NCERT Exemplar for Class 12 Maths Chapter 5 (Book Solutions)

Solved Examples 

Short Answer Questions 

1. Find the value of the constant k so that the function f defined below is continuous at x = 0, where

$f\left( x \right)=\left\{ \begin{align} & \dfrac{1-cos~4x}{8{{x}^{2}}},\text{ at } x\ne 0 \\ & k,\text{ at } x = 0 \\ \end{align} \right.$

Ans: Here, It is given that the function f is continuous at x = 0. Therefore, $\underset{x\to 0}{\mathop{lim}}\,$f(x) = f(0)

\[\Rightarrow \underset{x\to 0}{\mathop{lim}}\,\dfrac{1-cos~4x}{8{{x}^{2}}} = k\]

\[\Rightarrow \underset{x\to 0}{\mathop{lim}}\,\dfrac{2si{{n}^{2}}2x}{8{{x}^{2}}} = k \]

\[\Rightarrow \underset{x\to 0}{\mathop{lim}}\,{{(\dfrac{sin2x}{2{{x}^{{}}}})}^{2}} = k \]

\[\Rightarrow k = 1 \]

Hence the value of k = 1 

2. Discuss the continuity of the function f(x) = sin x. cos x.

Ans: As we know that sin x and cos x are continuous functions and the product of two continuous functions is always a continuous function.

Hence, f(x) = sin x . cos x is a continuous function.

3. $f\left( x \right)=\left\{ \begin{align} & \dfrac{{{x}^{3}}+{{x}^{2}}-16x+20}{{{(x-2)}^{2}}}, x\ne 2 \\ & k,\text{ }x=2 \\ \end{align} \right.$ is continuous at x =2, Find the value of k .

Ans: Given f (2) = k. 

Now, $\underset{x\to {{2}^{-}}}{\mathop{lim}}\,$f(x) = $\underset{x\to {{2}^{+}}}{\mathop{lim}}\,$f(x) = $\underset{x\to {{2}^{{}}}}{\mathop{lim}}\,\dfrac{{{x}^{3~}}~+~{{x}^{2~}}-16x~+~20}{{{(x-2)}^{2}}}$

\[\Rightarrow \] $\underset{x\to {{2}^{{}}}}{\mathop{lim}}\,\dfrac{\left( x+5 \right){{(x-2)}^{2}}}{{{(x-2)}^{2}}}$

\[\Rightarrow \]$\underset{x\to {{2}^{{}}}}{\mathop{lim}}\,\left( x+5 \right)$ = 7

Now since f is continuous at x = 2, we have

$\underset{x\to {{2}^{{}}}}{\mathop{lim}}\,$f(x) = f(2)

 k = 7

4. Show that the function f defined by $f\left( x \right)=\left\{ \begin{align} & x\sin \dfrac{1}{x},\text{ }x\ne 0 \\ & 0,\text{ }x=0 \\ \end{align} \right.$ is continuous at x = 0

Ans: Left hand limit at x = 0 is given by 

$\underset{x\to {{0}^{-}}}{\mathop{lim}}\,$f(x) = $\underset{x\to {{0}^{-}}}{\mathop{lim}}\,$xsin$\dfrac{1}{x}$ = 0

Similarly , $\underset{x\to {{0}^{+}}}{\mathop{lim}}\,$f(x) = $\underset{x\to {{0}^{+}}}{\mathop{lim}}\,$xsin$\dfrac{1}{x}$ = 0

Also f(0) = 0

Hence $\underset{x\to {{0}^{-}}}{\mathop{lim}}\,$f(x) = $\underset{x\to {{0}^{+}}}{\mathop{lim}}\,$f(x) = f(0) 

\[\Rightarrow \] f is continuous at x = 0

5. Given f(x) = $\dfrac{1}{x-1}$. Find the points of discontinuity of the composite function y = $f\left[f(x)\right]$. 

Ans: We know that f (x) = $\dfrac{1}{x-1}$ is discontinuous at x = 1 Now, for x ≠ 1,

f (f (x)) = f$\left(\dfrac{1}{x-1}\right)$ = $\dfrac{1}{\dfrac{1}{x-1}-1}$ = $\dfrac{x-1}{2-x}$

which is discontinuous at x = 2. 

Therefore, the points of discontinuity are x = 1 and x = 2. 

6. Let f(x) = x|x|, for all x ∈ R. Discuss the derivability of f(x) at x = 0

Ans: We may rewrite f as $f\left( x \right)=\left\{ \begin{align} & {{x}^{2}},\text{if}\text{ }x\ge 0 \\ & -{{x}^{2}},\text{ }if\text{ }x < 0 \\ \end{align} \right.$ 

We know that 

Lf ′ (0) = $\underset{x\to {{0}^{-}}}{\mathop{lim}}\,\dfrac{f\left( 0-h \right)~-~f\left( 0 \right)}{h}$ = $\underset{x\to {{0}^{-}}}{\mathop{lim}}\,\dfrac{-{{h}^{2~}}-0~}{-h}$ = $\underset{x\to {{0}^{-}}}{\mathop{lim}}\,$-h = 0

Rf ′ (0) = $\underset{x\to {{0}^{+}}}{\mathop{lim}}\,\dfrac{f\left( 0+h \right)~-~f\left( 0 \right)}{h}$ = $\underset{x\to {{0}^{+}}}{\mathop{lim}}\,\dfrac{{{h}^{2~}}-0~}{h}$ = $\underset{x\to {{0}^{+}}}{\mathop{lim}}\,$h = 0

Now clearly, the left hand derivative and right hand derivative both are equal, Therefore , f is differentiable at x = 0.

7. Differentiate $\sqrt{tan\sqrt{x}}$ w.r.t x

Ans: Let us consider y = $\sqrt{tan\sqrt{x}}$ 

$\dfrac{dy}{dx}$ = $\dfrac{1}{2\sqrt{tan\sqrt{x}}}\dfrac{d}{dx}$tan$\sqrt{x}$

=$\dfrac{1}{2\sqrt{tan\sqrt{x}}}se{{c}^{2}}{{\sqrt{x}}^{{}}}\dfrac{d}{dx}{{\sqrt{x}}^{{}}}$

 =$\dfrac{1}{2\sqrt{tan\sqrt{x}}}se{{c}^{2}}{{\sqrt{x}}^{{}}}{{\dfrac{1}{2\sqrt{x}}}^{{}}}$

=$\dfrac{sec^{2}\sqrt{x}}{4\sqrt{x}\sqrt{tan{\sqrt{x}}}}$

8. If y = tan(x + y), find $\dfrac{dy}{dx}$ .

Ans: Here, y = tan (x + y).

On differentiating both sides w.r.t. x, we have

$\dfrac{dy}{dx}$ = $\dfrac{d}{dx}$tan(x+y)

$\Rightarrow \dfrac{dy}{dx}$ = $se{{c}^{2}}\left( x+y \right){{~}^{{}}}\dfrac{d}{dx}$(x+y)

$\Rightarrow \dfrac{dy}{dx}$ = (1+$\dfrac{dy}{dx}$) $se{{c}^{2}}\left( x+y \right){{~}^{{}}}$

$\Rightarrow \dfrac{dy}{dx}$ = $se{{c}^{2}}\left( x+y \right){{~}^{{}}}$+ $\dfrac{dy}{dx}se{{c}^{2}}\left( x+y \right){{~}^{{}}}$

$\Rightarrow \dfrac{dy}{dx}$ - $\dfrac{dy}{dx}se{{c}^{2}}\left( x+y \right){{~}^{{}}}$ = $se{{c}^{2}}\left( x+y \right)~$

$\Rightarrow (1-se{{c}^{2}}\left( x+y \right)~)\dfrac{dy}{dx}$ = $se{{c}^{2}}\left( x+y \right)~$

$\Rightarrow \dfrac{dy}{dx}$ = $\dfrac{se{{c}^{2}}\left( x+y \right)~}{1-se{{c}^{2}}\left( x+y \right)~}$ = - co$se{{c}^{2}}\left( x+y \right)$

Hence, $\dfrac{dy}{dx}$ = - co$se{{c}^{2}}\left( x+y \right)$

9. If ${{e}^{x}}^{{}}$ + ${{e}^{y}}^{{}}$ = ${{e}^{x+y}}^{{}}$, Prove that $\dfrac{dy}{dx}$ = - ${{e}^{y-x}}^{{}}$.

Ans: Here we have , ${{e}^{x}}^{{}}$+ ${{e}^{y}}^{{}}$ = ${{e}^{x+y}}^{{}}$.

On Differentiating both sides w.r.t x , we get 

$\dfrac{d}{dx}$ $({{e}^{x}}^{{}}$+ ${{e}^{y}}^{{}})$ = $\dfrac{d}{dx}{{e}^{x+y}}$

$\Rightarrow {{e}^{x}}$ + ${{e}^{y}}^{{}}\dfrac{dy}{dx}$ = ${{e}^{x+y}}^{{}}\dfrac{d}{dx}$(x+y)

$\Rightarrow {{e}^{x}}$ + ${{e}^{y}}^{{}}\dfrac{dy}{dx}$ = (1+$\dfrac{dy}{dx}$)${{e}^{x+y}}^{{}}$

$\Rightarrow {{e}^{x}}$ + ${{e}^{y}}^{{}}\dfrac{dy}{dx}$ = $~{{e}^{x+y}}^{{}}$ + ${{e}^{x+y}}^{{}}\dfrac{dy}{dx}$

$\Rightarrow({{e}^{y}}$ - ${{e}^{x+y}}^{{}}$)$\dfrac{dy}{dx}$ = ${{e}^{x+y}}^{{}}$- ${{e}^{x}}$ 

$\Rightarrow \dfrac{dy}{dx}$ = $\dfrac{{{e}^{x+y}}^{{}}-~{{e}^{x}}~}{{{e}^{y}}~-~{{e}^{x+y}}^{{}}}$ 

$\Rightarrow\dfrac{dy}{dx}$=$\dfrac{{{e}^{x~}}^{{}}+~{{e}^{y}}~-~{{e}^{x}}~}{{{e}^{y}}~-~{{e}^{x~}}-~{{e}^{y}}{{~}^{{}}}^{{}}}$ $\left[\because {{e}^{x}}^{{}}+{{e}^{y}}^{{}} = {{e}^{x+y}}^{{}}\right]$

$\Rightarrow \dfrac{dy}{dx}$ = -${{e}^{y-x}}^{{}}$

Hence Proved.

10. Find $\dfrac{dy}{dx}$ if y = $ta{{n}^{-1}}{{\dfrac{\left( 3x~-{{x}^{3}} \right)}{1-3{{x}^{2}}}}^{{}}}$, - $\dfrac{1}{\sqrt{3}}~$<x <$\dfrac{1}{\sqrt{3}}$

Ans: Let us put x = tanθ ⇒ θ = $ta{{n}^{-1}}$x 

so, y = $ta{{n}^{-1}}{{\dfrac{\left( 3~tan\theta ~~-~ta{{n}^{3}}\theta {{~}^{{}}} \right)}{1-3~ta{{n}^{2}}\theta ~}}^{{}}}$

We know that tan3θ = $\dfrac{\left( 3~tan\theta ~~-~ta{{n}^{3}}\theta {{~}^{{}}} \right)}{1-3~ta{{n}^{2}}\theta ~}$

$\Rightarrow y = ta{{n}^{-1}}tan3{{\theta }^{{}}}$ 

$\Rightarrow y = 3\theta $  $\left( -\dfrac{\pi }{2}<3\theta <\dfrac{\pi }{2} \right)$

$\Rightarrow y = 3ta{{n}^{-1}}$x 

Now on differentiating both sides w.r.t x 

$\Rightarrow \dfrac{dy}{dx}$ = $\dfrac{d}{dx}$(3$ta{{n}^{-1}}$x)

$\Rightarrow \dfrac{dy}{dx}$ = 3$\dfrac{1}{1+{{x}^{2}}}$

Hence, $\dfrac{dy}{dx}$ = $\dfrac{3}{1+{{x}^{2}}}$

11. If y = $si{{n}^{-1}}$$\left\{ x\sqrt{1-x~}~-\sqrt{x}\sqrt{1-{{x}^{2}}} \right\}$ and 0 < x < 1, then find $\dfrac{dy}{dx}$.

Ans: we have 

y = $si{{n}^{-1}}${x$\sqrt{1-x~}~-\sqrt{x}\sqrt{1-{{x}^{2}}}$ } ; 0<x<1

Now, let's put x = sinA and $\sqrt{x}$ = sinB

y = $si{{n}^{-1}}$ {sinA$\sqrt{1-si{{n}^{2}}B~}~-sinB\sqrt{1-si{{n}^{2}}A}$ } 

= $si{{n}^{-1}}$ {sinAcosB - sinBcosA }

= $si{{n}^{-1}}$ { sin(A - B) } = A – B

y = $si{{n}^{-1}}$x – $si{{n}^{-1}}\sqrt{x}$

On Differentiating w.r.t to x , we get

$\dfrac{dy}{dx}$ = $\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ - $\dfrac{1}{\sqrt{1-{{\sqrt{{{x}^{2}}}}^{{}}}}}$.$\dfrac{d}{dx}$($\sqrt{x})$

= $\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ - $\dfrac{1}{2\sqrt{x}~.\sqrt{1-x}}$

Therefore , $\dfrac{dy}{dx}$ = $\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ - $\dfrac{1}{2\sqrt{x}~.\sqrt{1-x}}$

12. If x = $a se{{c}^{3}}\theta ~$ and y = $a ta{{n}^{3}}\theta ~$, find $\dfrac{dy}{dx}$ at $\theta $ = $\dfrac{\pi }{3}$

Ans: we have x = a$se{{c}^{3}}\theta ~$ and y = a $ta{{n}^{3}}\theta ~$

On Differentiating both sides w.r.t $\theta $

$\dfrac{dx}{d\theta }$ = 3a$se{{c}^{2}}\theta ~$sec$\theta $.tan$\theta $ 

and $\dfrac{dy}{d\theta }$ = 3a$ta{{n}^{2}}\theta ~se{{c}^{2}}\theta $

Now, $\dfrac{dy}{dx}$ = $\dfrac{\dfrac{dy}{d\theta }}{\dfrac{dx}{d\theta }}$ = $\dfrac{3ata{{n}^{2}}\theta ~se{{c}^{2}}\theta }{3ase{{c}^{2}}\theta ~sec\theta .tan\theta ~}$ = sin$\theta $

At $\theta $ = $\dfrac{\pi }{3}$

$\dfrac{dy}{dx}$ = sin $\dfrac{\pi }{3}$ = $\dfrac{\sqrt{3}}{2}$

Hence, 

$\dfrac{dy}{dx}$ = $\dfrac{\sqrt{3}}{2}$ , at $\theta $ = $\dfrac{\pi }{3}$

13. If ${{x}^{y}}$ = ${{e}^{\left( x-y \right)}}$, Prove that $\dfrac{dy}{dx}$ = $\dfrac{log~x~}{{{(1+logx)}^{2}}}$

Ans: we have ,

${{x}^{y}}$ = ${{e}^{\left( x-y \right)}}$

Taking log on both the sides we get ,

log(${{x}^{y}}$) = log$({{e}^{x-y}})$

⇒y log x = (x-y)loge      

⇒y log x = (x-y).1

⇒y (log x +1) = x 

$\Rightarrow y = \dfrac{x}{1+logx}$

Using Quotient Rule , we get 

$\Rightarrow \dfrac{dy}{dx}$ = $\dfrac{\left( 1+logx \right).\dfrac{d}{dx}\left( x \right)-x.\dfrac{d}{dx}\left( 1+logx \right)}{{{(1+logx)}^{2}}}$

$\Rightarrow \dfrac{dy}{dx}$ = $~\dfrac{\left( 1+logx \right)- 1}{{{(1+logx)}^{2}}}$ 

$\Rightarrow \dfrac{dy}{dx}$ = $\dfrac{log~x~}{{{(1+logx)}^{2}}}$

Hence Proved.

14. If y = tanx + secx, prove that $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ = $\dfrac{cos x}{{{(1-sinx)}^{2}}}$ 

Ans: we have, 

y = tanx + secx

On differentiating both sides w.r.t x , we get 

$\dfrac{dy}{dx}$ = $\dfrac{d}{dx}$ ( tanx ) + $\dfrac{d}{dx}$ ( secx )

$\Rightarrow \dfrac{dy}{dx}$ = $se{{c}^{2}}x$ + secx.tanx 

$\Rightarrow \dfrac{dy}{dx}$ = $\dfrac{1}{co{{s}^{2}}x}$ + $\dfrac{sinx}{co{{s}^{2}}x}$ = $ \dfrac{1+sinx}{co{{s}^{2}}x}$ 

$\Rightarrow \dfrac{dy}{dx}$ = $\dfrac{1~+~sinx}{co{{s}^{2}}x}$ 

$\Rightarrow \dfrac{dy}{dx}$ = $\dfrac{1~+~sinx}{1-si{{n}^{2}}x}$ 

$\Rightarrow \dfrac{dy}{dx}$ = $\dfrac{1~+~sinx}{\left( 1+si{{n}^{{}}}x \right)\left( 1-sinx \right)}$ 

$\Rightarrow \dfrac{dy}{dx}$ = $\dfrac{1~}{\left( 1-sinx \right)}$ 

Again Differentiating both sides w.r.t x

$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ = $\dfrac{-\left( -cosx \right)}{{{(1-sinx)}^{2}}}$ 

$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ = $\dfrac{cosx}{{{(1-sinx)}^{2}}}$ 

Hence Proved .

15. If f(x) = |cosx|, find f’ ($\dfrac{3\pi }{4}$)

Ans: we know that cosx < 0 when $\dfrac{\pi }{2}~$< x < $\pi $

 |cosx| = -cosx 

so , f(x) = -cosx 

 f’(x) = sinx 

Therefore, f’ $\left( \dfrac{3\pi }{4} \right)$ = sin$\dfrac{3\pi }{4}$

  f’ $\left( \dfrac{3\pi }{4} \right)$ = $\dfrac{1}{\sqrt{2}}$

Hence Proved 

16. If f(x) = |cosx-sinx |, find f’($\dfrac{\pi }{6}$)

Ans: we know that cosx > sinx when 0 < x <$\dfrac{\pi }{4}$

 cos x – sin x > 0 

so, f(x) = cosx - sinx 

f’(x) = – sin x – cos x

Therefore, f’ ($\dfrac{\pi }{6}$) = -sin$\dfrac{\pi }{6}$ - cos$\dfrac{\pi }{6}$

f’ ($\dfrac{\pi }{6}$) = $\dfrac{-1}{2}$(1+$\sqrt{3}$)

Hence Proved 

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

Ans: we have,

f (x) = sin 2x in $\left[ 0,\dfrac{\pi }{2} \right]$ 

The given function is continuous in $\left[ 0,\dfrac{\pi }{2} \right]$ as f is a sine function.

f’(x) = 2cos2x , exists in $\left( 0,\dfrac{\pi }{2} \right)$, hence f is differentiable in $\left( 0,\dfrac{\pi }{2} \right)$ 

f(0) = sin 0 = 0 and f$\left( \dfrac{\pi }{2} \right)$= sin$\dfrac{\pi }{2}$ = 0 

f(0) = f$\left( \dfrac{\pi }{2} \right)$

Since, all the conditions of Rolle’s theorem are satisfied , so there exists a c ∈

$\left( 0,\dfrac{\pi }{2} \right)$ such that f’(c) = 0 

f’(x) = 2cos2x 

⇒f’(c) = 2cos2c = 0 

⇒cos2c = 0 

⇒2c = $\dfrac{\pi }{2}$

⇒c = $\dfrac{\pi }{4}$

18. Verify mean value theorem for the function f (x) = (x – 3) (x – 6) (x – 9) in $\left[3, 5\right]$. 

Ans: (i) Given function f(x) is continuous in $\left[3, 5\right]$ as the product of polynomial functions is a polynomial and all polynomial functions are continuous 

(ii) f ′(x) = 3${{x}^{2}}$ – 36x + 99 exists in (3, 5) and hence is derivable in (3, 5).

Thus conditions of the mean value theorem are satisfied. Hence, there exists at least one c ∈ (3, 5) such that

f’(c) = $\dfrac{f\left( 5 \right)~-~f\left( 3 \right)}{5-3}$ 

⇒3${{c}^{2}}$ – 36c + 99 = $\dfrac{8-0}{2}$ 

⇒3${{c}^{2}}$ – 36c + 99 = 4

⇒3${{c}^{2}}$ – 36c + 95 = 0

⇒c = 6±$\sqrt{\dfrac{13}{3}}$

⇒c $\ne ~6~+\sqrt{\dfrac{13}{3}}$

Therefore , c = $~6~-\sqrt{\dfrac{13}{3}}$

Long Answers (L.A.)

