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Questions:

Case Study Questions

10 x 4 = 40

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

Logarithmic differentiation is a powerful technique to differentiate functions of the form \(\begin{equation} f(x)=[u(x)]^{\nu(x)} \end{equation}\) ,where both u(x) and vex) are differentiable functions and f and u need to be positive functions.
Let function \(\begin{equation} y=f(x)=(u(x))^{v(x)} \end{equation}\),then \(\begin{equation} y^{\prime}=y\left[\frac{v(x)}{u(x)} u^{\prime}(x)+v^{\prime}(x) \cdot \log [u(x)]\right] \end{equation}\)
On the basis of above information, answer the following questions.
(i) Differentiate x^x w.r.t. x

(a) x^x(l + log x)

(b) x^x(l -log x)

(c) -x^x(1 + log x)

(d) x^xlogx

(ii) Differentiate x^x+ a^x+a^a w.r.t. x

(a) (1 + log x) + (a^xlog a + ax^a-1	(b) x^x(1 + log x) + log a + ax^a-1
(c) x^x(1 + log x) + x^a(log x + ax^a-1)	(d) ~(1 + log x) + aXlog a + ax^a-1

(iii) If x = e^x/y,then find \(\begin{equation} \frac{d y}{d x} \end{equation}\)

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

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

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

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

(iv) If = (2 - x)³ + 2x)⁵, then find \(\begin{equation} \frac{d y}{d x} \end{equation}\)

(a) \(\begin{equation} (2-x)^{3}(3+2 x)^{5}\left[\frac{15}{3+2 x}-\frac{8}{2-x}\right] \end{equation}\)	(b) \(\begin{equation} (2-x)^{3}(3+2 x)^{5}\left[\frac{15}{3+2 x}+\frac{3}{2-x}\right] \end{equation}\)
(c) \(\begin{equation} (2-x)^{3}(3+2 x)^{5}\left[\frac{10}{3+2 x}-\frac{3}{2-x}\right] \end{equation}\)	(d) \(\begin{equation} (2-x)^{3}(3+2 x)^{5} \cdot\left[\frac{10}{3+2 x}+\frac{3}{2-x}\right] \end{equation}\)

(v) If \(\begin{equation} y=x^{x} \cdot e^{(2 x+5)} \end{equation}\) then find \(\begin{equation} \frac{d y}{d x} \end{equation}\)

(a) x^xe^2x+5

(b) x^xe^2x+5(3-logx)

(c) x^xe^2x+5(1-logx)

(d) x^xe^2x+5 (3+logx)

(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

Derivative of y = f(x) w.r.t. x (if exists) is denoted by \(\begin{equation} \frac{d y}{d x} \end{equation}\) .or f(x) and is called the first order derivative of y If we take derivative of \(\begin{equation} \frac{d y}{d x} \end{equation}\) again,
then we get \(\begin{equation} \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d^{2} y}{d x^{2}} \end{equation}\) or f"(x) and is called the second order derivative of y.
Similarly \(\begin{equation} \frac{d}{d x}\left(\frac{d^{2} y}{d x^{2}}\right) \end{equation}\) is denoted and defined as \(\begin{equation} \frac{d^{3} y}{d x^{3}} \text { or } f^{\prime \prime \prime}(x) \end{equation}\) and is known as third order derivative of y.
Based on the above information, answer the following questions.
(i) If \(\begin{equation} y=\tan ^{-1}\left(\frac{\log \left(e / x^{2}\right)}{\log \left(e x^{2}\right)}\right)+\tan ^{-1}\left(\frac{3+2 \log x}{1-6 \log x}\right) \end{equation}\), then \(\begin{equation} \frac{d^{2} y}{d x^{2}} \end{equation}\) is equal to

(a) 2

(b) 1

(c) 0

(d) -1

(ii) If u = x² + y² and x = s + 3t, y = 2s - t, then \(\begin{equation} \frac{d^{2} u}{d s^{2}} \end{equation}\) is equal to

