Chapter 5

CONTINUITY AND DIFFERENTIABILITY

5.1 Overview

5.1.1 Continuity of a function at a point

Let $f$ be a real function on a subset of the real numbers and let $c$ be a point in the domain of $f$. Then $f$ is continuous at $c$ if

$ \lim _{x \to c} f(x)=f(c) $

More elaborately, if the left hand limit, right hand limit and the value of the function at $x=c$ exist and are equal to each other, i.e.,

$ \lim _{x \to c^{-}} f(x)=f(c)=\lim _{x \to c^{+}} f(x) $

then $f$ is said to be continuous at $x=c$.

5.1.2 Continuity in an interval

(i) $f$ is said to be continuous in an open interval $(a, b)$ if it is continuous at every point in this interval.

(ii) $f$ is said to be continuous in the closed interval $[a, b]$ if

• $f$ is continuous in $(a, b)$
• $\lim _{x \to a^{+}} f(x)=f(a)$
• $\lim _{x \to b^{-}} f(x)=f(b)$

5.1.3 Geometrical meaning of continuity

(i) Function $f$ will be continuous at $x=c$ if there is no break in the graph of the function at the point $(c, f(c))$.

(ii) In an interval, function is said to be continuous if there is no break in the graph of the function in the entire interval.

5.1.4 Discontinuity

The function $f$ will be discontinuous at $x=a$ in any of the following cases :

(i) $\lim _{x \to a^{-}} f(x)$ and $\lim _{x \to a^{+}} f(x)$ exist but are not equal.

(ii) $\lim _{x \to a^{-}} f(x)$ and $\lim _{x \to a^{+}} f(x)$ exist and are equal but not equal to $f(a)$.

(iii) $f(a)$ is not defined.

5.1.5 Continuity of some of the common functions

$ \hspace{30 mm} $ Function f(x) $\hspace{35 mm}$ Interval in which f is continuous

$\begin{matrix} \text{1. The constant function,} i.e. f(x)=c & \mathbf{R} \\ \text{2. The identity function,} i.e. f(x)=x & \mathbf{R} \\ \text{3. The polynomial function,} i.e. \newline f(x)=a_0 x^n+a_1 x^{n-1}+\ldots+a_{n-1} x+a_n & \mathbf{R} \\ \text{4. |x-a| } & (-\infty, \infty) \\ \text{5.} x^{-n}, \text{n is a positive integer} & (-\infty, \infty)- \lbrace 0 \rbrace \\ \text{6. p(x) / q(x), where p(x) and q(x) are polynomials in x} & \mathbf{R}-\lbrace x: q(x)=0\rbrace \\ \text{7.} \sin x, \cos x & \mathbf{R} \\ \text{8.} \tan x, \sec x & \mathbf{R}-\lbrace (2 n+1) \frac{\pi}{2}: n \in \mathbf{Z}\rbrace \\ \text{9.} \cot x, \operatorname{cosec} x & \mathbf{R}-\lbrace n \pi: n \in \mathbf{Z}\rbrace \\ \text{10.} e^{x} & \mathbf{R} \\ \text{11.} \log x & (0, \infty) \\ \text{12. The inverse trigonometric functions, i.e. } & \text{In their respective domains } \end{matrix}$

5.1.6 Continuity of composite functions

Let $f$ and $g$ be real valued functions such that ( $f \circ g$ ) is defined at $a$. If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then $(f \circ g)$ is continuous at $a$.

5.1.7 Differentiability

The function defined by $f^{\prime}(x)=\lim _{h \to 0} \frac{f(x+h)-f(x)}{h}$, wherever the limit exists, is defined to be the derivative of $f$ at $x$. In other words, we say that a function $f$ is differentiable at a point $c$ in its domain if both $\lim _{h \to 0^{-}} \frac{f(c+h)-f(c)}{h}$, called left hand derivative, denoted by $L f^{\prime}(c)$, and $\lim _{h \to 0^{+}} \frac{f(c+h)-f(c)}{h}$, called right hand derivative, denoted by $R f^{\prime}(c)$, are finite and equal.