19. If f (x) = $\dfrac{\sqrt{2}cosx~-1}{cotx-1}$ , x$\ne $ $\dfrac{\pi }{4}$, Find the value of f($\dfrac{\pi }{4}$) so that f(x) becomes continuous at x = $\dfrac{\pi }{4}$

Ans: we have f (x) = $\dfrac{\sqrt{2}cosx~-1}{cotx-1}$ , x$\ne $ $\dfrac{\pi }{4}$ 

$\underset{x\to ~~\dfrac{\pi }{4}}{\mathop{lim}}\,$f(x) = $\underset{x\to ~~\dfrac{\pi }{4}}{\mathop{lim}}\,\dfrac{\sqrt{2}cosx~-1}{cotx-1}$ 

= $\underset{x\to ~~\dfrac{\pi }{4}}{\mathop{lim}}\,\dfrac{\left( \sqrt{2}cosx~-1 \right)sinx}{cosx~-~sinx}$ 

= $\underset{x\to ~~\dfrac{\pi }{4}}{\mathop{lim}}\,\dfrac{\left( \sqrt{2}cosx~-1 \right)}{cosx~-~sinx}$ $\dfrac{\left( \sqrt{2}cosx~+1 \right)~\left( cosx~+sinx \right)sinx}{\left( \sqrt{2}cosx~+1 \right)\left( cosx~+sinx \right)}$

= $\underset{x\to ~~\dfrac{\pi }{4}}{\mathop{lim}}\,\dfrac{\left( 2co{{s}^{2}}x~-1 \right)\left( cosx~+sinx \right)}{\left( co{{s}^{2}}x~-~si{{n}^{2}}x \right)\left( \sqrt{2}cosx~+1 \right)}.sinx$ 

= $\underset{x\to ~~\dfrac{\pi }{4}}{\mathop{lim}}\,\dfrac{cos2x\left( cosx~+sinx \right)}{cos2x\left( \sqrt{2}cosx~+1 \right)}.sinx$ 

= $\underset{x\to ~~\dfrac{\pi }{4}}{\mathop{lim}}\,\dfrac{\left( cosx~+sinx \right)}{\left( \sqrt{2}cosx~+1 \right)}.sinx$ 

= $\dfrac{~\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} \right)}{\dfrac{1}{\sqrt{2}}.\sqrt{2}+1}\cdot \dfrac{1}{\sqrt{2}}$ = $\dfrac{1}{2}$

Therefore, $\underset{x\to ~~\dfrac{\pi }{4}}{\mathop{lim}}\,$f(x) = $\dfrac{1}{2}$

Hence , for f to be continuous at x = $\dfrac{\pi }{4}$ , f($\dfrac{\pi }{4}$) should be equal to $\dfrac{1}{2}$ 

i.e. f($\dfrac{\pi }{4}$) = $\dfrac{1}{2}$

20. Show that the function f given by $f\left( x \right)=\left\{ \begin{align} & \dfrac{{{e}^{\dfrac{1}{x}}~-1}}{{{e}^{\dfrac{1}{x}}~+1}},\text{ }if\text{ }x\ne 0 \\ & 0,\text{ }if\text{ }x=0 \\ \end{align} \right.$ is discontinuous at x =0.

Ans: The left hand limit of f at x = 0 is given by

$\underset{x\to ~~{{0}^{-}}}{\mathop{lim}}\,$f(x) = $\underset{x\to ~~{{0}^{-}}}{\mathop{lim}}\,\dfrac{{{e}^{\dfrac{1}{x}}}~-1}{{{e}^{\dfrac{1}{x}}}~+1}$ = $\dfrac{0-1}{0+1}$ = -1

Similarly,

$\underset{x\to ~~{{0}^{+}}}{\mathop{lim}}\,$f(x) = $\underset{x\to ~~{{0}^{+}}}{\mathop{lim}}\,\dfrac{{{e}^{\dfrac{1}{x}}}~-1}{{{e}^{\dfrac{1}{x}}}~+1}$ = $\dfrac{1-0}{1+0}$ = 1

Therefore,

$\underset{x\to ~~{{0}^{-}}}{\mathop{lim}}\,$f(x) $\ne $ $\underset{x\to ~~{{0}^{+}}}{\mathop{lim}}\,$f(x) , hence $\underset{x\to ~~{{0}^{{}}}}{\mathop{lim}}\,$f(x) does not exist .

f is discontinuous at x = 0

21. Let $f\left( x \right)=\left\{ \begin{align} & \dfrac{1-cos4x}{{x}^{2}},\text{ }if\text{ }x\le 0 \\ & a,\text{}if\text{ }x=0 \\ & \dfrac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4},\text{}if\text{ }x>0\\ \end{align} \right.$ For what value of a , f is continuous at x = 0 ?

Ans: Here f (0) = a Left hand limit of f at 0 is 

$\underset{x\to ~~{{0}^{-}}}{\mathop{lim}}\,$f(x) = $\underset{x\to ~~{{0}^{-}}}{\mathop{lim}}\,\dfrac{1-cos4x}{{{x}^{2}}}$ 

= $\underset{x\to ~~{{0}^{-}}}{\mathop{lim}}\,\dfrac{2si{{n}^{2}}2x}{{{x}^{2}}}$

= $~\underset{x\to ~~{{0}^{-}}}{\mathop{lim}}\,8.{{(\dfrac{si{{n}^{{}}}2x}{{{x}^{{}}}})}^{2}}$ 

= $8.{{(1)}^{2}}$ 

= 8

Now, $\underset{x\to ~~{{0}^{+}}}{\mathop{lim}}\,$f(x) = $\underset{x\to ~~{{0}^{+}}}{\mathop{lim}}\,\dfrac{\sqrt{x}}{\sqrt{16~+~\sqrt{x}}~-~4}$

= $\underset{x\to ~~{{0}^{+}}}{\mathop{lim}}\,\dfrac{\sqrt{x}\left( \sqrt{16~+~\sqrt{x}}~+4 \right)}{\left( \sqrt{16~+~\sqrt{x}}~-~4 \right)\left( \sqrt{16~+~\sqrt{x}}~+~4 \right)}$

= $\underset{x\to ~~{{0}^{+}}}{\mathop{lim}}\,\dfrac{\sqrt{x}\left( \sqrt{16~+~\sqrt{x}}~+4 \right)}{\left( \sqrt{16~+~\sqrt{x}}~-~4 \right)\left( \sqrt{16~+~\sqrt{x}}~+~4 \right)}$

= $\underset{x\to ~~{{0}^{+}}}{\mathop{lim}}\,\dfrac{\sqrt{x}\left( \sqrt{16~+~\sqrt{x}}~+4 \right)}{\left( 16~+\sqrt{x}~-~16 \right)}$

 = $\underset{x\to ~~{{0}^{+}}}{\mathop{lim}}\,\dfrac{\sqrt{x}\left( \sqrt{16~+~\sqrt{x}}~+4 \right)}{\sqrt{x}}$

$=\underset{x\to ~~{{0}^{+}}}{\mathop{lim}}\,(\sqrt{16~+~\sqrt{x}}~+4$) 

= 8

Therefore, $\underset{x\to ~~{{0}^{-}}}{\mathop{lim}}\,$f(x) $=$ $\underset{x\to ~~{{0}^{+}}}{\mathop{lim}}\,$f(x) = 8 , hence f is continuous at x = 0 only if a = 8.

22. Examine the differentiability of the function f defined by

$f\left( x \right)=\left\{ \begin{align} & 2x+3 ,,\text{ }if\text{ }-3 \le x < -2 \\ & x+1,\text{ }if\text{ }-2 \le x < 0 \\ & x+2, \text{ }if\text{ }0 \le x \le 1 \end{align} \right.$

Ans: The points where differentiability can't be directly predicted of f (x) are x = – 2 and x = 0.

Let us find Differentiability at x = – 2 and at x = 0 

Differentiability at x = – 2

L.H.D = f’(-2) = $\underset{h\to {{0}^{-}}}{\mathop{lim}}\,\dfrac{f\left( -2+h \right)~-~f\left( -2 \right)~}{h}$

= $\underset{h\to {{0}^{-}}}{\mathop{lim}}\,\dfrac{2\left( -2+h \right)~+3~-~\left( -2+1 \right)~}{h}$

= $\underset{h\to {{0}^{-}}}{\mathop{lim}}\,\dfrac{2h}{h}$

= 2

R.H.D = f’(-2) = $\underset{h\to {{0}^{+}}}{\mathop{lim}}\,\dfrac{f\left( -2+h \right)~-~f\left( -2 \right)~}{h}$

= $\underset{h\to {{0}^{+}}}{\mathop{lim}}\,\dfrac{-2+h+1~~-~\left( -2+1 \right)~}{h}$

= $\underset{h\to {{0}^{+}}}{\mathop{lim}}\,\dfrac{h}{h}$

= 1

Clearly, L.H.D ≠ R.H.D

f(x) is not differentiable at x = – 2

Now , Differentiability at x = 0

L.H.D = f’(0) = $\underset{h\to {{0}^{-}}}{\mathop{lim}}\,\dfrac{f\left( -2+h \right)~-~f\left( -2 \right)~}{h}$

= $\underset{h\to {{0}^{-}}}{\mathop{lim}}\,\dfrac{f\left( 0+h \right)~-~f\left( 0 \right)~}{h}$

= $\underset{h\to {{0}^{-}}}{\mathop{lim}}\,\dfrac{\left( 0+h+1 \right)~-~\left( 0+2 \right)~}{h}$

= $\underset{h\to {{0}^{-}}}{\mathop{lim}}\,\dfrac{h-1~}{h}$

= $\underset{h\to {{0}^{-}}}{\mathop{lim}}\,(1-\dfrac{1}{h}$) 

which does not exist. Hence f is not differentiable at x = 0.

Therefore, f(x) is not differentiable at points x = -2 and x = 0

23. Differentiate $~u~=~ta{{n}^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)$ w.r.t $v = ~co{{s}^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)$ , where x$\in \left( \dfrac{1}{\sqrt{2}},~1 \right)$.

Ans: Given,

$~u~=~ta{{n}^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)$ and v = $~co{{s}^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)$ 

we have, $~u~=~ta{{n}^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)$ 

Put x = sin$\theta $ , where $\dfrac{\pi }{4}$ < $\theta ~$< $\dfrac{\pi }{2}$

$~u~=~ta{{n}^{-1}}\left( \dfrac{\sqrt{1-si{{n}^{2}}\theta {{~}^{{}}}}}{sin\theta ~} \right)$ 

$\Rightarrow ~u~=~ta{{n}^{-1}}\left( \dfrac{cos\theta }{sin\theta ~} \right)$ 

$\Rightarrow ~u~=~ta{{n}^{-1}}\left( cot\theta \right)$ 

$\Rightarrow ~u~=~ta{{n}^{-1}}(tan\left( \dfrac{\pi }{2}-\theta \right)$ 

$\Rightarrow ~u~=~\left( \dfrac{\pi }{2}-\theta \right)$ 

$\Rightarrow ~u~=~\left( \dfrac{\pi }{2}-si{{n}^{-1}}x \right)$ 

On differentiating both sides w.r.t x

$\dfrac{du}{dx}$ = $\dfrac{-1}{\sqrt{1-{{x}^{2}}}}$    ….......…eqn (1)

Now, v = $~co{{s}^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)$

v = $\dfrac{\pi }{2}-~si{{n}^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)$

Again, Put x = sin$\theta $ , where $\dfrac{\pi }{4}$ < $\theta ~$< $\dfrac{\pi }{2}$

v = $~\dfrac{\pi }{2}-si{{n}^{-1}}\left( 2sin\theta ~\sqrt{1-~si{{n}^{2}}\theta ~} \right)$

⇒v = $~\dfrac{\pi }{2}-si{{n}^{-1}}\left( 2sin\theta cos\theta \right)$

⇒v = $~si{{n}^{-1}}\left( sin2\theta \right)$

⇒v = $2\theta $

⇒v = $~2si{{n}^{-1}}x$

On differentiating both sides w.r.t x

$\dfrac{dv}{d~x}$= $\dfrac{2}{\sqrt{1-{{x}^{2}}}}$     ……………..eqn(2)

Therefore, 

$\dfrac{du}{dv}$ = $\dfrac{\dfrac{du}{dx}}{\dfrac{dv}{d~x}}$ = $\dfrac{\dfrac{-1}{\sqrt{1-{{x}^{2}}}}}{\dfrac{2}{\sqrt{1-{{x}^{2}}}}~}$   …from eqn(1) & eqn(2)

$\Rightarrow \dfrac{du}{dv}$ = $\dfrac{-1}{2}$

Objective Type Questions

24. The function f (x) = $f\left( x \right)=\left\{ \begin{align} & \dfrac{sinx}{x} + cosx,\text{ }if\text{ }x \ne 0 \\ & k,\text{ }if\text{ }x=0 \\ \end{align} \right.$ is continuous at x = 0, then the value of k is

(A) 3

 (B) 2 

(C) 1 

(D) 1.5

Ans: Since the function f(x) is continuous at x = 0 

 L.H.L = R.H.L = f(0) 

\[\underset{x\to {{0}^{+}}}{\mathop{lim}}\,\left( \dfrac{sinx}{x}+cosx \right)\] + cosx = f(0)

Put x = 0 +h 

$\underset{h\to 0}{\mathop{lim}}\,\left( \dfrac{sinh}{h}+cosh \right)$ = k

Using identities : $\underset{x\to 0}{\mathop{lim}}\,\dfrac{sinx}{x}$ = 1 and $\underset{x\to 0}{\mathop{lim}}\,$cosx = 1

1 +1 = k

k = 2 

Therefore, the value of k = 2 

Hence, the correct option is (B)

25. The function f (x) =$\left[x\right]$, where $\left[x\right]$ denotes the greatest integer function, is continuous at 

(A) 4

 (B) – 2 

(C) 1 

(D) 1.5

Ans: The greatest integer function f(x) = $\left[x\right]$ is discontinuous at all integral values of x. 

Hence the correct option is (D)

26. The number of points at which the function f (x) = $\dfrac{1}{x-\left[ x \right]}$ is not continuous is

(A) 1 

(B) 2 

(C) 3 

(D) none of these

Ans: As we know that $\dfrac{1}{x-\left[ x \right]} = 0$, when x is an integer therefore f (x) is discontinuous for all x ∈ Z.

Hence, the correct option is (D)

27. The function given by f (x) = tanx is discontinuous on the set 

(A) \[\left\{ \text{n }\!\!\pi\!\!\text{ :n}\in \text{Z} \right\}\]

(B) \[\left\{ \text{2n }\!\!\pi\!\!\text{ :n}\in \text{Z} \right\}\]

(C) \[\left\{ \left( \text{2n+1} \right)\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{:n}\in \text{Z} \right\}\]

(D) \[\left\{ \dfrac{\text{n }\!\!\pi\!\!\text{ }}{\text{2}}\text{:n}\in \text{Z} \right\}\]

Ans: f (x) = tanx is always discontinuous on the set \[\left\{ \left( \text{2n+1} \right)\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{:n}\in \text{Z} \right\}\]

Hence the correct option is (C)

28. Let f (x)= |cosx|. Then,

(A) f is everywhere differentiable.

(B) f is everywhere continuous but not differentiable at x = nπ, n ∈Z .

(C) f is everywhere continuous but not differentiable at x = (2n+1)$\dfrac{\pi}{2}$ , n ∈Z

(D) none of these

Ans: We have 

f (x)= |cosx| 

Here, f is everywhere continuous but not differentiable at x = (2n+1)$\dfrac{\pi}{2}$ , n ∈Z

Hence the correct option is (C)

29. The function f (x) = |x| + |x – 1| is

(A) continuous at x = 0 as well as at x = 1.

(B) continuous at x = 1 but not at x = 0. 

(C) discontinuous at x = 0 as well as at x = 1. 

(D) continuous at x = 0 but not at x = 1.

Ans: f (x) = |x| + |x – 1| is continuous at x = 0 as well as at x = 1.

Hence the correct option is (A).

30. The value of k which makes the function defined by $f\left( x \right)=\left\{ \begin{align} & Sin\dfrac{1}{x},\text{ }if\text{ }x\ne 0 \\ & k,\text{ }if\text{ }x=0 \\ \end{align} \right.$, continuous at x=0 is 

(A) 8

(B) 1 

(C) –1

(D) none of these

Ans: As we know that $\underset{x\to {{0}^{{}}}}{\mathop{lim}}\,$sin$\dfrac{1}{x}$ does not exist 

Hence the correct option is (D)

31. The set of points where the functions f given by f (x) = |x – 3|cos x is differentiable is

(A) R 

(B) R – {3} 

(C) (0, ∞) 

(D) none of these

Ans: The set of points where the functions f given by f (x) = |x – 3| cosx is differentiable is R – {3} 

Hence the correct option is (B)

32. Differential coefficient of sec ($ta{{n}^{-1}}$x) w.r.t. x is

(A) $\dfrac{x}{\sqrt{1+{{x}^{2}}}}$

(B) $\dfrac{x}{1+{{x}^{2}}}$

(C) $x\sqrt{1+{{x}^{2}}}$

 (D) $\dfrac{1}{1+{{x}^{2}}}$

Ans: Let y = sec ($ta{{n}^{-1}}$x)

On Differentiating both sides w.r.t x 

$\dfrac{dy}{dx}$ = $\dfrac{d}{dx}~sec~\left( ta{{n}^{-1}}x \right)$

 = $~sec~\left( ta{{n}^{-1}}x \right)$. tan$\left( ta{{n}^{-1}}x \right)$.$\dfrac{d}{dx}$($ta{{n}^{-1}}x$)

= x.$~sec~\left( ta{{n}^{-1}}x \right)$.$\dfrac{1}{1+{{x}^{2}}}$

Clearly the differential coefficient of sec ($ta{{n}^{-1}}$x) is $\dfrac{x}{1+{{x}^{2}}}$

Hence the correct option is (B)

33.If u = $si{{n}^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)$ and v = $ta{{n}^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ , then $\dfrac{du}{dv}$ is 

(A) $\dfrac{1}{2}$

(B) x 

(C) $\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}}$

(D) 1

Ans: Given u = $si{{n}^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)$ and v = $ta{{n}^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ 

we have , 

u = $si{{n}^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)$

Put x = sinθ 

⇒u = $si{{n}^{-1}}\left( \dfrac{2sin\theta ~}{1+si{{n}^{2}}{{\theta }^{{}}}} \right)$

⇒u = $si{{n}^{-1}}\left( sin2\theta \right)$

⇒u = $2\theta $

$\Rightarrow \dfrac{du}{d\theta }$ = 2        .. eqn (1)

Now, we have ,

v = $ta{{n}^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$

put v = tan$\theta $

v = $ta{{n}^{-1}}\left( \dfrac{2~tan\theta }{1-~ta{{n}^{2}}{{\theta }^{{}}}} \right)$

⇒v = $ta{{n}^{-1}}\left( tan2\theta \right)$

⇒v= $2\theta $

$\Rightarrow \dfrac{dv}{d\theta }$ = 2        ...…eqn(2)

Dividing equation (1) by (2)

$\dfrac{\dfrac{du}{d\theta }}{\dfrac{dv}{d\theta }}$ = $\dfrac{2}{2~}$ = 1

$\Rightarrow \dfrac{du}{dv}$ = 1 

Hence the correct option is (D).

34. The value of c in Rolle’s Theorem for the function f (x) = ${{e}^{x}}$ sinx, x∈$\left[0, π \right]$ is

(A) $\dfrac{\pi }{6}$

(B) $\dfrac{\pi }{4}$

(C) $\dfrac{\pi }{2}$

(D) $\dfrac{3\pi }{4}$

Ans: We Know that by Rolle's Theorem that a function f(x) is continuous on [a,b] and differentiable in (a,b) then there exists c $\in $(a,b) such that f’(c) = 0 

We have , 

f (x) = ${{e}^{x}}$ sinx, x∈$\left[0, π \right]$ is

f’(x) = ${{e}^{x}}$ sinx + ${{e}^{x}}$ cosx

f’(c) = ${{e}^{c}}$(sinx + cosx) = 0 

Now we know that ${{e}^{c}}$ $\ne 0$

 tanc = -1 

c = $\dfrac{3\pi }{4}$

Hence option (D) is the correct answer.

35. The value of c in Mean value theorem for the function f (x) = x (x – 2), x ∈$\left[1, 2 \right]$ is

(A) $\dfrac{3}{2~}$

(B) $\dfrac{2}{3~}$

(C) $\dfrac{1}{2~}$

 (D) $\dfrac{3}{2~}$

Ans: We have,

 f (x) = x (x – 2)

 f (x) = ${{x}^{2}}-2x$

Since the polynomial function is always continuous and differentiable 

f (x) = ${{x}^{2}}-2x$ is always continuous on $\left[1,2\right]$ and differentiable in (1,2)

Now since, f(x) satisfies both the conditions of Mean Value Theorem there exist a real number c $\in $ (1,2) such that 

f’(c) = $\left(\dfrac{f\left( 2 \right)~-f\left( 1 \right)}{2-1}\right)$

⇒f’(c) = $\left(\dfrac{f\left( 2 \right)~-f\left( 1 \right)}{1}\right)$

⇒f’(c) = f(2) -f(1)

⇒f’(c) = f(2) -f(1)

we have f(x) = ${{x}^{2}}-2x$

f’(x) = 2x -2 

⇒f’(c) = 2c -2 

⇒f(2) -f(1) = 2c -2 

$\left({{2}^{2}}-2\times 2\right)$ - $\left({{1}^{2}}-2\times 1\right)$ = 2c -2 

⇒0 -(-1) = 2c -2 

⇒2c = 1 +2 

⇒c = $\dfrac{3}{2~}$

Clearly c = $\dfrac{3}{2~}\in \left( 1,2 \right)$

Hence option (A) is the correct answer

36. Match the following  

Ans: \[\text{ A}\to \text{c, B}\to \text{d, C}\to \text{a, D}\to \text{b}\]

Fill in the blanks in each of the Examples 37 to 41

37. The number of points at which the function f (x) =$\dfrac{1}{log\left| x \right|~}$ is discontinuous ________.

Ans: The given function is discontinuous at x = 0, ± 1 and hence the number of points of discontinuity is 3.

Therefore, The number of points at which the function f (x) =$\dfrac{1}{log\left| x \right|~}$ is discontinuous is 3

38.If f(x) = ax+1 , x $\ge 1~$and f(x) = x+2, x $<1~$ is continuous, then a should be equal to _______.

Ans: If f(x) = ax+1 , x $\ge 1~$and f(x) = x+2, x $<1~$

is continuous, then a should be equal to 2.

39. The derivative of ${log}_{10}{x}$ w.r.t. x is ________.

Ans: The derivative of ${log}_{10}{x}$ w.r.t. x is ${log}_{10}{e}$.$\dfrac{1}{x~}$

40. If y = $se{{c}^{-1}}\dfrac{\sqrt{x+1}}{\sqrt{x-1}}$ + $si{{n}^{-1}}\dfrac{\sqrt{x-1}}{\sqrt{x+1}}$ , then $\dfrac{dy}{dx}$ is equal to ________

Ans: y = $se{{c}^{-1}}\dfrac{\sqrt{x+1}}{\sqrt{x-1}}$ + $si{{n}^{-1}}\dfrac{\sqrt{x-1}}{\sqrt{x+1}}$, then $\dfrac{\text{dy}}{\text{dx}}$ is equal to 0

41. The derivative of sin x w.r.t. cos x is ________

Ans: The derivative of sin x w.r.t. cos x is -cotx

State whether the statements are True or False in each of the Exercises 42 to 46.

