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Questions:

Case Study Questions

5 x 4 = 20

If a real valued function f(x) is finitely derivable at any point of its domain, it is necessarily continuous at that point. But its converse need not be true.
For example, every polynomial. constant function are both continuous as well as differentiable and inverse trigonometric functions are continuous and differentiable in its domains etc.
Based on the above information, answer the following questions.
(i) If \(\begin{equation} f(x)=\left\{\begin{array}{l} x, \text { for } x \leq 0 \\ 0, \text { for } x>0 \end{array}\right. \end{equation}\) , then at x = 0

(a) f(x) is differentiable and continuous	(b) j(x) is neither continuous nor differentiable
(c) f(x) is continuous but not differentiable	(d) none of these

(ii) If \(\begin{equation} f(x)=|x-1|, x \in R \end{equation}\) ,then at x= 1

(a) f(x) is not continuous	(b) f(x) is continuous but not differentiable
(c) f(x) is continuous and differentiable	(d) none of these

(iii) f(x) = x³ is

(a) continuous but not differentiable at x = 3	(b) continuous and differentiable at x = 3
(c) neither continuous nor differentiable at x = 3	(d) none of these

(iv) f(x) = [sin x], then which of the following is true?

(a) j(x) is continuous and differentiable at x = o.	(b) j(x) is discontinuous at x = o.
(c) j(x) is continuous at x = 0 but not differentiable	(d) fix) is differentiable but not continuous at \(\begin{equation} x=\pi / 2 \end{equation}\)

(v) If f(x) = sin^-1x, \(\begin{equation} -1 \leq x \leq 1 \end{equation}\), then

(a) f(x) is both continuous and differentiable	(b) f(x) is neither continuous nor differentiable.
(c) f(x) is continuous but not differentiable	(d) None of these

The function f(x) will be discontinuous at x = a if f(x) has
(a) Discontinuity of first kind \(\begin{equation} \lim _{h \rightarrow 0} f(a-h) \text { and } \lim _{h \rightarrow 0} f(a+h) \end{equation}\) both exist but are not equal. If is also known as irremovable discontinuity.
(b) Discontinuity of second kind: If none of the limits \(\begin{equation} \lim _{h \rightarrow 0} f(a-h) \text { and } \lim _{h \rightarrow 0} f(a+h) \end{equation}\) exist.
(c) Removable discontinuity: \(\begin{equation} \lim _{h \rightarrow 0} f(a-h) \text { and } \lim _{h \rightarrow 0} f(a+h) \end{equation}\) both exist and equal but not equal to f(a).
Based on the above information, answer the following questions.
(i) If \(\begin{equation} f(x)=\left\{\begin{array}{ll} \frac{x^{2}-9}{x-3}, & \text { for } x \neq 3 \\ 4, & \text { for } x=3 \end{array}\right. \end{equation}\) ,then at x= 3

(a) f has removable discontinuity	(b) f is continuous
(c) f has irremovable discontinuity	(d) none of these

(ii) Let \(\begin{equation} f(x)=\left\{\begin{array}{ll} x+2, & \text { if } x \leq 4 \\ x+4, & \text { if } x>4 \end{array}\right. \end{equation}\) ,then at x = 4

(a) f is continuous	(b) f has removable discontinuity
(c) f has irremovable discontinuity	(d) none of these

(iii) Consider the function f(x) defined \(\begin{equation} f(x)=\left\{\begin{array}{l} \frac{x^{2}-4}{x-2} \\ 5 \end{array}\right. \end{equation}\), for \(\begin{equation} x \neq 2 \end{equation}\)

(a) f has removable discontinuity	(b) f has irremovable discontinuity
(c) f is continuous	(d) f is continuous if f(2) = 3

(iv) If \(\begin{equation} f(x)=\left\{\begin{array}{cc} \frac{x-|x|}{x}, & if\ x \neq 0 \\ 2, & if\ x=0 \end{array}\right. \end{equation}\) ,then x = 0

a) f is continuous	(b) f has removable discontinuity
(c) f has irremovable discontinuity	(d) none of these

(v) If \(\begin{equation} f^{\prime}(x)=\left\{\begin{array}{cl} \frac{e^{x}-1}{\log (1+2 x)}, & \text { if } x \neq 0 \\ 7, & \text { if } x=0 \end{array}\right. \end{equation}\), then at x = 0

(a) f is continuous if f(0) = 2	(b) f is continuous
(c) f has irremovable discontinuity	(d) f has removable discontinuity

