mathrm{that}\phantom{\rule{0ex}{0ex}}\quad \frac{{\mathrm{d}}^{2}\mathrm{y}}{\mathrm{d}{\mathrm{x}}^{2}}-\left(\mathrm{m}+\mathrm{n}\right)\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{mny}=0

Continuity and Differentiability

Medium

Q.

{If}\phantom{\rule{0ex}{0ex}}\mathrm{y}=3\mathrm{cos}\left(\mathrm{logx}\right)+\quad 4\mathrm{sin}\left(\mathrm{logx}\right),\quad \mathrm{show}\phantom{\rule{0ex}{0ex}}\quad \mathrm{that}\phantom{\rule{0ex}{0ex}}\quad {\mathrm{x}}^{2}{\mathrm{y}}_{2}+\mathrm{x}{\mathrm{y}}_{1}+\mathrm{y}=0

Continuity and Differentiability

Medium

Q.

{If }\phantom{\rule{0ex}{0ex}}\mathrm{y}=\mathrm{co}{\mathrm{s}}^{-1}\mathrm{x}.\phantom{\rule{0ex}{0ex}}\quad \mathrm{Find}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{d}}^{2}\mathrm{y}}{\mathrm{d}{\mathrm{x}}^{2}}\quad \phantom{\rule{0ex}{0ex}}\mathrm{in}\phantom{\rule{0ex}{0ex}}\quad \mathrm{term}\quad \mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{o}\quad \mathrm{f}\phantom{\rule{0ex}{0ex}}\mathrm{y}\phantom{\rule{0ex}{0ex}}\quad \mathrm{alone}.

Continuity and Differentiability

Medium

Q.

{If}\phantom{\rule{0ex}{0ex}}\quad \mathrm{y}=5\mathrm{cosx}-3\phantom{\rule{0ex}{0ex}}\mathrm{sinx},\phantom{\rule{0ex}{0ex}}\quad \mathrm{prove}\phantom{\rule{0ex}{0ex}}\quad \mathrm{that}\phantom{\rule{0ex}{0ex}}\quad \frac{{\mathrm{d}}^{2}\mathrm{y}}{\mathrm{d}{\mathrm{x}}^{2}}+\mathrm{y}=0

Continuity and Differentiability

Medium

Q.

Find the second order derivative of the function:
sin (log x)

Continuity and Differentiability

Medium

Q.

\begin{array}{l}\mathrm{Find}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{second}\mathrm{\hspace{0.33em}}\mathrm{order}\mathrm{\hspace{0.33em}}\mathrm{derivative}\mathrm{\hspace{0.33em}}\mathrm{of}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{function}:\\ {\mathrm{e}}^{6\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{cos}3\mathrm{x}\end{array}

Continuity and Differentiability

Easy

Q.

Examine the continuity of the function
f(x) = 2x² – 1 at x = 3.

Continuity and Differentiability

Medium

Q.

\begin{array}{l}\mathrm{Find}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{second}\mathrm{\hspace{0.33em}}\mathrm{order}\mathrm{\hspace{0.33em}}\mathrm{derivative}\mathrm{\hspace{0.33em}}\mathrm{of}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{function}:\\ {\mathrm{e}}^{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{sin}5\mathrm{x}\end{array}

Continuity and Differentiability

Medium

Q.

\begin{array}{l}\mathrm{Find}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{second}\mathrm{\hspace{0.33em}}\mathrm{order}\mathrm{\hspace{0.33em}}\mathrm{derivative}\mathrm{\hspace{0.33em}}\mathrm{of}\mathrm{\hspace{0.33em}}\mathrm{the}\mathrm{\hspace{0.33em}}\mathrm{function}:\\ {\mathrm{x}}^{3}\mathrm{\hspace{0.17em}}\mathrm{logx}\end{array}

Continuity and Differentiability

Medium

Q.

Find all the points of discontinuity of f, where f is defined by

f(x)={\begin{array}{l} \frac{| x |}{x}, if x \neq 0 \\ 0, if x = 0 \end{array}

Continuity and Differentiability

Medium

Q.

