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

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

NCERT Exemplar for Class 12 Maths - Continuity And Differentiability - Free PDF Download

Free PDF download of NCERT Exemplar for Class 12 Maths Chapter 5 - Continuity And Differentiability solved by expert Maths teachers on Vedantu.com as per NCERT (CBSE) Book guidelines. All Chapter 5 - Continuity And Differentiability Exercise questions with solutions to help you to revise the complete syllabus and score more marks in your Examinations.

NCERT Exemplar for Class 12 Maths Chapter 5 - Continuity And Differentiability (Book Solutions) is useful for students as it allows them to become familiar with different types of questions and, as a result, develop problem-solving skills. Students can use the NCERT Exemplar Solutions, which are provided subject-by-subject, to help them solve the Exercise questions in each Chapter. All of the answers are created following the most recent CBSE criteria for students to score well.

NCERT Exemplar for Class 12 Maths Chapter 5 - Continuity And Differentiability (Book Solutions) is an important Chapter as it establishes the groundwork for Differential Calculus. Continuity, Differentiability, Algebra of Continuous Functions, Derivatives of Composite Functions, Implicit Functions, and Inverse Trigonometric Functions, Exponential and Logarithmic Functions, Logarithmic Differentiation, Derivatives of Functions in Parametric forms, second-order Derivative, and mean value theorem are some of the topics covered in this Chapter. The answers to the Chapter's Exercise problems are also available on the Vedantu website and app in PDF format.

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Access NCERT Exemplar Solutions for Class 12 Mathematics Chapter 5 - Continuity and Differentiability

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

$f (x) = {\begin{aligned} \frac{1 - c o s 4 x}{8 x^{2}}, at x \neq 0 \\ k, at x = 0 \end{aligned}$

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

$\Rightarrow \underset{x \to 0}{l i m} \frac{1 - c o s 4 x}{8 x^{2}} = k$

$\Rightarrow \underset{x \to 0}{l i m} \frac{2 s i n^{2} 2 x}{8 x^{2}} = k$

$\Rightarrow \underset{x \to 0}{l i m} {(\frac{s i n 2 x}{2 x^{}})}^{2} = k$

$\Rightarrow k = 1$

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. $f (x) = {\begin{aligned} \frac{x^{3} + x^{2} - 16 x + 20}{{(x - 2)}^{2}}, x \neq 2 \\ k, x = 2 \end{aligned}$ is continuous at x =2, Find the value of k .

Ans: Given f (2) = k.

Now, $\underset{x \to 2^{-}}{l i m}$ f(x) = $\underset{x \to 2^{+}}{l i m}$ f(x) = $\underset{x \to 2^{}}{l i m} \frac{x^{3} + x^{2} - 16 x + 20}{{(x - 2)}^{2}}$

$\Rightarrow$ $\underset{x \to 2^{}}{l i m} \frac{(x + 5) {(x - 2)}^{2}}{{(x - 2)}^{2}}$

$\Rightarrow$ $\underset{x \to 2^{}}{l i m} (x + 5)$ = 7

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

$\underset{x \to 2^{}}{l i m}$ f(x) = f(2)

k = 7

4. Show that the function f defined by $f (x) = {\begin{aligned} x \sin \frac{1}{x}, x \neq 0 \\ 0, x = 0 \end{aligned}$ is continuous at x = 0

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

$\underset{x \to 0^{-}}{l i m}$ f(x) = $\underset{x \to 0^{-}}{l i m}$ xsin $\frac{1}{x}$ = 0

Similarly , $\underset{x \to 0^{+}}{l i m}$ f(x) = $\underset{x \to 0^{+}}{l i m}$ xsin $\frac{1}{x}$ = 0

Also f(0) = 0

Hence $\underset{x \to 0^{-}}{l i m}$ f(x) = $\underset{x \to 0^{+}}{l i m}$ f(x) = f(0)

$\Rightarrow$ f is continuous at x = 0

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

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

f (f (x)) = f $(\frac{1}{x - 1})$ = $\frac{1}{\frac{1}{x - 1} - 1}$ = $\frac{x - 1}{2 - x}$

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 $f (x) = {\begin{aligned} x^{2}, if x \geq 0 \\ - x^{2}, i f x < 0 \end{aligned}$

We know that

Lf ′ (0) = $\underset{x \to 0^{-}}{l i m} \frac{f (0 - h) - f (0)}{h}$ = $\underset{x \to 0^{-}}{l i m} \frac{- h^{2} - 0}{- h}$ = $\underset{x \to 0^{-}}{l i m}$ -h = 0

Rf ′ (0) = $\underset{x \to 0^{+}}{l i m} \frac{f (0 + h) - f (0)}{h}$ = $\underset{x \to 0^{+}}{l i m} \frac{h^{2} - 0}{h}$ = $\underset{x \to 0^{+}}{l i m}$ h = 0

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

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

Ans: Let us consider y = $\sqrt{t a n \sqrt{x}}$

$\frac{d y}{d x}$ = $\frac{1}{2 \sqrt{t a n \sqrt{x}}} \frac{d}{d x}$ tan $\sqrt{x}$

= $\frac{1}{2 \sqrt{t a n \sqrt{x}}} s e c^{2} {\sqrt{x}}^{} \frac{d}{d x} {\sqrt{x}}^{}$

= $\frac{1}{2 \sqrt{t a n \sqrt{x}}} s e c^{2} {\sqrt{x}}^{} {\frac{1}{2 \sqrt{x}}}^{}$

= $\frac{s e c^{2} \sqrt{x}}{4 \sqrt{x} \sqrt{t a n \sqrt{x}}}$

8. If y = tan(x + y), find $\frac{d y}{d x}$ .

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

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

$\frac{d y}{d x}$ = $\frac{d}{d x}$ tan(x+y)

$\Rightarrow \frac{d y}{d x}$ = $s e c^{2} (x + y)^{} \frac{d}{d x}$ (x+y)

$\Rightarrow \frac{d y}{d x}$ = (1+ $\frac{d y}{d x}$ ) $s e c^{2} (x + y)^{}$

$\Rightarrow \frac{d y}{d x}$ = $s e c^{2} (x + y)^{}$ + $\frac{d y}{d x} s e c^{2} (x + y)^{}$

$\Rightarrow \frac{d y}{d x}$ - $\frac{d y}{d x} s e c^{2} (x + y)^{}$ = $s e c^{2} (x + y)$

$\Rightarrow (1 - s e c^{2} (x + y)) \frac{d y}{d x}$ = $s e c^{2} (x + y)$

$\Rightarrow \frac{d y}{d x}$ = $\frac{s e c^{2} (x + y)}{1 - s e c^{2} (x + y)}$ = - co $s e c^{2} (x + y)$

Hence, $\frac{d y}{d x}$ = - co $s e c^{2} (x + y)$

9. If ${e^{x}}^{}$ + ${e^{y}}^{}$ = ${e^{x + y}}^{}$ , Prove that $\frac{d y}{d x}$ = - ${e^{y - x}}^{}$ .

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

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

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

$\Rightarrow e^{x}$ + ${e^{y}}^{} \frac{d y}{d x}$ = ${e^{x + y}}^{} \frac{d}{d x}$ (x+y)

$\Rightarrow e^{x}$ + ${e^{y}}^{} \frac{d y}{d x}$ = (1+ $\frac{d y}{d x}$ ) ${e^{x + y}}^{}$

$\Rightarrow e^{x}$ + ${e^{y}}^{} \frac{d y}{d x}$ = ${e^{x + y}}^{}$ + ${e^{x + y}}^{} \frac{d y}{d x}$

$\Rightarrow (e^{y}$ - ${e^{x + y}}^{}$ ) $\frac{d y}{d x}$ = ${e^{x + y}}^{}$ - $e^{x}$

$\Rightarrow \frac{d y}{d x}$ = $\frac{{e^{x + y}}^{} - e^{x}}{e^{y} - {e^{x + y}}^{}}$

$\Rightarrow \frac{d y}{d x}$ = $\frac{{e^{x}}^{} + e^{y} - e^{x}}{e^{y} - e^{x} - e^{y} {^{}}^{}}$ $[∵ {e^{x}}^{} + {e^{y}}^{} = {e^{x + y}}^{}]$

$\Rightarrow \frac{d y}{d x}$ = - ${e^{y - x}}^{}$

Hence Proved.

10. Find $\frac{d y}{d x}$ if y = $t a n^{- 1} {\frac{(3 x - x^{3})}{1 - 3 x^{2}}}^{}$ , - $\frac{1}{\sqrt{3}}$ <x < $\frac{1}{\sqrt{3}}$

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

so, y = $t a n^{- 1} {\frac{(3 t a n θ - t a n^{3} θ^{})}{1 - 3 t a n^{2} θ}}^{}$

We know that tan3θ = $\frac{(3 t a n θ - t a n^{3} θ^{})}{1 - 3 t a n^{2} θ}$

$\Rightarrow y = t a n^{- 1} t a n 3 θ^{}$

$\Rightarrow y = 3 θ$ $(- \frac{π}{2} < 3 θ < \frac{π}{2})$

$\Rightarrow y = 3 t a n^{- 1}$ x

Now on differentiating both sides w.r.t x

$\Rightarrow \frac{d y}{d x}$ = $\frac{d}{d x}$ (3 $t a n^{- 1}$ x)

$\Rightarrow \frac{d y}{d x}$ = 3 $\frac{1}{1 + x^{2}}$

Hence, $\frac{d y}{d x}$ = $\frac{3}{1 + x^{2}}$

11. If y = $s i n^{- 1}$ ${x \sqrt{1 - x} - \sqrt{x} \sqrt{1 - x^{2}}}$ and 0 < x < 1, then find $\frac{d y}{d x}$ .

Ans: we have

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

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

y = $s i n^{- 1}$ {sinA $\sqrt{1 - s i n^{2} B} - s i n B \sqrt{1 - s i n^{2} A}$ }

= $s i n^{- 1}$ {sinAcosB - sinBcosA }

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

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

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

$\frac{d y}{d x}$ = $\frac{1}{\sqrt{1 - x^{2}}}$ - $\frac{1}{\sqrt{1 - {\sqrt{x^{2}}}^{}}}$ . $\frac{d}{d x}$ ( $\sqrt{x})$

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

Therefore , $\frac{d y}{d x}$ = $\frac{1}{\sqrt{1 - x^{2}}}$ - $\frac{1}{2 \sqrt{x} . \sqrt{1 - x}}$

12. If x = $a s e c^{3} θ$ and y = $a t a n^{3} θ$ , find $\frac{d y}{d x}$ at $θ$ = $\frac{π}{3}$

Ans: we have x = a $s e c^{3} θ$ and y = a $t a n^{3} θ$

On Differentiating both sides w.r.t $θ$

$\frac{d x}{d θ}$ = 3a $s e c^{2} θ$ sec $θ$ .tan $θ$

and $\frac{d y}{d θ}$ = 3a $t a n^{2} θ s e c^{2} θ$

Now, $\frac{d y}{d x}$ = $\frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$ = $\frac{3 a t a n^{2} θ s e c^{2} θ}{3 a s e c^{2} θ s e c θ . t a n θ}$ = sin $θ$

At $θ$ = $\frac{π}{3}$

$\frac{d y}{d x}$ = sin $\frac{π}{3}$ = $\frac{\sqrt{3}}{2}$

Hence,

$\frac{d y}{d x}$ = $\frac{\sqrt{3}}{2}$ , at $θ$ = $\frac{π}{3}$

13. If $x^{y}$ = $e^{(x - y)}$ , Prove that $\frac{d y}{d x}$ = $\frac{l o g x}{{(1 + l o g x)}^{2}}$

Ans: we have ,

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

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

$\Rightarrow y = \frac{x}{1 + l o g x}$

Using Quotient Rule , we get

$\Rightarrow \frac{d y}{d x}$ = $\frac{(1 + l o g x) . \frac{d}{d x} (x) - x . \frac{d}{d x} (1 + l o g x)}{{(1 + l o g x)}^{2}}$

$\Rightarrow \frac{d y}{d x}$ = $\frac{(1 + l o g x) - 1}{{(1 + l o g x)}^{2}}$

$\Rightarrow \frac{d y}{d x}$ = $\frac{l o g x}{{(1 + l o g x)}^{2}}$

Hence Proved.

14. If y = tanx + secx, prove that $\frac{d^{2} y}{d x^{2}}$ = $\frac{c o s x}{{(1 - s i n x)}^{2}}$

Ans: we have,

y = tanx + secx

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

$\frac{d y}{d x}$ = $\frac{d}{d x}$ ( tanx ) + $\frac{d}{d x}$ ( secx )

$\Rightarrow \frac{d y}{d x}$ = $s e c^{2} x$ + secx.tanx

$\Rightarrow \frac{d y}{d x}$ = $\frac{1}{c o s^{2} x}$ + $\frac{s i n x}{c o s^{2} x}$ = $\frac{1 + s i n x}{c o s^{2} x}$

$\Rightarrow \frac{d y}{d x}$ = $\frac{1 + s i n x}{c o s^{2} x}$

$\Rightarrow \frac{d y}{d x}$ = $\frac{1 + s i n x}{1 - s i n^{2} x}$

$\Rightarrow \frac{d y}{d x}$ = $\frac{1 + s i n x}{(1 + s i n^{} x) (1 - s i n x)}$

$\Rightarrow \frac{d y}{d x}$ = $\frac{1}{(1 - s i n x)}$

Again Differentiating both sides w.r.t x

$\frac{d^{2} y}{d x^{2}}$ = $\frac{- (- c o s x)}{{(1 - s i n x)}^{2}}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}$ = $\frac{c o s x}{{(1 - s i n x)}^{2}}$

Hence Proved .

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

Ans: we know that cosx < 0 when $\frac{π}{2}$ < x < $π$

|cosx| = -cosx

so , f(x) = -cosx

f’(x) = sinx

Therefore, f’ $(\frac{3 π}{4})$ = sin $\frac{3 π}{4}$

f’ $(\frac{3 π}{4})$ = $\frac{1}{\sqrt{2}}$

Hence Proved

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

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

cos x – sin x > 0

so, f(x) = cosx - sinx

f’(x) = – sin x – cos x

Therefore, f’ ( $\frac{π}{6}$ ) = -sin $\frac{π}{6}$ - cos $\frac{π}{6}$

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

Hence Proved

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

Ans: we have,

f (x) = sin 2x in $[0, \frac{π}{2}]$

The given function is continuous in $[0, \frac{π}{2}]$ as f is a sine function.

f’(x) = 2cos2x , exists in $(0, \frac{π}{2})$ , hence f is differentiable in $(0, \frac{π}{2})$

f(0) = sin 0 = 0 and f $(\frac{π}{2})$ = sin $\frac{π}{2}$ = 0

f(0) = f $(\frac{π}{2})$

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

$(0, \frac{π}{2})$ such that f’(c) = 0

f’(x) = 2cos2x

⇒f’(c) = 2cos2c = 0

⇒cos2c = 0

⇒2c = $\frac{π}{2}$

⇒c = $\frac{π}{4}$

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

Ans: (i) Given function f(x) is continuous in $[3, 5]$ 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) = $\frac{f (5) - f (3)}{5 - 3}$

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

⇒3 $c^{2}$ – 36c + 99 = 4

⇒3 $c^{2}$ – 36c + 95 = 0

⇒c = 6± $\sqrt{\frac{13}{3}}$

⇒c $\neq 6 + \sqrt{\frac{13}{3}}$

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

Long Answers (L.A.)