Let \(\begin{equation} f: A \rightarrow B \end{equation}\) and \(\begin{equation} g: B \rightarrow C \end{equation}\) be two functions defined on non-empty sets A, B, C,
then \(\begin{equation} \text { gof }: A \rightarrow C \end{equation}\) be is called the composition off and g defined as, \(\begin{equation} g o f(x)=g\{f(x)\} \forall x \in A \end{equation}\) .
Consider the functions \(\begin{equation} f(x)=\left\{\begin{array}{ll} \sin x, & x \geq 0 \\ 1-\cos x, & x \leq 0 \end{array}, g(x)=e^{x}\right. \end{equation}\) and
then answer the following questions.
(i) The function gof(x) is defined as

(a) \(\begin{equation} g o f(x)=\left\{\begin{array}{ll} e^{x} & , x \geq 0 \\ 1-e^{\cos x} & , x \leq 0 \end{array}\right. \end{equation}\)	(b) \(\begin{equation} \operatorname{gof}(x)=\left\{\begin{array}{ll} e^{\sin x} & , x \leq 0 \\ e^{1-\cos x} & , x \geq 0 \end{array}\right. \end{equation}\)
(c) \(\begin{equation} g o f(x)=\left\{\begin{array}{ll} e^{\sin x} & , x \leq 0 \\ 1-e^{\cos x} & , x \geq 0 \end{array}\right. \end{equation}\)	(d) \(\begin{equation} g o f(x)=\left\{\begin{array}{ll} e^{\sin x} & , x \geq 0 \\ e^{1-\cos x} & , x \leq 0 \end{array}\right. \end{equation}\)

(ii) \(\begin{equation} \frac{d}{d x}\{\operatorname{gof}(x)\}= \end{equation}\)

(a) \(\begin{equation} [g o f(x)]^{\prime}=\left\{\begin{array}{ll} \cos x \cdot e^{\sin x} & , x \geq 0 \\ e^{1-\cos x} \cdot \sin x & , x \leq 0 \end{array}\right. \end{equation}\)	(b) \(\begin{equation} [g o f(x)]^{\prime}=\left\{\begin{array}{ll} \cos x \cdot e^{\sin x} & , x \geq 0 \\ -\sin x \cdot e^{1-\cos x} & , x \leq 0 \end{array}\right. \end{equation}\)
(c) \(\begin{equation} [g o f(x)]^{\prime}=\left\{\begin{array}{ll} \cos x \cdot e^{\sin x} & , x \geq 0 \\ \sin x \cdot(1-\cos x) & , x \leq 0 \end{array}\right. \end{equation}\)	(d) \(\begin{equation} [g o f(x)]^{\prime}=\left\{\begin{array}{ll} \cos x \cdot e^{\sin x} & , x \geq 0 \\ (1-\sin x) \cdot e^{1-\cos x} & , x \leq 0 \end{array}\right. \end{equation}\)

(iii) R.H.D. of gof(x) at x = 0 is

(a) 0

(b) 1

(c) -1

(d) 2

(iv) L.H.D. of gof(x) at x = 0 is

(a) 0

(b) 1

(c) -1

(d) 2

(v) The value of \(\begin{equation} f^{\prime}(x) \text { at } x=\frac{\pi}{4} \end{equation}\) is

(a) 1/9

(b) \(\begin{equation} 1 / \sqrt{2} \end{equation}\)

(c) 1/2

(d) not defined

Let x = f(t) and y = get) be parametric forms with t as a parameter,
then \(\begin{equation} \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{g^{\prime}(t)}{f^{\prime}(t)} \end{equation}\) ,where \(\begin{equation} f^{\prime}(t) \neq 0 \end{equation}\) \(\begin{equation} \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{g^{\prime}(t)}{f^{\prime}(t)} \end{equation}\) where \(\begin{equation} f^{\prime}(t) \neq 0 \end{equation}\).
(i) The derivative off (tanx) w.r.t. \(\begin{equation} g(\sec x) \text { at } x=\frac{\pi}{4} \end{equation}\) ,where f'(1) and \(\begin{equation} g^{\prime}(\sqrt{2})=4 \end{equation}\) is

(a) \(\begin{equation} \frac{1}{\sqrt{2}} \end{equation}\)

(b) \(\begin{equation} \sqrt{2} \end{equation}\)

(c) 1

(d) 0

(ii) The derivate of \(\begin{equation} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \end{equation}\) ,with respect to \(\begin{equation} \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \end{equation}\) is

(a) -1

(b) 1

(c) 2

(d) 4

(iii) The derivative of \(\begin{equation} e^{x^{3}} \end{equation}\) with respect to log x is

(a) \(\begin{equation} e^{x^{3}} \end{equation}\)

(b) \(\begin{equation} 3 x^{2} 2 e^{x^{3}} \end{equation}\)

(c) \(\begin{equation} 3 x^{3} e^{x^{3}} \end{equation}\)