Find all the points of discontinuity of f, where f is defined by

f(x)={\begin{array}{l} | x | + 3, if x \leq - 3 \\ - 2 x, if - 3 < x < 3 \\ 6 x + 2, if x \geq 3 \end{array}

Continuity and Differentiability

Difficult

Q.

Find all the points of discontinuity of f, where f is defined by

Continuity and Differentiability

Medium

Q.

\begin{array}{l} Is the function f defined by \\ f (x) = \{\begin{cases} x, if x \leq 1 \\ 5, if x > 1 \end{cases} \\ continuous at x = 0 ? Atx = 1 ? Atx = 2 ? \end{array}

Continuity and Differentiability

Easy

Q.

\begin{array}{l} P r o v e t h a t t h e f u n c t i o n f (x) = x^{n} i s c o n t i n u o u s a t x = n, \\ w h e r e n i s a p o s i t i v e i n t e g e r . \end{array}

Continuity and Differentiability

Easy

Q.

\begin{array}{l} Examine the following functions for continuity . \\ (a) f (x) = x - 5 (b) f (x) = \frac{1}{x - 5}, x \neq 3 \\ (c) f (x) = \frac{x^{2} - 25}{x + 5}, x \neq - 5 (d) f (x) = |x - 5| \end{array}

Continuity and Differentiability

Medium

Q.

\begin{array}{l}\mathrm{VerifyRollesTheoremforthe}function\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\\ \mathrm{f}\left(\mathrm{x}\right)\phantom{\rule{0ex}{0ex}}=\phantom{\rule{0ex}{0ex}}{\mathrm{x}}^{2}+2\mathrm{x}-8,\phantom{\rule{0ex}{0ex}}\mathrm{x}\phantom{\rule{0ex}{0ex}}\in \phantom{\rule{0ex}{0ex}}\left[-4,2\right].\end{array}

The chapter covers important topics such as limits of functions, continuity at a point, differentiability, derivatives of composite, implicit and inverse functions, exponential and logarithmic functions, logarithmic differentiation, and derivatives of parametric functions.

These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability are prepared strictly as per the latest CBSE syllabus and NCERT guidelines. All questions are solved in a step-by-step and exam-oriented manner, making it easier for students to understand complex concepts and score confidently in board examinations as well as competitive exams like JEE.

1) Discontinuity points (piecewise function)

$f(x)= \begin{cases} 2x+3,\; x\le 2\\ 2x-3,\; x>2 \end{cases}$

$f (x) = {2 x + 3, x \leq 22 x - 3, x > 2$

At $x=2$

$x = 2$ :
LHL $=2(2)+3=7$

$= 2 (2) + 3 = 7$
RHL $=2(2)-3=1$

$= 2 (2) - 3 = 1$
LHL $\ne$

$\neq =$ RHL ⇒ Discontinuous at $x=2$

$x = 2$ .

✅ Answer: $x=2$

$x = 2$

2) Second derivative of $x^3\log x$

$x^{3} log x$ (log = ln)

Let $y=x^3\ln x$

$y = x^{3} ln x$

$y' = 3x^2\ln x + x^2 = x^2(3\ln x+1)$

$y^{'} = 3 x^{2} ln x + x^{2} = x^{2} (3 ln x + 1)$ $y'' = 6x\ln x + 5x = x(6\ln x+5)$