(i) The function $y=f(x)$ is said to be differentiable in an open interval $(a, b)$ if it is differentiable at every point of $(a, b)$

(ii) The function $y=f(x)$ is said to be differentiable in the closed interval $[a, b]$ if $R f^{\prime}(a)$ and $L f^{\prime}(b)$ exist and $f^{\prime}(x)$ exists for every point of $(a, b)$.

(iii) Every differentiable function is continuous, but the converse is not true

5.1.8 Algebra of derivatives

If $u, v$ are functions of $x$, then

(i) $\frac{d(u \pm v)}{d x}=\frac{d u}{d x} \pm \frac{d v}{d x}$

(ii) $\frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x}$

(iii) $\frac{d}{d x} (\frac{u}{v})=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$

5.1.9 Chain rule is a rule to differentiate composition of functions. Let $f=v o u$. If $t=u(x)$ and both $\frac{d t}{d x}$ and $\frac{d v}{d t}$ exist then $\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}$

5.1.10 Following are some of the standard derivatives (in appropriate domains)

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

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

3. $\frac{d}{d x}(\tan ^{-1} x)=\frac{1}{1+x^{2}}$

4. $\frac{d}{d x}(\cot ^{-1} x)=\frac{-1}{1+x^{2}}$

5. $\frac{d}{d x}(\sec ^{-1} x)=\frac{1}{|x| \sqrt{x^{2}-1}},|x|>1$

6. $\frac{d}{d x}(cosec^{-1} x)=\frac{-1}{|x| \sqrt{x^{2}-1}},|x|>1$

5.1.11 Exponential and logarithmic functions

(i) The exponential function with positive base $b>1$ is the function $y=f(x)=b^{x}$. Its domain is $\mathbf{R}$, the set of all real numbers and range is the set of all positive real numbers. Exponential function with base 10 is called the common exponential function and with base $e$ is called the natural exponential function.

(ii) Let $b>1$ be a real number. Then we say logarithm of $a$ to base $b$ is $x$ if $b^{x}=a$, Logarithm of $a$ to the base $b$ is denoted by $\log _{b} a$. If the base $b=10$, we say it is common logarithm and if $b=e$, then we say it is natural logarithms. $\log x$ denotes the logarithm function to base $e$. The domain of logarithm function is $\mathbf{R}^{+}$, the set of all positive real numbers and the range is the set of all real numbers.

(iii) The properties of logarithmic function to any base $b>1$ are listed below:

1. $\log _{b}(x y)=\log _{b} x+\log _{b} y$

2. $\log _{b} (\frac{x}{y})=\log _{b} x-\log _{b} y$

3. $\log _{b} x^{n}=n \log _{b} x$

4. $\log _{b} x=\frac{\log _{c} x}{\log _{c} b}$, where $c>1$

5. $\log _{b} x=\frac{1}{\log _{x} b}$

6. $\log _{b} b=1$ and $\log _{b} 1=0$

(iv) The derivative of $e^{x}$ w.r.t., $x$ is $e^{x}$, i.e. $\frac{d}{d x}(e^{x})=e^{x}$. The derivative of $\log x$ w.r.t., $x$ is $\frac{1}{x}$; i.e. $\frac{d}{d x}(\log x)=\frac{1}{x}$.

5.1.12 Logarithmic differentiation is a powerful technique to differentiate functions of the form $f(x)=(u(x))^{v(x)}$, where both $f$ and $u$ need to be positive functions for this technique to make sense.

5.1.13 Differentiation of a function with respect to another function

Let $u=f(x)$ and $v=g(x)$ be two functions of $x$, then to find derivative of $f(x)$ w.r.t. to $g(x)$, i.e., to find $\frac{d u}{d v}$, we use the formula

$ \frac{d u}{d v}=\frac{d u}{\frac{d v}{d x}} $

5.1.14 Second order derivative

$\frac{d}{d x} \frac{d y}{d x}=\frac{d^{2} y}{d x^{2}}$ is called the second order derivative of $y$ w.r.t. $x$. It is denoted by $y^{\prime}$ or $y_2$, if $y=f(x)$.