19. If f (x) = $\frac{\sqrt{2} c o s x - 1}{c o t x - 1}$ , x $\neq$ $\frac{π}{4}$ , Find the value of f( $\frac{π}{4}$ ) so that f(x) becomes continuous at x = $\frac{π}{4}$

Ans: we have f (x) = $\frac{\sqrt{2} c o s x - 1}{c o t x - 1}$ , x $\neq$ $\frac{π}{4}$

$\underset{x \to \frac{π}{4}}{l i m}$ f(x) = $\underset{x \to \frac{π}{4}}{l i m} \frac{\sqrt{2} c o s x - 1}{c o t x - 1}$

= $\underset{x \to \frac{π}{4}}{l i m} \frac{(\sqrt{2} c o s x - 1) s i n x}{c o s x - s i n x}$

= $\underset{x \to \frac{π}{4}}{l i m} \frac{(\sqrt{2} c o s x - 1)}{c o s x - s i n x}$ $\frac{(\sqrt{2} c o s x + 1) (c o s x + s i n x) s i n x}{(\sqrt{2} c o s x + 1) (c o s x + s i n x)}$

= $\underset{x \to \frac{π}{4}}{l i m} \frac{(2 c o s^{2} x - 1) (c o s x + s i n x)}{(c o s^{2} x - s i n^{2} x) (\sqrt{2} c o s x + 1)} . s i n x$

= $\underset{x \to \frac{π}{4}}{l i m} \frac{c o s 2 x (c o s x + s i n x)}{c o s 2 x (\sqrt{2} c o s x + 1)} . s i n x$

= $\underset{x \to \frac{π}{4}}{l i m} \frac{(c o s x + s i n x)}{(\sqrt{2} c o s x + 1)} . s i n x$

= $\frac{(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})}{\frac{1}{\sqrt{2}} . \sqrt{2} + 1} \cdot \frac{1}{\sqrt{2}}$ = $\frac{1}{2}$

Therefore, $\underset{x \to \frac{π}{4}}{l i m}$ f(x) = $\frac{1}{2}$

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

i.e. f( $\frac{π}{4}$ ) = $\frac{1}{2}$

20. Show that the function f given by $f (x) = {\begin{aligned} \frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1}, i f x \neq 0 \\ 0, i f x = 0 \end{aligned}$ is discontinuous at x =0.

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

$\underset{x \to 0^{-}}{l i m}$ f(x) = $\underset{x \to 0^{-}}{l i m} \frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1}$ = $\frac{0 - 1}{0 + 1}$ = -1

Similarly,

$\underset{x \to 0^{+}}{l i m}$ f(x) = $\underset{x \to 0^{+}}{l i m} \frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1}$ = $\frac{1 - 0}{1 + 0}$ = 1

Therefore,

$\underset{x \to 0^{-}}{l i m}$ f(x) $\neq$ $\underset{x \to 0^{+}}{l i m}$ f(x) , hence $\underset{x \to 0^{}}{l i m}$ f(x) does not exist .

f is discontinuous at x = 0

21. Let $f (x) = {\begin{aligned} \frac{1 - c o s 4 x}{x^{2}}, i f x \leq 0 \\ a, i f x = 0 \\ \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, i f x > 0 \end{aligned}$ For what value of a , f is continuous at x = 0 ?

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

$\underset{x \to 0^{-}}{l i m}$ f(x) = $\underset{x \to 0^{-}}{l i m} \frac{1 - c o s 4 x}{x^{2}}$

= $\underset{x \to 0^{-}}{l i m} \frac{2 s i n^{2} 2 x}{x^{2}}$

= $\underset{x \to 0^{-}}{l i m} 8. {(\frac{s i n^{} 2 x}{x^{}})}^{2}$

= $8. {(1)}^{2}$

= 8

Now, $\underset{x \to 0^{+}}{l i m}$ f(x) = $\underset{x \to 0^{+}}{l i m} \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}$

= $\underset{x \to 0^{+}}{l i m} \frac{\sqrt{x} (\sqrt{16 + \sqrt{x}} + 4)}{(\sqrt{16 + \sqrt{x}} - 4) (\sqrt{16 + \sqrt{x}} + 4)}$

= $\underset{x \to 0^{+}}{l i m} \frac{\sqrt{x} (\sqrt{16 + \sqrt{x}} + 4)}{(16 + \sqrt{x} - 16)}$

= $\underset{x \to 0^{+}}{l i m} \frac{\sqrt{x} (\sqrt{16 + \sqrt{x}} + 4)}{\sqrt{x}}$

$= \underset{x \to 0^{+}}{l i m} (\sqrt{16 + \sqrt{x}} + 4$ )

= 8

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

22. Examine the differentiability of the function f defined by

$f (x) = {\begin{aligned} 2 x + 3,, i f - 3 \leq x < - 2 \\ x + 1, i f - 2 \leq x < 0 \\ x + 2, i f 0 \leq x \leq 1 \end{aligned}$

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) = $\underset{h \to 0^{-}}{l i m} \frac{f (- 2 + h) - f (- 2)}{h}$

= $\underset{h \to 0^{-}}{l i m} \frac{2 (- 2 + h) + 3 - (- 2 + 1)}{h}$

= $\underset{h \to 0^{-}}{l i m} \frac{2 h}{h}$

= 2

R.H.D = f’(-2) = $\underset{h \to 0^{+}}{l i m} \frac{f (- 2 + h) - f (- 2)}{h}$

= $\underset{h \to 0^{+}}{l i m} \frac{- 2 + h + 1 - (- 2 + 1)}{h}$

= $\underset{h \to 0^{+}}{l i m} \frac{h}{h}$

= 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) = $\underset{h \to 0^{-}}{l i m} \frac{f (- 2 + h) - f (- 2)}{h}$

= $\underset{h \to 0^{-}}{l i m} \frac{f (0 + h) - f (0)}{h}$

= $\underset{h \to 0^{-}}{l i m} \frac{(0 + h + 1) - (0 + 2)}{h}$

= $\underset{h \to 0^{-}}{l i m} \frac{h - 1}{h}$

= $\underset{h \to 0^{-}}{l i m} (1 - \frac{1}{h}$ )

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 $u = t a n^{- 1} (\frac{\sqrt{1 - x^{2}}}{x})$ w.r.t $v = c o s^{- 1} (2 x \sqrt{1 - x^{2}})$ , where x $\in (\frac{1}{\sqrt{2}}, 1)$ .

Ans: Given,

$u = t a n^{- 1} (\frac{\sqrt{1 - x^{2}}}{x})$ and v = $c o s^{- 1} (2 x \sqrt{1 - x^{2}})$

we have, $u = t a n^{- 1} (\frac{\sqrt{1 - x^{2}}}{x})$

Put x = sin $θ$ , where $\frac{π}{4}$ < $θ$ < $\frac{π}{2}$

$u = t a n^{- 1} (\frac{\sqrt{1 - s i n^{2} θ^{}}}{s i n θ})$

$\Rightarrow u = t a n^{- 1} (\frac{c o s θ}{s i n θ})$

$\Rightarrow u = t a n^{- 1} (c o t θ)$

$\Rightarrow u = t a n^{- 1} (t a n (\frac{π}{2} - θ)$

$\Rightarrow u = (\frac{π}{2} - θ)$

$\Rightarrow u = (\frac{π}{2} - s i n^{- 1} x)$

On differentiating both sides w.r.t x

$\frac{d u}{d x}$ = $\frac{- 1}{\sqrt{1 - x^{2}}}$ ….......…eqn (1)

Now, v = $c o s^{- 1} (2 x \sqrt{1 - x^{2}})$

v = $\frac{π}{2} - s i n^{- 1} (2 x \sqrt{1 - x^{2}})$

Again, Put x = sin $θ$ , where $\frac{π}{4}$ < $θ$ < $\frac{π}{2}$

v = $\frac{π}{2} - s i n^{- 1} (2 s i n θ \sqrt{1 - s i n^{2} θ})$

⇒v = $\frac{π}{2} - s i n^{- 1} (2 s i n θ c o s θ)$

⇒v = $s i n^{- 1} (s i n 2 θ)$

⇒v = $2 θ$

⇒v = $2 s i n^{- 1} x$

On differentiating both sides w.r.t x

$\frac{d v}{d x}$ = $\frac{2}{\sqrt{1 - x^{2}}}$ ……………..eqn(2)

Therefore,

$\frac{d u}{d v}$ = $\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$ = $\frac{\frac{- 1}{\sqrt{1 - x^{2}}}}{\frac{2}{\sqrt{1 - x^{2}}}}$ …from eqn(1) & eqn(2)

$\Rightarrow \frac{d u}{d v}$ = $\frac{- 1}{2}$

Objective Type Questions

24. The function f (x) = $f (x) = {\begin{aligned} \frac{s i n x}{x} + c o s x, i f x \neq 0 \\ k, i f x = 0 \end{aligned}$ is continuous at x = 0, then the value of k is

(A) 3

(B) 2

(D) 1.5

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

L.H.L = R.H.L = f(0)

$\underset{x \to 0^{+}}{l i m} (\frac{s i n x}{x} + c o s x)$ + cosx = f(0)

Put x = 0 +h

$\underset{h \to 0}{l i m} (\frac{s i n h}{h} + c o s h)$ = k

Using identities : $\underset{x \to 0}{l i m} \frac{s i n x}{x}$ = 1 and $\underset{x \to 0}{l i m}$ cosx = 1

1 +1 = k

k = 2

Therefore, the value of k = 2

Hence, the correct option is (B)

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

(A) 4

(B) – 2

(D) 1.5

Ans: The greatest integer function f(x) = $[x]$ 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) = $\frac{1}{x - [x]}$ is not continuous is

(A) 1

(B) 2

(D) none of these

Ans: As we know that $\frac{1}{x - [x]} = 0$ , 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) ${n π :n \in Z}$

(B) ${2n π :n \in Z}$

(D) ${\frac{n π}{2} :n \in Z}$

Ans: f (x) = tanx is always discontinuous on the set ${(2n+1) \frac{π}{2} :n \in Z}$

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 .

(D) none of these

Ans: We have

f (x)= |cosx|

Here, f is everywhere continuous but not differentiable at x = (2n+1) $\frac{π}{2}$ , 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.

(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 $f (x) = {\begin{aligned} S i n \frac{1}{x}, i f x \neq 0 \\ k, i f x = 0 \end{aligned}$ , continuous at x=0 is

(A) 8

(B) 1

(D) none of these

Ans: As we know that $\underset{x \to 0^{}}{l i m}$ sin $\frac{1}{x}$ 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}

(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 ( $t a n^{- 1}$ x) w.r.t. x is

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

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

(D) $\frac{1}{1 + x^{2}}$

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

On Differentiating both sides w.r.t x

$\frac{d y}{d x}$ = $\frac{d}{d x} s e c (t a n^{- 1} x)$

= $s e c (t a n^{- 1} x)$ . tan $(t a n^{- 1} x)$ . $\frac{d}{d x}$ ( $t a n^{- 1} x$ )

= x. $s e c (t a n^{- 1} x)$ . $\frac{1}{1 + x^{2}}$

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

Hence the correct option is (B)

33.If u = $s i n^{- 1} (\frac{2 x}{1 + x^{2}})$ and v = $t a n^{- 1} (\frac{2 x}{1 - x^{2}})$ , then $\frac{d u}{d v}$ is

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

(B) x

(D) 1

Ans: Given u = $s i n^{- 1} (\frac{2 x}{1 + x^{2}})$ and v = $t a n^{- 1} (\frac{2 x}{1 - x^{2}})$

we have ,

u = $s i n^{- 1} (\frac{2 x}{1 + x^{2}})$

Put x = sinθ

⇒u = $s i n^{- 1} (\frac{2 s i n θ}{1 + s i n^{2} θ^{}})$

⇒u = $s i n^{- 1} (s i n 2 θ)$

⇒u = $2 θ$

$\Rightarrow \frac{d u}{d θ}$ = 2 .. eqn (1)

Now, we have ,

v = $t a n^{- 1} (\frac{2 x}{1 - x^{2}})$

put v = tan $θ$

v = $t a n^{- 1} (\frac{2 t a n θ}{1 - t a n^{2} θ^{}})$

⇒v = $t a n^{- 1} (t a n 2 θ)$

⇒v= $2 θ$

$\Rightarrow \frac{d v}{d θ}$ = 2 ...…eqn(2)

Dividing equation (1) by (2)

$\frac{\frac{d u}{d θ}}{\frac{d v}{d θ}}$ = $\frac{2}{2}$ = 1

$\Rightarrow \frac{d u}{d v}$ = 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∈ $[0, π]$ is

(A) $\frac{π}{6}$

(B) $\frac{π}{4}$

(D) $\frac{3 π}{4}$

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∈ $[0, π]$ is

f’(x) = $e^{x}$ sinx + $e^{x}$ cosx

f’(c) = $e^{c}$ (sinx + cosx) = 0

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

tanc = -1

c = $\frac{3 π}{4}$

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 ∈ $[1, 2]$ is

(A) $\frac{3}{2}$

(B) $\frac{2}{3}$

(D) $\frac{3}{2}$

Ans: We have,

f (x) = x (x – 2)

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

Since the polynomial function is always continuous and differentiable

f (x) = $x^{2} - 2 x$ is always continuous on $[1, 2]$ 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) = $(\frac{f (2) - f (1)}{2 - 1})$

⇒f’(c) = $(\frac{f (2) - f (1)}{1})$

⇒f’(c) = f(2) -f(1)

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

f’(x) = 2x -2

⇒f’(c) = 2c -2

⇒f(2) -f(1) = 2c -2

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

⇒0 -(-1) = 2c -2

⇒2c = 1 +2

⇒c = $\frac{3}{2}$

Clearly c = $\frac{3}{2} \in (1, 2)$

Hence option (A) is the correct answer

36. Match the following

Column-I	Column-II
(A) If a function $f (x) = {\begin{aligned} \frac{s i n 3 x}{x}, i f x \neq 0 \\ \frac{k}{2}, i f x = 0 \end{aligned}$ is continuous at x = 0, then k is equal to	(a) \|x\|
(B) Every continuous function is differentiable	(b) True
(C) An example of a function which is continuous	(c) 6
(D) The identity function i.e. f (x) = x ∀ ∈x R is a continuous function.	(d) False

Ans: $A \to c, B \to d, C \to a, D \to b$

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

37. The number of points at which the function f (x) = $\frac{1}{l o g | x |}$ 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) = $\frac{1}{l o g | x |}$ is discontinuous is 3

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

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

is continuous, then a should be equal to 2.

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

Ans: The derivative of ${l o g}_{10} x$ w.r.t. x is ${l o g}_{10} e$ . $\frac{1}{x}$

40. If y = $s e c^{- 1} \frac{\sqrt{x + 1}}{\sqrt{x - 1}}$ + $s i n^{- 1} \frac{\sqrt{x - 1}}{\sqrt{x + 1}}$ , then $\frac{d y}{d x}$ is equal to ________

Ans: y = $s e c^{- 1} \frac{\sqrt{x + 1}}{\sqrt{x - 1}}$ + $s i n^{- 1} \frac{\sqrt{x - 1}}{\sqrt{x + 1}}$ , then $\frac{dy}{dx}$ 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 $\underset{x \to a^{+}}{l i m} f (x)$ and $\underset{x \to a^{_}}{l i m} f (x)$ is equal to f(a).

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

L.H.L = R.H.L = f(a)

$\underset{x \to a^{-}}{l i m} f (x)$ = $\underset{x \to a^{+}}{l i m} f (x)$ = 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 $f (x) = x^{3} + 2 x^{2} - 1$ at $x = 1$ .

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

$x = 1$ .

$R H L = f (x)$

$= \underset{x \to 1^{+}}{l i m} (x^{3} + 2 x^{2} - 1)$ ; here, $x = 1 + h, h \to 0$

$= \underset{h \to 0}{l i m} {(1 + h)}^{3} + 2 {(1 + h)}^{2} - 1$

$= 1 + 2 - 1$

$R H L = 2$

$L H L = f (x)$

$= \underset{x \to 1^{-}}{l i m} (x^{3} + 2 x^{2} - 1)$ ; here, $x = 1 - h, h \to 0$

$= \underset{h \to 0}{l i m} {(1 - h)}^{3} + 2 {(1 - h)}^{2} - 1$

$= 1 + 2 - 1$

$L H L = 2$

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

So, we have $L H L = R H L = f (x)$ at $x = 1$ .