42. For continuity, at x = a, each of $\underset{x\to {{a}^{+}}}{\mathop{lim}}\,f\left( x \right)$ and $\underset{x\to {{a}^{\_}}}{\mathop{lim}}\,f\left( x \right)$ is equal to f(a).

Ans: We know that for a continuous function at x = a 

L.H.L = R.H.L = f(a) 

$\underset{x\to {{a}^{-}}}{\mathop{lim}}\,f\left( x \right)$ = $\underset{x\to {{a}^{+}}}{\mathop{lim}}\,f\left( x \right)$ = f(a)

Hence the given statement is True.

43. y = |x – 1| is a continuous function .

Ans: Yes, y = |x – 1| is a continuous function .

Hence the given statement is True.

44. A continuous function can have some points where the limit does not exist.

Ans: A continuous function can never have some points where the limit does not exist. 

Hence the given statement is False .

45. |sinx| is a differentiable function for every value of x.

Ans: |sinx| is not a differentiable function for every value of x.

Hence the given statement is False

46. cos|x| is differentiable everywhere. 

Ans: Yes, cos |x| is differentiable everywhere.

Hence the given statement is True.

Exercise

Short Answer Type Questions

1. Examine the continuity of the function $f\left( x \right)={{x}^{3}}+2{{x}^{2}}-1$ at $x=1$.

Ans: Given that $f\left( x \right)={{x}^{3}}+2{{x}^{2}}-1$ we must check continuity at

$x=1$. 

$RHL=f\left( x \right)~$ 

$=\underset{x\to {{1}^{+}}}{\mathop{lim}}\,\left( {{x}^{3}}+2{{x}^{2}}-1 \right)$ ; here, $x=1+h,~h\to 0$ 

$=\underset{h\to 0}{\mathop{lim}}\,{{\left( 1+h \right)}^{3}}+2{{\left( 1+h \right)}^{2}}-1$

$=1+2-1$ 

$RHL=2$ 

$LHL=f\left( x \right)~$ 

$=\underset{x\to {{1}^{-}}}{\mathop{lim}}\,\left( {{x}^{3}}+2{{x}^{2}}-1 \right)$ ; here, $x=1-h,~h\to 0$ 

$=\underset{h\to 0}{\mathop{lim}}\,{{\left( 1-h \right)}^{3}}+2{{\left( 1-h \right)}^{2}}-1$

$=1+2-1$ 

$LHL=2$ 

$f\left( 1 \right)={{1}^{3}}+2\times {{1}^{2}}-1=2$ 

So, we have $LHL=RHL=f\left( x \right)$ at $x=1$.

Hence $f\left( x \right)$ is continuous at $x=1$. 

Find which of the functions in Exercises 2 to 10 is continuous or discontinuous at the indicated points:

2. $f\left( x \right)=\left\{ \begin{align} & 3x+5,\text{if}\text{ }x\ge 2 \\ & {{x}^{2}},\text{if}\text{ }x<2 \\ \end{align} \right.$ at $x=2$. 

Ans: Given that, $f\left( x \right)=\left\{ \begin{align} & 3x+5,\text{if}\text{ }x\ge 2 \\ & {{x}^{2}},\text{if}\text{ }x<2 \\ \end{align} \right.$ at $x=2$

$LHL=f\left( x \right)~$$=\underset{x\to {{2}^{-}}}{\mathop{lim}}\,{{x}^{2}}$ 

$={{2}^{2}}$ 

$LHL=4$ 

$RHL=f\left( x \right)~$$=\underset{x\to {{2}^{+}}}{\mathop{lim}}\,\left( 3x+5 \right)$ 

$=3\times 2+5$ 

$RHL=11$ 

$f\left( 2 \right)=3\times 2+5=11$ 

Here $LHL\ne RHL=f\left( x \right)$ at $x=2$ 

Hence $f\left( x \right)$ is discontinuous at $x=2$. 

3. $f\left( x \right)=\left\{ \begin{align} & \dfrac{1-cos~2x}{{{x}^{2}}},\text{ }if\text{ }x\ne 0 \\ & 5,\text{if}\text{ }x=0 \\ \end{align} \right.$ at $x=0$. 

Ans: Given that, $f\left( x \right)=\left\{ \begin{align} & \dfrac{1-cos~2x}{{{x}^{2}}},\text{ }if\text{ }x\ne 0 \\ & 5,\text{if}\text{ }x=0 \\ \end{align} \right.$ at $x=0$

$LHL=f\left( x \right)~$ 

$=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,\dfrac{1-cos~2x}{{{x}^{2}}}$ 

$=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,\dfrac{2x~}{{{x}^{2}}}$ 

$=2\times {{\left( 1 \right)}^{2}}$ 

$LHL=2$ 

$RHL=f\left( x \right)~$ 

$=\underset{x\to {{0}^{+}}}{\mathop{lim}}\,\dfrac{1-cos~2x}{{{x}^{2}}}$ 

$=\underset{x\to {{0}^{+}}}{\mathop{lim}}\,\dfrac{2x~}{{{x}^{2}}}$ 

$=2\times {{\left( 1 \right)}^{2}}$ 

$RHL=2$ 

$f\left( 0 \right)=5$ 

Here $LHL=RHL\ne f\left( x \right)$ at $x=0$ 

Hence $f\left( x \right)$ is discontinuous at $x=0$.

4. $f\left( x \right)=\left\{ \begin{align} & \dfrac{2{{x}^{2}}-3x-2}{x-2},\text{ }if\text{ }x\ne 2 \\ & 5,\text{if}\text{ }x=2 \\ \end{align} \right.$ at $x=2$. 

Ans: Given that, $f\left( x \right)=\left\{ \begin{align} & \dfrac{2{{x}^{2}}-3x-2}{x-2},\text{ }if\text{ }x\ne 2 \\ & 5,\text{if}\text{ }x=2 \\ \end{align} \right.$ at $x=2$

$LHL=f\left( x \right)~$ 

$=\underset{x\to {{2}^{-}}}{\mathop{lim}}\,\dfrac{2{{x}^{2}}-3x-2}{x-2}$ 

$=\underset{x\to {{2}^{-}}}{\mathop{lim}}\,\dfrac{\left( 2x+1 \right)\left( x-2 \right)}{x-2}$ 

$=\underset{x\to {{2}^{-}}}{\mathop{lim}}\,\left( 2x+1 \right)$

$=2\times 2+1$ 

$LHL~~=~5$ 

$RHL~=f\left( x \right)~$ 

$=\underset{x\to {{2}^{-}}}{\mathop{lim}}\,\dfrac{2{{x}^{2}}-3x-2}{x-2}$ 

$=\underset{x\to {{2}^{-}}}{\mathop{lim}}\,\dfrac{\left( 2x+1 \right)\left( x-2 \right)}{x-2}$ 

$=\underset{x\to {{2}^{-}}}{\mathop{lim}}\,\left( 2x+1 \right)$

$=2\times 2+1$ 

$RHL=5$

$f\left( 2 \right)=5$ 

 Here $LHL=RHL=f\left( x \right)$ at $x=2$ 

 Hence $f\left( x \right)$ is continuous at $x=2$.

5. $f\left( x \right)=\left\{ \begin{align} & \dfrac{\left| x-4 \right|}{2\left( x-4 \right)},\text{ }if\text{ }x\ne 4 \\ & 0,\text{if}\text{ }x=4 \\ \end{align} \right.$ at $x=4$. 

Ans: Given that, $f\left( x \right)=\left\{ \begin{align} & \dfrac{\left| x-4 \right|}{2\left( x-4 \right)},\text{ }if\text{ }x\ne 4 \\ & 0,\text{if}\text{ }x=4 \\ \end{align} \right.$ at $x=4$.

We know that $\left| x-4 \right|=\left\{ \begin{align} & -\left( x-4 \right),\text{if}\text{ }x< 4 \\ & \left( x-4 \right),\text{if}\text{ }x > 4 \\ \end{align} \right.$

$LHL=f\left( x \right)~$ 

$=\underset{x\to {{4}^{-}}}{\mathop{lim}}\,\dfrac{\left| x-4 \right|}{2\left( x-4 \right)}$ 

$=\underset{x\to {{4}^{-}}}{\mathop{lim}}\,\dfrac{-\left( x-4 \right)}{2\left( x-4 \right)}$ 

$LHL=-\dfrac{1}{2}$ 

$RHL=f\left( x \right)~$ 

$=\underset{x\to {{4}^{+}}}{\mathop{lim}}\,\dfrac{\left| x-4 \right|}{2\left( x-4 \right)}$ 

$=\underset{x\to {{4}^{+}}}{\mathop{lim}}\,\dfrac{\left( x-4 \right)}{2\left( x-4 \right)}$ 

$RHL=\dfrac{1}{2}$ 

$f\left( 4 \right)=0$ 

Here $LHL\ne RHL\ne f\left( x \right)$ at $x=4$ 

Hence $f\left( x \right)$ is discontinuous at $x=4$.

6. $f\left( x \right)=\left\{ \begin{align} & \left| x \right|cos~\left( \dfrac{1}{x} \right),\text{if}\text{ }x\ne 0 \\ & 0,\text{if}\text{ }x=0 \\ \end{align} \right.$ at $x=0$. 

Ans: Given that $f\left( x \right)=\left\{ \begin{align} & \left| x \right|cos~\left( \dfrac{1}{x} \right),\text{if}\text{ }x\ne 0 \\ & 0,\text{if}\text{ }x=0 \\ \end{align} \right.$ at $x=0$. 

$LHL=f\left( x \right)~$ 

$=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,\left| x \right|cos~\left( \dfrac{1}{x} \right)~$ 

$=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,\left( -x \right)cos~\left( \dfrac{1}{x} \right)~$ 

$=0\times \left( number~which~will~vary~from-1~to~1 \right)$ 

$LHL=0$ 

$RHL=f\left( x \right)~$ 

$=\underset{x\to {{0}^{+}}}{\mathop{lim}}\,\left| x \right|cos~\left( \dfrac{1}{x} \right)~$ 

$=\underset{x\to {{0}^{+}}}{\mathop{lim}}\,\left( x \right)cos~\left( \dfrac{1}{x} \right)~$ 

$=0\times \left( number~which~will~vary~from-1~to~1 \right)$ 

$RHL=0$ 

$f\left( 0 \right)=0$ 

Here $LHL=RHL=f\left( x \right)$ at $x=0$ 

Hence $f\left( x \right)$ is continuous at $x=0$.

7. $f\left( x \right)=\left\{ \begin{align} & \left| x-a \right|sin~\left( \dfrac{1}{x-a} \right),\text{ }if\text{ }x\ne 0 \\ & 0,\text{ }if\text{ }x=a \\ \end{align} \right.$ at $x=a$. 

Ans: Given that $f\left( x \right)=\left\{ \begin{align} & \left| x-a \right|sin~\left( \dfrac{1}{x-a} \right),\text{ }if\text{ }x\ne 0 \\ & 0,\text{ }if\text{ }x=a \\ \end{align} \right.$ at $x=a$. 

 $LHL=f\left( x \right)~$ 

$=\underset{x\to {{a}^{-}}}{\mathop{lim}}\,\left| x-a \right|sin~\left( \dfrac{1}{x-a} \right)~$ ; 

A. $\left| x-a \right|=\left\{ \begin{align} & -(x-a),\text{ }if\text{ }x < a \\ & x-a,\text{ }if\text{ }x \ge a \\ \end{align} \right.$

$=\underset{x\to {{a}^{-}}}{\mathop{lim}}\,\left( -\left( x-a \right) \right)sin~\left( \dfrac{1}{x-a} \right)~$. 

$=0\times \left( number~which~will~vary~from-1~to~1 \right)$ 

$LHL=0$ 

$RHL=f\left( x \right)~$ 

$=\underset{x\to {{a}^{+}}}{\mathop{lim}}\,\left| x-a \right|sin~\left( \dfrac{1}{x-a} \right)~$ ; 

B. $\left| x-a \right|=\left\{ \begin{align} & -\left( x-a \right),\text{ }if\text{ }x < a \\ & x-a,\text{ }if\text{ }x \ge a \\ \end{align} \right.$

$=\underset{x\to {{a}^{+}}}{\mathop{lim}}\,\left( \left( x-a \right) \right)sin~\left( \dfrac{1}{x-a} \right)~$ 

$=0\times \left( number~which~will~vary~from-1~to~1 \right)$ 

$RHL=0$ 

$f\left( a \right)=0$ 

Here $LHL=RHL=f\left( x \right)$ at $x=a$ 

Hence $f\left( x \right)$ is continuous at $x=a$.

8. $f\left( x \right)=\left\{ \begin{align} & \dfrac{{{e}^{\dfrac{1}{x}}}}{1+{{e}^{\dfrac{1}{x}}}},\text{ }if\text{ }x\ne 0 \\ & 0,\text{ }if\text{ }x=0 \\ \end{align} \right.$ at $x=0$. 

Ans: Given $f\left( x \right)=\left\{ \begin{align} & \dfrac{{{e}^{\dfrac{1}{x}}}}{1+{{e}^{\dfrac{1}{x}}}},\text{ }if\text{ }x\ne 0 \\ & 0,\text{ }if\text{ }x=0 \\ \end{align} \right.$ at $x=0$.

$LHL=f\left( x \right)~$ 

$=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,\dfrac{{{e}^{\dfrac{1}{x}}}}{1+{{e}^{\dfrac{1}{x}}}}$

$x\to {{0}^{-}}\Rightarrow x<0$

$\therefore \dfrac{1}{x}\to -\infty $ 

${{e}^{\dfrac{1}{x}}}\to 0$ 

$=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,\left( \dfrac{0}{1+0} \right)$ 

$LHL=0$ 

$RHL=f\left( x \right)~$$=\underset{x\to {{0}^{+}}}{\mathop{lim}}\,\dfrac{{{e}^{\dfrac{1}{x}}}}{1+{{e}^{\dfrac{1}{x}}}}$

$x\to {{0}^{+}}\Rightarrow x>0$

$\therefore \dfrac{1}{x}\to \infty $ 

${{e}^{\dfrac{1}{x}}}\to \infty $ 

$=\underset{x\to {{0}^{+}}}{\mathop{lim}}\,\dfrac{1}{1+\dfrac{1}{{{e}^{\dfrac{1}{x}}}}}~$

$=\left( \dfrac{1}{1+0} \right)$ 

$RHL=1$ 

$f\left( 0 \right)=0$ 

Here $LHL\ne RHL$ at $x=0$ 

Hence $f\left( x \right)$ is discontinuous at $x=0$.

9. $f\left( x \right)=\left\{ \begin{align} & \dfrac{{{x}^{2}}}{2},\text{ }if\text{ }0\le x\le 1\\ &2{{x}^{2}}-3x+\dfrac{3}{2},\text{ }if\text{ }1<x\le 2 \\ \end{align} \right.$ at $x=1$.

Ans: $f\left( x \right)=\left\{ \begin{align} & \dfrac{{{x}^{2}}}{2},\text{ }if\text{ }0\le x\le 1\\ &2{{x}^{2}}-3x+\dfrac{3}{2},\text{ }if\text{ }1<x\le 2 \\ \end{align} \right.$ at $x=1$. 

$LHL=f\left( x \right)~$ 

 $=\underset{x\to {{1}^{-}}}{\mathop{lim}}\,\dfrac{{{x}^{2}}}{2}$ 

 $=\underset{x\to {{1}^{-}}}{\mathop{lim}}\,\left( \dfrac{1}{2} \right)$ 

 $LHL=\dfrac{1}{2}$

 $RHL=f\left( x \right)~$ 

 $=\underset{x\to {{1}^{+}}}{\mathop{lim}}\,\left( 2{{x}^{2}}-3x+\dfrac{3}{2} \right)$ 

 $=\underset{x\to {{1}^{+}}}{\mathop{lim}}\,\left( 2\times {{1}^{2}}-3\times 1+\dfrac{3}{2} \right)$

 $=\left( \dfrac{3}{2}-1 \right)$

 $RHL=\dfrac{1}{2}$ 

 $f\left( 1 \right)=\dfrac{1}{2}$ 

 Here $LHL=RHL=f\left( 1 \right)$ at $x=1$ 

 Hence $f\left( x \right)$ is continuous at $x=1$.

10. $f\left( x \right)=\left| x \right|+\left| x-1 \right|$ at $x=1$.

Ans: Given $f\left( x \right)=\left| x \right|+\left| x-1 \right|$ at $x=1$ 

 $LHL=f\left( x \right)~$ 

 $=\underset{x\to {{1}^{-}}}{\mathop{lim}}\,\left| x \right|+\left| x-1 \right|$

As x < 1

$\therefore \left| x \right|=x,~~\left| x-1 \right|=-\left( x-1 \right)$ 

 $=\underset{x\to {{1}^{-}}}{\mathop{lim}}\,\left( x-x+1 \right)$ 

 $LHL=1$

 $RHL=f\left( x \right)~$ 

 $=\underset{x\to {{1}^{+}}}{\mathop{lim}}\,\left| x \right|+\left| x-1 \right|$

As x > 1

$\therefore \left| x \right|=x,~~\left| x-1 \right|=\left( x-1 \right)$ 

 $=\underset{x\to {{1}^{+}}}{\mathop{lim}}\,\left( 2x-1 \right)$

 $=\left( 2-1 \right)$

 $RHL=1$

 $f\left( 1 \right)=1$ 

 Here $LHL=RHL=f\left( 1 \right)$ at $x=1$ 

 Hence $f\left( x \right)$ is continuous at $x=1$.

Find the value of k in each of the Exercises 11 to 14 so that the function f is continuous at the indicated point: 

11. $f\left( x \right)=\left\{ \begin{align} & 3x-8,\text{ }if\text{ }x\le 5 \\ & 2k,\text{if}\text{ }x>5 \\ \end{align} \right.$ at $x=5$. 

Ans: Given $f\left( x \right)=\left\{ \begin{align} & 3x-8,\text{ }if\text{ }x\le 5 \\ & 2k,\text{if}\text{ }x>5 \\ \end{align} \right.$ at $x=5$ 

$f\left( x \right)$ will be continuous at $x=5$ if 

$LHL=RHL=f\left( 5 \right)$ 

$LHL=f\left( x \right)~$ 

 $=\underset{x\to {{5}^{-}}}{\mathop{lim}}\,\left( 3x-8 \right)$ 

 $=\underset{x\to {{5}^{-}}}{\mathop{lim}}\,\left( 15-8 \right)$ 

$LHL=7$

$RHL=f\left( x \right)~$ 

$=\underset{x\to {{5}^{+}}}{\mathop{lim}}\,\left( 2k \right)$ 

$RHL=2k$ 

$2k=7\Rightarrow k=\dfrac{7}{2}$ 

12. $f\left( x \right)=\left\{ \begin{align} & \dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16},\text{if}\text{ }x\ne 2 \\ & k,\text{if}\text{ }x=2 \\ \end{align} \right.$ at $x=2$. 

Ans: Given $f\left( x \right)=\left\{ \begin{align} & \dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16},\text{if}\text{ }x\ne 2 \\ & k,\text{if}\text{ }x=2 \\ \end{align} \right.$ at $x=2$

 $f\left( x \right)$ will be continuous at $x=2$ if 

$LHL=RHL=f\left( 2 \right)$ 

$LHL=f\left( x \right)~$ 

 $=\underset{x\to {{2}^{-}}}{\mathop{lim}}\,~\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}$

$=\underset{x\to {{2}^{-}}}{\mathop{lim}}\,~\dfrac{4\left( {{2}^{x}}-4 \right)}{\left( {{2}^{x}}-4 \right)\left( {{2}^{x}}+4 \right)}$ 

 $=\underset{x\to {{2}^{-}}}{\mathop{lim}}\,\left( \dfrac{4}{{{2}^{2}}+4} \right)$ 

 $LHL=\dfrac{1}{2}$

$RHL=f\left( x \right)~$ 

 $=\underset{x\to {{2}^{+}}}{\mathop{lim}}\,~\dfrac{{{2}^{x+2}}-16}{{{4}^{x}}-16}$

$=\underset{x\to {{2}^{+}}}{\mathop{lim}}\,~\dfrac{4\left( {{2}^{x}}-4 \right)}{\left( {{2}^{x}}-4 \right)\left( {{2}^{x}}+4 \right)}$ 

 $=\underset{x\to {{2}^{+}}}{\mathop{lim}}\,\left( \dfrac{4}{{{2}^{2}}+4} \right)$ 

 $RHL=\dfrac{1}{2}$

$k=\dfrac{1}{2}$ 

13. $f\left( x \right)=\left\{ \begin{align} & \dfrac{\sqrt{1+kx}-\sqrt{1-kx}}{x},\text{if}\text{ }-1\le x<0 \\ & \dfrac{2x+1}{x-1},\text{ }if\text{ }0\le x\le 1 \\ \end{align} \right.$ at $x=0$. 

Ans: Given $f\left( x \right)=\left\{ \begin{align} & \dfrac{\sqrt{1+kx}-\sqrt{1-kx}}{x},\text{if}\text{ }-1\le x<0 \\ & \dfrac{2x+1}{x-1},\text{ }if\text{ }0\le x\le 1 \\ \end{align} \right.$ at $x=0$ 

$f\left( x \right)$ will be continuous at $x=0$ if 

$LHL=RHL=f\left( 0 \right)$ 

$LHL=f\left( x \right)~$ 

$=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,\left( \dfrac{\sqrt{1+kx}-\sqrt{1-kx}}{x} \right)$ 

$=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,\left( \dfrac{\sqrt{1+kx}-\sqrt{1-kx}}{x} \right)\left( \dfrac{\sqrt{1+kx}+\sqrt{1-kx}}{\sqrt{1+kx}+\sqrt{1-kx}} \right)$ 

$=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,~\dfrac{\left( 1+kx \right)-\left( 1-kx \right)}{x\left( \sqrt{1+kx}+\sqrt{1-kx} \right)}$ 

$=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,~\dfrac{2kx}{x\left( \sqrt{1+kx}+\sqrt{1-kx} \right)}$ 

C. $=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,~\dfrac{2k}{\left( \sqrt{1+0}+\sqrt{1-0} \right)}$

 $LHL=k$

 $RHL=f\left( x \right)~$ 

 $=\underset{x\to {{0}^{+}}}{\mathop{lim}}\,\left( \dfrac{2x+1}{x-1} \right)$ 

$RHL=-1$ 

$k=-1$ 

14. $f\left( x \right)=\left\{ \begin{align} & \dfrac{1-cos~kx}{x~sin~x},\text{if}\text{ }x\ne 0 \\ & \dfrac{1}{2},\text{if}\text{ }x=0 \\ \end{align} \right.$ at $x=0$. 