$y^{''} = 6 x ln x + 5 x = x (6 ln x + 5)$

✅ Answer: $y''=x(6\log x+5)$

$y^{''} = x (6 log x + 5)$

3) Second derivative of $e^x\sin 5x$

$e^{x} sin 5 x$

$y=e^x\sin 5x$

$y = e^{x} sin 5 x$ $y' = e^x(\sin 5x + 5\cos 5x)$

$y^{'} = e^{x} (sin 5 x + 5 cos 5 x)$ $y'' = e^x(-24\sin 5x + 10\cos 5x)$

$y^{''} = e^{x} (- 24 sin 5 x + 10 cos 5 x)$

✅ Answer: $y''=e^x(10\cos 5x-24\sin 5x)$

$y^{''} = e^{x} (10 cos 5 x - 24 sin 5 x)$

4) Second derivative of $e^{6x}\cos 3x$

$e^{6 x} cos 3 x$

$y=e^{6x}\cos 3x$

$y = e^{6 x} cos 3 x$ $y' = e^{6x}(6\cos 3x - 3\sin 3x)$

$y^{'} = e^{6 x} (6 cos 3 x - 3 sin 3 x)$ $y'' = e^{6x}(27\cos 3x - 36\sin 3x)$

$y^{''} = e^{6 x} (27 cos 3 x - 36 sin 3 x)$

✅ Answer: $y''=e^{6x}(27\cos 3x-36\sin 3x)$

$y^{''} = e^{6 x} (27 cos 3 x - 36 sin 3 x)$

5) Second derivative of $\tan^{-1}x$

$tan^{-1} x$

$y=\tan^{-1}x$

$y = tan^{-1} x$ $y'=\frac1{1+x^2}$

$y^{'} = 1 + x ^{2} 1$ $y''=\frac{-2x}{(1+x^2)^2}$

$y^{''} = ( 1 + x ^{2} ) ^{2} -2 x$

✅ Answer: $\;y''=-\dfrac{2x}{(1+x^2)^2}$

$y^{''} = - ( 1 + x ^{2} ) ^{2} 2 x$

6) Second derivative of $\sin(\log x)$

$sin (log x)$

$y=\sin(\ln x)$

$y = sin (ln x)$ $y'=\frac{\cos(\ln x)}{x}$

$y^{'} = x cos ( ln x )$ $y''=-\frac{\cos(\ln x)+\sin(\ln x)}{x^2}$

$y^{''} = - x ^{2} cos ( ln x ) + sin ( ln x )$

✅ Answer: $\;y''=-\dfrac{\cos(\log x)+\sin(\log x)}{x^2}$

$y^{''} = - x ^{2} cos ( log x ) + sin ( log x )$

7) If $y=5\cos x-3\sin x$

$y = 5 cos x - 3 sin x$ , prove $y''+y=0$

$y^{''} + y = 0$

$y'=-5\sin x-3\cos x$

$y^{'} = - 5 sin x - 3 cos x$ $y''=-5\cos x+3\sin x = -(5\cos x-3\sin x)=-y$

$y^{''} = - 5 cos x + 3 sin x = - (5 cos x - 3 sin x) = - y$

So $y''+y=0$

$y^{''} + y = 0$ .

✅ Proved.

8) If $y=\cos^{-1}x$

$y = cos^{-1} x$ , find $y''$

$y^{''}$ in terms of $y$

$y$ alone

$y'=-\frac1{\sqrt{1-x^2}},\quad y''=-\frac{x}{(1-x^2)^{3/2}}$

$y^{'} = - 1 - x ^{2} 1, y^{''} = - ( 1 - x ^{2} ) ^{3/2} x$

Now $x=\cos y$

$x = cos y$ and $1-x^2=\sin^2 y$

$1 - x^{2} = sin^{2} y$ (since $y\in[0,\pi]$

$y \in [0, π]$ , $\sin y\ge0$

$sin y \geq 0$ )

$y''=-\frac{\cos y}{\sin^3 y}$

$y^{''} = - sin ^{3} y cos y$

✅ Answer: $\;y''=-\dfrac{\cos y}{\sin^3 y}$

$y^{''} = - sin ^{3} y cos y$

9) If $y=3\cos(\log x)+4\sin(\log x)$

$y = 3 cos (log x) + 4 sin (log x)$ , show $x^2y''+xy'+y=0$

$x^{2} y^{''} + x y^{'} + y = 0$

(standard result)
Compute $y'=\frac{-3\sin(\log x)+4\cos(\log x)}{x}$

$y^{'} = x -3 s i n ( l o g x ) + 4 c o s ( l o g x )$ ,
$y''=\frac{-7\cos(\log x)-\sin(\log x)}{x^2}$

$y^{''} = x ^{2} -7 c o s ( l o g x ) - s i n ( l o g x )$
Then:

$x^2y''+xy'+y=0$

$x^{2} y^{''} + x y^{'} + y = 0$

✅ Proved.