5.1.15 Rolle’s Theorem

Let $f:[a, b] \to \mathbf{R}$ be continuous on $[a, b]$ and differentiable on $(a, b)$, such that $f(a)$ $=f(b)$, where $a$ and $b$ are some real numbers. Then there exists at least one point $c$ in $(a, b)$ such that $f^{\prime}(c)=0$.

Geometrically Rolle’s theorem ensures that there is at least one point on the curve $y=f(x)$ at which tangent is parallel to $x$-axis (abscissa of the point lying in $(a, b)$ ).

5.1.16 Mean Value Theorem (Lagrange)

Let $f:[a, b] \to \mathbf{R}$ be a continuous function on $[a, b]$ and differentiable on $(a, b)$. Then there exists at least one point $c$ in $(a, b)$ such that $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$.

Geometrically, Mean Value Theorem states that there exists at least one point $c$ in $(a, b)$ such that the tangent at the point $(c, f(c))$ is parallel to the secant joining the points $(a, f(a)$ and $(b, f(b))$.

5.2 Solved Examples

Short Answer (S.A.)

Example 1 Find the value of the constant $k$ so that the function $f$ defined below is

continuous at $x=0$, where f(x)= $ \begin{cases} \frac{1-\cos 4 x}{8 x^{2}}, & x \neq 0. \\ k, & x=0 \end{cases} $

Solution It is given that the function $f$ is continuous at $x=0$. Therefore, $\lim _{x \to 0} f(x)=f(0)$

$ \begin{matrix} \Rightarrow & \lim _{x \to 0} \frac{1-\cos 4 x}{8 x^{2}}=k \\ \Rightarrow & \lim _{x \to 0} \frac{2 \sin ^{2} 2 x}{8 x^{2}}=k \\ \Rightarrow & \lim _{x \to 0} (\frac{\sin 2 x^{2}}{2 x})=k \\ \Rightarrow & k=1 \end{matrix} $

Thus, $f$ is continuous at $x=0$ if $k=1$.

Example 2 Discuss the continuity of the function $f(x)=\sin x \cdot \cos x$.

Solution Since $\sin x$ and $\cos x$ are continuous functions and product of two continuous function is a continuous function, therefore $f(x)=\sin x \cdot \cos x$ is a continuous function.

Example 3 If f(x)= $ \begin{cases} \frac{x^{3}+x^{2}-16 x+20}{(x-2)^{2}}, & x \neq 2 \\ k, & x=2 \end{cases} $ is continuous at $x=2$, find

the value of $k$.

Solution Given $f(2)=k$.

Now, $\lim _{x \to 2^{-}} f(x)=\lim _{x \to 2^{+}} f(x)=\lim _{x \to 2} \frac{x^{3}+x^{2}-16 x+20}{(x-2)^{2}}$

$ =\lim _{x \to 2} \frac{(x+5)(x-2)^{2}}{(x-2)^{2}}=\lim _{x \to 2}(x+5)=7 $

As $f$ is continuous at $x=2$, we have

$ \lim _{x \to 2} f(x)=f(2) $

$ \Rightarrow k=7 $

Example 4 Show that the function $f$ defined by

f(x)= $ \begin{cases} x \sin \frac{1}{x}, x \neq 0 \\ 0, x=0 \end{cases} $

is continuous at $x=0$.

Solution Left hand limit at $x=0$ is given by

$ \lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{-}} x \sin \frac{1}{x}=0 \quad[\text{ since, }-1<\sin \frac{1}{x}<1] $

Similarly $\lim _{x \to 0^{+}} f(x)=\lim _{x \to 0^{+}} x \sin \frac{1}{x}=0$. Moreover $f(0)=0$.