Ans: Given $f\left( x \right)=\left\{ \begin{align} & \dfrac{1-cos~kx}{x~sin~x},\text{if}\text{ }x\ne 0 \\ & \dfrac{1}{2},\text{if}\text{ }x=0 \\ \end{align} \right.$ at $x=0$.

 $f\left( x \right)$ will be continuous at $x=0$ if 

$LHL=RHL=f\left( 0 \right)$ 

$LHL=RHL=f\left( x \right)~$ 

 $=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,\left( \dfrac{1-cos~kx}{x~sin~x} \right)$ 

 $=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,\left( \dfrac{1-cos~kx}{{{k}^{2}}{{x}^{2}}\left( \dfrac{sin~x}{x} \right)} \right){{k}^{2}}$ 

$=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,\left( \dfrac{1}{2} \right){{k}^{2}}$ 

 $LHL=RHL=\dfrac{{{k}^{2}}}{2}$

 $f\left( 0 \right)=\dfrac{1}{2}$ 

$\dfrac{{{k}^{2}}}{2}=\dfrac{1}{2}\Rightarrow {{k}^{2}}=1$ 

$k=\pm 1$. 

15. Prove that the function $f$ defined by $f\left( x \right)=\left\{ \begin{align} & \dfrac{x}{\left| x \right|+2{{x}^{2}}},\text{if}\text{ }x\ne 0 \\ & k,\text{if}\text{ }x=0 \\ \end{align} \right.$ remains discontinuous at $x=0$, regardless of the choice of k. 

Ans: Given that, $f\left( x \right)=\left\{ \begin{align} & \dfrac{x}{\left| x \right|+2{{x}^{2}}},\text{if}\text{ }x\ne 0 \\ & k,\text{if}\text{ }x=0 \\ \end{align} \right.$. here to check continuity at $x=0$

 $LHL=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,~f\left( x \right)$

 $=\dfrac{x}{\left| x \right|+2{{x}^{2}}}~$ 

$\left| x \right|= \left\{ \begin{align} & -x, \text{ }x<0 \\ & x,\text{ }x\ge 0\\ \end{align} \right.$ 

 $=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,~\dfrac{x}{-x+2{{x}^{2}}}$ 

$=\underset{x\to {{0}^{-}}}{\mathop{lim}}\,~\dfrac{1}{-1+2x}$ 

$LHL=-1$ 

 $RHL=\underset{x\to {{0}^{+}}}{\mathop{lim}}\,~f\left( x \right)$

 $=\dfrac{x}{\left| x \right|+2{{x}^{2}}}~$ 

$\left| x \right|= \left\{ \begin{align} & -x, \text{ }x<0 \\ & x,\text{ }x\ge 0\\ \end{align} \right.$ 

 $=\underset{x\to {{0}^{+}}}{\mathop{lim}}\,~\dfrac{x}{x+2{{x}^{2}}}$. 

$=\underset{x\to {{0}^{+}}}{\mathop{lim}}\,~\dfrac{1}{1+2x}$ 

$RHL=1$ 

re $LHL\ne RHL$ 

Hence limit doesn’t exist, the discontinuity is irremovable so there is no point of value of function at $x=0$ for continuity. Proved. 

16. Find the values of a and b such that the function f defined by 

$f\left( x \right)=\left\{ \begin{align} & \dfrac{x-4}{\left| x-4 \right|}+a,\text{if}\text{ }x< 4 \\ & a+b,\text{if}\text{ }x=4 \\ &\dfrac{x-4}{\left| x-4 \right|}+b, \text{if}\text{ }x>4\\ \end{align} \right.$ is a continuous function at $x=4$. 

Ans: Given that $f\left( x \right)=\left\{ \begin{align} & \dfrac{x-4}{\left| x-4 \right|}+a,\text{if}\text{ }x< 4 \\ & a+b,\text{if}\text{ }x=4 \\ &\dfrac{x-4}{\left| x-4 \right|}+b, \text{if}\text{ }x>4\\ \end{align} \right.$

We must find a and b such that the function is continuous at $x=4$.

$LHL=\underset{x\to {{4}^{-}}}{\mathop{lim}}\,~f\left( x \right)$ 

 $=\dfrac{x-4}{\left| x-4 \right|}+a~$ 

$\left| x-4 \right|=\left\{ \begin{align} &-\left( x-4 \right),\text{}x<4\\ & \left(x-4\right),\text{ } x\ge 4\\ \end{align} \right.$ 

$=\underset{x\to {{4}^{-}}}{\mathop{lim}}\,~\dfrac{x-4}{-\left( x-4 \right)}+a$ 

$=\underset{x\to {{4}^{-}}}{\mathop{lim}}\,~\left( -1+a \right)$ 

$LHL=a-1$ 

$RHL=\underset{x\to {{4}^{-}}}{\mathop{lim}}\,~f\left( x \right)$$=\dfrac{x-4}{\left| x-4 \right|}+b~$ 

$\left| x-4 \right|=\left\{ \begin{align} &-\left( x-4 \right),\text{}x<4\\ & \left(x-4\right),\text{ } x\ge 4\\ \end{align} \right.$

$=\underset{x\to {{4}^{+}}}{\mathop{lim}}\,~\dfrac{x-4}{\left( x-4 \right)}+b$ 

$=\underset{x\to {{4}^{+}}}{\mathop{lim}}\,~\left( 1+b \right)$ 

$RHL=b+1$ 

$f\left( 4 \right)=a+b$ 

$LHL=RHL=f\left( 4 \right)$ 

$a-1=b+1=a+b$ 

$a=1,~b=-1$. 

17. Given the function $f\left( x \right)=\dfrac{1}{x+2}$. Find the points of discontinuity of the composite function $y=f\left( f\left( x \right) \right)$. 

Ans: Given that $f\left( x \right)=\dfrac{1}{x+2}$, 

 Now, $f\left( f\left( x \right) \right)=f\left( \dfrac{1}{x+2} \right)=\dfrac{1}{\dfrac{1}{x+2}+2}=\dfrac{x+2}{1+2x+4}=\dfrac{x+2}{2x+5}$ 

 Here $y=\dfrac{x+2}{2x+5}$ so, for y to be defined $2x+5\ne 0$ 

 $2x\ne -5\Rightarrow x\ne -\dfrac{5}{2}$ 

 Hence $f\left( f\left( x \right) \right)$ is discontinuous at $x=-\dfrac{5}{2}$. 

18. Find all points of discontinuity of the function $f\left( t \right)=\dfrac{1}{{{t}^{2}}+t-2}$, where $t=\dfrac{1}{x-1}$. 

Ans: Given, $f\left( t \right)=\dfrac{1}{{{t}^{2}}+t-2}$ and $t=\dfrac{1}{x-1}$ 

 $f\left( x \right)=\dfrac{1}{{{\left( \dfrac{1}{x-1} \right)}^{2}}+\left( \dfrac{1}{x-1} \right)-2}$ 

$=\dfrac{{{\left( x-1 \right)}^{2}}}{1+\left( x-1 \right)-2{{\left( x-1 \right)}^{2}}}$ 

$=\dfrac{{{\left( x-1 \right)}^{2}}}{-2\left( {{x}^{2}}-2x+1 \right)+x-1+1}$ 

$=\dfrac{{{\left( x-1 \right)}^{2}}}{-2{{x}^{2}}+5x-2}$ 

For $f\left( x \right)$ to be defined $-2{{x}^{2}}+5x-2\ne 0$ 

$-\left( 2{{x}^{2}}-4x-x+2 \right)\ne 0$ 

$\Rightarrow 2x\left( x-2 \right)-\left( x-2 \right)\ne 0$ 

$\Rightarrow \left( 2x-1 \right)\left( x-2 \right)\ne 0$ 

$\Rightarrow x\ne \dfrac{1}{2}~and~2$ 

Hence points of discontinuity for the function $f\left( x \right)$ are 

$x=\dfrac{1}{2}$ and $x=2$. 

19. Show that the function $f\left( x \right)=\left| sin~x~+cos~x~ \right|$ is continuous at $x=\pi $. 

Ans: Given $f\left( x \right)=\left| sin~x~+cos~x~ \right|$ to check continuity at $x=\pi $. 

 $f\left( x \right)~=\left| sin~x~+cos~x~ \right|~$ 

$=\left| sin~\pi ~+cos~\pi ~ \right|~$ 

$=\left| 0-1 \right|$ 

$f\left( x \right)~=1$ 

$f\left( \pi \right)=\left| sin~\pi ~+cos~\pi ~ \right|=\left| 0-1 \right|=1$ 

So, $f\left( x \right)$ is continuous at $x=\pi $. 

Hence proved. 

Examine the differentiability of function f, where f is defined by

20. $f\left( x \right)=\left\{ \begin{align} & x\left[ x \right],\text{if}\text{ }0\le x<2 \\ & \left( x-1 \right),\text{if}\text{ }2\le x<3 \\ \end{align} \right.$ at $x=2$. 

Ans: Given $f\left( x \right)=\left\{ \begin{align} & x\left[ x \right],\text{if}\text{ }0\le x<2 \\ & \left( x-1 \right),\text{if}\text{ }2\le x<3 \\ \end{align} \right.$ at $x=2$, 

$LHD=f'\left( {{2}^{-}} \right)$ 

$=\dfrac{f\left( 2-h \right)-f\left( 2 \right)}{-h}~$ 

$=\dfrac{\left( 2-h \right)\left[ 2-h \right]-2\left( 2-1 \right)}{-h}~$ 

$=\dfrac{\left( 2-h \right)1-2}{-h}~$ 

 $=\dfrac{-h}{-h}~$ 

$LHD=1$ 

$RHD=f'\left( {{2}^{+}} \right)$ 

$=\dfrac{f\left( 2+h \right)-f\left( 2 \right)}{h}~$ 

$=\dfrac{\left( 2+h-1 \right)\left( 2+h \right)-2\left( 2-1 \right)}{h}~$ 

$=\dfrac{{{h}^{2}}+3h+2-2}{h}~$ 

 $=\dfrac{{{h}^{2}}+3h}{h}~$ 

$RHD=3$ 

Here $LHD\ne RHD$ 

Hence function f is not differentiable at $x=2$.

21. $f\left( x \right)=\left\{ \begin{align} & {{x}^{2}}.sin\dfrac{1}{x},\text{if}\text{ }x\ne 0 \\ & 0,\text{if}\text{ }x=0 \\ \end{align} \right.$

Ans: Given $f\left( x \right)=\left\{ \begin{align} & {{x}^{2}}.sin\dfrac{1}{x},\text{if}\text{ }x\ne 0 \\ & 0,\text{if}\text{ }x=0 \\ \end{align} \right.$, 

 $LHD=f'\left( {{0}^{-}} \right)$ 

$=\dfrac{f\left( 0-h \right)-f\left( 0 \right)}{-h}~$ 

$=\dfrac{{{\left( 0-h \right)}^{2}}sin~\left( \dfrac{1}{0-h} \right)~-0}{-h}~$ 

$=\dfrac{-{{h}^{2}}sin~\left( \dfrac{1}{h} \right)~}{-h}~$ 

 $=hsin~\left( \dfrac{1}{h} \right)~~$; $\because 0\times \left( numbers~vary~between-1~to~1 \right)$ 

$LHD=0$ 

$RHD=f'\left( {{0}^{+}} \right)$ 

$=\dfrac{f\left( 0+h \right)-f\left( 0 \right)}{h}~$ 

$=\dfrac{{{\left( 0+h \right)}^{2}}sin~\left( \dfrac{1}{0+h} \right)~-0}{-h}~$ 

$=\dfrac{{{h}^{2}}sin~\left( \dfrac{1}{h} \right)~}{h}~$ 

 $=hsin~\left( \dfrac{1}{h} \right)~~$; $\because 0\times \left( numbers~vary~between-1~to~1 \right)$

$RHD=0$ Here $LHD=RHD$ 

Hence function f is differentiable at $x=0$.

22. $f\left( x \right)=\left\{ \begin{align} & 1+x,\text{ }if\text{ }x\le 2 \\ & 5-x,\text{ }if\text{ }x>2 \\ \end{align} \right.$ at $x=2$. 

Ans: Given $f\left( x \right)=\left\{ \begin{align} & 1+x,\text{ }if\text{ }x\le 2 \\ & 5-x,\text{ }if\text{ }x>2 \\ \end{align} \right.$ at $x=2$, 

 $LHD=f'\left( {{2}^{-}} \right)$ 

$=\dfrac{f\left( 2-h \right)-f\left( 2 \right)}{-h}~$ 

$=\dfrac{\left( 1+2-h \right)-3}{-h}~$ 

$=\dfrac{-h}{-h}~$ 

 $=1~$ 

$LHD=1$ 

$RHD=f'\left( {{2}^{+}} \right)$ 

$=\dfrac{f\left( 2+h \right)-f\left( 2 \right)}{h}~$ 

$=\dfrac{\left( 5-2-h \right)-3}{h}~$ 

$=\dfrac{-h}{h}~$ 

$=-1~$ 

$RHD=-1$ 

Here $LHD\ne RHD$ 

Hence function f is not differentiable at $x=2$.

23. Show that $f\left( x \right)=\left| x-5 \right|$ is continuous but not differentiable at $x=5$.

Ans: Given that $f\left( x \right)=\left| x-5 \right|$ 

 Here, $f\left( x \right)=\left| x-5 \right|=\left\{\begin{align} & -\left( x-5 \right),\text{ }x<5 \\ & x-5,\text{ } x\ge 5 \\ \end{align} \right.$ 

 Continuity at $x=5$ 

 $LHL=f\left( x \right)~$ 

 $=\underset{x\to {{5}^{-}}}{\mathop{lim}}\,\left( 5-x \right)$ 

 $=0$ 

 $RHL=f\left( x \right)~$ 

 $=\underset{x\to {{5}^{+}}}{\mathop{lim}}\,\left( x-5 \right)$ 

 $=0$

 $f\left( 5 \right)=0$ 

Here $LHL=RHL=f\left( 5 \right)$ 

Hence f is continuous at $x=5$. 

Differentiability at $x=5$ 

 $LHD={f}'\left( {{5}^{-}} \right)$ 

$=\underset{h\to 0}{\mathop{lim}}\,\left( \dfrac{1}{-h} \right)\left( f\left( 5-h \right)-f\left( 5 \right) \right)$ 

$=\underset{h\to 0}{\mathop{lim}}\,\left( \dfrac{1}{-h} \right)\left\{ \left| 5-h-5 \right|-0 \right\}$ 

$=\underset{h\to 0}{\mathop{lim}}\,\left( \dfrac{1}{-h} \right)h$ 

$LHD=-1$ 

$RHD={f}'\left( {{5}^{+}} \right)$ 

$=\underset{h\to 0}{\mathop{lim}}\,\left( \dfrac{1}{h} \right)\left( f\left( 5+h \right)-f\left( 5 \right) \right)$ 

$=\underset{h\to 0}{\mathop{lim}}\,\left( \dfrac{1}{h} \right)\left\{ \left| 5+h-5 \right|-0 \right\}$ 

$=\underset{h\to 0}{\mathop{lim}}\,\left( \dfrac{1}{h} \right)h$ 

$RHD=1$ 

Here, $LHD\ne RHD$ 

Hence f is not differentiable at $x=5$. 

24. A function $f:R\to R$ satisfies the equation $f\left( x+y \right)=f\left( x \right).f\left( y \right)$ for all $x,~y\in R,~f\left( x \right)\ne 0$. Suppose that the function is differentiable at $x=0$ and ${f}'\left( 0 \right)=2$, then prove that ${f}'\left( x \right)=2f\left( x \right)$. 

Ans: Given $f:R\to R$ satisfies the equation $f\left( x+y \right)=f\left( x \right).f\left( y \right)$.

 for all $x,~y\in R,~f\left( x \right)\ne 0$. 

${f}'\left( x \right)=\dfrac{1}{h}\left\{ f\left( x+h \right)-f\left( x \right) \right\}~$ 

 $=\dfrac{1}{h}\left\{ f\left( x \right).f\left( h \right)-f\left( x \right) \right\}~$ $\left[\because f\left( x+y \right)=f\left( x \right).f\left( y \right)\right]$ 

$=f\left( x \right)~\dfrac{1}{h}\left\{ f\left( h \right)-1 \right\}~$ 

By putting $x=0=y$, $f\left( x+y \right)=f\left( x \right).f\left( y \right)$ 

 $f\left( 0 \right)=f\left( 0 \right).f\left( 0 \right)\Rightarrow f\left( 0 \right)=0~or~1$ but $f\left( x \right)\ne 0$ 

So, $f\left( 0 \right)=1$

$f'\left( x \right)=f\left( x \right)~\dfrac{1}{h}\left\{ f\left( h \right)-f\left( 0 \right) \right\}~$ 

$=f\left( x \right).f'\left( 0 \right)$ 

 ${f}'\left( x \right)=2f\left( x \right)$ 

 Hence proved.

Differentiate each of the following w.r.to x (Exercise 25 to 43)

25. ${{2}^{cos^{2}x}}$ 

Ans: Given $y={{2}^{cos^{2}x}}$ 

By taking log to the base $e$.

 $ln~y~=ln~{{2}^{cos^{2}x}}~={cos^{2}x}~ln~2~$ 

By differentiating w.r.to x we have

$\dfrac{d}{dx}ln~y~=ln~2~\dfrac{d}{dx}{cos^{2}x}~$

$\Rightarrow \dfrac{1}{y}~\dfrac{dy}{dx}=\left(ln~2~ \right).\left(2cos~x\right)\left( -sin~x~ \right)$ 

$\Rightarrow \dfrac{dy}{dx}=-y.sin~2x~. ln~2~$ 

$\Rightarrow \dfrac{dy}{dx}=-{{2}^{cos^{2}x~}}.sin~2x~. ln~2~$ 

26. $\dfrac{{{8}^{x}}}{{{x}^{8}}}$ 

Ans: Given $y=\dfrac{{{8}^{x}}}{{{x}^{8}}}$ 

By taking log to base e. 