10) If $y=Ae^{mx}+Be^{nx}$

$y = A e^{m x} + B e^{n x}$ , show $y''-(m+n)y'+mny=0$

$y^{''} - (m + n) y^{'} + mn y = 0$

$y'=Am e^{mx}+Bn e^{nx},\quad y''=Am^2 e^{mx}+Bn^2 e^{nx}$

$y^{'} = A m e^{m x} + B n e^{n x}, y^{''} = A m^{2} e^{m x} + B n^{2} e^{n x}$

Substitute ⇒ both terms become 0.

✅ Proved.

11) If $y=500e^{7x}+600e^{-7x}$

$y = 500 e^{7 x} + 600 e^{-7 x}$ , show $y''=49y$

$y^{''} = 49 y$

$y''=24500e^{7x}+29400e^{-7x}=49(500e^{7x}+600e^{-7x})=49y$

$y^{''} = 24500 e^{7 x} + 29400 e^{-7 x} = 49 (500 e^{7 x} + 600 e^{-7 x}) = 49 y$

✅ Proved.

12) If $e^y(x+1)=1$

$e^{y} (x + 1) = 1$ , show $y''=(y')^2$

$y^{''} = (y^{'})^{2}$

From $e^y=\frac1{x+1}\Rightarrow y=-\ln(x+1)$

$e^{y} = x + 1 1 \Rightarrow y = - ln (x + 1)$

$y'=-\frac1{x+1},\quad y''=\frac1{(x+1)^2}$

$y^{'} = - x + 11, y^{''} = ( x + 1 ) ^{2} 1$ $(y')^2=\frac1{(x+1)^2}=y''$

$(y^{'})^{2} = ( x + 1 ) ^{2} 1 = y^{''}$

✅ Proved.

13) Verify Rolle’s theorem for $f(x)=x^2+2x-8$

$f (x) = x^{2} + 2 x - 8$ on $[-4,2]$

$[- 4, 2]$

Polynomial ⇒ continuous & differentiable.

$f(-4)=16-8-8=0,\quad f(2)=4+4-8=0$

$f (- 4) = 16 - 8 - 8 = 0, f (2) = 4 + 4 - 8 = 0$

So $f(-4)=f(2)$

$f (- 4) = f (2)$ .

$f'(x)=2x+2=0 \Rightarrow x=-1$

$f^{'} (x) = 2 x + 2 = 0 \Rightarrow x = - 1$

$-1\in(-4,2)$

$- 1 \in (- 4, 2)$

✅ Rolle’s theorem verified at $c=-1$

$c = - 1$

FAQs: NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

1. Is Chapter 5 Continuity and Differentiability important for the Class 12 board exam?

Yes, this is one of the most important and high-weightage chapters in Class 12 Maths. Questions based on continuity, differentiability, and derivatives are frequently asked in the board exams.

2. Are NCERT Solutions enough to score well in this chapter?

Yes. Practising all NCERT examples and exercise questions thoroughly is sufficient, as most CBSE board questions are directly or indirectly based on NCERT.

3. Is this chapter important for JEE preparation?

Absolutely. Continuity and Differentiability forms the core foundation of calculus and is extremely important for JEE Main and JEE Advanced.

4. Which topics require more practice in this chapter?

Students should practise more:

Continuity of functions
Differentiability at a point
Derivatives using chain rule
Logarithmic differentiation
Derivatives of inverse and implicit functions

5. Are these NCERT Solutions updated as per the latest CBSE syllabus?

Yes, these NCERT Solutions for Class 12 Maths Chapter 5 are fully updated according to the latest CBSE curriculum and NCERT textbook.

6. Can these solutions be used for last-minute revision?

Yes. The step-wise explanation and clear presentation make these solutions ideal for quick revision before exams.