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Questions:

Case Study Questions

5 x 4 = 20

If a relation between x and y is such that y cannot be expressed in terms of x, then y is called an implicit function of x.
When a given relation expresses y as an implicit function of x and we want to find \(\begin{equation} \frac{d y}{d x} \end{equation}\).then
we differentiate every term of the given relation w.r.t. x. remembering that a term in y is first differentiated w.r.t. y and then multiplied by \(\begin{equation} \frac{d y}{d x} \end{equation}\).
Based on the above information, find the value of \(\begin{equation} \frac{d y}{d x} \end{equation}\) in each of the following questions
(i) x³+x²y+xy²+y3=81

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

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

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

(ii) x^y = c- y

(a) \(\begin{equation} \frac{x-y}{(1+\log x)} \end{equation}\)

(b) \(\begin{equation} \frac{x+y}{(1+\log x)} \end{equation}\)

(d) \(\begin{equation} \frac{x+y}{x(1+\log x)} \end{equation}\)

(iii) e^siny = xy

(a) \(\begin{equation} \frac{-y}{x(y \cos y-1)} \end{equation}\)

(b) \(\begin{equation} \frac{y}{y \cos y-1} \end{equation}\)

(d) \(\begin{equation} \frac{y}{x(y \cos y-1)} \end{equation}\)

(iv) sin²x + cos²y = 1

(a) \(\begin{equation} \frac{\sin 2 y}{\sin 2 x} \end{equation}\)

(b) \(\begin{equation} -\frac{\sin 2 x}{\sin 2 y} \end{equation}\)

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

(d) \(\begin{equation} \frac{\sin 2 x}{\sin 2 y} \end{equation}\)

(v) \(\begin{equation} y=(\sqrt{x})^{\sqrt{x}} \end{equation}\)

(a) \(\begin{equation} \frac{-y^{2}}{x(2-y \log x)} \end{equation}\)

(b) \(\begin{equation} \frac{y^{2}}{2+y \log x} \end{equation}\)

(c) \(\begin{equation} \frac{y^{2}}{x(2+y \log x)} \end{equation}\)

(d) \(\begin{equation} \frac{y^{2}}{x(2-y \log x)} \end{equation}\)

If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then y = f[g(x)] is a differentiable function of x and \(\begin{equation} \frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x} \end{equation}\). This rule is also known as CHAIN RULE.
Based on the above information, find the derivative of functions w.r.t. x in the following questions
(i) \(\begin{equation} \cos \sqrt{x} \end{equation}\)

(a) \(\begin{equation} \frac{-\sin \sqrt{x}}{2 \sqrt{x}} \end{equation}\)

(b) \(\begin{equation} \frac{\sin \sqrt{x}}{2 \sqrt{x}} \end{equation}\)

(c) \(\begin{equation} \sin \sqrt{x} \end{equation}\)

(d) \(\begin{equation} -\sin \sqrt{x} \end{equation}\)

(ii) \(\begin{equation} 7^{x+\frac{1}{x}} \end{equation}\)

(a) \(\begin{equation} \left(\frac{x^{2}-1}{x^{2}}\right) \cdot 7^{x+\frac{1}{x}} \cdot \log 7 \end{equation}\)

(b) \(\begin{equation} \left(\frac{x^{2}+1}{x^{2}}\right) \cdot 7^{x+\frac{1}{x}} \cdot \log 7 \end{equation}\)

(c) \(\begin{equation} \left(\frac{x^{2}-1}{x^{2}}\right) \cdot 7^{x-\frac{1}{x}} \cdot \log 7 \end{equation}\)

(d) \(\begin{equation} \left(\frac{x^{2}+1}{x^{2}}\right) \cdot 7^{x-\frac{1}{x}} \cdot \log 7 \end{equation}\)

(iii) \(\begin{equation} \sqrt{\frac{1-\cos x}{1+\cos x}} \end{equation}\)

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

(b) \(\begin{equation} \frac{1}{x^{2}+b^{2}}+\frac{1}{x^{2}+a^{2}} \end{equation}\)

(c) \(\begin{equation} \frac{1}{x^{2}+b^{2}}-\frac{1}{x^{2}+a^{2}} \end{equation}\)

(d) none of these

(v) (d) :\(\begin{equation} \sec ^{-1} x+\operatorname{cosec}^{-1} \frac{x}{\sqrt{x^{2}-1}} \end{equation}\)