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

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

2. $f (x) = {\begin{aligned} 3 x + 5, if x \geq 2 \\ x^{2}, if x < 2 \end{aligned}$ at $x = 2$ .

Ans: Given that, $f (x) = {\begin{aligned} 3 x + 5, if x \geq 2 \\ x^{2}, if x < 2 \end{aligned}$ at $x = 2$

$L H L = f (x)$ $= \underset{x \to 2^{-}}{l i m} x^{2}$

$= 2^{2}$

$L H L = 4$

$R H L = f (x)$ $= \underset{x \to 2^{+}}{l i m} (3 x + 5)$

$= 3 \times 2 + 5$

$R H L = 11$

$f (2) = 3 \times 2 + 5 = 11$

Here $L H L \neq R H L = f (x)$ at $x = 2$

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

3. $f (x) = {\begin{aligned} \frac{1 - c o s 2 x}{x^{2}}, i f x \neq 0 \\ 5, if x = 0 \end{aligned}$ at $x = 0$ .

Ans: Given that, $f (x) = {\begin{aligned} \frac{1 - c o s 2 x}{x^{2}}, i f x \neq 0 \\ 5, if x = 0 \end{aligned}$ at $x = 0$

$L H L = f (x)$

$= \underset{x \to 0^{-}}{l i m} \frac{1 - c o s 2 x}{x^{2}}$

$= \underset{x \to 0^{-}}{l i m} \frac{2 x}{x^{2}}$

$= 2 \times {(1)}^{2}$

$L H L = 2$

$R H L = f (x)$

$= \underset{x \to 0^{+}}{l i m} \frac{1 - c o s 2 x}{x^{2}}$

$= \underset{x \to 0^{+}}{l i m} \frac{2 x}{x^{2}}$

$= 2 \times {(1)}^{2}$

$R H L = 2$

$f (0) = 5$

Here $L H L = R H L \neq f (x)$ at $x = 0$

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

4. $f (x) = {\begin{aligned} \frac{2 x^{2} - 3 x - 2}{x - 2}, i f x \neq 2 \\ 5, if x = 2 \end{aligned}$ at $x = 2$ .

Ans: Given that, $f (x) = {\begin{aligned} \frac{2 x^{2} - 3 x - 2}{x - 2}, i f x \neq 2 \\ 5, if x = 2 \end{aligned}$ at $x = 2$

$L H L = f (x)$

$= \underset{x \to 2^{-}}{l i m} \frac{2 x^{2} - 3 x - 2}{x - 2}$

$= \underset{x \to 2^{-}}{l i m} \frac{(2 x + 1) (x - 2)}{x - 2}$

$= \underset{x \to 2^{-}}{l i m} (2 x + 1)$

$= 2 \times 2 + 1$

$L H L = 5$

$R H L = f (x)$

$= \underset{x \to 2^{-}}{l i m} \frac{2 x^{2} - 3 x - 2}{x - 2}$

$= \underset{x \to 2^{-}}{l i m} \frac{(2 x + 1) (x - 2)}{x - 2}$

$= \underset{x \to 2^{-}}{l i m} (2 x + 1)$

$= 2 \times 2 + 1$

$R H L = 5$

$f (2) = 5$

Here $L H L = R H L = f (x)$ at $x = 2$

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

5. $f (x) = {\begin{aligned} \frac{| x - 4 |}{2 (x - 4)}, i f x \neq 4 \\ 0, if x = 4 \end{aligned}$ at $x = 4$ .

Ans: Given that, $f (x) = {\begin{aligned} \frac{| x - 4 |}{2 (x - 4)}, i f x \neq 4 \\ 0, if x = 4 \end{aligned}$ at $x = 4$ .

We know that $| x - 4 | = {\begin{aligned} - (x - 4), if x < 4 \\ (x - 4), if x > 4 \end{aligned}$

$L H L = f (x)$

$= \underset{x \to 4^{-}}{l i m} \frac{| x - 4 |}{2 (x - 4)}$

$= \underset{x \to 4^{-}}{l i m} \frac{- (x - 4)}{2 (x - 4)}$

$L H L = - \frac{1}{2}$

$R H L = f (x)$

$= \underset{x \to 4^{+}}{l i m} \frac{| x - 4 |}{2 (x - 4)}$

$= \underset{x \to 4^{+}}{l i m} \frac{(x - 4)}{2 (x - 4)}$

$R H L = \frac{1}{2}$

$f (4) = 0$

Here $L H L \neq R H L \neq f (x)$ at $x = 4$

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

6. $f (x) = {\begin{aligned} | x | c o s (\frac{1}{x}), if x \neq 0 \\ 0, if x = 0 \end{aligned}$ at $x = 0$ .

Ans: Given that $f (x) = {\begin{aligned} | x | c o s (\frac{1}{x}), if x \neq 0 \\ 0, if x = 0 \end{aligned}$ at $x = 0$ .

$L H L = f (x)$

$= \underset{x \to 0^{-}}{l i m} | x | c o s (\frac{1}{x})$

$= \underset{x \to 0^{-}}{l i m} (- x) c o s (\frac{1}{x})$

$= 0 \times (n u m b e r w h i c h w i l l v a r y f r o m - 1 t o 1)$

$L H L = 0$

$R H L = f (x)$

$= \underset{x \to 0^{+}}{l i m} | x | c o s (\frac{1}{x})$

$= \underset{x \to 0^{+}}{l i m} (x) c o s (\frac{1}{x})$

$= 0 \times (n u m b e r w h i c h w i l l v a r y f r o m - 1 t o 1)$

$R H L = 0$

$f (0) = 0$

Here $L H L = R H L = f (x)$ at $x = 0$

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

7. $f (x) = {\begin{aligned} | x - a | s i n (\frac{1}{x - a}), i f x \neq 0 \\ 0, i f x = a \end{aligned}$ at $x = a$ .

Ans: Given that $f (x) = {\begin{aligned} | x - a | s i n (\frac{1}{x - a}), i f x \neq 0 \\ 0, i f x = a \end{aligned}$ at $x = a$ .

$L H L = f (x)$

$= \underset{x \to a^{-}}{l i m} | x - a | s i n (\frac{1}{x - a})$ ;

A. $| x - a | = {\begin{aligned} - (x - a), i f x < a \\ x - a, i f x \geq a \end{aligned}$

$= \underset{x \to a^{-}}{l i m} (- (x - a)) s i n (\frac{1}{x - a})$ .

$= 0 \times (n u m b e r w h i c h w i l l v a r y f r o m - 1 t o 1)$

$L H L = 0$

$R H L = f (x)$

$= \underset{x \to a^{+}}{l i m} | x - a | s i n (\frac{1}{x - a})$ ;

B. $| x - a | = {\begin{aligned} - (x - a), i f x < a \\ x - a, i f x \geq a \end{aligned}$

$= \underset{x \to a^{+}}{l i m} ((x - a)) s i n (\frac{1}{x - a})$

$= 0 \times (n u m b e r w h i c h w i l l v a r y f r o m - 1 t o 1)$

$R H L = 0$

$f (a) = 0$

Here $L H L = R H L = f (x)$ at $x = a$

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

8. $f (x) = {\begin{aligned} \frac{e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}}, i f x \neq 0 \\ 0, i f x = 0 \end{aligned}$ at $x = 0$ .

Ans: Given $f (x) = {\begin{aligned} \frac{e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}}, i f x \neq 0 \\ 0, i f x = 0 \end{aligned}$ at $x = 0$ .

$L H L = f (x)$

$= \underset{x \to 0^{-}}{l i m} \frac{e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}}$

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

$∴ \frac{1}{x} \to - \infty$

$e^{\frac{1}{x}} \to 0$

$= \underset{x \to 0^{-}}{l i m} (\frac{0}{1 + 0})$

$L H L = 0$

$R H L = f (x)$ $= \underset{x \to 0^{+}}{l i m} \frac{e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}}$

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

$∴ \frac{1}{x} \to \infty$

$e^{\frac{1}{x}} \to \infty$

$= \underset{x \to 0^{+}}{l i m} \frac{1}{1 + \frac{1}{e^{\frac{1}{x}}}}$

$= (\frac{1}{1 + 0})$

$R H L = 1$

$f (0) = 0$

Here $L H L \neq R H L$ at $x = 0$

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

9. $f (x) = {\begin{aligned} \frac{x^{2}}{2}, i f 0 \leq x \leq 1 \\ 2 x^{2} - 3 x + \frac{3}{2}, i f 1 < x \leq 2 \end{aligned}$ at $x = 1$ .

Ans: $f (x) = {\begin{aligned} \frac{x^{2}}{2}, i f 0 \leq x \leq 1 \\ 2 x^{2} - 3 x + \frac{3}{2}, i f 1 < x \leq 2 \end{aligned}$ at $x = 1$ .

$L H L = f (x)$

$= \underset{x \to 1^{-}}{l i m} \frac{x^{2}}{2}$

$= \underset{x \to 1^{-}}{l i m} (\frac{1}{2})$

$L H L = \frac{1}{2}$

$R H L = f (x)$

$= \underset{x \to 1^{+}}{l i m} (2 x^{2} - 3 x + \frac{3}{2})$

$= \underset{x \to 1^{+}}{l i m} (2 \times 1^{2} - 3 \times 1 + \frac{3}{2})$

$= (\frac{3}{2} - 1)$

$R H L = \frac{1}{2}$

$f (1) = \frac{1}{2}$

Here $L H L = R H L = f (1)$ at $x = 1$

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

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

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

$L H L = f (x)$

$= \underset{x \to 1^{-}}{l i m} | x | + | x - 1 |$

As x < 1

$∴ | x | = x, | x - 1 | = - (x - 1)$

$= \underset{x \to 1^{-}}{l i m} (x - x + 1)$

$L H L = 1$

$R H L = f (x)$

$= \underset{x \to 1^{+}}{l i m} | x | + | x - 1 |$

As x > 1

$∴ | x | = x, | x - 1 | = (x - 1)$

$= \underset{x \to 1^{+}}{l i m} (2 x - 1)$

$= (2 - 1)$

$R H L = 1$

$f (1) = 1$

Here $L H L = R H L = f (1)$ at $x = 1$

Hence $f (x)$ 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. $f (x) = {\begin{aligned} 3 x - 8, i f x \leq 5 \\ 2 k, if x > 5 \end{aligned}$ at $x = 5$ .

Ans: Given $f (x) = {\begin{aligned} 3 x - 8, i f x \leq 5 \\ 2 k, if x > 5 \end{aligned}$ at $x = 5$

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

$L H L = R H L = f (5)$

$L H L = f (x)$

$= \underset{x \to 5^{-}}{l i m} (3 x - 8)$

$= \underset{x \to 5^{-}}{l i m} (15 - 8)$

$L H L = 7$

$R H L = f (x)$

$= \underset{x \to 5^{+}}{l i m} (2 k)$

$R H L = 2 k$

$2 k = 7 \Rightarrow k = \frac{7}{2}$

12. $f (x) = {\begin{aligned} \frac{2^{x + 2} - 16}{4^{x} - 16}, if x \neq 2 \\ k, if x = 2 \end{aligned}$ at $x = 2$ .

Ans: Given $f (x) = {\begin{aligned} \frac{2^{x + 2} - 16}{4^{x} - 16}, if x \neq 2 \\ k, if x = 2 \end{aligned}$ at $x = 2$

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

$L H L = R H L = f (2)$

$L H L = f (x)$

$= \underset{x \to 2^{-}}{l i m} \frac{2^{x + 2} - 16}{4^{x} - 16}$

$= \underset{x \to 2^{-}}{l i m} \frac{4 (2^{x} - 4)}{(2^{x} - 4) (2^{x} + 4)}$

$= \underset{x \to 2^{-}}{l i m} (\frac{4}{2^{2} + 4})$

$L H L = \frac{1}{2}$

$R H L = f (x)$

$= \underset{x \to 2^{+}}{l i m} \frac{2^{x + 2} - 16}{4^{x} - 16}$

$= \underset{x \to 2^{+}}{l i m} \frac{4 (2^{x} - 4)}{(2^{x} - 4) (2^{x} + 4)}$

$= \underset{x \to 2^{+}}{l i m} (\frac{4}{2^{2} + 4})$

$R H L = \frac{1}{2}$

$k = \frac{1}{2}$

13. $f (x) = {\begin{aligned} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}, if - 1 \leq x < 0 \\ \frac{2 x + 1}{x - 1}, i f 0 \leq x \leq 1 \end{aligned}$ at $x = 0$ .

Ans: Given $f (x) = {\begin{aligned} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}, if - 1 \leq x < 0 \\ \frac{2 x + 1}{x - 1}, i f 0 \leq x \leq 1 \end{aligned}$ at $x = 0$

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

$L H L = R H L = f (0)$

$L H L = f (x)$

$= \underset{x \to 0^{-}}{l i m} (\frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x})$

$= \underset{x \to 0^{-}}{l i m} (\frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}) (\frac{\sqrt{1 + k x} + \sqrt{1 - k x}}{\sqrt{1 + k x} + \sqrt{1 - k x}})$

$= \underset{x \to 0^{-}}{l i m} \frac{(1 + k x) - (1 - k x)}{x (\sqrt{1 + k x} + \sqrt{1 - k x})}$

$= \underset{x \to 0^{-}}{l i m} \frac{2 k x}{x (\sqrt{1 + k x} + \sqrt{1 - k x})}$

C. $= \underset{x \to 0^{-}}{l i m} \frac{2 k}{(\sqrt{1 + 0} + \sqrt{1 - 0})}$

$L H L = k$

$R H L = f (x)$

$= \underset{x \to 0^{+}}{l i m} (\frac{2 x + 1}{x - 1})$

$R H L = - 1$

$k = - 1$

14. $f (x) = {\begin{aligned} \frac{1 - c o s k x}{x s i n x}, if x \neq 0 \\ \frac{1}{2}, if x = 0 \end{aligned}$ at $x = 0$ .

Ans: Given $f (x) = {\begin{aligned} \frac{1 - c o s k x}{x s i n x}, if x \neq 0 \\ \frac{1}{2}, if x = 0 \end{aligned}$ at $x = 0$ .

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

$L H L = R H L = f (0)$

$L H L = R H L = f (x)$

$= \underset{x \to 0^{-}}{l i m} (\frac{1 - c o s k x}{x s i n x})$

$= \underset{x \to 0^{-}}{l i m} (\frac{1 - c o s k x}{k^{2} x^{2} (\frac{s i n x}{x})}) k^{2}$

$= \underset{x \to 0^{-}}{l i m} (\frac{1}{2}) k^{2}$

$L H L = R H L = \frac{k^{2}}{2}$

$f (0) = \frac{1}{2}$

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

$k = \pm 1$ .

15. Prove that the function $f$ defined by $f (x) = {\begin{aligned} \frac{x}{| x | + 2 x^{2}}, if x \neq 0 \\ k, if x = 0 \end{aligned}$ remains discontinuous at $x = 0$ , regardless of the choice of k.

Ans: Given that, $f (x) = {\begin{aligned} \frac{x}{| x | + 2 x^{2}}, if x \neq 0 \\ k, if x = 0 \end{aligned}$ . here to check continuity at $x = 0$

$L H L = \underset{x \to 0^{-}}{l i m} f (x)$

$= \frac{x}{| x | + 2 x^{2}}$

$| x | = {\begin{aligned} - x, x < 0 \\ x, x \geq 0 \end{aligned}$

$= \underset{x \to 0^{-}}{l i m} \frac{x}{- x + 2 x^{2}}$

$= \underset{x \to 0^{-}}{l i m} \frac{1}{- 1 + 2 x}$

$L H L = - 1$

$R H L = \underset{x \to 0^{+}}{l i m} f (x)$

$= \frac{x}{| x | + 2 x^{2}}$

$| x | = {\begin{aligned} - x, x < 0 \\ x, x \geq 0 \end{aligned}$

$= \underset{x \to 0^{+}}{l i m} \frac{x}{x + 2 x^{2}}$ .