$ln~y~=ln~\dfrac{{{8}^{x}}}{{{x}^{8}}}~=lnln~{{8}^{x}}~-lnln~{{x}^{8}}~$ 

$\Rightarrow ln~y~=x~ln~8~-8lnln~x~$ 

$\Rightarrow \dfrac{d}{dx}ln~y~=\dfrac{d}{dx}\left( x~ln~8~-8~ln~x~ \right)$ 

$\Rightarrow \dfrac{d}{dy}ln~y~\dfrac{dy}{dx}=\left(ln~8~-\dfrac{8}{x} \right)$ 

$\Rightarrow \dfrac{dy}{dx}=y\left(ln~8~-\dfrac{8}{x} \right)$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{{{8}^{x}}}{{{x}^{8}}}\left(ln~8~-\dfrac{8}{x} \right)$

27. $log~\left( x+\sqrt{{{x}^{2}}+a} \right)~$ 

Ans: Given that $y=log~\left( x+\sqrt{{{x}^{2}}+a} \right)~$ 

 $\dfrac{dy}{dx}=\dfrac{d}{dx}\left( log~\left( x+\sqrt{{{x}^{2}}+a} \right)~ \right)$ 

$=\dfrac{d}{d\left( x+\sqrt{{{x}^{2}}+a} \right)}log~\left( x+\sqrt{{{x}^{2}}+a} \right)~\dfrac{d}{dx}\left( x+\sqrt{{{x}^{2}}+a} \right)$ 

$=\dfrac{1}{\left( x+\sqrt{{{x}^{2}}+a} \right)}\left\{ 1+\dfrac{d}{d\left( {{x}^{2}}+a \right)}\sqrt{{{x}^{2}}+a}~\dfrac{d\left( {{x}^{2}}+a \right)}{dx} \right\}$ 

$=\dfrac{1}{\left( x+\sqrt{{{x}^{2}}+a} \right)}\left\{ 1+\dfrac{1}{2\sqrt{{{x}^{2}}+a}}~2x \right\}$ 

$=\dfrac{1}{\left( x+\sqrt{{{x}^{2}}+a} \right)}\left\{ \dfrac{x+\sqrt{{{x}^{2}}+a}}{\sqrt{{{x}^{2}}+a}}~ \right\}$ 

$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{{{x}^{2}}+a}}$ 

28. $log~\left[ log~\left( log~{{x}^{5}}~ \right)~ \right]~$ 

Ans: Given $y=log~\left[ log~\left( log~{{x}^{5}}~ \right)~ \right]~$ 

 $\dfrac{dy}{dx}=\dfrac{d}{dx}log~\left[ log~\left( log~{{x}^{5}}~ \right)~ \right]~$ 

$=\dfrac{d}{dlog~\left( log~{{x}^{5}}~ \right)~}log~\left[ log~\left( log~{{x}^{5}}~ \right)~ \right]$ 

$~~\dfrac{d}{dlog~{{x}^{5}}~}log~\left( log~{{x}^{5}}~ \right)~~5\dfrac{d}{dx}log~x~$

$=\dfrac{1}{log~\left( log~{{x}^{5}}~ \right)~}\times \dfrac{1}{log~{{x}^{5}}~}\times \dfrac{5}{x}$ 

$\dfrac{dy}{dx}=\dfrac{1}{x.log~{{x}^{5}}.log~\left( log~{{x}^{5}}~ \right)~}$ 

29. $sin~\sqrt{x}~+cos^{2}\sqrt{x}~$ 

Ans: Given $y=sin~\sqrt{x}~+cos^{2}\sqrt{x}~$ 

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( sin~\sqrt{x}~+cos^{2}\sqrt{x}~ \right)$ 

$=\dfrac{d}{d\sqrt{x}}sin~\sqrt{x}~~\dfrac{d\sqrt{x}}{dx}+\dfrac{d}{d\sqrt{x}}\sqrt{x}~~\dfrac{d\sqrt{x}}{dx}$ 

$=cos~\sqrt{x}~\dfrac{1}{2\sqrt{x}}-sin~2\sqrt{x}~\dfrac{1}{2\sqrt{x}}$ 

$=\dfrac{1}{2\sqrt{x}}\left( cos~\sqrt{x}~-sin~2\sqrt{x}~~ \right)$ 

30. $sin^{n}\left( a{{x}^{2}}+bx+c \right)~$ 

Ans: Given $y=sin^{n}\left( a{{x}^{2}}+bx+c \right)~$ 

 $\dfrac{dy}{dx}=\dfrac{d}{dx}sin^{n}\left( a{{x}^{2}}+bx+c \right)~$ 

$=\dfrac{d sin^{n}\left( a{{x}^{2}}+bx+c \right)~}{dsin~\left( a{{x}^{2}}+bx+c \right)~}\times \dfrac{dsin~\left( a{{x}^{2}}+bx+c \right)~}{d\left( a{{x}^{2}}+bx+c \right)}\times \dfrac{d\left( a{{x}^{2}}+bx+c \right)}{dx}$ 

$=n sin^{n-1}\left( a{{x}^{2}}+bx+c \right)~.cos\left( a{{x}^{2}}+bx+c \right).\left( 2ax+b \right)$ 

$\dfrac{dy}{dx}=n\left( 2ax+b \right).sin^{n-1}\left( a{{x}^{2}}+bx+c \right)~.cos~\left( a{{x}^{2}}+bx+c \right)~$ 

31. $cos~\left( tan~\sqrt{x+1}~ \right)~$ 

Ans: Given $y=cos~\left( tan~\sqrt{x+1}~ \right)~$ 

 $\dfrac{dy}{dx}=\dfrac{d}{dx}cos~\left( tan~\sqrt{x+1}~ \right)~$ 

 $=\dfrac{dcos~\left( tan~\sqrt{x+1}~ \right)~}{d\left( tan~\sqrt{x+1}~ \right)}\times \dfrac{d\left( tan~\sqrt{x+1}~ \right)}{d\sqrt{x+1}}\times \dfrac{d\sqrt{x+1}}{dx}$ 

 $=-sin~\left( tan~\sqrt{x+1}~ \right)~.\sqrt{x+1}.\dfrac{1}{2\sqrt{x+1}}$ 

$\dfrac{dy}{dx}=\dfrac{-1}{2\sqrt{x+1}}.sin~\left( tan~\sqrt{x+1}~ \right)~.\sqrt{x+1}~$ 

32. $sin~{{x}^{2}}~+sin^{2}x~+sin^{2}({{x}^{2}})~$ 

Ans: Given $y=sin~{{x}^{2}}~+sin^{2}x~+sin^{2}({{x}^{2}})~$ 

 $\dfrac{dy}{dx}=\dfrac{d}{dx}\left( sin~{{x}^{2}}~+x~+{{x}^{2}}~ \right)$ 

$=\dfrac{dsin~{{x}^{2}}~}{d{{x}^{2}}}\times \dfrac{d{{x}^{2}}}{dx}+\dfrac{d}{dx}\left(sin^{2}{x}\right)+\dfrac{d}{dx}{sin^{2}\left({x^2}\right)}$ 

 $\dfrac{dy}{dx}=2xcos~{{x}^{2}}~+sin~2x~+2xsin\left( 2{{x}^{2}} \right)~$ 

33. $sin^{-1}\left( \dfrac{1}{\sqrt{x+1}} \right)~$ 

Ans: Given $y=sin^{-1}\left( \dfrac{1}{\sqrt{x+1}} \right)~$ 

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left\{ sin^{-1}\left( \dfrac{1}{\sqrt{x+1}} \right)~ \right\}$

$=\dfrac{d~sin^{-1}\left( \dfrac{1}{\sqrt{x+1}} \right)~}{d\left( \dfrac{1}{\sqrt{x+1}} \right)}\times \dfrac{d\left( \dfrac{1}{\sqrt{x+1}} \right)}{d\sqrt{x+1}}\times \dfrac{d\sqrt{x+1}}{dx}$ 

$=\dfrac{1}{\sqrt{1-{{\left( \dfrac{1}{\sqrt{x+1}} \right)}^{2}}}}\times \left\{ -\dfrac{1}{{{\left( \sqrt{x+1} \right)}^{2}}} \right\}\times \dfrac{1}{2\sqrt{x+1}}$ 

$=-\dfrac{\sqrt{x+1}}{\sqrt{x+1-1}}\times \dfrac{1}{{{\left( \sqrt{x+1} \right)}^{2}}}\times \dfrac{1}{2\sqrt{x+1}}$ 

$\dfrac{dy}{dx}=-\dfrac{1}{2\sqrt{x}\left( x+1 \right)}$ 

34. ${{\left( sin~x~ \right)}^{cos~x~}}$ 

Ans: Given $y={{\left( sin~x~ \right)}^{cos~x~}}$ 

Taking log

$\Rightarrow log~y~=log~\left( {{\left( sin~x~ \right)}^{cos~x~}} \right)~=cos~x~log~sin~x~~$

$\Rightarrow \dfrac{d}{dx}log~y~=\dfrac{d}{dx}cos~x~log~sin~x~~$ 

$\Rightarrow \dfrac{d}{dy}log~y~.\dfrac{dy}{dx}=cos~x~\dfrac{dlog~sin~x~~}{dsin~x~}\times \dfrac{dsin~x~}{dx}+log~sin~x~.\dfrac{dcos~x~}{dx}$ 

$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=cos~x.\dfrac{1}{sin~x~}.cos~x~-sin~x~log~sin~x~~$ 

$\Rightarrow \dfrac{dy}{dx}={{\left( sin~x~ \right)}^{cos~x~}}\left( cos~x~.cot~x~-sin~x~log~sin~x~~ \right)$ 

35. $sin^{m}x~\cdot cos^{n}x~$ 

Ans: Given $y=sin^{m}x~\cdot cos^{n}x~$ 

 $\dfrac{dy}{dx}=\dfrac{d}{dx}\left( sin^{m}x~\cdot cos^{n}x~ \right)$ 

$=cos^{n}x~\dfrac{d}{dx}{sin^{m}x}+sin^{m}x~\dfrac{d}{dx}{cos^{n}x}~$ 

 $=cos^{n}x~\cdot m~sin^{m-1}x~cos~x+sin^{m}x \cdot n~~\left(cos^{n-1}x~ \right)~sin~x$

$=m~cos^{n+1}x\cdot sin^{m-1}x - n sin^{m+1}x \cdot \left(cos^{n-1}x~ \right)~$ 

36. ${{\left( x+1 \right)}^{2}}{{\left( x+2 \right)}^{3}}{{\left( x+3 \right)}^{4}}$ 

Ans: Given $y={{\left( x+1 \right)}^{2}}{{\left( x+2 \right)}^{3}}{{\left( x+3 \right)}^{4}}$

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left\{ {{\left( x+1 \right)}^{2}}{{\left( x+2 \right)}^{3}}{{\left( x+3 \right)}^{4}} \right\}$ $={{\left( x+2 \right)}^{3}}{{\left( x+3 \right)}^{4}}\dfrac{d}{dx}{{\left( x+1 \right)}^{2}}+{{\left( x+1 \right)}^{2}}{{\left( x+3 \right)}^{4}}~\dfrac{d}{dx}{{\left( x+2 \right)}^{3}}+{{\left( x+1 \right)}^{2}}{{\left( x+2 \right)}^{3}}\dfrac{d}{dx}{{\left( x+3 \right)}^{4}}$ 

$={{\left( x+2 \right)}^{3}}{{\left( x+3 \right)}^{4}}\cdot 2\left( x+1 \right)+{{\left( x+1 \right)}^{2}}{{\left( x+3 \right)}^{4}}\cdot 3{{\left( x+2 \right)}^{2}}+{{\left( x+1 \right)}^{2}}{{\left( x+2 \right)}^{3}}\cdot 4{{\left( x+3 \right)}^{3}}$ 

$=\left( x+1 \right){{\left( x+2 \right)}^{2}}{{\left( x+3 \right)}^{3}}\left\{ 2\left( x+2 \right)\left( x+3 \right)+3\left( x+1 \right)\left( x+3 \right)+4\left( x+1 \right)\left( x+2 \right) \right\}$ 

$=\left( x+1 \right){{\left( x+2 \right)}^{2}}{{\left( x+3 \right)}^{3}}\left\{ \left( 2+3+4 \right){{x}^{2}}+\left( 10+12+12 \right)x+12+9+8 \right\}$ 

$=\left( x+1 \right){{\left( x+2 \right)}^{2}}{{\left( x+3 \right)}^{3}}\left( 9{{x}^{2}}+34x+29 \right)$ 

37. $cos^{-1}\left( \dfrac{sin~x~+cos~x~}{\sqrt{2}} \right)~,-\dfrac{\pi }{4}<x<\dfrac{\pi }{4}$ 

Ans: Given $y=cos^{-1}\left( \dfrac{sin~x~+cos~x~}{\sqrt{2}} \right)~,-\dfrac{\pi }{4}<x<\dfrac{\pi }{4}$ 

 $=cos^{-1}\left( sin~\dfrac{\pi }{4}~sin~x~+cos~\dfrac{\pi }{4}~cos~x~ \right)~$ 

$=cos^{-1}\left( cos~\left( \dfrac{\pi }{4}-x \right)~ \right)~$ 

 $y=\dfrac{\pi }{4}-x$

 $\dfrac{dy}{dx}=-1$ 

38. $tan^{-1}\left( \sqrt{\dfrac{1-cos~x~}{1+cos~x~}} \right)~,-\dfrac{\pi }{4}<x<\dfrac{\pi }{4}$ 

Ans: Given $y=tan^{-1}\left( \sqrt{\dfrac{1-cos~x~}{1+cos~x~}} \right)~,-\dfrac{\pi }{4}<x<\dfrac{\pi }{4}$ 

 $=tan^{-1}\left( \sqrt{\dfrac{2sin^{2}\dfrac{x}{2}~}{2cos^{2}\dfrac{x}{2}~}} \right)~$ 

 $=tan^{-1}\left( \sqrt{tan^{2}\dfrac{x}{2}~} \right)~$ 

$=tan^{-1}\left| tan~\dfrac{x}{2}~ \right|~$ 

$y=\dfrac{x}{2}$ 

$\dfrac{dy}{dx}=\dfrac{1}{2}$ 

39. $tan^{-1}\left( sec~x~+tan~x~ \right)~,-\dfrac{\pi }{2}<x<\dfrac{\pi }{2}$ 

Ans: Given $y= tan^{-1}\left( sec~x~+tan~x~ \right)~,-\dfrac{\pi }{2}<x<\dfrac{\pi }{2}$ 

 $=tan^{-1}\left[\dfrac{\left( 1+sin~x~ \right)}{cos~x~}\right]$ 

$=tan^{-1}\left[\dfrac{1-cos~\left( \dfrac{\pi }{2}+x \right)~}{sin~\left( \dfrac{\pi }{2}+x \right)~}~\right]$ 

$=tan^{-1}\left[\dfrac{2sin^{2}\left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)~}{2sin~\left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)~cos~\left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)~}~\right]$ 

 $=tan^{-1}\left[\left( tan~\left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)~ \right)~\right]$ 

 $y=\dfrac{\pi }{4}+\dfrac{x}{2}$ 

$\dfrac{dy}{dx}=\dfrac{1}{2}$ 

40. $tan^{-1}\left( \dfrac{acos~x~-bsin~x~}{bcos~x~+asin~x~} \right)~,-\dfrac{\pi }{2}< x < \dfrac{\pi }{2}~and~\dfrac{a}{b}tan~x~ > -1$ 

Ans: Given $y=tan^{-1}\left( \dfrac{acos~x~-bsin~x~}{bcos~x~+asin~x~} \right)~,-\dfrac{\pi }{2}<x < \dfrac{\pi }{2}~and~\dfrac{a}{b}tan~x~ > -1$ 

$=tan^{-1}\left(\dfrac{\dfrac{acos~x~}{bcos~x~}-\dfrac{bsin~x~}{bcos~x~}}{\dfrac{bcos~x~}{bcos~x~}+\dfrac{asin~x~}{bcos~x~}} \right)~$ 

$=tan^{-1}\left( \dfrac{\dfrac{a}{b}-tan~x~}{1+\dfrac{a}{b}tan~x~} \right)~$ 

$=tan^{-1}\left[tan\left( \dfrac{a}{b} \right)~-x\right]$ 

$=\left( \dfrac{a}{b} \right)~-x$

$\dfrac{dy}{dx}=-$1 

41. $sec^{-1}\left( \dfrac{1}{4{{x}^{3}}-3x} \right),~0<x<\dfrac{1}{\sqrt{2}}~$ 

Ans: Given $y=sec^{-1}\left( \dfrac{1}{4{{x}^{3}}-3x} \right),~0<x<\dfrac{1}{\sqrt{2}}~$ 

 Let, $x=cos~t~$ 

 $y=sec^{-1}\left( \dfrac{1}{4cos^{3}~t~-3cos~t~} \right)~$ 

 $=sec^{-1}\left( \dfrac{1}{cos~3t~} \right)~$ 

$=sec^{-1}\left( sec~3t~ \right)~$ 

$=3t$ 

$y=3~cos^{-1}~x$ 

$\dfrac{dy}{dx}=3\times \dfrac{-1}{\sqrt{1-{{x}^{2}}}}=\dfrac{-3}{\sqrt{1-{{x}^{2}}}}$ 

42. $tan^{-1}\left( \dfrac{3{{a}^{2}}x-{{x}^{3}}}{{{a}^{3}}-3a{{x}^{2}}} \right)~,-\dfrac{1}{\sqrt{3}}<\dfrac{x}{a}<\dfrac{1}{\sqrt{3}}$ 

Ans: Given $y=tan^{-1}\left( \dfrac{3{{a}^{2}}x-{{x}^{3}}}{{{a}^{3}}-3a{{x}^{2}}} \right)~,-\dfrac{1}{\sqrt{3}}<\dfrac{x}{a}<\dfrac{1}{\sqrt{3}}$ 

$\Rightarrow y=tan^{-1}\left( \dfrac{3\cdot \dfrac{x}{a}-{{\left( \dfrac{x}{a} \right)}^{3}}}{1-3{{\left( \dfrac{x}{a} \right)}^{2}}} \right)~$ 

Let, $\dfrac{x}{a}=tan~t~$ 

$\Rightarrow y=tan^{-1}\left( tan~3t~ \right)~$    $\left[\because tan~3t~=\dfrac{3tan~t~-tan^{3}t~}{1-3tan^{2}{t}~}\right]$ 

$\Rightarrow y=3t=3\cdot tan^{-1} \dfrac{x}{a}~$ 

$\Rightarrow \dfrac{dy}{dx}=3\cdot \dfrac{1}{a\left( 1+{{\left( \dfrac{x}{a} \right)}^{2}} \right)}$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{3a}{{{a}^{2}}+{{x}^{2}}}$ 

43. $tan^{-1}\left( \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)~,~-1<x<1,~x\ne 0$ 

Ans:Given $y=tan^{-1}\left( \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)~,~-1<x<1,~x\ne 0$ 

 Let, ${{x}^{2}}=cos~2t~$ 

$y=tan^{-1}\left(\dfrac{\sqrt{1+cos~2t~}+\sqrt{1-cos~2t~}}{\sqrt{1+cos~2t~}-\sqrt{1-cos~2t~}} \right)~$ 

$=tan^{-1}\left( \dfrac{\sqrt{2}cos~t~+\sqrt{2}sin~t~}{\sqrt{2}cos~t~-\sqrt{2}sin~t~} \right)~$ 

$=tan^{-1}\left( \dfrac{1+tan~t~}{1-tan~t~} \right)~$ 

$=tan^{-1}\left[tan\left(\dfrac{\pi}{4}+t\right)\right]~=\dfrac{\pi }{4}+t$ 

$=\dfrac{\pi }{4}+\dfrac{1}{2}{cos^{-1}{{x}^{2}}}$ 

$\dfrac{dy}{dx}=\dfrac{1}{2}\times \dfrac{-2x}{\sqrt{1-{{x}^{4}}}}$ 

$\dfrac{dy}{dx}=\dfrac{-x}{\sqrt{1-{{x}^{4}}}}$ 

Find $\dfrac{dy}{dx}$ of each of the functions expressed in parametric form in Exercises from 44 to 48.

44. $x=t+\dfrac{1}{t},~y=t-\dfrac{1}{t}$ 

Ans: Given $x=t+\dfrac{1}{t},~y=t-\dfrac{1}{t}$ 

$\Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( t+\dfrac{1}{t} \right),~\dfrac{dy}{dt}=\dfrac{d}{dt}\left( t-\dfrac{1}{t} \right)$ 

$\Rightarrow\dfrac{dx}{dt}=1-\dfrac{1}{{{t}^{2}}},~\dfrac{dy}{dt}=1+\dfrac{1}{{{t}^{2}}}$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ 

$=\dfrac{1+\dfrac{1}{{{t}^{2}}}}{1-\dfrac{1}{{{t}^{2}}}}$ 

$=\dfrac{{{t}^{2}}+1}{{{t}^{2}}-1}$ 

$x+y=2t\Rightarrow t=\dfrac{x+y}{2}$ 

$\dfrac{dy}{dx}=\dfrac{{{\left( x+y \right)}^{2}}+4}{{{\left( x+y \right)}^{2}}-4}$ 

45. $x={{e}^{\theta }}\left( \theta +\dfrac{1}{\theta } \right),~y={{e}^{\theta }}\left( \theta - \dfrac{1}{\theta } \right)$ 

Ans: Given $x={{e}^{\theta }}\left( \theta +\dfrac{1}{\theta } \right),~y={{e}^{-\theta }}\left( \theta -\dfrac{1}{\theta } \right)$ 

 $\Rightarrow \dfrac{dx}{d\theta }=\dfrac{d}{d\theta }\left\{ {{e}^{\theta }}\left( \theta +\dfrac{1}{\theta } \right) \right\},~\dfrac{dy}{d\theta }=\dfrac{d}{d\theta }\left\{ {{e}^{-\theta }}\left( \theta -\dfrac{1}{\theta } \right) \right\}$ 

$\Rightarrow \dfrac{dx}{d\theta }={{e}^{\theta }}\left( \theta +\dfrac{1}{\theta } \right)+{{e}^{\theta }}\left( 1-\dfrac{1}{{{\theta }^{2}}} \right),~\dfrac{dy}{d\theta }=-{{e}^{-\theta }}\left( \theta -\dfrac{1}{\theta } \right)+{{e}^{-\theta }}\left( 1+\dfrac{1}{{{\theta }^{2}}} \right)$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{dy/d\theta }{dx/d\theta }$ 

 $=\dfrac{-{{e}^{-\theta }}\left( \theta -\dfrac{1}{\theta } \right)+{{e}^{-\theta }}\left( 1+\dfrac{1}{{{\theta }^{2}}} \right)}{{{e}^{\theta }}\left( \theta +\dfrac{1}{\theta } \right)+{{e}^{\theta }}\left( 1-\dfrac{1}{{{\theta }^{2}}} \right)}$ 

$={{e}^{-2\theta }}\left( \dfrac{\dfrac{-\left( {{\theta }^{2}}-1 \right)}{\theta }+\dfrac{\left( {{\theta }^{2}}+1 \right)}{{{\theta }^{2}}}}{\dfrac{\left( {{\theta }^{2}}+1 \right)}{\theta }+\dfrac{\left( {{\theta }^{2}}-1 \right)}{{{\theta }^{2}}}} \right)$ 

$\dfrac{dy}{dx}={{e}^{-2\theta }}\left( \dfrac{-{{\theta }^{3}}+{{\theta }^{2}}+\theta +1}{{{\theta }^{3}}+{{\theta }^{2}}+\theta -1} \right)$ 

46. $x=3~cos~\theta ~-2~cos^{3}\theta ~,~y=3sin~\theta ~-2sin^{3}\theta ~$ 

Ans: Given $x=3cos~\theta ~-2\theta ~,~y=3sin~\theta ~-2\theta ~$ 

$\Rightarrow \dfrac{d}{d\theta }x=\dfrac{d}{d\theta }\left( 3cos~\theta ~-2\theta ~ \right),~\dfrac{d}{d\theta }y=\dfrac{d}{d\theta }\left( 3sin~\theta ~-2\theta ~ \right)$ 

$\Rightarrow \dfrac{dx}{d\theta }=-3sin~\theta ~-6~cos^{2}\theta ~~\left( -sin~\theta ~ \right),\dfrac{dy}{d\theta }=3cos~\theta ~-6~sin^{2}\theta ~cos~\theta ~$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{dy/d\theta }{dx/d\theta }$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{3cos~\theta ~-6~sin^{2}\theta ~cos~\theta ~}{-3sin~\theta ~+6sin~\theta cos^{2}\theta ~} =\dfrac{3cos~\theta ~}{3sin~\theta ~}\times \dfrac{1-2\theta ~}{2\theta ~-1}$ 

$\Rightarrow \dfrac{dy}{dx}=cot~\theta ~$ 

47. $sin~x~=\dfrac{2t}{1+{{t}^{2}}}~,tan~y~=\dfrac{2t}{1-{{t}^{2}}}$ 

Ans: Given $sin~x~=\dfrac{2t}{1+{{t}^{2}}}~,tan~y~=\dfrac{2t}{1-{{t}^{2}}}$ 

 Let $t=tan~\theta ~$ 

 $\Rightarrow sin~x~=\dfrac{2tan~\theta ~~}{1+\theta ~~}~,tan~y~=\dfrac{2tan~\theta ~~}{1-\theta ~}$ 

 $\Rightarrow sin~x~=sin~2\theta ~~,tan~y~=tan~2\theta ~$ 

 $\Rightarrow x=2\theta ,~~y=2\theta $ 

 $\Rightarrow y=x$ 

 $\Rightarrow \dfrac{dy}{dx}=1$ 

48. $x=\dfrac{1+log~t~}{{{t}^{2}}}~,~y=\dfrac{3+2log~t~}{t}$. 