(d) \(\begin{equation} 3 x^{2} e^{x^{3}}+3 x \end{equation}\)

(iv) The derivative of \(\begin{equation} \cos ^{-1}\left(2 x^{2}-1\right) \end{equation}\) w.r.t. cos^-1x is

(a) 2

(b) \(\begin{equation} \frac{-1}{2 \sqrt{1-x^{2}}} \end{equation}\)

(c) \(\begin{equation} \frac{2}{x} \end{equation}\)

(d) 1 -x²

(v) If \(\begin{equation} y=\frac{1}{4} u^{4} \end{equation}\) and \(\begin{equation} u=\frac{2}{3} x^{3}+5 \end{equation}\) then \(\begin{equation} \frac{d y}{d x}= \end{equation}\)

(a) \(\begin{equation} \frac{2}{27} x^{2}\left(2 x^{3}+15\right)^{3} \end{equation}\)

(b) \(\begin{equation} \frac{2}{7} x^{2}\left(2 x^{3}+15\right)^{3} \end{equation}\)

(c) \(\begin{equation} \frac{2}{27} x\left(2 x^{3}+5\right)^{3} \end{equation}\)

(d) \(\begin{equation} \frac{2}{7}\left(2 x^{3}+15\right)^{3} \end{equation}\)

Let f(x) be a real valued function, then its
Left Hand Derivative (L.H.D.) : \(\begin{equation} \mathrm{L} f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h} \end{equation}\)
Right Hand Derivative (R.H.D.) : \(\begin{equation} \mathrm{Rf}^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \end{equation}\)
Also, a function jfx) is said to be differentiable at x = a if its L.H.D. and R.H.D. at x = a exist and are equal
For the function \(\begin{equation} f(x)=\left\{\begin{array}{l} |x-3|, x \geq 1 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4}, x<1 \end{array}\right. \end{equation}\) answer the following questions
(i) R.H.D. of f(x) at x = 1is

(a) 1

(b) -1

(c) 0

(d) 2

(ii) L.H.D. of f(x) at x = 1 is

(a) 1

(b) -1

(c) 0

(d) 2

(iii) f(x) is non-differentiable at

(a) x = 1

(b) x = 2

(c) x = 3

(d) x = 4

(iv) Find the value of f'(2).

(a) 1

(b) 2

(c) 3

(d) -1

(v) The value of f'( -1) is

(a) 2

(b) 1

(c) -2

(d) -1

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Answers Key:

Case Study Questions

5 x 4 = 20

(i) (c)
(ii) (b)
(iii) (b)
(iv) (b)
(v) (a)

(i) (a) : f(3) = 4
\(\begin{equation} \lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}=\lim _{x \rightarrow 3} \frac{(x+3)(x-3)}{(x-3)} \end{equation}\)
\(\begin{equation} =\lim _{x \rightarrow 3}(x+3)=6 \because \lim _{x \rightarrow 3} f(x) \neq f(3) \end{equation}\)
\(\therefore\) f(x) has removable discontinuity at x = 3.
(ii) (c) : \(\begin{equation} \lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4}(x+2)=4+2=6 \end{equation}\)
\(\begin{equation} \lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4}(x+4)=4+4=8 \end{equation}\)
\(\begin{equation} \therefore \quad \lim _{x \rightarrow 4^{-}} f(x) \neq \lim _{x \rightarrow 4^{+}} f(x) \end{equation}\)
\(\begin{equation} \therefore f(x) \end{equation}\) has an irremovable discontinuity at x = 4.
(iii) (a) : \(\begin{equation} \lim _{x \rightarrow 2} f(x)=\lim _{x \rightarrow 2} \frac{\left(x^{-4}-4\right)}{(x-2)}=\lim _{x \rightarrow 2}(x+2)=4 \end{equation}\)
\(\begin{equation} \therefore f(x) \end{equation}\) has removable discontinuity at x = 2.
(iv) (c) : f(0)=2
\(\begin{equation} \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{x+x}{x}=2 \end{equation}\)
\(\begin{equation} \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} \frac{x-x}{x}=0 \end{equation}\)
\(\begin{equation} \because \lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x) \end{equation}\)
\(\begin{equation} \therefore f(x) \end{equation}\) has an irremovable discontinuity at x = 0.
(v) (d) : f(0) = 7
\(\begin{equation} \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{e^{x}-1}{\log (1+2 x)}=\lim _{x \rightarrow 0} \frac{\frac{\left(\frac{e^{x}-1}{x}\right)}{\log (1+2 x)}{2 x} \cdot 2}=\frac{1}{2} \end{equation}\)
\(\begin{equation} \because \ \lim _{x \rightarrow 0} f(x) \neq f(0) \end{equation}\)
\(\begin{equation} \therefore f(x) \end{equation}\) has removable discontinuity at x = 0.