$= \underset{x \to 0^{+}}{l i m} \frac{1}{1 + 2 x}$

$R H L = 1$

re $L H L \neq R H L$

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

$f (x) = {\begin{aligned} \frac{x - 4}{| x - 4 |} + a, if x < 4 \\ a + b, if x = 4 \\ \frac{x - 4}{| x - 4 |} + b, if x > 4 \end{aligned}$ is a continuous function at $x = 4$ .

Ans: Given that $f (x) = {\begin{aligned} \frac{x - 4}{| x - 4 |} + a, if x < 4 \\ a + b, if x = 4 \\ \frac{x - 4}{| x - 4 |} + b, if x > 4 \end{aligned}$

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

$L H L = \underset{x \to 4^{-}}{l i m} f (x)$

$= \frac{x - 4}{| x - 4 |} + a$

$| x - 4 | = {\begin{aligned} - (x - 4), x < 4 \\ (x - 4), x \geq 4 \end{aligned}$

$= \underset{x \to 4^{-}}{l i m} \frac{x - 4}{- (x - 4)} + a$

$= \underset{x \to 4^{-}}{l i m} (- 1 + a)$

$L H L = a - 1$

$R H L = \underset{x \to 4^{-}}{l i m} f (x)$ $= \frac{x - 4}{| x - 4 |} + b$

$| x - 4 | = {\begin{aligned} - (x - 4), x < 4 \\ (x - 4), x \geq 4 \end{aligned}$

$= \underset{x \to 4^{+}}{l i m} \frac{x - 4}{(x - 4)} + b$

$= \underset{x \to 4^{+}}{l i m} (1 + b)$

$R H L = b + 1$

$f (4) = a + b$

$L H L = R H L = f (4)$

$a - 1 = b + 1 = a + b$

$a = 1, b = - 1$ .

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

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

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

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

$2 x \neq - 5 \Rightarrow x \neq - \frac{5}{2}$

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

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

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

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

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

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

$= \frac{{(x - 1)}^{2}}{- 2 x^{2} + 5 x - 2}$

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

$- (2 x^{2} - 4 x - x + 2) \neq 0$

$\Rightarrow 2 x (x - 2) - (x - 2) \neq 0$

$\Rightarrow (2 x - 1) (x - 2) \neq 0$

$\Rightarrow x \neq \frac{1}{2} a n d 2$

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

$x = \frac{1}{2}$ and $x = 2$ .

19. Show that the function $f (x) = | s i n x + c o s x |$ is continuous at $x = π$ .

Ans: Given $f (x) = | s i n x + c o s x |$ to check continuity at $x = π$ .

$f (x) = | s i n x + c o s x |$

$= | s i n π + c o s π |$

$= | 0 - 1 |$

$f (x) = 1$

$f (π) = | s i n π + c o s π | = | 0 - 1 | = 1$

So, $f (x)$ is continuous at $x = π$ .

Hence proved.

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

20. $f (x) = {\begin{aligned} x [x], if 0 \leq x < 2 \\ (x - 1), if 2 \leq x < 3 \end{aligned}$ at $x = 2$ .

Ans: Given $f (x) = {\begin{aligned} x [x], if 0 \leq x < 2 \\ (x - 1), if 2 \leq x < 3 \end{aligned}$ at $x = 2$ ,

$L H D = f^{'} (2^{-})$

$= \frac{f (2 - h) - f (2)}{- h}$

$= \frac{(2 - h) [2 - h] - 2 (2 - 1)}{- h}$

$= \frac{(2 - h) 1 - 2}{- h}$

$= \frac{- h}{- h}$

$L H D = 1$

$R H D = f^{'} (2^{+})$

$= \frac{f (2 + h) - f (2)}{h}$

$= \frac{(2 + h - 1) (2 + h) - 2 (2 - 1)}{h}$

$= \frac{h^{2} + 3 h + 2 - 2}{h}$

$= \frac{h^{2} + 3 h}{h}$

$R H D = 3$

Here $L H D \neq R H D$

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

21. $f (x) = {\begin{aligned} x^{2} . s i n \frac{1}{x}, if x \neq 0 \\ 0, if x = 0 \end{aligned}$

Ans: Given $f (x) = {\begin{aligned} x^{2} . s i n \frac{1}{x}, if x \neq 0 \\ 0, if x = 0 \end{aligned}$ ,

$L H D = f^{'} (0^{-})$

$= \frac{f (0 - h) - f (0)}{- h}$

$= \frac{{(0 - h)}^{2} s i n (\frac{1}{0 - h}) - 0}{- h}$

$= \frac{- h^{2} s i n (\frac{1}{h})}{- h}$

$= h s i n (\frac{1}{h})$ ; $∵ 0 \times (n u m b e r s v a r y b e t w e e n - 1 t o 1)$

$L H D = 0$

$R H D = f^{'} (0^{+})$

$= \frac{f (0 + h) - f (0)}{h}$

$= \frac{{(0 + h)}^{2} s i n (\frac{1}{0 + h}) - 0}{- h}$

$= \frac{h^{2} s i n (\frac{1}{h})}{h}$

$= h s i n (\frac{1}{h})$ ; $∵ 0 \times (n u m b e r s v a r y b e t w e e n - 1 t o 1)$

$R H D = 0$ Here $L H D = R H D$

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

22. $f (x) = {\begin{aligned} 1 + x, i f x \leq 2 \\ 5 - x, i f x > 2 \end{aligned}$ at $x = 2$ .

Ans: Given $f (x) = {\begin{aligned} 1 + x, i f x \leq 2 \\ 5 - x, i f x > 2 \end{aligned}$ at $x = 2$ ,

$L H D = f^{'} (2^{-})$

$= \frac{f (2 - h) - f (2)}{- h}$

$= \frac{(1 + 2 - h) - 3}{- h}$

$= \frac{- h}{- h}$

$= 1$

$L H D = 1$

$R H D = f^{'} (2^{+})$

$= \frac{f (2 + h) - f (2)}{h}$

$= \frac{(5 - 2 - h) - 3}{h}$

$= \frac{- h}{h}$

$= - 1$

$R H D = - 1$

Here $L H D \neq R H D$

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

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

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

Here, $f (x) = | x - 5 | = {\begin{aligned} - (x - 5), x < 5 \\ x - 5, x \geq 5 \end{aligned}$

Continuity at $x = 5$

$L H L = f (x)$

$= \underset{x \to 5^{-}}{l i m} (5 - x)$

$= 0$

$R H L = f (x)$

$= \underset{x \to 5^{+}}{l i m} (x - 5)$

$= 0$

$f (5) = 0$

Here $L H L = R H L = f (5)$

Hence f is continuous at $x = 5$ .

Differentiability at $x = 5$

$L H D = f^{'} (5^{-})$

$= \underset{h \to 0}{l i m} (\frac{1}{- h}) (f (5 - h) - f (5))$

$= \underset{h \to 0}{l i m} (\frac{1}{- h}) {| 5 - h - 5 | - 0}$

$= \underset{h \to 0}{l i m} (\frac{1}{- h}) h$

$L H D = - 1$

$R H D = f^{'} (5^{+})$

$= \underset{h \to 0}{l i m} (\frac{1}{h}) (f (5 + h) - f (5))$

$= \underset{h \to 0}{l i m} (\frac{1}{h}) {| 5 + h - 5 | - 0}$

$= \underset{h \to 0}{l i m} (\frac{1}{h}) h$

$R H D = 1$

Here, $L H D \neq R H D$

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

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

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

for all $x, y \in R, f (x) \neq 0$ .

$f^{'} (x) = \frac{1}{h} {f (x + h) - f (x)}$

$= \frac{1}{h} {f (x) . f (h) - f (x)}$ $[∵ f (x + y) = f (x) . f (y)]$

$= f (x) \frac{1}{h} {f (h) - 1}$

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

$f (0) = f (0) . f (0) \Rightarrow f (0) = 0 o r 1$ but $f (x) \neq 0$

So, $f (0) = 1$

$f^{'} (x) = f (x) \frac{1}{h} {f (h) - f (0)}$

$= f (x) . f^{'} (0)$

$f^{'} (x) = 2 f (x)$

Hence proved.

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

25. $2^{c o s^{2} x}$

Ans: Given $y = 2^{c o s^{2} x}$

By taking log to the base $e$ .

$l n y = l n 2^{c o s^{2} x} = c o s^{2} x l n 2$

By differentiating w.r.to x we have

$\frac{d}{d x} l n y = l n 2 \frac{d}{d x} c o s^{2} x$

$\Rightarrow \frac{1}{y} \frac{d y}{d x} = (l n 2) . (2 c o s x) (- s i n x)$

$\Rightarrow \frac{d y}{d x} = - y . s i n 2 x . l n 2$

$\Rightarrow \frac{d y}{d x} = - 2^{c o s^{2} x} . s i n 2 x . l n 2$

26. $\frac{8^{x}}{x^{8}}$

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

By taking log to base e.

$l n y = l n \frac{8^{x}}{x^{8}} = l n l n 8^{x} - l n l n x^{8}$

$\Rightarrow l n y = x l n 8 - 8 l n l n x$

$\Rightarrow \frac{d}{d x} l n y = \frac{d}{d x} (x l n 8 - 8 l n x)$

$\Rightarrow \frac{d}{d y} l n y \frac{d y}{d x} = (l n 8 - \frac{8}{x})$

$\Rightarrow \frac{d y}{d x} = y (l n 8 - \frac{8}{x})$

$\Rightarrow \frac{d y}{d x} = \frac{8^{x}}{x^{8}} (l n 8 - \frac{8}{x})$

27. $l o g (x + \sqrt{x^{2} + a})$

Ans: Given that $y = l o g (x + \sqrt{x^{2} + a})$

$\frac{d y}{d x} = \frac{d}{d x} (l o g (x + \sqrt{x^{2} + a}))$

$= \frac{d}{d (x + \sqrt{x^{2} + a})} l o g (x + \sqrt{x^{2} + a}) \frac{d}{d x} (x + \sqrt{x^{2} + a})$

$= \frac{1}{(x + \sqrt{x^{2} + a})} {1 + \frac{d}{d (x^{2} + a)} \sqrt{x^{2} + a} \frac{d (x^{2} + a)}{d x}}$

$= \frac{1}{(x + \sqrt{x^{2} + a})} {1 + \frac{1}{2 \sqrt{x^{2} + a}} 2 x}$

$= \frac{1}{(x + \sqrt{x^{2} + a})} {\frac{x + \sqrt{x^{2} + a}}{\sqrt{x^{2} + a}}}$

$\frac{d y}{d x} = \frac{1}{\sqrt{x^{2} + a}}$

28. $l o g [l o g (l o g x^{5})]$

Ans: Given $y = l o g [l o g (l o g x^{5})]$

$\frac{d y}{d x} = \frac{d}{d x} l o g [l o g (l o g x^{5})]$

$= \frac{d}{d l o g (l o g x^{5})} l o g [l o g (l o g x^{5})]$

$\frac{d}{d l o g x^{5}} l o g (l o g x^{5}) 5 \frac{d}{d x} l o g x$

$= \frac{1}{l o g (l o g x^{5})} \times \frac{1}{l o g x^{5}} \times \frac{5}{x}$

$\frac{d y}{d x} = \frac{1}{x . l o g x^{5} . l o g (l o g x^{5})}$

29. $s i n \sqrt{x} + c o s^{2} \sqrt{x}$

Ans: Given $y = s i n \sqrt{x} + c o s^{2} \sqrt{x}$

$\frac{d y}{d x} = \frac{d}{d x} (s i n \sqrt{x} + c o s^{2} \sqrt{x})$

$= \frac{d}{d \sqrt{x}} s i n \sqrt{x} \frac{d \sqrt{x}}{d x} + \frac{d}{d \sqrt{x}} \sqrt{x} \frac{d \sqrt{x}}{d x}$

$= c o s \sqrt{x} \frac{1}{2 \sqrt{x}} - s i n 2 \sqrt{x} \frac{1}{2 \sqrt{x}}$

$= \frac{1}{2 \sqrt{x}} (c o s \sqrt{x} - s i n 2 \sqrt{x})$

30. $s i n^{n} (a x^{2} + b x + c)$

Ans: Given $y = s i n^{n} (a x^{2} + b x + c)$

$\frac{d y}{d x} = \frac{d}{d x} s i n^{n} (a x^{2} + b x + c)$

$= \frac{d s i n^{n} (a x^{2} + b x + c)}{d s i n (a x^{2} + b x + c)} \times \frac{d s i n (a x^{2} + b x + c)}{d (a x^{2} + b x + c)} \times \frac{d (a x^{2} + b x + c)}{d x}$

$= n s i n^{n - 1} (a x^{2} + b x + c) . c o s (a x^{2} + b x + c) . (2 a x + b)$

$\frac{d y}{d x} = n (2 a x + b) . s i n^{n - 1} (a x^{2} + b x + c) . c o s (a x^{2} + b x + c)$

31. $c o s (t a n \sqrt{x + 1})$

Ans: Given $y = c o s (t a n \sqrt{x + 1})$

$\frac{d y}{d x} = \frac{d}{d x} c o s (t a n \sqrt{x + 1})$

$= \frac{d c o s (t a n \sqrt{x + 1})}{d (t a n \sqrt{x + 1})} \times \frac{d (t a n \sqrt{x + 1})}{d \sqrt{x + 1}} \times \frac{d \sqrt{x + 1}}{d x}$

$= - s i n (t a n \sqrt{x + 1}) . \sqrt{x + 1} . \frac{1}{2 \sqrt{x + 1}}$

$\frac{d y}{d x} = \frac{- 1}{2 \sqrt{x + 1}} . s i n (t a n \sqrt{x + 1}) . \sqrt{x + 1}$

32. $s i n x^{2} + s i n^{2} x + s i n^{2} (x^{2})$

Ans: Given $y = s i n x^{2} + s i n^{2} x + s i n^{2} (x^{2})$

$\frac{d y}{d x} = \frac{d}{d x} (s i n x^{2} + x + x^{2})$

$= \frac{d s i n x^{2}}{d x^{2}} \times \frac{d x^{2}}{d x} + \frac{d}{d x} (s i n^{2} x) + \frac{d}{d x} s i n^{2} (x^{2})$

$\frac{d y}{d x} = 2 x c o s x^{2} + s i n 2 x + 2 x s i n (2 x^{2})$

33. $s i n^{- 1} (\frac{1}{\sqrt{x + 1}})$

Ans: Given $y = s i n^{- 1} (\frac{1}{\sqrt{x + 1}})$

$\frac{d y}{d x} = \frac{d}{d x} {s i n^{- 1} (\frac{1}{\sqrt{x + 1}})}$

$= \frac{d s i n^{- 1} (\frac{1}{\sqrt{x + 1}})}{d (\frac{1}{\sqrt{x + 1}})} \times \frac{d (\frac{1}{\sqrt{x + 1}})}{d \sqrt{x + 1}} \times \frac{d \sqrt{x + 1}}{d x}$

$= \frac{1}{\sqrt{1 - {(\frac{1}{\sqrt{x + 1}})}^{2}}} \times {- \frac{1}{{(\sqrt{x + 1})}^{2}}} \times \frac{1}{2 \sqrt{x + 1}}$

$= - \frac{\sqrt{x + 1}}{\sqrt{x + 1 - 1}} \times \frac{1}{{(\sqrt{x + 1})}^{2}} \times \frac{1}{2 \sqrt{x + 1}}$

$\frac{d y}{d x} = - \frac{1}{2 \sqrt{x} (x + 1)}$

34. ${(s i n x)}^{c o s x}$

Ans: Given $y = {(s i n x)}^{c o s x}$

Taking log

$\Rightarrow l o g y = l o g ({(s i n x)}^{c o s x}) = c o s x l o g s i n x$

$\Rightarrow \frac{d}{d x} l o g y = \frac{d}{d x} c o s x l o g s i n x$

$\Rightarrow \frac{d}{d y} l o g y . \frac{d y}{d x} = c o s x \frac{d l o g s i n x}{d s i n x} \times \frac{d s i n x}{d x} + l o g s i n x . \frac{d c o s x}{d x}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x} = c o s x . \frac{1}{s i n x} . c o s x - s i n x l o g s i n x$