(a) 12

(b) 32

(c) 36

(d) 10

(iii) If f(x) = 2 log sin x, then f"(x) is equal to

(a) 2cosec³x

(b) 2 cot² x - 4x² cosec² x2

(c) 2x cotx²

(d) -2 cosec²x

(iv) If f(x) = e^xsin x, then f"'(x) =

(a) 2tf(sin x + cos x)

(b) 2e^x(cos x - sin x)

(c) 2e^x(sin x - cos x)

(d) 0

(v) If y² = ax²+ bx + c, then \(\begin{equation} \frac{d}{d x}\left(y^{3} y_{2}\right)= \end{equation}\)

(a) 1

(b) -1

(c) \(\begin{equation} \frac{4 a c-b^{2}}{a^{2}} \end{equation}\)

(d) 0

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

10.

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

10 x 4 = 40

(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 \(\begin{equation} x=\sec \theta \Rightarrow \theta=\sec ^{-1} x \end{equation}\)
\(\begin{equation} \therefore \quad y=\sec ^{-1}(\sec \theta)+\operatorname{cosec}^{-1}\left(\frac{\sec \theta}{\sqrt{\sec ^{2} \theta-1}}\right) \end{equation}\)
\(\begin{equation} =\theta+\sin ^{-1}\left[\sqrt{1-\cos ^{2} \theta}\right] \end{equation}\)
\(\begin{equation} =\theta+\sin ^{-1}(\sin \theta)=\theta+\theta=2 \theta=2 \sec ^{-1} x \end{equation}\)
\(\begin{equation} \therefore \quad \frac{d y}{d x}=2 \frac{d}{d x}\left(\sec ^{-1} x\right)=2 \times \frac{1}{|x| \sqrt{x^{2}-1}}=\frac{2}{|x| \sqrt{x^{2}-1}} \end{equation}\)

(i) (a) : Let \(\begin{equation} y=x^{x} \Rightarrow \log y=x \log x \end{equation}\)
\(\begin{equation} \Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}(x \log x) \Rightarrow \frac{d y}{d x}=x^{x}\left[1 \times \log x+x \times \frac{1}{x}\right] \end{equation}\)
(ii) (d)
(iii) (d) : Given \(\begin{equation} x=e^{x / y} \Rightarrow \log x=\frac{x}{v} \log e \Rightarrow y \log x=x \end{equation}\)
\(\begin{equation} \Rightarrow y \frac{1}{x}+(\log x) \frac{d y}{d x}=1 \end{equation}\)
\(\begin{equation} \Rightarrow \frac{d y}{d x}=\left(1-\frac{y}{x}\right) \frac{1}{\log x} \Rightarrow \frac{1}{x \log x}(x-y) \end{equation}\)
(iv) (c) : y = (2 - x)³ (3 + 2x)⁵ '
\(\begin{equation} \Rightarrow \log y=\log (2-x)^{3}+\log (3+2 x)^{5} \end{equation}\)
= 3 log (2 - x) + 5log (3 + 2x)
\(\begin{equation} \Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{3 \times(-1)}{2-x}+\frac{5}{3+2 x} \times(2) \end{equation}\)
\(\begin{equation} \Rightarrow \frac{d y}{d x}=(2-x)^{3}(3+2 x)^{5}\left[\frac{10}{3+2 x}-\frac{3}{2-x}\right] \end{equation}\)
(v) (d) : y = x^x.e^{(2x + 5)}
\(\begin{equation} \Rightarrow \log y=x \log x+(2 x+5) \end{equation}\)
\(\begin{equation} \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\left(x \cdot \frac{1}{x}+\log x\right)+2 \end{equation}\)
\(\begin{equation} \Rightarrow \frac{d y}{d x}=x^{x} \cdot e^{2 x+5} \cdot(3+\log x) \end{equation}\)