Ans: Given that $x=\dfrac{1+log~t~}{{{t}^{2}}}~,~y=\dfrac{3+2log~t~}{t}$ 

$\Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( \dfrac{1+log~t~}{{{t}^{2}}} \right)~,\dfrac{dy}{dt}=\dfrac{d}{dt}\left( \dfrac{3+2log~t~}{t} \right)$ 

$\Rightarrow \dfrac{dx}{dt}=\dfrac{{{t}^{2}}\left( \dfrac{1}{t} \right)-\left( 1+log~t~ \right)2t}{{{t}^{4}}}~,~\dfrac{dy}{dt}=\dfrac{t\left( \dfrac{2}{t} \right)-\left( 3+2log~t~ \right)}{{{t}^{2}}}$ 

$\Rightarrow \dfrac{dx}{dt}=-\dfrac{1+2log~t~}{{{t}^{3}}}~,~\dfrac{dy}{dt}=-\dfrac{1+2log~t~}{{{t}^{2}}}$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ 

$\Rightarrow \dfrac{dy}{dx}=\left( -\dfrac{1+2log~t~}{{{t}^{2}}} \right)\times \left( -\dfrac{{{t}^{3}}}{1+2log~t~} \right)$ 

$\Rightarrow \dfrac{dy}{dx}=t$ 

49. If $x={{e}^{cos~2t~}}$ and $y={{e}^{sin~2t}}$, then prove that $\dfrac{dy}{dx}=-\dfrac{y~log~x}{x~log~y}$. 

Ans: Given that $x={{e}^{cos~2t~}}\Rightarrow cos~2t~=log~x~$ 

 $y={{e}^{sin~2t~}}\Rightarrow sin~2t~=log~y~$ 

 Squaring and adding we have

 $\Rightarrow 2t~+2t~={{\left( log~x~ \right)}^{2}}+{{\left( log~y~ \right)}^{2}}$ 

 $\Rightarrow {{\left( log~x~ \right)}^{2}}+{{\left( log~y~ \right)}^{2}}=1$ 

 By differentiating w.r.to x we have

 $\Rightarrow \dfrac{d}{dx}{{\left( log~x~ \right)}^{2}}+\dfrac{d}{dx}{{\left( log~y~ \right)}^{2}}=\dfrac{d}{dx}1$

 $\Rightarrow \dfrac{2log~x~}{x}+\dfrac{2log~y~}{y}\dfrac{dy}{dx}=0$ 

 $\Rightarrow \dfrac{dy}{dx}=-\dfrac{ylog~x~}{xlog~y~}$ 

50. If $x=asin~2t~~\left( 1+cos~2t~ \right)$ and $y=bcos~2t~\left( 1-cos~2t~ \right)$, show that ${{\left( \dfrac{dy}{dx} \right)}_{at~t=\dfrac{\pi }{4}}}=\dfrac{b}{a}$ . 

Ans: Given $x=asin~2t~~\left( 1+cos~2t~ \right)$ and $y=bcos~2t~\left( 1-cos~2t~ \right)$ 

$\Rightarrow x=a\left( sin~2t~+sin~2t~cos~2t~ \right),~y=b\left( cos~2t~-2t~ \right)$ 

$\Rightarrow x=a\left( sin~2t~+\dfrac{1}{2}sin~4t~ \right),~y=b\left( cos~2t~-\dfrac{1}{2}-\dfrac{1}{2}cos~4t~ \right)$ 

$\Rightarrow \dfrac{dx}{dt}=a\left( 2cos~2t~+2cos~4t~ \right),\dfrac{dy}{dt}=b\left( -2sin~2t~+2sin~4t~ \right)$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{b\left( -2sin~2t~+2sin~4t~ \right)}{a\left( 2cos~2t~+2cos~4t~ \right)}$ 

At $t=\dfrac{\pi }{4}$ 

$\Rightarrow {{\left( \dfrac{dy}{dx} \right)}_{at~t=\dfrac{\pi }{4}}}=\dfrac{b\left( -sin~\dfrac{\pi }{2}~~+sin~\pi ~ \right)}{a\left( cos~\dfrac{\pi }{2}~+cos~\pi ~ \right)}$ 

$\Rightarrow {{\left( \dfrac{dy}{dx} \right)}_{at~t=\dfrac{\pi }{4}}}=\dfrac{b\left( -1 \right)}{a\left( -1 \right)}=\dfrac{b}{a}$ 

 Hence proved. 

51. If $x=3sin~t~-sin~3t~,~y=3cos~t~-cos~3t~$, find $\dfrac{dy}{dx}~at~t=\dfrac{\pi }{3}$. 

Ans: Given $x=3sin~t~-sin~3t~,~y=3cos~t~-cos~3t~$ 

 $x=4t~,~y=3cos~t~-cos~3t~$ 

 $\Rightarrow \dfrac{dx}{dt}=12t~cos~t~,~\dfrac{dy}{dt}=-3sin~t~+3sin~3t~$ 

 $\Rightarrow \dfrac{dy}{dx}=\dfrac{-3sin~t~+3sin~3t~}{12t~cos~t~}$ 

$at~t=\dfrac{\pi }{3}$ 

$\Rightarrow {{\left( \dfrac{dy}{dx} \right)}_{at~t=\dfrac{\pi }{3}}}=\dfrac{-sin~\dfrac{\pi }{3}~+sin~\pi ~}{4\dfrac{\pi }{3}~cos~\dfrac{\pi }{3}~}=\dfrac{-\dfrac{\sqrt{3}}{2}}{4\times {{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}\times \dfrac{1}{2}}$ 

$\Rightarrow {{\left( \dfrac{dy}{dx} \right)}_{at~t=\dfrac{\pi }{3}}}=-\dfrac{1}{\sqrt{3}}$ 

52. Differentiate $\dfrac{x}{sin~x~}$ w.r.to $sin~x~$. 

Ans: Given functions are $f\left( x \right)=\dfrac{x}{sin~x~}~,~g\left( x \right)=sin~x~$ 

Differentiation of f w.r.to g 

$\dfrac{df\left( x \right)}{dg\left( x \right)}=\dfrac{df\left( x \right)/dx}{dg\left( x \right)/dx}$ 

$\Rightarrow \dfrac{df\left( x \right)}{dx}=\dfrac{d}{dx}\left( \dfrac{x}{sin~x~} \right)=\dfrac{sin~x~-xcos~x~}{sin^{2}x~}$ 

$\Rightarrow \dfrac{dg\left( x \right)}{dx}=\dfrac{d}{dx}sin~x~=cos~x~$ 

$\Rightarrow \dfrac{df\left( x \right)}{dg\left( x \right)}=\left( \dfrac{sin~x~-xcos~x~}{sin^{2}x~} \right)\left( \dfrac{1}{cos~x~} \right)$ 

$\Rightarrow \dfrac{df\left( x \right)}{dg\left( x \right)}=\dfrac{tan~x~-x}{sin^{2}x~}$ 

53. Differentiate $tan^{-1}\left( \dfrac{\sqrt{1+{{x}^{2}}}-1}{x} \right)~$ w.r.to $tan^{-1}x~$ when $x\ne 0$. 

Ans: Given functions are $f\left( x \right)=tan^{-1}\left( \dfrac{\sqrt{1+{{x}^{2}}}-1}{x} \right)~~,~g\left( x \right)=tan^{-1}x~$ 

 Differentiation of f w.r.to g 

 $\dfrac{df\left( x \right)}{dg\left( x \right)}=\dfrac{df\left( x \right)/dx}{dg\left( x \right)/dx}$ 

 $\dfrac{df\left( x \right)}{dx}=\dfrac{d}{dx}tan^{-1}\left( \dfrac{\sqrt{1+{{x}^{2}}}-1}{x} \right)~,~\dfrac{dg\left( x \right)}{dx}=\dfrac{d}{dx}tan^{-1}x~$

 For f , let $x=tan~t~$ 

 $\left( \dfrac{\sqrt{1+tan^{2}t~}-1}{tan~t~} \right)~=\left( \dfrac{\left( sec~t~-1 \right)cos~t~}{sin~t~} \right)~$ 

 $=\left( \dfrac{1-cos~t~}{sin~t~} \right)~$ 

 $=\left( tan~\dfrac{t}{2}~ \right)~$ 

 $=\dfrac{t}{2}=\dfrac{1}{2}tan^{-1}x~$ 

 $\dfrac{df\left( x \right)}{dx}=\dfrac{d}{dx}\left( \dfrac{\sqrt{1+{{x}^{2}}}-1}{x} \right)~$ 

 $=\dfrac{d}{dx}\left( \dfrac{1}{2}tan^{-1}x~ \right)$ 

$=\dfrac{1}{2\left( 1+{{x}^{2}} \right)}$ 

 $\dfrac{dg\left( x \right)}{dx}=\dfrac{d}{dx}tan^{-1}x~=\dfrac{1}{1+{{x}^{2}}}$ 

$\dfrac{df\left( x \right)}{dg\left( x \right)}=\dfrac{df\left( x \right)/dx}{dg\left( x \right)/dx}=\dfrac{\dfrac{1}{2\left( 1+{{x}^{2}} \right)}}{\dfrac{1}{1+{{x}^{2}}}~}=\dfrac{1}{2}$ 

Find $\dfrac{dy}{dx}$ when x and y are connected by the relation given in each of the exercise 54 to 57.

54. $sin~\left( xy \right)~+\dfrac{x}{y}={{x}^{2}}-y$ 

Ans: 

 Given $sin~\left( xy \right)~+\dfrac{x}{y}={{x}^{2}}-y$ 

 Differentiating w.r.to x we have

$\Rightarrow \dfrac{d}{dx}sin~\left( xy \right)~+\dfrac{d}{dx}\left( \dfrac{x}{y} \right)=\dfrac{d}{dx}{{x}^{2}}-\dfrac{dy}{dx}$ 

$\Rightarrow cos~\left( xy \right)~\left\{ y+x\dfrac{dy}{dx} \right\}+\dfrac{y-x\dfrac{dy}{dx}}{{{y}^{2}}}=2x-\dfrac{dy}{dx}$ 

$\Rightarrow {{y}^{3}}cos~\left( xy \right)~+cos~\left( xy \right)~{{y}^{2}}x\dfrac{dy}{dx}+y-x\dfrac{dy}{dx}=2x{{y}^{2}}-{{y}^{2}}\dfrac{dy}{dx}$ 

$\Rightarrow \left( cos~\left( xy \right)~{{y}^{2}}x-x+{{y}^{2}} \right)\dfrac{dy}{dx}=2x{{y}^{2}}-y-{{y}^{3}}cos~\left( xy \right)~$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{2x{{y}^{2}}-y-{{y}^{3}}cos~\left( xy \right)~}{cos~\left( xy \right)~{{y}^{2}}x-x+{{y}^{2}}}$ 

55. $sec~\left( x+y \right)~=xy$ 

Ans: Given $sec~\left( x+y \right)~=xy$ 

$\Rightarrow \dfrac{d}{dx}sec~\left( x+y \right)~=\dfrac{d}{dx}\left( xy \right)$ 

$\Rightarrow sec~\left( x+y \right)~tan~\left( x+y \right)~\left\{ 1+\dfrac{dy}{dx} \right\}=y+x\dfrac{dy}{dx}$ 

$\Rightarrow sec~\left( x+y \right)~tan~\left( x+y \right)~+sec~\left( x+y \right)~tan~\left( x+y \right)~\dfrac{dy}{dx}=y+x\dfrac{dy}{dx}$ 

$\Rightarrow sec~\left( x+y \right)~tan~\left( x+y \right)~\dfrac{dy}{dx}-x\dfrac{dy}{dx}~~=y-sec~\left( x+y \right)~tan~\left( x+y \right)~$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{y-sec~\left( x+y \right)~tan~\left( x+y \right)~}{sec~\left( x+y \right)~tan~\left( x+y \right)~-x}$ 

56. $tan^{-1}\left( {{x}^{2}}+{{y}^{2}} \right)~=a$. 

Ans: Given $tan^{-1}\left( {{x}^{2}}+{{y}^{2}} \right)~=a$ 

 $\therefore {{x}^{2}}+{{y}^{2}}=tan~a~$ 

 Differentiating w.r.to x we get

 $\Rightarrow 2x+2y\dfrac{dy}{dx}=0$ 

 $\Rightarrow \dfrac{dy}{dx}=-\dfrac{x}{y}$ 

57. ${{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=xy$. 

Ans: Given ${{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=xy$ 

 ${{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}-xy=0$ 

 Differentiating w.r.to x we have

 $\Rightarrow 4{{x}^{3}}+4{{y}^{3}}\dfrac{dy}{dx}+4x{{y}^{2}}+4{{x}^{2}}y\dfrac{dy}{dx}-y-x\dfrac{dy}{dx}=0$ 

 $\Rightarrow \left( 4{{y}^{3}}+4{{x}^{2}}y-x \right)\dfrac{dy}{dx}=y-4x{{y}^{2}}-4{{x}^{3}}$ 

 $\Rightarrow\dfrac{dy}{dx}=\dfrac{4{{y}^{3}}+4{{x}^{2}}y-x}{4{{y}^{3}}+4{{x}^{2}}y-x}$ 

58. If $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$, then show that $\dfrac{dy}{dx}\dfrac{dx}{dy}=1$. 

Ans: Given $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ 

 Differentiating w.r.to x we have

 $2ax+2h\left( y+x\dfrac{dy}{dx} \right)+2by\dfrac{dy}{dx}+2g+2f\dfrac{dy}{dx}=0$ 

 $\Rightarrow \left( hx+by+f \right)\dfrac{dy}{dx}=-\left( ax+hy+g \right)~$

 $\Rightarrow \dfrac{dy}{dx}=-\dfrac{ax+hy+g}{hx+by+f}$ 

 Now differentiating w.r.to y we have

 $2ax\dfrac{dx}{dy}+2h\left( x+y\dfrac{dx}{dy} \right)+2by+2g\dfrac{dx}{dy}+2f=0$ 

 $\Rightarrow \left( ax+hy+g \right)\dfrac{dx}{dy}=-\left( 2hx+by+f \right)$ 

 $\Rightarrow \dfrac{dx}{dy}=-\dfrac{hx+by+f}{ax+hy+g}$ 

 Now, $\dfrac{dy}{dx}\cdot \dfrac{dx}{dy}=1$. 

 Hence proved. 

59. If $x={{e}^{\dfrac{x}{y}}}$, prove that $\dfrac{dy}{dx}=\dfrac{x-y}{xlog~x~}$ 

Ans: Given that $x={{e}^{\dfrac{x}{y}}}$ 

 Taking log, we have

 $log~x~=\dfrac{x}{y}$ 

 $\Rightarrow y~log~x~=x$ 

 Differentiating w.r.to x we have

 $\dfrac{y}{x}+log~x~\dfrac{dy}{dx}=1$ 

 $\Rightarrow log~x~\dfrac{dy}{dx}=1-\dfrac{y}{x}=\dfrac{x-y}{x}$ 

 $\Rightarrow \dfrac{dy}{dx}=\dfrac{x-y}{xlog~x~}$ 

 Hence proved. 

60. If ${{y}^{x}}={{e}^{y-x}}$, prove that $\dfrac{dy}{dx}=\dfrac{{{\left( 1+log~y~ \right)}^{2}}}{log~y~}$. 

Ans: Given ${{y}^{x}}={{e}^{y-x}}$

 Taking log, we have

 $xlog~y~=y-x$ 

 Differentiating w.r.to x we have

 $log~y~+\dfrac{x}{y}\dfrac{dy}{dx}=\dfrac{dy}{dx}-1$ 

 $\Rightarrow 1+log~y~=\left( 1-\dfrac{x}{y} \right)\dfrac{dy}{dx}$ 

 $\Rightarrow \dfrac{dy}{dx}=\dfrac{y\left( 1+log~y~ \right)}{y-x}$ 

 Now $xlog~y~=y-x$ 

 $\Rightarrow log~y~=\dfrac{y}{x}-1$ 

 $\Rightarrow 1+log~y~=\dfrac{y}{x}$ 

 $\Rightarrow 1-\dfrac{x}{y}=1-\dfrac{1}{1+log~y~~}=\dfrac{1+log~y~-1}{1+log~y~}=\dfrac{log~y~}{1+log~y~}$ 

 $\Rightarrow \dfrac{dy}{dx}=\dfrac{\left( 1+log~y~ \right)}{1-\dfrac{x}{y}}=\dfrac{\left( 1+log~y~ \right)}{\dfrac{log~y~}{1+log~y~}~}$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{{{\left( 1+log~y~ \right)}^{2}}}{log~y~}$ 

 Hence proved.

61. If $y={{\left( cos~x~ \right)}^{{{\left( cos~x~ \right)}^{x~)\ldots \infty }}}}$, show that $\dfrac{dy}{dx}=\dfrac{{{y}^{2}}tan~x~}{ylog~cos~x~~-1}$. 

Ans: Given $y={{\left( cos~x~ \right)}^{{{\left( cos~x~ \right)}^{x~)\ldots \infty }}}}$ 

 $y={{\left( cos~x~ \right)}^{y}}$ 

 $\Rightarrow log~y~=ylog~cos~x~~$ 

 $\Rightarrow \dfrac{d}{dx}log~y~=\dfrac{d}{dx}\left( ylog~cos~x~~ \right)$ 

 $\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=log~cos~x~~\dfrac{dy}{dx}+y\dfrac{1}{cos~x~}\left( -sin~x~ \right)$ 

 $\Rightarrow \dfrac{dy}{dx}=ylog~cos~x~~\dfrac{dy}{dx}-{{y}^{2}}tan~x~$ 

 $\Rightarrow {{y}^{2}}tan~x~=\left( ylog~cos~x~~-1 \right)\dfrac{dy}{dx}$ 

 $\Rightarrow \dfrac{dy}{dx}=\dfrac{{{y}^{2}}tan~x~}{ylog~cos~x~~-1}$ 

 Hence proved. 

62. If $xsin~\left( a+y \right)~+sin~a~cos~\left( a+y \right)~=0$, prove that $\dfrac{dy}{dx}=\dfrac{sin^{2}\left( a+y \right)~}{sin~a~}$. 

Ans: Given $xsin~\left( a+y \right)~+sin~a~cos~\left( a+y \right)~=0$

$\dfrac{d}{dx}\left( xsin~\left( a+y \right)~+sin~a~cos~\left( a+y \right)~ \right)=0$

$\Rightarrow sin~\left( a+y \right)~+xcos~\left( a+y \right)~\dfrac{dy}{dx}-sin~a~sin~\left( a+y \right)~\dfrac{dy}{dx}=0$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{sin~\left( a+y \right)~}{sin~a~sin~\left( a+y \right)~-xcos~\left( a+y \right)~}$ 

$\Rightarrow x=-\dfrac{sin~a~cos~\left( a+y \right)~}{sin~\left( a+y \right)~}$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{sin~\left( a+y \right)~}{sin~a~sin~\left( a+y \right)~-\left( -\dfrac{sin~a~cos~\left( a+y \right)~}{sin~\left( a+y \right)~}~ \right)cos~\left( a+y \right)~}$ 

 $=\dfrac{sin^{2}\left( a+y \right)~}{sin~a~\left\{ sin^{2}\left( a+y \right)~+cos^{2}\left( a+y \right)~ \right\}}$ 

 $\dfrac{dy}{dx}=\dfrac{sin^{2}\left( a+y \right)~}{sin~a~}$ 

 Hence proved. 

63. If $\sqrt{1-{{x}^{2}}}+\sqrt{1-{{y}^{2}}}=a\left( x-y \right)$, prove that $\dfrac{dy}{dx}=\sqrt{\dfrac{1-{{y}^{2}}}{1-{{x}^{2}}}}$. 

Ans: Given $\sqrt{1-{{x}^{2}}}+\sqrt{1-{{y}^{2}}}=a\left( x-y \right)$ 

 Let $x=sin~p~,~y=sin~q~$ 

 $1-{{x}^{2}}=1-p~=p~$ 

 $\Rightarrow 1-{{y}^{2}}=1-q~=q~$ 

 $\Rightarrow cos~p~+cos~q~=a\left( sin~p~-sin~q~ \right)$ 

 $\Rightarrow 2cos~\dfrac{p+q}{2}~~cos~\dfrac{p-q}{2}~=a\left( 2cos~\dfrac{p+q}{2}~~sin~\dfrac{p-q}{2}~ \right)$ 

 $\Rightarrow cot~\dfrac{p-q}{2}~=a$ 

 $\Rightarrow p-q=2a~$ 

 $\Rightarrow x~-y~=2a~$ 

 $\Rightarrow \dfrac{d}{dx}\left( x~-y~ \right)=2\dfrac{d}{dx}a~$ 

$\Rightarrow \dfrac{1}{\sqrt{1-{{x}^{2}}}}-\dfrac{1}{\sqrt{1-{{y}^{2}}}}\dfrac{dy}{dx}=0$ 

 $\Rightarrow \dfrac{dy}{dx}=\dfrac{\sqrt{1-{{y}^{2}}}}{\sqrt{1-{{x}^{2}}}}$ 

 Hence prove. 

64. If $y= tan^{-1}x~$, find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ in terms of y alone. 

Ans: Given $y= tan^{-1}x~$ 

 $x=tan~y~$ 

 Differentiating w.r.to x 

 $1= sec^{2}y~\dfrac{dy}{dx}$ 

 $\Rightarrow \dfrac{dy}{dx}= cos^{2}y~$ 

 Again, $\dfrac{d}{dx}\dfrac{dy}{dx}=\dfrac{d}{dx} cos^{2}y~$ 

$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-sin~2y~\dfrac{dy}{dx}=-sin~2y~cos^{2}y~$

Verify Rolle's theorem for each of the functions in Exercise 65 to 69. 