$\Rightarrow \frac{d y}{d x} = {(s i n x)}^{c o s x} (c o s x . c o t x - s i n x l o g s i n x)$

35. $s i n^{m} x \cdot c o s^{n} x$

Ans: Given $y = s i n^{m} x \cdot c o s^{n} x$

$\frac{d y}{d x} = \frac{d}{d x} (s i n^{m} x \cdot c o s^{n} x)$

$= c o s^{n} x \frac{d}{d x} s i n^{m} x + s i n^{m} x \frac{d}{d x} c o s^{n} x$

$= c o s^{n} x \cdot m s i n^{m - 1} x c o s x + s i n^{m} x \cdot n (c o s^{n - 1} x) s i n x$

$= m c o s^{n + 1} x \cdot s i n^{m - 1} x - n s i n^{m + 1} x \cdot (c o s^{n - 1} x)$

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

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

$\frac{d y}{d x} = \frac{d}{d x} {{(x + 1)}^{2} {(x + 2)}^{3} {(x + 3)}^{4}}$ $= {(x + 2)}^{3} {(x + 3)}^{4} \frac{d}{d x} {(x + 1)}^{2} + {(x + 1)}^{2} {(x + 3)}^{4} \frac{d}{d x} {(x + 2)}^{3} + {(x + 1)}^{2} {(x + 2)}^{3} \frac{d}{d x} {(x + 3)}^{4}$

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

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

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

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

37. $c o s^{- 1} (\frac{s i n x + c o s x}{\sqrt{2}}), - \frac{π}{4} < x < \frac{π}{4}$

Ans: Given $y = c o s^{- 1} (\frac{s i n x + c o s x}{\sqrt{2}}), - \frac{π}{4} < x < \frac{π}{4}$

$= c o s^{- 1} (s i n \frac{π}{4} s i n x + c o s \frac{π}{4} c o s x)$

$= c o s^{- 1} (c o s (\frac{π}{4} - x))$

$y = \frac{π}{4} - x$

$\frac{d y}{d x} = - 1$

38. $t a n^{- 1} (\sqrt{\frac{1 - c o s x}{1 + c o s x}}), - \frac{π}{4} < x < \frac{π}{4}$

Ans: Given $y = t a n^{- 1} (\sqrt{\frac{1 - c o s x}{1 + c o s x}}), - \frac{π}{4} < x < \frac{π}{4}$

$= t a n^{- 1} (\sqrt{\frac{2 s i n^{2} \frac{x}{2}}{2 c o s^{2} \frac{x}{2}}})$

$= t a n^{- 1} (\sqrt{t a n^{2} \frac{x}{2}})$

$= t a n^{- 1} | t a n \frac{x}{2} |$

$y = \frac{x}{2}$

$\frac{d y}{d x} = \frac{1}{2}$

39. $t a n^{- 1} (s e c x + t a n x), - \frac{π}{2} < x < \frac{π}{2}$

Ans: Given $y = t a n^{- 1} (s e c x + t a n x), - \frac{π}{2} < x < \frac{π}{2}$

$= t a n^{- 1} [\frac{(1 + s i n x)}{c o s x}]$

$= t a n^{- 1} [\frac{1 - c o s (\frac{π}{2} + x)}{s i n (\frac{π}{2} + x)}]$

$= t a n^{- 1} [\frac{2 s i n^{2} (\frac{π}{4} + \frac{x}{2})}{2 s i n (\frac{π}{4} + \frac{x}{2}) c o s (\frac{π}{4} + \frac{x}{2})}]$

$= t a n^{- 1} [(t a n (\frac{π}{4} + \frac{x}{2}))]$

$y = \frac{π}{4} + \frac{x}{2}$

$\frac{d y}{d x} = \frac{1}{2}$

40. $t a n^{- 1} (\frac{a c o s x - b s i n x}{b c o s x + a s i n x}), - \frac{π}{2} < x < \frac{π}{2} a n d \frac{a}{b} t a n x > - 1$

Ans: Given $y = t a n^{- 1} (\frac{a c o s x - b s i n x}{b c o s x + a s i n x}), - \frac{π}{2} < x < \frac{π}{2} a n d \frac{a}{b} t a n x > - 1$

$= t a n^{- 1} (\frac{\frac{a c o s x}{b c o s x} - \frac{b s i n x}{b c o s x}}{\frac{b c o s x}{b c o s x} + \frac{a s i n x}{b c o s x}})$

$= t a n^{- 1} (\frac{\frac{a}{b} - t a n x}{1 + \frac{a}{b} t a n x})$

$= t a n^{- 1} [t a n (\frac{a}{b}) - x]$

$= (\frac{a}{b}) - x$

$\frac{d y}{d x} = -$ 1

41. $s e c^{- 1} (\frac{1}{4 x^{3} - 3 x}), 0 < x < \frac{1}{\sqrt{2}}$

Ans: Given $y = s e c^{- 1} (\frac{1}{4 x^{3} - 3 x}), 0 < x < \frac{1}{\sqrt{2}}$

Let, $x = c o s t$

$y = s e c^{- 1} (\frac{1}{4 c o s^{3} t - 3 c o s t})$

$= s e c^{- 1} (\frac{1}{c o s 3 t})$

$= s e c^{- 1} (s e c 3 t)$

$= 3 t$

$y = 3 c o s^{- 1} x$

$\frac{d y}{d x} = 3 \times \frac{- 1}{\sqrt{1 - x^{2}}} = \frac{- 3}{\sqrt{1 - x^{2}}}$

42. $t a n^{- 1} (\frac{3 a^{2} x - x^{3}}{a^{3} - 3 a x^{2}}), - \frac{1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}$

Ans: Given $y = t a n^{- 1} (\frac{3 a^{2} x - x^{3}}{a^{3} - 3 a x^{2}}), - \frac{1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}$

$\Rightarrow y = t a n^{- 1} (\frac{3 \cdot \frac{x}{a} - {(\frac{x}{a})}^{3}}{1 - 3 {(\frac{x}{a})}^{2}})$

Let, $\frac{x}{a} = t a n t$

$\Rightarrow y = t a n^{- 1} (t a n 3 t)$ $[∵ t a n 3 t = \frac{3 t a n t - t a n^{3} t}{1 - 3 t a n^{2} t}]$

$\Rightarrow y = 3 t = 3 \cdot t a n^{- 1} \frac{x}{a}$

$\Rightarrow \frac{d y}{d x} = 3 \cdot \frac{1}{a (1 + {(\frac{x}{a})}^{2})}$

$\Rightarrow \frac{d y}{d x} = \frac{3 a}{a^{2} + x^{2}}$

43. $t a n^{- 1} (\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}), - 1 < x < 1, x \neq 0$

Ans:Given $y = t a n^{- 1} (\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}), - 1 < x < 1, x \neq 0$

Let, $x^{2} = c o s 2 t$

$y = t a n^{- 1} (\frac{\sqrt{1 + c o s 2 t} + \sqrt{1 - c o s 2 t}}{\sqrt{1 + c o s 2 t} - \sqrt{1 - c o s 2 t}})$

$= t a n^{- 1} (\frac{\sqrt{2} c o s t + \sqrt{2} s i n t}{\sqrt{2} c o s t - \sqrt{2} s i n t})$

$= t a n^{- 1} (\frac{1 + t a n t}{1 - t a n t})$

$= t a n^{- 1} [t a n (\frac{π}{4} + t)] = \frac{π}{4} + t$

$= \frac{π}{4} + \frac{1}{2} c o s^{- 1} x^{2}$

$\frac{d y}{d x} = \frac{1}{2} \times \frac{- 2 x}{\sqrt{1 - x^{4}}}$

$\frac{d y}{d x} = \frac{- x}{\sqrt{1 - x^{4}}}$

Find $\frac{d y}{d x}$ of each of the functions expressed in parametric form in Exercises from 44 to 48.

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

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

$\Rightarrow \frac{d x}{d t} = \frac{d}{d t} (t + \frac{1}{t}), \frac{d y}{d t} = \frac{d}{d t} (t - \frac{1}{t})$

$\Rightarrow \frac{d x}{d t} = 1 - \frac{1}{t^{2}}, \frac{d y}{d t} = 1 + \frac{1}{t^{2}}$

$\Rightarrow \frac{d y}{d x} = \frac{d y / d t}{d x / d t}$

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

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

$x + y = 2 t \Rightarrow t = \frac{x + y}{2}$

$\frac{d y}{d x} = \frac{{(x + y)}^{2} + 4}{{(x + y)}^{2} - 4}$

45. $x = e^{θ} (θ + \frac{1}{θ}), y = e^{θ} (θ - \frac{1}{θ})$

Ans: Given $x = e^{θ} (θ + \frac{1}{θ}), y = e^{- θ} (θ - \frac{1}{θ})$

$\Rightarrow \frac{d x}{d θ} = \frac{d}{d θ} {e^{θ} (θ + \frac{1}{θ})}, \frac{d y}{d θ} = \frac{d}{d θ} {e^{- θ} (θ - \frac{1}{θ})}$

$\Rightarrow \frac{d x}{d θ} = e^{θ} (θ + \frac{1}{θ}) + e^{θ} (1 - \frac{1}{θ^{2}}), \frac{d y}{d θ} = - e^{- θ} (θ - \frac{1}{θ}) + e^{- θ} (1 + \frac{1}{θ^{2}})$

$\Rightarrow \frac{d y}{d x} = \frac{d y / d θ}{d x / d θ}$

$= \frac{- e^{- θ} (θ - \frac{1}{θ}) + e^{- θ} (1 + \frac{1}{θ^{2}})}{e^{θ} (θ + \frac{1}{θ}) + e^{θ} (1 - \frac{1}{θ^{2}})}$

$= e^{- 2 θ} (\frac{\frac{- (θ^{2} - 1)}{θ} + \frac{(θ^{2} + 1)}{θ^{2}}}{\frac{(θ^{2} + 1)}{θ} + \frac{(θ^{2} - 1)}{θ^{2}}})$

$\frac{d y}{d x} = e^{- 2 θ} (\frac{- θ^{3} + θ^{2} + θ + 1}{θ^{3} + θ^{2} + θ - 1})$

46. $x = 3 c o s θ - 2 c o s^{3} θ, y = 3 s i n θ - 2 s i n^{3} θ$

Ans: Given $x = 3 c o s θ - 2 θ, y = 3 s i n θ - 2 θ$

$\Rightarrow \frac{d}{d θ} x = \frac{d}{d θ} (3 c o s θ - 2 θ), \frac{d}{d θ} y = \frac{d}{d θ} (3 s i n θ - 2 θ)$

$\Rightarrow \frac{d x}{d θ} = - 3 s i n θ - 6 c o s^{2} θ (- s i n θ), \frac{d y}{d θ} = 3 c o s θ - 6 s i n^{2} θ c o s θ$

$\Rightarrow \frac{d y}{d x} = \frac{d y / d θ}{d x / d θ}$

$\Rightarrow \frac{d y}{d x} = \frac{3 c o s θ - 6 s i n^{2} θ c o s θ}{- 3 s i n θ + 6 s i n θ c o s^{2} θ} = \frac{3 c o s θ}{3 s i n θ} \times \frac{1 - 2 θ}{2 θ - 1}$

$\Rightarrow \frac{d y}{d x} = c o t θ$

47. $s i n x = \frac{2 t}{1 + t^{2}}, t a n y = \frac{2 t}{1 - t^{2}}$

Ans: Given $s i n x = \frac{2 t}{1 + t^{2}}, t a n y = \frac{2 t}{1 - t^{2}}$

Let $t = t a n θ$

$\Rightarrow s i n x = \frac{2 t a n θ}{1 + θ}, t a n y = \frac{2 t a n θ}{1 - θ}$

$\Rightarrow s i n x = s i n 2 θ, t a n y = t a n 2 θ$

$\Rightarrow x = 2 θ, y = 2 θ$

$\Rightarrow y = x$

$\Rightarrow \frac{d y}{d x} = 1$

48. $x = \frac{1 + l o g t}{t^{2}}, y = \frac{3 + 2 l o g t}{t}$ .

Ans: Given that $x = \frac{1 + l o g t}{t^{2}}, y = \frac{3 + 2 l o g t}{t}$

$\Rightarrow \frac{d x}{d t} = \frac{d}{d t} (\frac{1 + l o g t}{t^{2}}), \frac{d y}{d t} = \frac{d}{d t} (\frac{3 + 2 l o g t}{t})$

$\Rightarrow \frac{d x}{d t} = \frac{t^{2} (\frac{1}{t}) - (1 + l o g t) 2 t}{t^{4}}, \frac{d y}{d t} = \frac{t (\frac{2}{t}) - (3 + 2 l o g t)}{t^{2}}$

$\Rightarrow \frac{d x}{d t} = - \frac{1 + 2 l o g t}{t^{3}}, \frac{d y}{d t} = - \frac{1 + 2 l o g t}{t^{2}}$

$\Rightarrow \frac{d y}{d x} = \frac{d y / d t}{d x / d t}$

$\Rightarrow \frac{d y}{d x} = (- \frac{1 + 2 l o g t}{t^{2}}) \times (- \frac{t^{3}}{1 + 2 l o g t})$

$\Rightarrow \frac{d y}{d x} = t$

49. If $x = e^{c o s 2 t}$ and $y = e^{s i n 2 t}$ , then prove that $\frac{d y}{d x} = - \frac{y l o g x}{x l o g y}$ .

Ans: Given that $x = e^{c o s 2 t} \Rightarrow c o s 2 t = l o g x$

$y = e^{s i n 2 t} \Rightarrow s i n 2 t = l o g y$

Squaring and adding we have

$\Rightarrow 2 t + 2 t = {(l o g x)}^{2} + {(l o g y)}^{2}$

$\Rightarrow {(l o g x)}^{2} + {(l o g y)}^{2} = 1$

By differentiating w.r.to x we have

$\Rightarrow \frac{d}{d x} {(l o g x)}^{2} + \frac{d}{d x} {(l o g y)}^{2} = \frac{d}{d x} 1$

$\Rightarrow \frac{2 l o g x}{x} + \frac{2 l o g y}{y} \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{y l o g x}{x l o g y}$

50. If $x = a s i n 2 t (1 + c o s 2 t)$ and $y = b c o s 2 t (1 - c o s 2 t)$ , show that ${(\frac{d y}{d x})}_{a t t = \frac{π}{4}} = \frac{b}{a}$ .

Ans: Given $x = a s i n 2 t (1 + c o s 2 t)$ and $y = b c o s 2 t (1 - c o s 2 t)$

$\Rightarrow x = a (s i n 2 t + s i n 2 t c o s 2 t), y = b (c o s 2 t - 2 t)$

$\Rightarrow x = a (s i n 2 t + \frac{1}{2} s i n 4 t), y = b (c o s 2 t - \frac{1}{2} - \frac{1}{2} c o s 4 t)$

$\Rightarrow \frac{d x}{d t} = a (2 c o s 2 t + 2 c o s 4 t), \frac{d y}{d t} = b (- 2 s i n 2 t + 2 s i n 4 t)$

$\Rightarrow \frac{d y}{d x} = \frac{b (- 2 s i n 2 t + 2 s i n 4 t)}{a (2 c o s 2 t + 2 c o s 4 t)}$

At $t = \frac{π}{4}$

$\Rightarrow {(\frac{d y}{d x})}_{a t t = \frac{π}{4}} = \frac{b (- s i n \frac{π}{2} + s i n π)}{a (c o s \frac{π}{2} + c o s π)}$

$\Rightarrow {(\frac{d y}{d x})}_{a t t = \frac{π}{4}} = \frac{b (- 1)}{a (- 1)} = \frac{b}{a}$

Hence proved.