(i) (d)
(ii) (a)
(iii) (b) ,
(iv) (a)
(v) (b)

(i) (a) : Now, \(\begin{equation} \frac{d f(\tan x)}{d g(\sec x)}=\frac{f^{\prime}(\tan x) \sec ^{2} x}{g^{\prime}(\sec x) \sec x \tan x} \end{equation}\)
\(\begin{equation} =\frac{f^{\prime}(\tan x) \sec x}{g^{\prime}(\sec x) \tan x} \end{equation}\)
\(\begin{equation} \therefore\left[\frac{d f(\tan x)}{d g(\sec x)}\right]_{x=\pi / 4}=\frac{f^{\prime}(1) \sqrt{2}}{g^{\prime}(\sqrt{2}) \cdot 1}=\frac{2 \sqrt{2}}{4 \cdot 1}=\frac{1}{\sqrt{2}} \end{equation}\)
(ii) (b)
(iii) (c) : Let \(\begin{equation} y=e^{x^{3}}, z=\log x \end{equation}\)
Differentiating w.r.t. x, we get
\(\begin{equation} \frac{d y}{d x}=e^{x^{3}}\left(3 x^{2}\right)=3 x^{2} e^{x^{3}} \end{equation}\) and \(\begin{equation} \frac{d z}{d x}=\frac{1}{x} \end{equation}\)
\(\begin{equation} \therefore \ \frac{d y}{d z}=\frac{\frac{d x}{d x}}{\frac{d z}{d x}}=\frac{3 x^{2} e^{x^{3}}}{\left(\frac{1}{x}\right)}=3 x^{3} e^{x^{3}} \end{equation}\)
(iv) (a): Let y = cos^-1(2x² - 1) = 2cos^-1x
Differentiating w.r.t. cos'" x, we get
\(\begin{equation} \frac{d y}{d\left(\cos ^{-1} x\right)}=\frac{2 d\left(\cos ^{-1} x\right)}{d\left(\cos ^{-1} x\right)}=2 \end{equation}\)
(v) (a) : We have \(\begin{equation} y=\frac{1}{4} u^{4} \Rightarrow \frac{d y}{d u}=\frac{1}{4} \cdot 4 u^{3}=u^{3} \end{equation}\)
and \(\begin{equation} u=\frac{2}{3} x^{3}+5 \Rightarrow \frac{d u}{d x}=\frac{2}{3} \cdot 3 x^{2}=2 x^{2} \end{equation}\)
\(\begin{equation} \therefore \ \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}=u^{3} \cdot 2 x^{2}=\left(\frac{2}{3} x^{3}+5\right)^{3}\left(2 x^{2}\right) \end{equation}\)
\(\begin{equation} =\frac{2}{27} x^{2}\left(2 x^{3}+15\right)^{3} \end{equation}\)

we have,\(\begin{equation} f(x)=\left\{\begin{array}{ll} x-3 & , x \geq 3 \\ 3-x & , 1 \leq x<3 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4} & , x<1 \end{array}\right. \end{equation}\)
(i) (b) : \(\begin{equation} \mathrm{R} f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \end{equation}\)
\(\begin{equation} =\lim _{h \rightarrow 0} \frac{3-(1+h)-2}{h}=\lim _{h \rightarrow 0}-\frac{h}{h}=-1 \end{equation}\)
(ii) (b) : \(\begin{equation} \mathrm{L}_{\mathrm{s}}^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \end{equation}\)
\(\begin{equation} =\lim _{h \rightarrow 0} \frac{-1}{h}\left[\frac{(1-h)^{2}}{4}-\frac{3(1-h)}{2}+\frac{13}{4}-2\right] \end{equation}\)
\(\begin{equation} =\lim _{h \rightarrow 0}\left(\frac{1+h^{2}-2 h-6+6 h+13-8}{-4 h}\right) \end{equation}\)
\(\begin{equation} =\lim _{h \rightarrow 0}\left(\frac{h^{2}+4 h}{-4 h}\right)=-1 \end{equation}\)
(iii) (c) : Since, R.H.D. at x = 3 is 1 and L.H.D. at x = 3 is-1
\(\therefore\) f(x) is non-differentiable at x = 3.
(iv) (d)
(v) (c) : From above, we have
\(\begin{equation} f^{\prime}(x)=\frac{x}{2}-\frac{3}{2}, x<1 \end{equation}\)
\(\begin{equation} \therefore f^{\prime}(-1)=\frac{-1}{2}-\frac{3}{2}=-2 \end{equation}\)