65. $f\left( x \right)=x{{\left( x-1 \right)}^{2}}$ in $\left[ 0,~1 \right]$. 

Ans: Given $f\left( x \right)=x{{\left( x-1 \right)}^{2}}$ 

 Function f is polynomial, so 

i. it is continuous and 

ii. differentiable for all real x. 

iii. $f\left( 0 \right)=0$ and $f\left( 1 \right)=0$ 

 All three conditions for Rolle’s theorem are verified, so 

 $\exists c\in \left[ 0,~1 \right]$ such that ${f}'\left( c \right)=0$. 

 ${f}'\left( x \right)={{\left( x-1 \right)}^{2}}+2x\left( x-1 \right)$ 

 $=3{{x}^{2}}-4x+1$ 

 $\Rightarrow 3{{c}^{2}}-4c+1=0$ 

 $\Rightarrow 3{{c}^{2}}-3c-c+1=0$ 

 $\Rightarrow 3c\left( c-1 \right)-\left( c-1 \right)=0$ 

 $\Rightarrow \left( 3c-1 \right)\left( c-1 \right)=0$ 

 $\Rightarrow c=1,\dfrac{1}{3}$ 

 $\Rightarrow \dfrac{1}{3}\in \left[ 0,~1 \right]$ 

 Hence Rolle’s theorem is verified.

66. $f\left( x \right)=sin^{4}x~+cos^{4}x~$ in $\left[ 0,\dfrac{\pi }{2} \right]$. 

Ans: Given $f\left( x \right)=sin^{4}x~+cos^{4}x~~$ 

 Function f is sine and cosine function, so 

i. it is continuous and 

ii. differentiable for all real x. 

iii. $f\left( 0 \right)=1$ and $f\left( \dfrac{\pi }{2} \right)=1$ 

 All three conditions for Rolle’s theorem are verified, so 

 $\exists c\in \left[ 0,\dfrac{\pi }{2} \right]$ such that ${f}'\left( c \right)=0$. 

 ${f}'\left( x \right)=\dfrac{d}{dx}\left( sin^{4}x~+cos^{4}x~~ \right)$ 

 $=4sin^{3}x~cos~x~-4cos^{3}x~sin~x~$ 

 $=4sin~x~cos~x~\left( sin^{2}x~- cos^{2}x~ \right)$ 

 $=-2sin~2x~cos~2x~$ 

 $-2sin~2c~cos~2c~=0$ 

 $sin~2c~=0\Rightarrow 2c=0,~\pi $ 

 $c=0,\dfrac{\pi }{2}$ 

 $cos~2c~=0\Rightarrow 2c=\dfrac{\pi }{2}$ 

 $c=\dfrac{\pi }{4}\in \left[ 0,~~\dfrac{\pi }{2} \right]$ 

 Hence Rolle’s theorem is verified.

67. $f\left( x \right)=log~\left( {{x}^{2}}+2 \right)~-log~3~$ in $\left[ -1,~1 \right]$ 

Ans: 

 Given $f\left( x \right)=log~\left( {{x}^{2}}+2 \right)~-log~3~$ 

 Function f is logarithmic function and ${{x}^{2}}+2>0~\forall x\in R$, so 

i. it is continuous and 

ii. differentiable for all real x. 

iii. $f\left( -1 \right)=0$ and $f\left( 1 \right)=0$ 

 All three conditions for Rolle’s theorem are verified, so 

 $\exists c\in \left[ -1,~1 \right]$ such that ${f}'\left( c \right)=0$. 

 ${f}'\left( x \right)=\dfrac{d}{dx}\left( log~\left( {{x}^{2}}+2 \right)~-log~3~ \right)$ 

 $=\dfrac{2x}{{{x}^{2}}+2}$ 

 $\dfrac{2c}{{{c}^{2}}+1}=0\Rightarrow c=0\in \left[ -1,~1 \right]$ 

 Hence Rolle’s theorem is verified.

68. $f\left( x \right)=x\left( x+3 \right){{e}^{-\dfrac{x}{2}}}$ in $\left[ -3,~0 \right]$. 

Ans: Given $f\left( x \right)=x\left( x+3 \right){{e}^{-\dfrac{x}{2}}}$ 

 Function f is polynomial with exponential function, so 

i. it is continuous and 

ii. differentiable for all real x. 

iii. $f\left( -3 \right)=0$ and $f\left( 0 \right)=0$ 

 All three conditions for Rolle’s theorem are verified, so 

 $\exists c\in \left[ -3,~0 \right]$ such that ${f}'\left( c \right)=0$. 

 $\Rightarrow {f}'\left( x \right)=\dfrac{d}{dx}\left( x\left( x+3 \right){{e}^{-\dfrac{x}{2}}} \right)$ 

 $=\left( x+3 \right){{e}^{-\dfrac{x}{2}}}+x{{e}^{-\dfrac{x}{2}}}-\dfrac{1}{2}x\left( x+3 \right){{e}^{-\dfrac{x}{2}}}$ 

 $=\dfrac{1}{2}\left( {{x}^{2}}+7x+6 \right){{e}^{-\dfrac{x}{2}}}$ 

 $\Rightarrow f'\left( c \right)=\dfrac{1}{2}\left( {{c}^{2}}+7c+6 \right){{e}^{-\dfrac{c}{2}}}=0$ 

 $\Rightarrow \therefore {{c}^{2}}+c+6c+6=0$ 

 $\Rightarrow c\left( c+1 \right)+6\left( c+1 \right)=0$ 

 $\Rightarrow \left( c+1 \right)\left( c+6 \right)=0$ 

 $\Rightarrow c=-6,~-1$ 

 $\Rightarrow c=-1\in \left[ -3,~0 \right]$ 

 Hence Rolle’s theorem is verified.

69. $f\left( x \right)=\sqrt{4-{{x}^{2}}}$ in $\left[ -2,~2 \right]$ 

Ans: Given $f\left( x \right)=\sqrt{4-{{x}^{2}}}$ 

 Function f is defined $\forall x\in \left[ -2,~2 \right]$, so 

i. it is continuous in $\left[ -2,~2 \right]$ and 

ii. differentiable in$\left( -2,~2 \right)$. 

iii. $f\left( -2 \right)=0$ and $f\left( 2 \right)=0$ 

 All three conditions for Rolle’s theorem are verified, so 

 $\exists c\in \left[ -2,~2 \right]$ such that ${f}'\left( c \right)=0$. 

 ${f}'\left( x \right)=\dfrac{d}{dx}\sqrt{4-{{x}^{2}}}$ 

 $=-\dfrac{x}{\sqrt{4-{{x}^{2}}}}$ 

 ${f}'\left( c \right)=-\dfrac{c}{\sqrt{4-{{c}^{2}}}}=0$ 

 $c=0\in \left( -2,~2 \right)$ 

 Hence Rolle’s theorem is verified.

70. Discuss the applicability of Rolle’s theorem on the function given by 

$f\left( x \right)=\left\{ \begin{align} &{{x}^{2}}+1,\text{if}\text{ }0\le x \le 1 \\ & 3-x,\text{if}\text{ }1\le x \le 2 \\ \end{align} \right.$

Ans: Given $f\left( x \right)=\left\{ \begin{align} &{{x}^{2}}+1,\text{if}\text{ }0\le x \le 1 \\ & 3-x,\text{if}\text{ }1\le x \le 2 \\ \end{align} \right.$

 At $x=1$ 

 $LHL=f\left( x \right)~$ 

 $=\underset{x\to {{1}^{-}}}{\mathop{lim}}\,\left( {{x}^{2}}+1 \right)$ 

 $=1+1=2$ 

 $RHL=\underset{x\to {{1}^{+}}}{\mathop{lim}}\,~f\left( x \right)$ 

 $=\underset{x\to {{1}^{+}}}{\mathop{lim}}\,\left( 3-x \right)$ 

 $=3-1=2$

 $LHL=RHL=f\left( 1 \right)$ 

 So, function is continuous.

 $LHD={f}'\left( {{1}^{-}} \right)={{\left( 2x \right)}_{at~x=1}}=2$ 

 $RHD={f}'\left( {{1}^{+}} \right)={{\left( -x \right)}_{at~x=1}}=-1$ 

 $LHD\ne RHD$. So, function is not differentiable.

 Hence Rolle’s theorem is not applicable in $\left[ 0,~2 \right]$. 

71. Find the points on the curve $y=cos~x~-1$ in $\left[ 0,~2\pi \right]$, where the tangents are parallel to the x-axis.

Ans: Given $y=cos~x~-1$

Tangent is parallel to the x-axis then the derivative must be zero at those points. So,$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( cos~x~-1 \right)=sin~x~=0$

 $x=n\pi ;~n\in Z$ 

 But in$\left( 0,~2\pi \right)$, $x=\pi $ 

 Point is $y=cos~\pi ~-1=-2$ 

 $\left( \pi ,~-2 \right)$.

 Hence the point is $\left( \pi ,~-2 \right)$. 

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

Ans: Given $y=x\left( x-4 \right),~x\in \left[ 0,~~4 \right]$, 

 Function f is polynomial, so 

i. it is continuous and 

ii. differentiable for all real x. 

iii. $f\left( 0 \right)=0$ and $f\left( 4 \right)=0$ 

 All three conditions for Rolle’s theorem are verified, so 

 $\exists c\in \left[ 0,~4 \right]$ such that ${f}'\left( c \right)=0$. 

 ${f}'\left( x \right)=\dfrac{d}{dx}\left( {{x}^{2}}-4x \right)$ 

 $=2x-4$ 

 ${f}'\left( c \right)=2x-4=0$ 

 $c=2\in \left( 0,~4 \right)$ 

 At $x=2$ function has a tangent parallel to the x-axis.

 Again, $y=2\left( 2-4 \right)=-4$ 

 Hence the point is $\left( 2,~-4 \right)$. 

Verify mean value theorem for each of the functions given Exercises 73 to 76.

73. $f\left( x \right)=\dfrac{1}{4x-1},~x\in \left[ 1,~4 \right]$ 

Ans: Given $f\left( x \right)=\dfrac{1}{4x-1},~x\in \left[ 1,~4 \right]$ 

For mean value theorem function must be continuous and differentiable in the given domain.

Here, the function $f$ is discontinuous at $x=\dfrac{1}{4}$. 

Again ${f}'\left( x \right)=-\dfrac{4}{{{\left( 4x-1 \right)}^{2}}}$, so $f$ is not differentiable at $x=\dfrac{1}{4}$. 

Both conditions are satisfied for the mean value theorem.

So, there exists a real number $c\in \left( 1,~4 \right)$ such that

${f}'\left( c \right)=\dfrac{f\left( 4 \right)-f\left( 1 \right)}{4-1}$ 

 $\Rightarrow -\dfrac{4}{{{\left( 4c-1 \right)}^{2}}}=\dfrac{1}{3}\left( \dfrac{1}{4\times 4-1}-\dfrac{1}{4\times 1-1} \right)$ 

 $\Rightarrow \dfrac{4}{{{\left( 4c-1 \right)}^{2}}}=-\dfrac{1}{3}\left( \dfrac{1}{15}-\dfrac{1}{3} \right)$ 

 $\Rightarrow {{\left( \dfrac{2}{4c-1} \right)}^{2}}=\dfrac{1}{3}\left( \dfrac{5-1}{15} \right)=\dfrac{4}{45}$ 

 $\Rightarrow \dfrac{2}{4c-1}=\pm \dfrac{2}{3\sqrt{5}}$ 

 $\Rightarrow 4c-1=\pm 3\sqrt{5}$ 

 $\Rightarrow c=\dfrac{1}{4}\left( 1\pm 3\sqrt{5}~ \right)$ 

 $\Rightarrow c=\dfrac{1}{4}\left( 1+3\sqrt{5}~ \right)\in \left( 1,~4 \right)$ 

 Hence the mean value theorem is verified. 

74. $f\left( x \right)={{x}^{3}}-2{{x}^{2}}-x+3,~x\in \left[ 0,~1 \right]$. 

Ans: Given $f\left( x \right)={{x}^{3}}-2{{x}^{2}}-x+3,~x\in \left[ 0,~1 \right]$ 

Since, $f\left( x \right)$ is polynomial and the polynomial function is continuous and differentiable for all real x. 

So there exists a real number $c\in \left( 0,~1 \right)$ such that

${f}'\left( c \right)=\dfrac{f\left( 1 \right)-f\left( 0 \right)}{1-0}$ 

$\Rightarrow {f}'\left( x \right)=3{{x}^{2}}-4x-1$ 

$\Rightarrow 3{{c}^{2}}-4c-1=\left( 1-2-1+3 \right)-\left( 3 \right)$ 

$\Rightarrow 3{{c}^{2}}-4c-1=-2$ 

$\Rightarrow 3{{c}^{2}}-3c-c+1=0$ 

$\Rightarrow 3c\left( c-1 \right)-\left( c-1 \right)=0$ 

$\Rightarrow \left( 3c-1 \right)\left( c-1 \right)=0$ 

$\Rightarrow c=1,\dfrac{1}{3}$ 

$\Rightarrow c = \dfrac{1}{3} \in \left( 0, 1 \right)$ 

 Hence the mean value theorem is verified. 

75. $f\left( x \right)=sin~x~-sin~2x~,~x\in \left[ 0,~\pi \right]$ 

Ans: Given that $f\left( x \right)=sin~x~-sin~2x~,~x\in \left[ 0,~\pi \right]$ 

Since sin function is continuous and differentiable for all real x, so there exists a real number $c\in \left( 0,~\pi \right)$ such that

${f}'\left( c \right)=\dfrac{f\left( \pi \right)-f\left( 0 \right)}{\pi -0}$ 

$\Rightarrow {f}'\left( x \right)=cos~x~-2cos~2x~$ 

$\Rightarrow cos~c~-2 cos~2c~=\dfrac{1}{\pi }\left( 0-0 \right)=0$ 

$\Rightarrow cos~c~-2\left( 2c~-1 \right)=0$ 

 $\Rightarrow 4c~-cos~c~-2=0$ 

 $\Rightarrow cos~c~=\dfrac{1\pm \sqrt{1+4\times 4\times 2}}{8}=\dfrac{1\pm \sqrt{33}}{8}$ 

 $\Rightarrow c=\left( \dfrac{1\pm \sqrt{33}}{8} \right)~\in \left( 0,~\pi \right)$ 

Hence the mean value theorem is verified.

76. $f\left( x \right)=\sqrt{25-{{x}^{2}}},~x\in \left[ 1,~5 \right]$. 

Ans: Given $f\left( x \right)=\sqrt{25-{{x}^{2}}},~x\in \left[ 1,~5 \right]$ 

 $f\left( x \right)$ is continuous when $25-{{x}^{2}}\ge 0$ 

 So, $x\in \left[ -5,~5 \right]$ 

 $\therefore f\left( x \right)$ is continuous in $\left[ 1,~5 \right]$ 

 Now, ${f}'\left( x \right)=-\dfrac{x}{\sqrt{25-{{x}^{2}}}}$ so, it is differentiable in $\left( 1,~5 \right)$ 

 It satisfies both the conditions of the mean value theorem. 

 There exists a real number $c\in \left( 1,~5 \right)$ such that

 ${f}'\left( c \right)=\dfrac{f\left( 5 \right)-f\left( 1 \right)}{5-1}$ 

 $\Rightarrow -\dfrac{c}{\sqrt{25-{{c}^{2}}}}=\dfrac{1}{4}\left( \sqrt{25-{{5}^{2}}}-\sqrt{25-{{1}^{2}}} \right)$ 

 $\Rightarrow \dfrac{c}{\sqrt{25-{{c}^{2}}}}=-\dfrac{1}{4}\left( -2\sqrt{6} \right)$ 

 Squaring both sides, we have

 $\dfrac{{{c}^{2}}}{25-{{c}^{2}}}=\dfrac{6}{4}=\dfrac{3}{2}$ 

 $\Rightarrow 2{{c}^{2}}=75-3{{c}^{2}}$ 

 $\Rightarrow 5{{c}^{2}}=75$ 

 $\Rightarrow {{c}^{2}}=15$ 

 $\Rightarrow c=\pm \sqrt{15}$ 

 $\Rightarrow c=\sqrt{15}\in \left( 1,~5 \right)$ 

 Hence, the mean value theorem is verified. 

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

Ans: Given $y={{\left( x-3 \right)}^{2}}$ 

Function is polynomial so, it will be continuous and differentiable for all real x. 

Mean value theorem is applicable. Points $\left( 3,~0 \right)$ and $\left( 4,~1 \right)$ are on curve so, there exists a real number $c\in \left( 3,~4 \right)$ such that

 ${f}'\left( c \right)=\dfrac{f\left( 4 \right)-f\left( 3 \right)}{4-3}$ 

 $\Rightarrow 2\left( c-3 \right)=\dfrac{1-0}{1}=1$ 

 $\Rightarrow c-3=\dfrac{1}{2}$ 

 $\Rightarrow c=3+\dfrac{1}{2}=\dfrac{7}{2}$ 

 $\Rightarrow y={{\left( \dfrac{7}{2}-3 \right)}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{1}{4}$ 

 Hence the point is $\left( \dfrac{7}{2},~\dfrac{1}{4} \right)$.

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

Ans:

 Given $y=2{{x}^{2}}-5x+3$ and points are $A\left( 1,~0 \right)~and~B\left( 2,~1 \right)$. 

 y is a polynomial, so it is continuous and differentiable for all real x. Mean value theorem is applicable. Points $\left( 1,~0 \right)$ and $\left( 2,~1 \right)$ are on curve so, there exists a real number $c\in \left( 1,~2 \right)$ such that

 ${f}'\left( c \right)=\dfrac{f\left( 2 \right)-f\left( 1 \right)}{2-1}$ 

 $\Rightarrow 4c-5=\dfrac{1-0}{1}$ 

 $\Rightarrow 4c=6\Rightarrow c=\dfrac{3}{2}\in \left( 1,~2 \right)$

There is a tangent to the curve at $x=\dfrac{3}{2}$ where the tangent is parallel to the chord AB. 

Hence proved.

For the point, $y=2\times {{\left( \dfrac{3}{2} \right)}^{2}}-5\times \dfrac{3}{2}+3=\dfrac{9}{2}-\dfrac{15}{2}+\dfrac{6}{2}=0$ 

Point is $\left( \dfrac{3}{2},~0 \right)$. 

Long Answer (L.A.) 

79. Find the values of p and q so that $f\left( x \right)=\left\{ \begin{align} & {{x}^{2}}+3x+p,\text{if }\text{ }x\le 1 \\ & qx+2,\text{if }\text{ }x>1 \\ \end{align} \right.$ is differentiable at $x=1$. 

Ans:

Given that $f\left( x \right)=\left\{ \begin{align} & {{x}^{2}}+3x+p,\text{if }\text{ }x\le 1 \\ & qx+2,\text{if }\text{ }x>1 \\ \end{align} \right.$ at $x=1$. 

For differentiability function must be continuous. So, for continuity 

 $LHL=f\left( x \right)~$ 

 $=\underset{x\to {{1}^{-}}}{\mathop{lim}}\,\left( {{x}^{2}}+3x+p \right)$ 

 $LHL=4+p$

 $RHL=f\left( x \right)~$ 

 $=\underset{x\to {{1}^{+}}}{\mathop{lim}}\,\left( qx+2 \right)$ 

 $RHL=q+2$

 $f\left( 1 \right)=p+4$ 

 $LHL=RHL=f\left( 1 \right)$ 

 $p+4=q+2$ 

 $p=q-2$ 

Now for differentiability 

 $LHD={f}'\left( {{1}^{-}} \right)={{\left( 2x+3 \right)}_{at~x=1}}=5$ 

 $RHD={f}'\left( {{1}^{+}} \right)=q$ 

For differentiability $LHD=RHD$ 

$q=5$ 

$p=5-2=3$ 

Hence the value of $p=3$ and $q=5$. 

80. If ${{x}^{m}}{{y}^{n}}={{\left( x+y \right)}^{m+n}}$, prove that 

(i). $\dfrac{dy}{dx}=\dfrac{y}{x}$

Ans: Given ${{x}^{m}}{{y}^{n}}={{\left( x+y \right)}^{m+n}}$ 

${{x}^{m}}{{y}^{n}}={{\left( x+y \right)}^{m+n}}$ 

$\dfrac{d}{dx}{{x}^{m}}{{y}^{n}}=\dfrac{d}{dx}{{\left( x+y \right)}^{m+n}}$

$x^{m}n{y^{n-1}}\dfrac{dy}{dx}+mx^{m-1}y^{n}=(m+n)(x+y)^{m+n-1}\left(1+\dfrac{dy}{dx}\right)$ 

$\left( {{x}^{m}}~n{{y}^{n-1}}-\left( m+n \right){{\left( x+y \right)}^{m+n-1}} \right)\dfrac{dy}{dx}=~\left( m+n \right){{\left( x+y \right)}^{m+n-1}}-m{{x}^{m-1}}~{{y}^{n}}$ 

$\dfrac{dy}{dx}=\dfrac{\left( m+n \right){{\left( x+y \right)}^{m+n-1}}-m{{x}^{m-1}}~{{y}^{n}}}{\left( {{x}^{m}}~n{{y}^{n-1}}-\left( m+n \right){{\left( x+y \right)}^{m+n-1}} \right)}$ 

$=\dfrac{{{x}^{m}}{{y}^{n}}\left( m+n \right){{\left( x+y \right)}^{-1}}-m{{x}^{m-1}}~{{y}^{n}}}{\left( {{x}^{m}}~n{{y}^{n-1}}-{{x}^{m}}{{y}^{n}}\left( m+n \right){{\left( x+y \right)}^{-1}} \right)}$ 

$=\dfrac{\left( m+n \right){{\left( x+y \right)}^{-1}}-m{{x}^{-1}}}{\left( ~n{{y}^{-1}}-\left( m+n \right){{\left( x+y \right)}^{-1}} \right)}$ 

$=\dfrac{x\left( m+n \right)-m\left( x+y \right)}{\left( ~n\left( x+y \right)-\left( m+n \right)y \right)}\cdot \dfrac{y}{x}$ 

$=\dfrac{mx+nx-mx-my}{nx+ny-my-ny}\cdot \dfrac{y}{x}$ 

$=\dfrac{nx-my}{nx-my}\cdot \dfrac{y}{x}$ 

$\dfrac{dy}{dx}=\dfrac{y}{x}$ 

Hence proved.