51. If $x = 3 s i n t - s i n 3 t, y = 3 c o s t - c o s 3 t$ , find $\frac{d y}{d x} a t t = \frac{π}{3}$ .

Ans: Given $x = 3 s i n t - s i n 3 t, y = 3 c o s t - c o s 3 t$

$x = 4 t, y = 3 c o s t - c o s 3 t$

$\Rightarrow \frac{d x}{d t} = 12 t c o s t, \frac{d y}{d t} = - 3 s i n t + 3 s i n 3 t$

$\Rightarrow \frac{d y}{d x} = \frac{- 3 s i n t + 3 s i n 3 t}{12 t c o s t}$

$a t t = \frac{π}{3}$

$\Rightarrow {(\frac{d y}{d x})}_{a t t = \frac{π}{3}} = \frac{- s i n \frac{π}{3} + s i n π}{4 \frac{π}{3} c o s \frac{π}{3}} = \frac{- \frac{\sqrt{3}}{2}}{4 \times {(\frac{\sqrt{3}}{2})}^{2} \times \frac{1}{2}}$

$\Rightarrow {(\frac{d y}{d x})}_{a t t = \frac{π}{3}} = - \frac{1}{\sqrt{3}}$

52. Differentiate $\frac{x}{s i n x}$ w.r.to $s i n x$ .

Ans: Given functions are $f (x) = \frac{x}{s i n x}, g (x) = s i n x$

Differentiation of f w.r.to g

$\frac{d f (x)}{d g (x)} = \frac{d f (x) / d x}{d g (x) / d x}$

$\Rightarrow \frac{d f (x)}{d x} = \frac{d}{d x} (\frac{x}{s i n x}) = \frac{s i n x - x c o s x}{s i n^{2} x}$

$\Rightarrow \frac{d g (x)}{d x} = \frac{d}{d x} s i n x = c o s x$

$\Rightarrow \frac{d f (x)}{d g (x)} = (\frac{s i n x - x c o s x}{s i n^{2} x}) (\frac{1}{c o s x})$

$\Rightarrow \frac{d f (x)}{d g (x)} = \frac{t a n x - x}{s i n^{2} x}$

53. Differentiate $t a n^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x})$ w.r.to $t a n^{- 1} x$ when $x \neq 0$ .

Ans: Given functions are $f (x) = t a n^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x}), g (x) = t a n^{- 1} x$

Differentiation of f w.r.to g

$\frac{d f (x)}{d g (x)} = \frac{d f (x) / d x}{d g (x) / d x}$

$\frac{d f (x)}{d x} = \frac{d}{d x} t a n^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x}), \frac{d g (x)}{d x} = \frac{d}{d x} t a n^{- 1} x$

For f , let $x = t a n t$

$(\frac{\sqrt{1 + t a n^{2} t} - 1}{t a n t}) = (\frac{(s e c t - 1) c o s t}{s i n t})$

$= (\frac{1 - c o s t}{s i n t})$

$= (t a n \frac{t}{2})$

$= \frac{t}{2} = \frac{1}{2} t a n^{- 1} x$

$\frac{d f (x)}{d x} = \frac{d}{d x} (\frac{\sqrt{1 + x^{2}} - 1}{x})$

$= \frac{d}{d x} (\frac{1}{2} t a n^{- 1} x)$

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

$\frac{d g (x)}{d x} = \frac{d}{d x} t a n^{- 1} x = \frac{1}{1 + x^{2}}$

$\frac{d f (x)}{d g (x)} = \frac{d f (x) / d x}{d g (x) / d x} = \frac{\frac{1}{2 (1 + x^{2})}}{\frac{1}{1 + x^{2}}} = \frac{1}{2}$

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

54. $s i n (x y) + \frac{x}{y} = x^{2} - y$

Ans:

Given $s i n (x y) + \frac{x}{y} = x^{2} - y$

Differentiating w.r.to x we have

$\Rightarrow \frac{d}{d x} s i n (x y) + \frac{d}{d x} (\frac{x}{y}) = \frac{d}{d x} x^{2} - \frac{d y}{d x}$

$\Rightarrow c o s (x y) {y + x \frac{d y}{d x}} + \frac{y - x \frac{d y}{d x}}{y^{2}} = 2 x - \frac{d y}{d x}$

$\Rightarrow y^{3} c o s (x y) + c o s (x y) y^{2} x \frac{d y}{d x} + y - x \frac{d y}{d x} = 2 x y^{2} - y^{2} \frac{d y}{d x}$

$\Rightarrow (c o s (x y) y^{2} x - x + y^{2}) \frac{d y}{d x} = 2 x y^{2} - y - y^{3} c o s (x y)$

$\Rightarrow \frac{d y}{d x} = \frac{2 x y^{2} - y - y^{3} c o s (x y)}{c o s (x y) y^{2} x - x + y^{2}}$

55. $s e c (x + y) = x y$

Ans: Given $s e c (x + y) = x y$

$\Rightarrow \frac{d}{d x} s e c (x + y) = \frac{d}{d x} (x y)$

$\Rightarrow s e c (x + y) t a n (x + y) {1 + \frac{d y}{d x}} = y + x \frac{d y}{d x}$

$\Rightarrow s e c (x + y) t a n (x + y) + s e c (x + y) t a n (x + y) \frac{d y}{d x} = y + x \frac{d y}{d x}$

$\Rightarrow s e c (x + y) t a n (x + y) \frac{d y}{d x} - x \frac{d y}{d x} = y - s e c (x + y) t a n (x + y)$

$\Rightarrow \frac{d y}{d x} = \frac{y - s e c (x + y) t a n (x + y)}{s e c (x + y) t a n (x + y) - x}$

56. $t a n^{- 1} (x^{2} + y^{2}) = a$ .

Ans: Given $t a n^{- 1} (x^{2} + y^{2}) = a$

$∴ x^{2} + y^{2} = t a n a$

Differentiating w.r.to x we get

$\Rightarrow 2 x + 2 y \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{x}{y}$

57. ${(x^{2} + y^{2})}^{2} = x y$ .

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

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

Differentiating w.r.to x we have

$\Rightarrow 4 x^{3} + 4 y^{3} \frac{d y}{d x} + 4 x y^{2} + 4 x^{2} y \frac{d y}{d x} - y - x \frac{d y}{d x} = 0$

$\Rightarrow (4 y^{3} + 4 x^{2} y - x) \frac{d y}{d x} = y - 4 x y^{2} - 4 x^{3}$

$\Rightarrow \frac{d y}{d x} = \frac{4 y^{3} + 4 x^{2} y - x}{4 y^{3} + 4 x^{2} y - x}$

58. If $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ , then show that $\frac{d y}{d x} \frac{d x}{d y} = 1$ .

Ans: Given $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$

Differentiating w.r.to x we have

$2 a x + 2 h (y + x \frac{d y}{d x}) + 2 b y \frac{d y}{d x} + 2 g + 2 f \frac{d y}{d x} = 0$

$\Rightarrow (h x + b y + f) \frac{d y}{d x} = - (a x + h y + g)$

$\Rightarrow \frac{d y}{d x} = - \frac{a x + h y + g}{h x + b y + f}$

Now differentiating w.r.to y we have

$2 a x \frac{d x}{d y} + 2 h (x + y \frac{d x}{d y}) + 2 b y + 2 g \frac{d x}{d y} + 2 f = 0$

$\Rightarrow (a x + h y + g) \frac{d x}{d y} = - (2 h x + b y + f)$

$\Rightarrow \frac{d x}{d y} = - \frac{h x + b y + f}{a x + h y + g}$

Now, $\frac{d y}{d x} \cdot \frac{d x}{d y} = 1$ .

Hence proved.

59. If $x = e^{\frac{x}{y}}$ , prove that $\frac{d y}{d x} = \frac{x - y}{x l o g x}$

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

Taking log, we have

$l o g x = \frac{x}{y}$

$\Rightarrow y l o g x = x$

Differentiating w.r.to x we have

$\frac{y}{x} + l o g x \frac{d y}{d x} = 1$

$\Rightarrow l o g x \frac{d y}{d x} = 1 - \frac{y}{x} = \frac{x - y}{x}$

$\Rightarrow \frac{d y}{d x} = \frac{x - y}{x l o g x}$

Hence proved.

60. If $y^{x} = e^{y - x}$ , prove that $\frac{d y}{d x} = \frac{{(1 + l o g y)}^{2}}{l o g y}$ .

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

Taking log, we have

$x l o g y = y - x$

Differentiating w.r.to x we have

$l o g y + \frac{x}{y} \frac{d y}{d x} = \frac{d y}{d x} - 1$

$\Rightarrow 1 + l o g y = (1 - \frac{x}{y}) \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x} = \frac{y (1 + l o g y)}{y - x}$

Now $x l o g y = y - x$

$\Rightarrow l o g y = \frac{y}{x} - 1$

$\Rightarrow 1 + l o g y = \frac{y}{x}$

$\Rightarrow 1 - \frac{x}{y} = 1 - \frac{1}{1 + l o g y} = \frac{1 + l o g y - 1}{1 + l o g y} = \frac{l o g y}{1 + l o g y}$

$\Rightarrow \frac{d y}{d x} = \frac{(1 + l o g y)}{1 - \frac{x}{y}} = \frac{(1 + l o g y)}{\frac{l o g y}{1 + l o g y}}$

$\Rightarrow \frac{d y}{d x} = \frac{{(1 + l o g y)}^{2}}{l o g y}$

Hence proved.

61. If $y = {(c o s x)}^{{(c o s x)}^{x) \dots \infty}}$ , show that $\frac{d y}{d x} = \frac{y^{2} t a n x}{y l o g c o s x - 1}$ .

Ans: Given $y = {(c o s x)}^{{(c o s x)}^{x) \dots \infty}}$

$y = {(c o s x)}^{y}$

$\Rightarrow l o g y = y l o g c o s x$

$\Rightarrow \frac{d}{d x} l o g y = \frac{d}{d x} (y l o g c o s x)$

$\Rightarrow \frac{1}{y} \frac{d y}{d x} = l o g c o s x \frac{d y}{d x} + y \frac{1}{c o s x} (- s i n x)$

$\Rightarrow \frac{d y}{d x} = y l o g c o s x \frac{d y}{d x} - y^{2} t a n x$

$\Rightarrow y^{2} t a n x = (y l o g c o s x - 1) \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x} = \frac{y^{2} t a n x}{y l o g c o s x - 1}$

Hence proved.

62. If $x s i n (a + y) + s i n a c o s (a + y) = 0$ , prove that $\frac{d y}{d x} = \frac{s i n^{2} (a + y)}{s i n a}$ .

Ans: Given $x s i n (a + y) + s i n a c o s (a + y) = 0$

$\frac{d}{d x} (x s i n (a + y) + s i n a c o s (a + y)) = 0$

$\Rightarrow s i n (a + y) + x c o s (a + y) \frac{d y}{d x} - s i n a s i n (a + y) \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = \frac{s i n (a + y)}{s i n a s i n (a + y) - x c o s (a + y)}$

$\Rightarrow x = - \frac{s i n a c o s (a + y)}{s i n (a + y)}$

$\Rightarrow \frac{d y}{d x} = \frac{s i n (a + y)}{s i n a s i n (a + y) - (- \frac{s i n a c o s (a + y)}{s i n (a + y)}) c o s (a + y)}$

$= \frac{s i n^{2} (a + y)}{s i n a {s i n^{2} (a + y) + c o s^{2} (a + y)}}$

$\frac{d y}{d x} = \frac{s i n^{2} (a + y)}{s i n a}$

Hence proved.

63. If $\sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = a (x - y)$ , prove that $\frac{d y}{d x} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}$ .

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

Let $x = s i n p, y = s i n q$

$1 - x^{2} = 1 - p = p$

$\Rightarrow 1 - y^{2} = 1 - q = q$

$\Rightarrow c o s p + c o s q = a (s i n p - s i n q)$

$\Rightarrow 2 c o s \frac{p + q}{2} c o s \frac{p - q}{2} = a (2 c o s \frac{p + q}{2} s i n \frac{p - q}{2})$

$\Rightarrow c o t \frac{p - q}{2} = a$

$\Rightarrow p - q = 2 a$

$\Rightarrow x - y = 2 a$

$\Rightarrow \frac{d}{d x} (x - y) = 2 \frac{d}{d x} a$

$\Rightarrow \frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - y^{2}}} \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = \frac{\sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}}$

Hence prove.

64. If $y = t a n^{- 1} x$ , find $\frac{d^{2} y}{d x^{2}}$ in terms of y alone.

Ans: Given $y = t a n^{- 1} x$

$x = t a n y$

Differentiating w.r.to x

$1 = s e c^{2} y \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x} = c o s^{2} y$

Again, $\frac{d}{d x} \frac{d y}{d x} = \frac{d}{d x} c o s^{2} y$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - s i n 2 y \frac{d y}{d x} = - s i n 2 y c o s^{2} y$

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

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

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

Function f is polynomial, so

i. it is continuous and

ii. differentiable for all real x.

iii. $f (0) = 0$ and $f (1) = 0$

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

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

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

$= 3 x^{2} - 4 x + 1$

$\Rightarrow 3 c^{2} - 4 c + 1 = 0$

$\Rightarrow 3 c^{2} - 3 c - c + 1 = 0$

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

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

$\Rightarrow c = 1, \frac{1}{3}$

$\Rightarrow \frac{1}{3} \in [0, 1]$

Hence Rolle’s theorem is verified.

66. $f (x) = s i n^{4} x + c o s^{4} x$ in $[0, \frac{π}{2}]$ .

Ans: Given $f (x) = s i n^{4} x + c o s^{4} x$

Function f is sine and cosine function, so

i. it is continuous and

ii. differentiable for all real x.

iii. $f (0) = 1$ and $f (\frac{π}{2}) = 1$

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

$\exists c \in [0, \frac{π}{2}]$ such that $f^{'} (c) = 0$ .

$f^{'} (x) = \frac{d}{d x} (s i n^{4} x + c o s^{4} x)$

$= 4 s i n^{3} x c o s x - 4 c o s^{3} x s i n x$

$= 4 s i n x c o s x (s i n^{2} x - c o s^{2} x)$

$= - 2 s i n 2 x c o s 2 x$

$- 2 s i n 2 c c o s 2 c = 0$

$s i n 2 c = 0 \Rightarrow 2 c = 0, π$

$c = 0, \frac{π}{2}$

$c o s 2 c = 0 \Rightarrow 2 c = \frac{π}{2}$

$c = \frac{π}{4} \in [0, \frac{π}{2}]$

Hence Rolle’s theorem is verified.

67. $f (x) = l o g (x^{2} + 2) - l o g 3$ in $[- 1, 1]$

Ans:

Given $f (x) = l o g (x^{2} + 2) - l o g 3$

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. $f (- 1) = 0$ and $f (1) = 0$

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

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

$f^{'} (x) = \frac{d}{d x} (l o g (x^{2} + 2) - l o g 3)$

$= \frac{2 x}{x^{2} + 2}$

$\frac{2 c}{c^{2} + 1} = 0 \Rightarrow c = 0 \in [- 1, 1]$

Hence Rolle’s theorem is verified.

68. $f (x) = x (x + 3) e^{- \frac{x}{2}}$ in $[- 3, 0]$ .

Ans: Given $f (x) = x (x + 3) e^{- \frac{x}{2}}$

Function f is polynomial with exponential function, so

i. it is continuous and

ii. differentiable for all real x.

iii. $f (- 3) = 0$ and $f (0) = 0$

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

$\exists c \in [- 3, 0]$ such that $f^{'} (c) = 0$ .