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) (c) : Given, \(\begin{equation} y=\tan ^{-1}\left(\frac{\log \left(\frac{e}{x^{2}}\right)}{\log e x^{2}}\right)+\tan ^{-1}\left(\frac{3+2 \log x}{1-6 \log x}\right) \end{equation}\)
\(\begin{equation} =\tan ^{-1}\left(\frac{1-\log x^{2}}{1+\log x^{2}}\right)+\tan ^{-1}\left(\frac{3+2 \log x}{1-6 \log x}\right) \end{equation}\)
\(\begin{equation} =\tan ^{-1}(1)-\tan ^{-1}(2 \log x)+\tan ^{-1}(3)+\tan ^{-1}(2 \log x) \end{equation}\)
\(\begin{equation} \Rightarrow y=\tan ^{-1}(1)+\tan ^{-1}(3) \end{equation}\)
\(\begin{equation} \Rightarrow \ \frac{d y}{d x}=0 \Rightarrow \frac{d^{2} y}{d x^{2}}=0 \end{equation}\)
(ii) (d) : Given, \(\begin{equation} x=s+3 t, y=2 s-t \Rightarrow \frac{d x}{d s}=1, \frac{d y}{d s}=2 \end{equation}\)
Now, \(\begin{equation} u=x^{2}+y^{2} \Rightarrow \frac{d u}{d s}=2 x \frac{d x}{d s}+2 y \frac{d y}{d s}=2 x+4 y \end{equation}\)
\(\begin{equation} \Rightarrow \ \frac{d^{2} u}{d s^{2}}=2\left(\frac{d x}{d s}\right)+4\left(\frac{d y}{d s}\right) \Rightarrow \frac{d^{2} u}{d s^{2}}=2(1)+4(2)=10 \end{equation}\)
(iii) (d) : We have, j(x) = 2 log sin x
\(\begin{equation} \Rightarrow f^{\prime}(x)=2 \cdot \frac{1}{\sin x} \cdot \cos x=2 \cot x \Rightarrow f^{\prime \prime}(x)=-2 \operatorname{cosec}^{2} x \end{equation}\)
(iv) (b) : We have, f(x) = e^xsinx
\(\begin{equation} \Rightarrow f^{\prime}(x)=e^{x} \cos x+e^{x} \sin x=e^{x}(\cos x+\sin x) \end{equation}\)
\(\begin{equation} \Rightarrow f^{\prime \prime}(x)=e^{x}(\cos x-\sin x)+e^{x}(\cos x+\sin x)=2 e^{x} \cos x \end{equation}\)
\(\begin{equation} \Rightarrow f^{\prime \prime \prime}(x)=2\left[e^{x} \cos x-e^{x} \sin x\right]=2 e^{x}[\cos x-\sin x] \end{equation}\)
(v) (d) : Given y² = ax² + bx + c
\(\begin{equation} \Rightarrow 2 y y_{1}=2 a x+b \end{equation}\)
\(\begin{equation} \Rightarrow 2 y y_{2}+y_{1}\left(2 y_{1}\right)=2 a \end{equation}\)
\(\begin{equation} \Rightarrow y y_{2}=a-y_{1}^{2} \Rightarrow y y_{2}=a-\left(\frac{2 a x+b}{2 y}\right)^{2} \end{equation}\) (Using (i))
\(\begin{equation} =\frac{4 y^{2} a-\left(4 a^{2} x^{2}+b^{2}+4 a b x\right)}{4 y^{2}} \end{equation}\)
\(\begin{equation} \Rightarrow y^{3} y_{2}=\frac{4 a\left(a x^{2}+b x+c\right)-\left(4 a^{2} x^{2}+b^{2}+4 a b x\right)}{4} \end{equation}\)
\(\begin{equation} =\frac{4 a c-b^{2}}{4} \end{equation}\)
\(\begin{equation} \Rightarrow \frac{d}{d x}\left(y^{3} y_{2}\right)=0 \end{equation}\)

(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 \(\begin{equation} u=\frac{2}{3} x^{3}+5 \Rightarrow \frac{d u}{d x}=\frac{2}{3} \cdot 3 x^{2}=2 x^{2} \end{equation}\)
\(\begin{equation} \therefore \ \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}=u^{3} \cdot 2 x^{2}=\left(\frac{2}{3} x^{3}+5\right)^{3}\left(2 x^{2}\right) \end{equation}\)
\(\begin{equation} =\frac{2}{27} x^{2}\left(2 x^{3}+15\right)^{3} \end{equation}\)

10.

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