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

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

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

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

(a) A function f(x) is said to be continuous in an open interval (a, b), if it is continuous at every point in this interval.
(b) A function f(x) is said to be continuous in the closed interval [a, b], if f(x) is continuous in (a, b) and \(\begin{equation} \lim _{h \rightarrow 0} f(a+h)=f(a) \text { and } \lim _{h \rightarrow 0} f(b-h)=f(b) \end{equation}\)
If function \(\begin{equation} f(x)=\left\{\begin{array}{ll} \frac{\sin (a+1) x+\sin x}{x} & , x<0 \\ c & , x=0 \\ \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x^{3 / 2}} & , x>0 \end{array}\right. \end{equation}\) is continuous at x = 0, then answer the following questions.
(i) The value of a is

(a) -3/2

(b) 0

(c) 1/2

(d) -1/2

(ii) The value of b is

(a) 1

(b) -1

(c) 0

(d) any real number

(iii) The value of c is

(a) 1

(b) 1/2

(c) -1

(d) -1/2

(iv) The value of a + c is

(a) 1

(b) 0

(c) -1

(d) -2

(v) The value oi c - a is

(a) 1

(b) 0

(c) -1

(d) 2

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 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) (b) : x³ + x²y+ xy²+y³= 81
\(\begin{equation} \Rightarrow 3 x^{2}+x^{2} \frac{d y}{d x}+2 x y+2 x y \frac{d y}{d x}+y^{2}+3 y^{2} \frac{d y}{d x}=0 \end{equation}\)
\(\begin{equation} \Rightarrow \left(x^{2}+2 x y+3 y^{2}\right) \frac{d y}{d x}=-3 x^{2}-2 x y-y^{2} \end{equation}\)
\(\begin{equation} \Rightarrow \frac{d y}{d x}=\frac{-\left(3 x^{2}+2 x y+y^{2}\right)}{x^{2}+2 x y+3 y^{2}} \end{equation}\)
(ii) (c) : x^y = e^x-y⇒y log x = x - y
\(\begin{equation} \Rightarrow y \times \frac{1}{x}+\log x \cdot \frac{d y}{d x}=1-\frac{d y}{d x} \end{equation}\)
\(\begin{equation} \Rightarrow \frac{d y}{d x}[\log x+1]=1-\frac{y}{x} \Rightarrow \frac{d y}{d x}=\frac{x-y}{x[1+\log x]} \end{equation}\)
(iii) (d) : \(\begin{equation} e^{\sin y}=x y \Rightarrow \sin y=\log x+\log y \end{equation}\)
\(\begin{equation} \Rightarrow \cos y \frac{d y}{d x}=\frac{1}{x}+\frac{1}{y} \frac{d y}{d x} \Rightarrow \frac{d y}{d x}\left[\cos y-\frac{1}{y}\right]=\frac{1}{x} \end{equation}\)
\(\begin{equation} \Rightarrow \frac{d y}{d x}=\frac{y}{x(y \cos y-1)} \end{equation}\)
(iv) (d) : sin²x + cos²y = 1
\(\begin{equation} \Rightarrow \quad 2 \sin x \cos x+2 \cos y\left(-\sin y \frac{d y}{d x}\right)=0 \end{equation}\)
\(\begin{equation} \Rightarrow \frac{d y}{d x}=\frac{-\sin 2 x}{-\sin 2 y}=\frac{\sin 2 x}{\sin 2 y} \end{equation}\)
(v) (d) : \(\begin{equation} y=(\sqrt{x})^{\sqrt{x}} \quad \Rightarrow y=(\sqrt{x})^{y} \end{equation}\)
\(\begin{equation} \Rightarrow \log y=y(\log \sqrt{x}) \Rightarrow \log y=\frac{1}{2}(y \log x) \end{equation}\)
\(\begin{equation} \Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1}{2}\left[y \times \frac{1}{x}+\log x\left(\frac{d y}{d x}\right)\right] \end{equation}\)
\(\begin{equation} \Rightarrow \frac{d y}{d x}\left\{\frac{1}{y}-\frac{1}{2} \log x\right\}=\frac{1}{2} \frac{y}{x} \end{equation}\)
\(\begin{equation} \Rightarrow \frac{d y}{d x}=\frac{y}{2 x} \times \frac{2 y}{(2-y \log x)}=\frac{y^{2}}{x(2-y \log x)} \end{equation}\)