$\Rightarrow f^{'} (x) = \frac{d}{d x} (x (x + 3) e^{- \frac{x}{2}})$

$= (x + 3) e^{- \frac{x}{2}} + x e^{- \frac{x}{2}} - \frac{1}{2} x (x + 3) e^{- \frac{x}{2}}$

$= \frac{1}{2} (x^{2} + 7 x + 6) e^{- \frac{x}{2}}$

$\Rightarrow f^{'} (c) = \frac{1}{2} (c^{2} + 7 c + 6) e^{- \frac{c}{2}} = 0$

$\Rightarrow∴ c^{2} + c + 6 c + 6 = 0$

$\Rightarrow c (c + 1) + 6 (c + 1) = 0$

$\Rightarrow (c + 1) (c + 6) = 0$

$\Rightarrow c = - 6, - 1$

$\Rightarrow c = - 1 \in [- 3, 0]$

Hence Rolle’s theorem is verified.

69. $f (x) = \sqrt{4 - x^{2}}$ in $[- 2, 2]$

Ans: Given $f (x) = \sqrt{4 - x^{2}}$

Function f is defined $\forall x \in [- 2, 2]$ , so

i. it is continuous in $[- 2, 2]$ and

ii. differentiable in $(- 2, 2)$ .

iii. $f (- 2) = 0$ and $f (2) = 0$

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

$\exists c \in [- 2, 2]$ such that $f^{'} (c) = 0$ .

$f^{'} (x) = \frac{d}{d x} \sqrt{4 - x^{2}}$

$= - \frac{x}{\sqrt{4 - x^{2}}}$

$f^{'} (c) = - \frac{c}{\sqrt{4 - c^{2}}} = 0$

$c = 0 \in (- 2, 2)$

Hence Rolle’s theorem is verified.

70. Discuss the applicability of Rolle’s theorem on the function given by

$f (x) = {\begin{aligned} x^{2} + 1, if 0 \leq x \leq 1 \\ 3 - x, if 1 \leq x \leq 2 \end{aligned}$

Ans: Given $f (x) = {\begin{aligned} x^{2} + 1, if 0 \leq x \leq 1 \\ 3 - x, if 1 \leq x \leq 2 \end{aligned}$

At $x = 1$

$L H L = f (x)$

$= \underset{x \to 1^{-}}{l i m} (x^{2} + 1)$

$= 1 + 1 = 2$

$R H L = \underset{x \to 1^{+}}{l i m} f (x)$

$= \underset{x \to 1^{+}}{l i m} (3 - x)$

$= 3 - 1 = 2$

$L H L = R H L = f (1)$

So, function is continuous.

$L H D = f^{'} (1^{-}) = {(2 x)}_{a t x = 1} = 2$

$R H D = f^{'} (1^{+}) = {(- x)}_{a t x = 1} = - 1$

$L H D \neq R H D$ . So, function is not differentiable.

Hence Rolle’s theorem is not applicable in $[0, 2]$ .

71. Find the points on the curve $y = c o s x - 1$ in $[0, 2 π]$ , where the tangents are parallel to the x-axis.

Ans: Given $y = c o s x - 1$

Tangent is parallel to the x-axis then the derivative must be zero at those points. So, $\frac{d y}{d x} = \frac{d}{d x} (c o s x - 1) = s i n x = 0$

$x = n π; n \in Z$

But in $(0, 2 π)$ , $x = π$

Point is $y = c o s π - 1 = - 2$

$(π, - 2)$ .

Hence the point is $(π, - 2)$ .

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

Ans: Given $y = x (x - 4), x \in [0, 4]$ ,

Function f is polynomial, so

i. it is continuous and

ii. differentiable for all real x.

iii. $f (0) = 0$ and $f (4) = 0$

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

$\exists c \in [0, 4]$ such that $f^{'} (c) = 0$ .

$f^{'} (x) = \frac{d}{d x} (x^{2} - 4 x)$

$= 2 x - 4$

$f^{'} (c) = 2 x - 4 = 0$

$c = 2 \in (0, 4)$

At $x = 2$ function has a tangent parallel to the x-axis.

Again, $y = 2 (2 - 4) = - 4$

Hence the point is $(2, - 4)$ .

Verify mean value theorem for each of the functions given Exercises 73 to 76.

73. $f (x) = \frac{1}{4 x - 1}, x \in [1, 4]$

Ans: Given $f (x) = \frac{1}{4 x - 1}, x \in [1, 4]$

For mean value theorem function must be continuous and differentiable in the given domain.

Here, the function $f$ is discontinuous at $x = \frac{1}{4}$ .

Again $f^{'} (x) = - \frac{4}{{(4 x - 1)}^{2}}$ , so $f$ is not differentiable at $x = \frac{1}{4}$ .

Both conditions are satisfied for the mean value theorem.

So, there exists a real number $c \in (1, 4)$ such that

$f^{'} (c) = \frac{f (4) - f (1)}{4 - 1}$

$\Rightarrow - \frac{4}{{(4 c - 1)}^{2}} = \frac{1}{3} (\frac{1}{4 \times 4 - 1} - \frac{1}{4 \times 1 - 1})$

$\Rightarrow \frac{4}{{(4 c - 1)}^{2}} = - \frac{1}{3} (\frac{1}{15} - \frac{1}{3})$

$\Rightarrow {(\frac{2}{4 c - 1})}^{2} = \frac{1}{3} (\frac{5 - 1}{15}) = \frac{4}{45}$

$\Rightarrow \frac{2}{4 c - 1} = \pm \frac{2}{3 \sqrt{5}}$

$\Rightarrow 4 c - 1 = \pm 3 \sqrt{5}$

$\Rightarrow c = \frac{1}{4} (1 \pm 3 \sqrt{5})$

$\Rightarrow c = \frac{1}{4} (1 + 3 \sqrt{5}) \in (1, 4)$

Hence the mean value theorem is verified.

74. $f (x) = x^{3} - 2 x^{2} - x + 3, x \in [0, 1]$ .

Ans: Given $f (x) = x^{3} - 2 x^{2} - x + 3, x \in [0, 1]$

Since, $f (x)$ is polynomial and the polynomial function is continuous and differentiable for all real x.

So there exists a real number $c \in (0, 1)$ such that

$f^{'} (c) = \frac{f (1) - f (0)}{1 - 0}$

$\Rightarrow f^{'} (x) = 3 x^{2} - 4 x - 1$

$\Rightarrow 3 c^{2} - 4 c - 1 = (1 - 2 - 1 + 3) - (3)$

$\Rightarrow 3 c^{2} - 4 c - 1 = - 2$

$\Rightarrow 3 c^{2} - 3 c - c + 1 = 0$

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

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

$\Rightarrow c = 1, \frac{1}{3}$

$\Rightarrow c = \frac{1}{3} \in (0, 1)$

Hence the mean value theorem is verified.

75. $f (x) = s i n x - s i n 2 x, x \in [0, π]$

Ans: Given that $f (x) = s i n x - s i n 2 x, x \in [0, π]$

Since sin function is continuous and differentiable for all real x, so there exists a real number $c \in (0, π)$ such that

$f^{'} (c) = \frac{f (π) - f (0)}{π - 0}$

$\Rightarrow f^{'} (x) = c o s x - 2 c o s 2 x$

$\Rightarrow c o s c - 2 c o s 2 c = \frac{1}{π} (0 - 0) = 0$

$\Rightarrow c o s c - 2 (2 c - 1) = 0$

$\Rightarrow 4 c - c o s c - 2 = 0$

$\Rightarrow c o s c = \frac{1 \pm \sqrt{1 + 4 \times 4 \times 2}}{8} = \frac{1 \pm \sqrt{33}}{8}$

$\Rightarrow c = (\frac{1 \pm \sqrt{33}}{8}) \in (0, π)$

Hence the mean value theorem is verified.

76. $f (x) = \sqrt{25 - x^{2}}, x \in [1, 5]$ .

Ans: Given $f (x) = \sqrt{25 - x^{2}}, x \in [1, 5]$

$f (x)$ is continuous when $25 - x^{2} \geq 0$

So, $x \in [- 5, 5]$

$∴ f (x)$ is continuous in $[1, 5]$

Now, $f^{'} (x) = - \frac{x}{\sqrt{25 - x^{2}}}$ so, it is differentiable in $(1, 5)$

It satisfies both the conditions of the mean value theorem.

There exists a real number $c \in (1, 5)$ such that

$f^{'} (c) = \frac{f (5) - f (1)}{5 - 1}$

$\Rightarrow - \frac{c}{\sqrt{25 - c^{2}}} = \frac{1}{4} (\sqrt{25 - 5^{2}} - \sqrt{25 - 1^{2}})$

$\Rightarrow \frac{c}{\sqrt{25 - c^{2}}} = - \frac{1}{4} (- 2 \sqrt{6})$

Squaring both sides, we have

$\frac{c^{2}}{25 - c^{2}} = \frac{6}{4} = \frac{3}{2}$

$\Rightarrow 2 c^{2} = 75 - 3 c^{2}$

$\Rightarrow 5 c^{2} = 75$

$\Rightarrow c^{2} = 15$

$\Rightarrow c = \pm \sqrt{15}$

$\Rightarrow c = \sqrt{15} \in (1, 5)$

Hence, the mean value theorem is verified.

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

Ans: Given $y = {(x - 3)}^{2}$

Function is polynomial so, it will be continuous and differentiable for all real x.

Mean value theorem is applicable. Points $(3, 0)$ and $(4, 1)$ are on curve so, there exists a real number $c \in (3, 4)$ such that

$f^{'} (c) = \frac{f (4) - f (3)}{4 - 3}$

$\Rightarrow 2 (c - 3) = \frac{1 - 0}{1} = 1$

$\Rightarrow c - 3 = \frac{1}{2}$

$\Rightarrow c = 3 + \frac{1}{2} = \frac{7}{2}$

$\Rightarrow y = {(\frac{7}{2} - 3)}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4}$

Hence the point is $(\frac{7}{2}, \frac{1}{4})$ .

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

Ans:

Given $y = 2 x^{2} - 5 x + 3$ and points are $A (1, 0) a n d B (2, 1)$ .

y is a polynomial, so it is continuous and differentiable for all real x. Mean value theorem is applicable. Points $(1, 0)$ and $(2, 1)$ are on curve so, there exists a real number $c \in (1, 2)$ such that

$f^{'} (c) = \frac{f (2) - f (1)}{2 - 1}$

$\Rightarrow 4 c - 5 = \frac{1 - 0}{1}$

$\Rightarrow 4 c = 6 \Rightarrow c = \frac{3}{2} \in (1, 2)$

There is a tangent to the curve at $x = \frac{3}{2}$ where the tangent is parallel to the chord AB.

Hence proved.

For the point, $y = 2 \times {(\frac{3}{2})}^{2} - 5 \times \frac{3}{2} + 3 = \frac{9}{2} - \frac{15}{2} + \frac{6}{2} = 0$

Point is $(\frac{3}{2}, 0)$ .

Long Answer (L.A.)

79. Find the values of p and q so that $f (x) = {\begin{aligned} x^{2} + 3 x + p, if x \leq 1 \\ q x + 2, if x > 1 \end{aligned}$ is differentiable at $x = 1$ .

Ans:

Given that $f (x) = {\begin{aligned} x^{2} + 3 x + p, if x \leq 1 \\ q x + 2, if x > 1 \end{aligned}$ at $x = 1$ .

For differentiability function must be continuous. So, for continuity

$L H L = f (x)$

$= \underset{x \to 1^{-}}{l i m} (x^{2} + 3 x + p)$

$L H L = 4 + p$

$R H L = f (x)$

$= \underset{x \to 1^{+}}{l i m} (q x + 2)$

$R H L = q + 2$

$f (1) = p + 4$

$L H L = R H L = f (1)$

$p + 4 = q + 2$

$p = q - 2$

Now for differentiability

$L H D = f^{'} (1^{-}) = {(2 x + 3)}_{a t x = 1} = 5$

$R H D = f^{'} (1^{+}) = q$

For differentiability $L H D = R H D$

$q = 5$

$p = 5 - 2 = 3$

Hence the value of $p = 3$ and $q = 5$ .

80. If $x^{m} y^{n} = {(x + y)}^{m + n}$ , prove that

(i). $\frac{d y}{d x} = \frac{y}{x}$

Ans: Given $x^{m} y^{n} = {(x + y)}^{m + n}$

$x^{m} y^{n} = {(x + y)}^{m + n}$

$\frac{d}{d x} x^{m} y^{n} = \frac{d}{d x} {(x + y)}^{m + n}$

$x^{m} n y^{n - 1} \frac{d y}{d x} + m x^{m - 1} y^{n} = (m + n) (x + y)^{m + n - 1} (1 + \frac{d y}{d x})$

$(x^{m} n y^{n - 1} - (m + n) {(x + y)}^{m + n - 1}) \frac{d y}{d x} = (m + n) {(x + y)}^{m + n - 1} - m x^{m - 1} y^{n}$

$\frac{d y}{d x} = \frac{(m + n) {(x + y)}^{m + n - 1} - m x^{m - 1} y^{n}}{(x^{m} n y^{n - 1} - (m + n) {(x + y)}^{m + n - 1})}$

$= \frac{x^{m} y^{n} (m + n) {(x + y)}^{- 1} - m x^{m - 1} y^{n}}{(x^{m} n y^{n - 1} - x^{m} y^{n} (m + n) {(x + y)}^{- 1})}$

$= \frac{(m + n) {(x + y)}^{- 1} - m x^{- 1}}{(n y^{- 1} - (m + n) {(x + y)}^{- 1})}$

$= \frac{x (m + n) - m (x + y)}{(n (x + y) - (m + n) y)} \cdot \frac{y}{x}$

$= \frac{m x + n x - m x - m y}{n x + n y - m y - n y} \cdot \frac{y}{x}$

$= \frac{n x - m y}{n x - m y} \cdot \frac{y}{x}$

$\frac{d y}{d x} = \frac{y}{x}$

Hence proved.

(ii). $\frac{d^{2} y}{d x^{2}} = 0$

Ans: we have $\frac{d y}{d x} = \frac{y}{x}$

$\frac{d}{d x} \frac{d y}{d x} = \frac{d}{d x} (\frac{y}{x})$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{x \frac{d y}{d x} - y}{x^{2}} = \frac{x \cdot \frac{y}{x} - y}{x^{2}}$

$\frac{d^{2} y}{d x^{2}} = 0$

Hence proved.

81. If $x = s i n t a n d y = s i n p t$ , prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0$ .

Ans: Given that $x = s i n t a n d y = s i n p t$

$\frac{d x}{d t} = c o s t, \frac{d y}{d t} = p c o s p t$

$\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{p c o s p t}{c o s t}$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d t} (\frac{p c o s p t}{c o s t}) \frac{1}{d x / d t}$

$= p \frac{c o s t (- p s i n p t) - c o s p t (- s i n t)}{t} \cdot \frac{1}{c o s t}$

$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} = p (1 - t) \frac{c o s p t s i n t - p s i n p t c o s t}{t}$

$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} = p (c o s p t t a n t - p s i n p t)$

$\Rightarrow x \frac{d y}{d x} = s i n t \frac{p c o s p t}{c o s t} = p c o s p t t a n t$

$\Rightarrow p^{2} y = p^{2} s i n p t$

$L H S = (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y$

$= p (c o s p t t a n t - p s i n p t) - p c o s p t t a n t + p^{2} s i n p t$

$= 0$

Hence proved.

82. Find $\frac{d y}{d x}$ , if $y = x^{t a n x} + \sqrt{\frac{x^{2} + 1}{2}}$

Ans: Given $y = x^{t a n x} + \sqrt{\frac{x^{2} + 1}{2}}$

$p = x^{t a n x}$

Taking log,

$l o g p = t a n x l o g x$

$\Rightarrow \frac{d}{d x} l o g p = \frac{d}{d x} (t a n x l o g x)$

$\Rightarrow \frac{1}{p} \frac{d p}{d x} = \frac{t a n x}{x} + x l o g x$

$\Rightarrow \frac{d}{d x} (x^{t a n x}) = x^{t a n x} (\frac{t a n x}{x} + s e c^{2} x l o g x)$

$\Rightarrow \frac{d}{d x} (\sqrt{\frac{x^{2} + 1}{2}}) = \frac{1}{\sqrt{2}} (\frac{x}{\sqrt{x^{2} + 1}})$

$\Rightarrow \frac{d y}{d x} = \frac{d}{d x} (x^{t a n x}) + \frac{d}{d x} (\sqrt{\frac{x^{2} + 1}{2}})$

$\Rightarrow \frac{d y}{d x} = x^{t a n x} (\frac{t a n x}{x} + x l o g x) + \frac{1}{\sqrt{2}} (\frac{x}{\sqrt{x^{2} + 1}})$

Choose the correct answers from the given four options in each of the Exercises 83 to 96.