(ii). $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0$

Ans: we have $\dfrac{dy}{dx}=\dfrac{y}{x}$ 

$\dfrac{d}{dx}\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \dfrac{y}{x} \right)$ 

$\Rightarrow\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{x\dfrac{dy}{dx}-y}{{{x}^{2}}}=\dfrac{x\cdot \dfrac{y}{x}-y}{{{x}^{2}}}$ 

$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0$ 

Hence proved.

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

Ans: Given that $x=sin~t~~and~y=sin~pt~$ 

$\dfrac{dx}{dt}=cos~t~,\dfrac{dy}{dt}=pcos~pt~$ 

$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{pcos~pt~}{cos~t~}$ 

$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dt}\left( \dfrac{pcos~pt~}{cos~t~} \right)\dfrac{1}{dx/dt}$ 

$=p\dfrac{cos~t~\left( -p~sin~pt~ \right)-cos~pt~\left( -sin~t~ \right)}{t~}\cdot \dfrac{1}{cos~t~}$ 

$\Rightarrow \left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=p\left( 1-t~ \right)\dfrac{cos~pt~sin~t~-p sin~pt~cos~t~}{t~}$ 

$\Rightarrow \left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=p\left( cos~pt~tan~t~-p~sin~pt~ \right)$ 

 $\Rightarrow x\dfrac{dy}{dx}=sin~t~\dfrac{pcos~pt~}{cos~t~}=pcos~pt~tan~t~$ 

 $\Rightarrow {{p}^{2}}y={{p}^{2}}sin~pt~$ 

 $LHS=\left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}+{{p}^{2}}y$ 

 $=p\left( cos~pt~tan~t~-p~sin~pt~ \right)-pcos~pt~tan~t~+{{p}^{2}}sin~pt~$ 

 $=0$ 

 Hence proved. 

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

Ans: Given $y={{x}^{tan~x~}}+\sqrt{\dfrac{{{x}^{2}}+1}{2}}$ 

$p={{x}^{tan~x~}}$ 

Taking log, 

$log~p~=tan~x~log~x~$ 

$\Rightarrow \dfrac{d}{dx}log~p~=\dfrac{d}{dx}\left( tan~x~log~x~ \right)$ 

$\Rightarrow \dfrac{1}{p}\dfrac{dp}{dx}=\dfrac{tan~x~}{x}+x~log~x~$ 

$\Rightarrow \dfrac{d}{dx}\left( {{x}^{tan~x~}} \right)={{x}^{tan~x~}}\left( \dfrac{tan~x~}{x}+sec^{2}x~log~x~ \right)$ 

$\Rightarrow \dfrac{d}{dx}\left( \sqrt{\dfrac{{{x}^{2}}+1}{2}} \right)=\dfrac{1}{\sqrt{2}}\left( \dfrac{x}{\sqrt{{{x}^{2}}+1}} \right)$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{tan~x~}} \right)+\dfrac{d}{dx}\left( \sqrt{\dfrac{{{x}^{2}}+1}{2}} \right)$ 

$\Rightarrow \dfrac{dy}{dx}={{x}^{tan~x~}}\left( \dfrac{tan~x~}{x}+x~log~x~ \right)+\dfrac{1}{\sqrt{2}}\left( \dfrac{x}{\sqrt{{{x}^{2}}+1}} \right)$ 

Choose the correct answers from the given four options in each of the Exercises 83 to 96. 

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

A. $f\left( x \right)+g\left( x \right)$ 

B. $f\left( x \right)-g\left( x \right)$

C. $f\left( x \right)\cdot g\left( x \right)$ 

D. $\dfrac{g\left( x \right)}{f\left( x \right)}$ 

Ans: The correct answer is option (D).

Given that $f\left( x \right)=2x$ and $g\left( x \right)=\dfrac{{{x}^{2}}}{2}+1$.

We can observe that both the functions are continuous so, 

$f+g,~f-g~and~f\cdot g$ are continuous but $\dfrac{g}{f}$ is not continuous at the point where f is zero. 

Hence $\dfrac{g\left( x \right)}{f\left( x \right)}$ can be discontinuous. 

Option (D) is correct.

84. The function $f\left( x \right)=\dfrac{4-{{x}^{2}}}{4x-{{x}^{3}}}$ 

A. Discontinuous at only one point

B. Discontinuous at exactly two points

C. Discontinuous at exactly three points

D. None of these

Ans: The correct answer is option (C).

Given $f\left( x \right)=\dfrac{4-{{x}^{2}}}{4x-{{x}^{3}}}$ 

For points of discontinuity $4x-{{x}^{3}}=0$ 

$x\left( 4-{{x}^{2}} \right)=0$ 

$\Rightarrow x\left( 2-x \right)\left( 2+x \right)=0$ 

$\Rightarrow x=0,~-2,~2$ 

Hence $f\left( x \right)$ is discontinuous at exactly three points. 

Option (C) is correct. 

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

A. $R$ 

B. $R-\left\{ \dfrac{1}{2} \right\}$ 

C. $\left( 0,~\infty \right)$ 

D. None of these.

Ans: The correct answer is option B.

Given function is $f\left( x \right)=\left| 2x-1 \right|sin~x~$

$f\left( x \right)=\left\{ \begin{align} &-\left( 2x-1 \right)sin~x,\text{ }if\text{ }x\le \dfrac{1}{2} \\ & \left( 2x-1 \right)sin~x,\text{ }if\text{ }x\ge \dfrac{1}{2} \\ \end{align} \right.$

We know that modulus function is continuous everywhere but not differentiable at $2x-1=0\Rightarrow x=\dfrac{1}{2}$ 

Hence, $f\left( x \right)$ is differentiable in $R-\left\{ \dfrac{1}{2} \right\}$.

86. The function $f\left( x \right)=cot~x~$ is discontinuous on the set

A. $\left\{ x=n\pi :n\in Z \right\}$ 

B. $\left\{ x=2n\pi :n\in Z \right\}$

C. $\left\{ x=\left( 2n+1 \right)\dfrac{\pi }{2}:n\in Z \right\}$

D. $\left\{ x=\dfrac{n\pi }{2}:n\in Z \right\}$

Ans: The correct answer is option A.

Given $f\left( x \right)=cot~x~$

We know that $f\left( x \right)=cot~x~$ is continuous at $x\in R-\left\{ n\pi ,~n\in Z \right\}$ 

Hence, $f\left( x \right)=cot~x~$ is discontinuous at $\left\{ x=n\pi :~n\in Z \right\}$.

87. The function $f\left( x \right)={{e}^{\left| x \right|}}$ 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

Ans: The correct answer is option (A).

Given $f\left( x \right)={{e}^{\left| x \right|}}$ 

$f\left( x \right)={{e}^{\left| x \right|}}=\left\{ \begin{align} & {{e}^{-x}},\text{if}\text{ }x\le 0 \\ & {{e}^{x}},\text{if}\text{ }x\ge 0 \\ \end{align} \right.$ 

$LHL={{e}^{0}}=1$ 

$RHL={{e}^{0}}=1$ 

 $LHL=RHL=f\left( 0 \right)$, function is continuous. 

$f\left( x \right)={{e}^{\left| x \right|}}=\left\{ \begin{align} & {{e}^{-x}},\text{if}\text{ }x\le 0 \\ & {{e}^{x}},\text{if}\text{ }x\ge 0 \\ \end{align} \right.$ 

$LHD=-{{e}^{0}}=-1$ 

$RHD={{e}^{0}}=1$ 

$LHD\ne RHD$, function is not differentiable.

Hence continuous everywhere but not differentiable at $x=0$ 

Option (A) is correct.

88. If $f\left( x \right)={{x}^{2}}sin~\dfrac{1}{x}~$, where $x\ne 0$, 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

Ans: The correct answer is option (A).

Given $f\left( x \right)={{x}^{2}}sin~\dfrac{1}{x}~$, where $x\ne 0$

$f\left( x \right)~=\underset{x\to 0}{\mathop{lim}}\,~{{x}^{2}}sin~\dfrac{1}{x}~$ 

$=~0\times \left(\text{number varying between -1 to 1} \right)$ 

$=0$ 

Hence the value of $f\left( x \right)$ at $x=0$, so that $f\left( x \right)$ be continuous at $x=0$, is 0. 

Option (A) is correct. 

89. If $f\left( x \right)=\left\{ \begin{align} & mx+1,\text{ }if\text{ }x\le \dfrac{\pi }{2} \\ & sin~x~+n,\text{ }if\text{ }x >\dfrac{\pi }{2} \\ \end{align} \right.$ 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. $m=n=\dfrac{\pi }{2}$

Ans: The correct answer is option (C). 

Given $f\left( x \right)=\left\{ \begin{align} & mx+1,\text{ }if\text{ }x\le \dfrac{\pi }{2} \\ & sin~x~+n,\text{ }if\text{ }x >\dfrac{\pi }{2} \\ \end{align} \right.$

 $LHL=f\left( x \right)~$ 

 $=\underset{x\to {{\dfrac{\pi }{2}}^{-}}}{\mathop{lim}}\,\left( mx+1 \right)$ 

 $=\dfrac{m\pi }{2}+1$ 

 $RHL=\underset{x\to {{\dfrac{\pi }{2}}^{+}}}{\mathop{lim}}\,f\left( x \right)$ 

 $=\underset{x\to {{\dfrac{\pi }{2}}^{+}}}{\mathop{lim}}\,\left( sin~x~+n \right)$ 

 $=n+1$ 

 For $f\left( x \right)$ to be continuous 

 $LHL=RHL$ 

 $\dfrac{m\pi }{2}+1=n+1$ 

 $n=\dfrac{m\pi }{2}$ 

 Option (C) is correct. 

90. Let $f\left( x \right)=\left| sin~x~ \right|$, then 

A. f is everywhere differentiable

B. f is everywhere continuous but not differentiable at $x=n\pi ,~n\in Z$.

C. f is everywhere continuous but not differentiable at $x=\left( 2n+1 \right)\dfrac{\pi }{2},~n\in Z$.

D. None of these.

Ans: The correct answer is option (B).

Given $f\left( x \right)=\left| sin~x~ \right|$ 

$f\left( x \right)=\left| sin~x~ \right|= \left\{ \begin{align} & -sin~x,\text{if}\text{ }x\le 0 \\ & sin~x,\text{ }if\text{ }x\ge 0 \\ \end{align} \right.$

Here we observe that $f$ is continuous.

$f\left( x \right)=\left\{ \begin{align} & -cos~x,\text{ }if\text{ }sin~x~\le 0 \\ & cos~x,\text{ }if\text{ }sin~x~ \ge 0 \\ \end{align} \right.$

At $x=n\pi ,~n\in Z$

$LHD={f}'\left( n\pi \right)=-cos~n\pi ~=-{{\left( -1 \right)}^{n}}$ 

$RHD={f}'\left( n\pi \right)=cos~n\pi ~={{\left( -1 \right)}^{n}}$ 

$LHD\ne RHD$ 

Hence $f$ is continuous but not differentiable at $x=n\pi ,~n\in Z$. 

Option (B) is correct. 

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

A. $\dfrac{4{{x}^{3}}}{1-{{x}^{4}}}$ 

B. $\dfrac{-4x}{1-{{x}^{4}}}$ 

C. $\dfrac{1}{4-{{x}^{4}}}$ 

D. $\dfrac{-4{{x}^{3}}}{1-{{x}^{4}}}$

Ans: The correct answer is option (B).

 Given $y= log\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)= log\left( 1-{{x}^{2}} \right)~- log~\left( 1+{{x}^{2}} \right)~$ 

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( log\left( 1-{{x}^{2}} \right) \right)-\dfrac{d}{dx}\left(log\left( 1+{{x}^{2}} \right)\right)$ $=\dfrac{-2x}{1-{{x}^{2}}}-\dfrac{2x}{1+{{x}^{2}}}$ 

$=-2x\left( \dfrac{1-{{x}^{2}}+1+{{x}^{2}}}{\left( 1-{{x}^{2}} \right)\left( 1+{{x}^{2}} \right)} \right)$ 

$=-\dfrac{4x}{1-{{x}^{4}}}$ 

Option (B) is correct. 

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. $\dfrac{sin~x~}{2y-1}$ 

Ans: The correct answer is option (A).

 Given $y=\sqrt{sin~x~+y}$ 

${{y}^{2}}=sin~x~+y$ 

$\Rightarrow {{y}^{2}}-y=sin~x~$ 

$\Rightarrow \dfrac{d}{dx}\left( {{y}^{2}}-y \right)=\dfrac{d}{dx}sin~x~$ 

$\Rightarrow \left( 2y-1 \right)\dfrac{dy}{dx}=cos~x~$ 

$\Rightarrow \dfrac{dy}{dx}=\dfrac{cos~x~}{2y-1}$ 

Option (A) is correct.

93. The derivative of $cos^{-1}\left( 2{{x}^{2}}-1 \right)~$ w.r.t $cos^{-1}x~$. 

A. 2 

B. $\dfrac{-1}{2\sqrt{1-{{x}^{2}}}}$ 

C. $\dfrac{2}{x}$ 

D. $1-{{x}^{2}}$ 

Ans: The correct answer is option (A)

 We must find derivative of $cos^{-1}\left( 2{{x}^{2}}-1 \right)~$ w.r.to $cos^{-1}x~$

 Let $t=cos^{-1}x~\Rightarrow x= cos~t~$

 $cos^{-1}\left( 2{{x}^{2}}-1 \right)~=cos^{-1}\left( 2cos^2{t}~-1 \right)~$ 

 $= cos^{-1}\left(cos~2t~ \right)~$ 

 $cos^{-1}\left( 2{{x}^{2}}-1 \right)~=2t$ 

 $\dfrac{d cos^{-1}\left( 2{{x}^{2}}-1 \right)~}{d cos^{-1}x~}=\dfrac{d\left( 2t \right)}{dt}=2$ 

 Option (A) is correct. 

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

A. $\dfrac{3}{2}$ 

B. $\dfrac{3}{4t}$ 

C. $\dfrac{3}{2t}$ 

D. $\dfrac{3}{4}$ 

Ans: The correct answer is option (B).

Given $x={{t}^{2}},~y={{t}^{3}}$ 

$\dfrac{dx}{dt}=2t,\dfrac{dy}{dt}=3{{t}^{2}}$ 

$\Rightarrow\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3{{t}^{2}}}{2t}=\dfrac{3t}{2}$ 

$\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( \dfrac{3t}{2} \right)=\dfrac{d}{dt}\left( \dfrac{3t}{2} \right)\cdot \dfrac{1}{dx/dt}$ 

$\Rightarrow\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{3}{2}\times \dfrac{1}{2t}=\dfrac{3}{4t}$ 

 Option (B) is correct.

95. The value of c in Rolle’s theorem for the function $f\left( x \right)={{x}^{3}}-3x$ in the interval $\left[ 0,~\sqrt{3} \right]$ is

A. 1 

B. $-1$ 

C. $\dfrac{3}{2}$ 

D. $\dfrac{1}{3}$ 

Ans: The correct answer is option (A).

Given that $f\left( x \right)={{x}^{3}}-3x$ 

Function is continuous and differentiable for all real x and $f\left( 0 \right)=0=f\left( \sqrt{3} \right)$. 

So, Rolle’s theorem is applicable here.

${f}'\left( x \right)=3{{x}^{2}}-3$ 

$\Rightarrow {f}'\left( c \right)=0\Rightarrow 3\left( {{c}^{2}}-1 \right)=0$

$\Rightarrow {{c}^{2}}=1\Rightarrow c=\pm 1$ 

$\Rightarrow c=1\in \left( 0,~\sqrt{3} \right)$ 

Option (A) is correct.

96. For the function $f\left( x \right)=x+\dfrac{1}{x}~,~x\in \left[ 1,~3 \right]$, the value of c for mean value theorem is

A. 1 

B. $\sqrt{3}$ 

C. 2 

D. None of these

Ans: The correct answer is option (B).

Given $f\left( x \right)=x+\dfrac{1}{x}$ is continuous in $x\in \left[ 1,~3 \right]$ 

${f}'\left( x \right)=1-\dfrac{1}{{{x}^{2}}}$ is differentiable in $x\in \left( 1,~3 \right)$ 

Mean value theorem is applicable here. 

${f}'\left( c \right)=1-\dfrac{1}{{{c}^{2}}}=\dfrac{f\left( 3 \right)-f\left( 1 \right)}{3-1}$ 

$\Rightarrow 1-\dfrac{1}{{{c}^{2}}}=\dfrac{3+\dfrac{1}{3}-1-1}{2}$ 

$\Rightarrow\dfrac{1}{{{c}^{2}}}=1-\dfrac{\dfrac{4}{3}}{2}=\dfrac{2-\dfrac{4}{3}}{2}=\dfrac{2}{6}=\dfrac{1}{3}$ 

$\Rightarrow {{c}^{2}}=3\Rightarrow c=\pm \sqrt{3}$ 

$\Rightarrow c=\sqrt{3}\in \left( 1,~3 \right)$ 

Option (B) is correct. 

Fill in the blanks in each of the exercises 97 to 101:

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

Ans: Function $\left| x-1 \right|+\left| x-2 \right|$ is continuous everywhere but it is not differentiable exactly at two points and the points are $x=0,~x=1$. 

98. Derivative f ${{x}^{2}}$ w.r.to ${{x}^{3}}$ is __________. 

Ans: To find derivative f ${{x}^{2}}$ w.r.to ${{x}^{3}}$ 

$\dfrac{d{{x}^{2}}}{d{{x}^{3}}}=\dfrac{\dfrac{d{{x}^{2}}}{dx}}{\dfrac{d{{x}^{3}}}{dx}}$

$=\dfrac{2x}{3{{x}^{2}}}$ 

$=\dfrac{2}{3x}$ 

99. If $f\left( x \right)=\left|cos~x~ \right|$, then $f'\left( \dfrac{\pi }{4} \right)=\_\_\_\_\_\_\_\_\_\_$. 

Ans: Given $f\left( x \right)=\left|cos~x~ \right|$ 

At $x=\dfrac{\pi }{4},~f\left( x \right)= cos~x~$ 

$\Rightarrow {f}'\left( x \right)=-sin~x~$ 

$\Rightarrow {f}'\left( \dfrac{\pi }{4} \right)=-sin~\left( \dfrac{\pi }{4} \right)~=-\dfrac{1}{\sqrt{2}}$ 

100. If $f\left( x \right)=\left|cos~x~- sin~x~ \right|$, then ${f}'\left( \dfrac{\pi }{3} \right)=\_\_\_\_\_\_\_\_\_$. 

Ans: Given $f\left( x \right)=\left|cos~x~- sin~x~ \right|$ 

$f\left( x \right)=\left| cos~x~- sin~x~ \right|$

$=\left\{ \begin{align} & sin~x~-cos~x,\text{ }if\text{ }sin~x~\ge cos~x \\ & -sin~x~+cos~x,\text{ }if\text{ }sin~x~\le cos~x \\ \end{align} \right.$ 

At $x=\dfrac{\pi }{3},~sin~x~\ge cos~x~$ 

$\Rightarrow f\left( x \right)=sin~x~-cos~x~$ 

$\Rightarrow {f}'\left( x \right)=cos~x~+sin~x~$ 

$\Rightarrow {f}'\left( \dfrac{\pi }{3} \right)=cos~\dfrac{\pi }{3}~+sin~\dfrac{\pi }{3}~=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}$ 

$\Rightarrow {f}'\left( \dfrac{\pi }{3} \right)=\dfrac{1+\sqrt{3}}{2}$ 

101. For the curve $\sqrt{x}+\sqrt{y}=1,\dfrac{dy}{dx}~at~\left( \dfrac{1}{4},~~\dfrac{1}{4} \right)$ is ________.

Ans: Given curve is $\sqrt{x}+\sqrt{y}=1$ 

 $\dfrac{d}{dx}\left( \sqrt{x}+\sqrt{y} \right)=\dfrac{d}{dx}1$ 

 $\Rightarrow \dfrac{1}{2\sqrt{x}}+\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}=0$ 

 $\Rightarrow \dfrac{dy}{dx}=-\dfrac{\sqrt{y}}{\sqrt{x}}$ 

 At $\left( \dfrac{1}{4},~~\dfrac{1}{4} \right)$, $\dfrac{dy}{dx}=-1$ 

State True or False for the following statements in each of the Exercise 102 to 106.

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

Ans: Given function $f\left( x \right)=\left| x-1 \right|$ is continuous in $\left[ 0,~2 \right]$ and not differentiable at$x=1\in \left( 0,~2 \right)$.

Hence the given statement is False. 

103. If $f$ is continuous on its domain D, then $\left| f \right|$ is also continuous on D.

Ans: The given statement is True. 

104. The composition of two continuous functions is a continuous function.

Ans: The given statement is True.

105. Trigonometric and inverse trigonometric functions are differentiable in their respective domain

Ans: The given statement is True.

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

Ans: The given statement is False. 

The Various Sub Topics Covered in the NCERT Exemplar for Class 12 Maths Chapter 5 - Continuity and Differentiability (Book Solutions) Includes

• Introduction (5.1)

• 5.2 Consistency

• 5.2.1 Continuous Functions Algebra

• Differentiability (5.3)

• 5.3.1 Composite Function derivatives

• 5.3.2 Implicit Function derivatives

• 5.4 Exponential and logarithmic Functions

• 5.5 Differentiation on a logarithmic scale

• 5.6 Function derivatives in parametric forms

• Second-order derivative (5.7)

• 5.8 Theorem of Mean Value