(i) (a) : Let \(\begin{equation} y=\cos \sqrt{x} \end{equation}\)
\(\begin{equation} \therefore \quad \frac{d y}{d x}=\frac{d}{d x}(\cos \sqrt{x})=-\sin \sqrt{x} \cdot \frac{d}{d x}(\sqrt{x}) \end{equation}\)
\(\begin{equation} =-\sin \sqrt{x} \times \frac{1}{2 \sqrt{x}}=\frac{-\sin \sqrt{x}}{2 \sqrt{x}} \end{equation}\)
(ii) (a) : Let \(\begin{equation} y=7^{x+\frac{1}{x}} \quad \therefore \quad \frac{d y}{d x}=\frac{d}{d x}\left(7^{x+\frac{1}{x}}\right) \end{equation}\)
\(\begin{equation} =7^{x+\frac{1}{x}} \cdot \log 7 \cdot \frac{d}{d x}\left(x+\frac{1}{x}\right)=7^{x+\frac{1}{x}} \cdot \log 7 \cdot\left(1-\frac{1}{x^{2}}\right) \end{equation}\)
\(\begin{equation} =\left(\frac{x^{2}-1}{x^{2}}\right) \cdot 7^{x+\frac{1}{x}} \cdot \log 7 \end{equation}\)
(iii) (a) : Let \(\begin{equation} y=\sqrt{\frac{1-\cos x}{1+\cos x}}=\sqrt{\frac{1-1+2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}-1+1}}=\tan \left(\frac{x}{2}\right) \end{equation}\)
\(\begin{equation} \therefore \frac{d y}{d x}=\sec ^{2} \frac{x}{2} \cdot \frac{1}{2}=\frac{1}{2} \sec ^{2} \frac{x}{2} \end{equation}\)
(iv) (b) : Let \(\begin{equation} y=\frac{1}{b} \tan ^{-1}\left(\frac{x}{b}\right)+\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right) \end{equation}\)
\(\begin{equation} \therefore \quad \frac{d y}{d x}=\frac{1}{b} \times \frac{1}{1+\frac{x^{2}}{b^{2}}} \times \frac{1}{b}+\frac{1}{a} \times \frac{1}{1+\frac{x^{2}}{a^{2}}} \times \frac{1}{a} \end{equation}\)
\(\begin{equation} =\frac{1}{b^{2}+x^{2}}+\frac{1}{a^{2}+x^{2}} \end{equation}\)
(v) (d) : Let \(\begin{equation} y=\sec ^{-1} x+\operatorname{cosec}^{-1} \frac{x}{\sqrt{x^{2}-1}} \end{equation}\)
Put \(\begin{equation} x=\sec \theta \Rightarrow \theta=\sec ^{-1} x \end{equation}\)
\(\begin{equation} \therefore \quad y=\sec ^{-1}(\sec \theta)+\operatorname{cosec}^{-1}\left(\frac{\sec \theta}{\sqrt{\sec ^{2} \theta-1}}\right) \end{equation}\)
\(\begin{equation} =\theta+\sin ^{-1}\left[\sqrt{1-\cos ^{2} \theta}\right] \end{equation}\)
\(\begin{equation} =\theta+\sin ^{-1}(\sin \theta)=\theta+\theta=2 \theta=2 \sec ^{-1} x \end{equation}\)
\(\begin{equation} \therefore \quad \frac{d y}{d x}=2 \frac{d}{d x}\left(\sec ^{-1} x\right)=2 \times \frac{1}{|x| \sqrt{x^{2}-1}}=\frac{2}{|x| \sqrt{x^{2}-1}} \end{equation}\)

L.H.L (at x = 0) = \(\begin{equation} \lim _{x \rightarrow 0} \frac{\sin (a+1) x+\sin x}{x}\left(\frac{0}{0} \text { form }\right) \end{equation}\)
Using L' Hospital rule, we get
L.H.L.(at x = 0)
\(\begin{equation} =\lim _{x \rightarrow 0}(a+1) \cos (a+1) x+\cos x=a+2 \end{equation}\)
R.H.L \(\begin{equation} \text { (at } x=0)=\lim _{x \rightarrow 0} \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x^{3 / 2}}=\lim _{x \rightarrow 0} \frac{\sqrt{1+b x}-1}{b x} \end{equation}\)
\(\begin{equation} =\lim _{x \rightarrow 0} \frac{1}{\sqrt{1+b x}+1}=\frac{1}{2} \end{equation}\)
Since,f(x) is continuous at x = 0.
\(\therefore\) From (i) and (ii), we get
\(\begin{equation} a+2=c=\frac{1}{2} \Rightarrow a=-\frac{3}{2}, c=\frac{1}{2} \end{equation}\)
Also, value of b does not affect the continuity of f(x), so b can be any real number.
(i) (a)
(ii) (d)
(iii) (b)
(iv) (c) : \(\begin{equation} a+c=-\frac{3}{2}+\frac{1}{2}=-1 \end{equation}\)
(v) (d) : \(\begin{equation} c-a=\frac{1}{2}+\frac{3}{2}=2 \end{equation}\)

(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.

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
\(\begin{equation} f^{\prime}(x)=\frac{x}{2}-\frac{3}{2}, x<1 \end{equation}\)
\(\begin{equation} \therefore f^{\prime}(-1)=\frac{-1}{2}-\frac{3}{2}=-2 \end{equation}\)