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

A. $f (x) + g (x)$

B. $f (x) - g (x)$

C. $f (x) \cdot g (x)$

D. $\frac{g (x)}{f (x)}$

Ans: The correct answer is option (D).

Given that $f (x) = 2 x$ and $g (x) = \frac{x^{2}}{2} + 1$ .

We can observe that both the functions are continuous so,

$f + g, f - g a n d f \cdot g$ are continuous but $\frac{g}{f}$ is not continuous at the point where f is zero.

Hence $\frac{g (x)}{f (x)}$ can be discontinuous.

Option (D) is correct.

84. The function $f (x) = \frac{4 - x^{2}}{4 x - x^{3}}$

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 $f (x) = \frac{4 - x^{2}}{4 x - x^{3}}$

For points of discontinuity $4 x - x^{3} = 0$

$x (4 - x^{2}) = 0$

$\Rightarrow x (2 - x) (2 + x) = 0$

$\Rightarrow x = 0, - 2, 2$

Hence $f (x)$ is discontinuous at exactly three points.

Option (C) is correct.

85. The set of points where the function f given by $f (x) = | 2 x - 1 | s i n x$ is differentiable is

A. $R$

B. $R - {\frac{1}{2}}$

C. $(0, \infty)$

D. None of these.

Ans: The correct answer is option B.

Given function is $f (x) = | 2 x - 1 | s i n x$

$f (x) = {\begin{aligned} - (2 x - 1) s i n x, i f x \leq \frac{1}{2} \\ (2 x - 1) s i n x, i f x \geq \frac{1}{2} \end{aligned}$

We know that modulus function is continuous everywhere but not differentiable at $2 x - 1 = 0 \Rightarrow x = \frac{1}{2}$

Hence, $f (x)$ is differentiable in $R - {\frac{1}{2}}$ .

86. The function $f (x) = c o t x$ is discontinuous on the set

A. ${x = n π : n \in Z}$

B. ${x = 2 n π : n \in Z}$

C. ${x = (2 n + 1) \frac{π}{2} : n \in Z}$

D. ${x = \frac{n π}{2} : n \in Z}$

Ans: The correct answer is option A.

Given $f (x) = c o t x$

We know that $f (x) = c o t x$ is continuous at $x \in R - {n π, n \in Z}$

Hence, $f (x) = c o t x$ is discontinuous at ${x = n π : n \in Z}$ .

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

A. Continuous everywhere but not differentiable at $x = 0$

B. Continuous and differentiable everywhere

C. Not continuous at $x = 0$

D. None of these

Ans: The correct answer is option (A).

Given $f (x) = e^{| x |}$

$f (x) = e^{| x |} = {\begin{aligned} e^{- x}, if x \leq 0 \\ e^{x}, if x \geq 0 \end{aligned}$

$L H L = e^{0} = 1$

$R H L = e^{0} = 1$

$L H L = R H L = f (0)$ , function is continuous.

$f (x) = e^{| x |} = {\begin{aligned} e^{- x}, if x \leq 0 \\ e^{x}, if x \geq 0 \end{aligned}$

$L H D = - e^{0} = - 1$

$R H D = e^{0} = 1$

$L H D \neq R H D$ , function is not differentiable.

Hence continuous everywhere but not differentiable at $x = 0$

Option (A) is correct.

88. If $f (x) = x^{2} s i n \frac{1}{x}$ , where $x \neq 0$ , 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 $f (x) = x^{2} s i n \frac{1}{x}$ , where $x \neq 0$

$f (x) = \underset{x \to 0}{l i m} x^{2} s i n \frac{1}{x}$

$= 0 \times (number varying between -1 to 1)$

$= 0$

Hence the value of $f (x)$ at $x = 0$ , so that $f (x)$ be continuous at $x = 0$ , is 0.

Option (A) is correct.

89. If $f (x) = {\begin{aligned} m x + 1, i f x \leq \frac{π}{2} \\ s i n x + n, i f x > \frac{π}{2} \end{aligned}$ is continuous at $x = \frac{π}{2}$ then

A. $m = 1, n = 0$

B. $m = \frac{n π}{2} + 1$

C. $n = \frac{m π}{2}$

D. $m = n = \frac{π}{2}$

Ans: The correct answer is option (C).

Given $f (x) = {\begin{aligned} m x + 1, i f x \leq \frac{π}{2} \\ s i n x + n, i f x > \frac{π}{2} \end{aligned}$

$L H L = f (x)$

$= \underset{x \to {\frac{π}{2}}^{-}}{l i m} (m x + 1)$

$= \frac{m π}{2} + 1$

$R H L = \underset{x \to {\frac{π}{2}}^{+}}{l i m} f (x)$

$= \underset{x \to {\frac{π}{2}}^{+}}{l i m} (s i n x + n)$

$= n + 1$

For $f (x)$ to be continuous

$L H L = R H L$

$\frac{m π}{2} + 1 = n + 1$

$n = \frac{m π}{2}$

Option (C) is correct.

90. Let $f (x) = | s i n x |$ , then

A. f is everywhere differentiable

B. f is everywhere continuous but not differentiable at $x = n π, n \in Z$ .

C. f is everywhere continuous but not differentiable at $x = (2 n + 1) \frac{π}{2}, n \in Z$ .

D. None of these.

Ans: The correct answer is option (B).

Given $f (x) = | s i n x |$

$f (x) = | s i n x | = {\begin{aligned} - s i n x, if x \leq 0 \\ s i n x, i f x \geq 0 \end{aligned}$

Here we observe that $f$ is continuous.

$f (x) = {\begin{aligned} - c o s x, i f s i n x \leq 0 \\ c o s x, i f s i n x \geq 0 \end{aligned}$

At $x = n π, n \in Z$

$L H D = f^{'} (n π) = - c o s n π = - {(- 1)}^{n}$

$R H D = f^{'} (n π) = c o s n π = {(- 1)}^{n}$

$L H D \neq R H D$

Hence $f$ is continuous but not differentiable at $x = n π, n \in Z$ .

Option (B) is correct.

91. If $y = l o g (\frac{1 - x^{2}}{1 + x^{2}})$ then $\frac{d y}{d x}$ is equal to

A. $\frac{4 x^{3}}{1 - x^{4}}$

B. $\frac{- 4 x}{1 - x^{4}}$

C. $\frac{1}{4 - x^{4}}$

D. $\frac{- 4 x^{3}}{1 - x^{4}}$

Ans: The correct answer is option (B).

Given $y = l o g (\frac{1 - x^{2}}{1 + x^{2}}) = l o g (1 - x^{2}) - l o g (1 + x^{2})$

$\frac{d y}{d x} = \frac{d}{d x} (l o g (1 - x^{2})) - \frac{d}{d x} (l o g (1 + x^{2}))$ $= \frac{- 2 x}{1 - x^{2}} - \frac{2 x}{1 + x^{2}}$

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

$= - \frac{4 x}{1 - x^{4}}$

Option (B) is correct.

92. If $y = \sqrt{s i n x + y}$ , then $\frac{d y}{d x}$ is equal to

A. $\frac{c o s x}{2 y - 1}$

B. $\frac{c o s x}{1 - 2 y}$

C. $\frac{s i n x}{1 - 2 y}$

D. $\frac{s i n x}{2 y - 1}$

Ans: The correct answer is option (A).

Given $y = \sqrt{s i n x + y}$

$y^{2} = s i n x + y$

$\Rightarrow y^{2} - y = s i n x$

$\Rightarrow \frac{d}{d x} (y^{2} - y) = \frac{d}{d x} s i n x$

$\Rightarrow (2 y - 1) \frac{d y}{d x} = c o s x$

$\Rightarrow \frac{d y}{d x} = \frac{c o s x}{2 y - 1}$

Option (A) is correct.

93. The derivative of $c o s^{- 1} (2 x^{2} - 1)$ w.r.t $c o s^{- 1} x$ .

A. 2

B. $\frac{- 1}{2 \sqrt{1 - x^{2}}}$

C. $\frac{2}{x}$

D. $1 - x^{2}$

Ans: The correct answer is option (A)

We must find derivative of $c o s^{- 1} (2 x^{2} - 1)$ w.r.to $c o s^{- 1} x$

Let $t = c o s^{- 1} x \Rightarrow x = c o s t$

$c o s^{- 1} (2 x^{2} - 1) = c o s^{- 1} (2 c o s^{2} t - 1)$

$= c o s^{- 1} (c o s 2 t)$

$c o s^{- 1} (2 x^{2} - 1) = 2 t$

$\frac{d c o s^{- 1} (2 x^{2} - 1)}{d c o s^{- 1} x} = \frac{d (2 t)}{d t} = 2$

Option (A) is correct.

94. If $x = t^{2}, y = t^{3}$ , then $\frac{d^{2} y}{d x^{2}}$ is

A. $\frac{3}{2}$

B. $\frac{3}{4 t}$

C. $\frac{3}{2 t}$

D. $\frac{3}{4}$

Ans: The correct answer is option (B).

Given $x = t^{2}, y = t^{3}$

$\frac{d x}{d t} = 2 t, \frac{d y}{d t} = 3 t^{2}$

$\Rightarrow \frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{3 t^{2}}{2 t} = \frac{3 t}{2}$

$\Rightarrow \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\frac{3 t}{2}) = \frac{d}{d t} (\frac{3 t}{2}) \cdot \frac{1}{d x / d t}$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{3}{2} \times \frac{1}{2 t} = \frac{3}{4 t}$

Option (B) is correct.

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

A. 1

B. $- 1$

C. $\frac{3}{2}$

D. $\frac{1}{3}$

Ans: The correct answer is option (A).

Given that $f (x) = x^{3} - 3 x$

Function is continuous and differentiable for all real x and $f (0) = 0 = f (\sqrt{3})$ .

So, Rolle’s theorem is applicable here.

$f^{'} (x) = 3 x^{2} - 3$

$\Rightarrow f^{'} (c) = 0 \Rightarrow 3 (c^{2} - 1) = 0$

$\Rightarrow c^{2} = 1 \Rightarrow c = \pm 1$

$\Rightarrow c = 1 \in (0, \sqrt{3})$

Option (A) is correct.

96. For the function $f (x) = x + \frac{1}{x}, x \in [1, 3]$ , 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 $f (x) = x + \frac{1}{x}$ is continuous in $x \in [1, 3]$

$f^{'} (x) = 1 - \frac{1}{x^{2}}$ is differentiable in $x \in (1, 3)$

Mean value theorem is applicable here.

$f^{'} (c) = 1 - \frac{1}{c^{2}} = \frac{f (3) - f (1)}{3 - 1}$

$\Rightarrow 1 - \frac{1}{c^{2}} = \frac{3 + \frac{1}{3} - 1 - 1}{2}$

$\Rightarrow \frac{1}{c^{2}} = 1 - \frac{\frac{4}{3}}{2} = \frac{2 - \frac{4}{3}}{2} = \frac{2}{6} = \frac{1}{3}$

$\Rightarrow c^{2} = 3 \Rightarrow c = \pm \sqrt{3}$

$\Rightarrow c = \sqrt{3} \in (1, 3)$

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 $| x - 1 | + | x - 2 |$ 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 $x^{3}$

$\frac{d x^{2}}{d x^{3}} = \frac{\frac{d x^{2}}{d x}}{\frac{d x^{3}}{d x}}$

$= \frac{2 x}{3 x^{2}}$

$= \frac{2}{3 x}$

99. If $f (x) = | c o s x |$ , then $f^{'} (\frac{π}{4}) =__________$ .

Ans: Given $f (x) = | c o s x |$

At $x = \frac{π}{4}, f (x) = c o s x$

$\Rightarrow f^{'} (x) = - s i n x$

$\Rightarrow f^{'} (\frac{π}{4}) = - s i n (\frac{π}{4}) = - \frac{1}{\sqrt{2}}$

100. If $f (x) = | c o s x - s i n x |$ , then $f^{'} (\frac{π}{3}) =_________$ .

Ans: Given $f (x) = | c o s x - s i n x |$

$f (x) = | c o s x - s i n x |$

$= {\begin{aligned} s i n x - c o s x, i f s i n x \geq c o s x \\ - s i n x + c o s x, i f s i n x \leq c o s x \end{aligned}$

At $x = \frac{π}{3}, s i n x \geq c o s x$

$\Rightarrow f (x) = s i n x - c o s x$

$\Rightarrow f^{'} (x) = c o s x + s i n x$

$\Rightarrow f^{'} (\frac{π}{3}) = c o s \frac{π}{3} + s i n \frac{π}{3} = \frac{1}{2} + \frac{\sqrt{3}}{2}$

$\Rightarrow f^{'} (\frac{π}{3}) = \frac{1 + \sqrt{3}}{2}$

101. For the curve $\sqrt{x} + \sqrt{y} = 1, \frac{d y}{d x} a t (\frac{1}{4}, \frac{1}{4})$ is ________.

Ans: Given curve is $\sqrt{x} + \sqrt{y} = 1$

$\frac{d}{d x} (\sqrt{x} + \sqrt{y}) = \frac{d}{d x} 1$

$\Rightarrow \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{\sqrt{y}}{\sqrt{x}}$

At $(\frac{1}{4}, \frac{1}{4})$ , $\frac{d y}{d x} = - 1$

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 $f (x) = | x - 1 | i n [0, 2]$ .

Ans: Given function $f (x) = | x - 1 |$ is continuous in $[0, 2]$ and not differentiable at $x = 1 \in (0, 2)$ .

Hence the given statement is False.

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

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

NCERT Exemplar for Class 12 Maths - Continuity And Differentiability - Free PDF Download

Access NCERT Exemplar Solutions for Class 12 Mathematics Chapter 5 - Continuity and Differentiability

Solved Examples

Short Answer Questions

Long Answers (L.A.)

Objective Type Questions

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

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

Exercise

Short Answer Type Questions

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

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

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

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

Find $\frac{d y}{d x}$ of each of the functions expressed in parametric form in Exercises from 44 to 48.

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

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

Verify mean value theorem for each of the functions given Exercises 73 to 76.

Long Answer (L.A.)

Choose the correct answers from the given four options in each of the Exercises 83 to 96.

Fill in the blanks in each of the exercises 97 to 101:

State True or False for the following statements in each of the Exercise 102 to 106.

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

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

NCERT Exemplar for Class 12 Maths - Continuity And Differentiability - Free PDF Download

Access NCERT Exemplar Solutions for Class 12 Mathematics Chapter 5 - Continuity and Differentiability

Solved Examples

Short Answer Questions

Long Answers (L.A.)

Objective Type Questions

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

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

Exercise

Short Answer Type Questions

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

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

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

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

Find dydx of each of the functions expressed in parametric form in Exercises from 44 to 48.

Find dydx when x and y are connected by the relation given in each of the exercise 54 to 57.

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

Verify mean value theorem for each of the functions given Exercises 73 to 76.

Long Answer (L.A.)

Choose the correct answers from the given four options in each of the Exercises 83 to 96.

Fill in the blanks in each of the exercises 97 to 101:

State True or False for the following statements in each of the Exercise 102 to 106.

The Various Sub Topics Covered in the NCERT Exemplar for Class 12 Maths Chapter 5 - Continuity and Differentiability (Book Solutions) Includes

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

Find $\frac{d y}{d x}$ of each of the functions expressed in parametric form in Exercises from 44 to 48.

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