Thus $\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{+}} f(x)=f(0)$. Hence $f$ is continuous at $x=0$

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

Solution We know that $f(x)=\frac{1}{x-1}$ is discontinuous at $x=1$

Now, for $x \neq 1$,

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

which is discontinuous at $x=2$.

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

Example 6 Let $f(x)=x|x|$, for all $x \in \mathbf{R}$. Discuss the derivability of $f(x)$ at $x=0$

Solution We may rewrite $f$ as f(x)= $\begin{cases}x^{2} & \text{, if } x \geq 0 \\ -x^{2} & \text{, if } x \lt 0 \end{cases}$

Now $L f^{\prime}(0)=\lim _{h \to 0^{-}} \frac{f(0+h)-f(0)}{h}=\lim _{h \to 0^{-}} \frac{-h^{2}-0}{h}=\lim _{h \to 0^{-}}-h=0$

Now Rf $(0)=\lim _{h \to 0^{+}} \frac{f(0+h)-f(0)}{h}=\lim _{h \to 0^{+}} \frac{h^{2}-0}{h}=\lim _{h \to 0^{-}} h=0$

Since the left hand derivative and right hand derivative both are equal, hence $f$ is differentiable at $x=0$.

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

Solution Let $y=\sqrt{\tan \sqrt{x}}$. Using chain rule, we have

$ \begin{aligned} & \frac{d y}{d x}=\frac{1}{2 \sqrt{\tan \sqrt{x}}} \cdot \frac{d}{d x}(\tan \sqrt{x}) \\ & =\frac{1}{2 \sqrt{\tan \sqrt{x}}} \cdot \sec ^{2} \sqrt{x} \frac{d}{d x}(\sqrt{x}) \\ & =\frac{1}{2 \sqrt{\tan \sqrt{x}}}(\sec ^{2} \sqrt{x}) \bigg (\frac{1}{2 \sqrt{x}} \bigg ) \\ & =\frac{(\sec ^{2} \sqrt{x})}{4 \sqrt{x} \sqrt{\tan \sqrt{x}}} . \end{aligned} $

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

Solution Given $y=\tan (x+y)$. differentiating both sides w.r.t. $x$, we have

$ \begin{aligned} \frac{d y}{d x}= & \sec ^{2}(x+y) \frac{d}{d x}(x+y) \\ & =\sec ^{2}(x+y) \bigg(1+\frac{d y}{d x} \bigg ) \end{aligned} $

or $\quad[1-\sec ^{2}(x+y] \frac{d y}{d x}=\sec ^{2}(x+y).$

Therefore, $\frac{d y}{d x}=\frac{\sec ^{2}(x+y)}{1-\sec ^{2}(x+y)}=-cosec^{2}(x+y)$.

Example 9 If $e^{x}+e^{y}=e^{x+y}$, prove that

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

Solution Given that $e^{x}+e^{y}=e^{x+y}$. Differentiating both sides w.r.t. $x$, we have

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

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

which implies that $\frac{d y}{d x}=\frac{e^{x+y}-e^{x}}{e^{y}-e^{x+y}}=\frac{e^{x}+e^{y}-e^{x}}{e^{y}-e^{x}-e^{y}}= - e^{y-x}$.

Example 10 Find $\frac{d y}{d x}$, if $y=\tan ^{-1} \bigg (\frac{3 x-x^{3}}{1-3 x^{2}} \bigg ) ,-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$.

Solution Put $x=\tan \theta$, where $\frac{-\pi}{6}<\theta<\frac{\pi}{6}$.

Therefore, $\quad y=\tan ^{-1} \bigg (\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta} \bigg )$

$ =\tan ^{-1}(\tan 3 \theta) $

$ \begin{aligned} & =3 \theta \quad \text{ (because } \frac{-\pi}{2}<3 \theta<\frac{\pi}{2} \text{ ) } \\ & =3 \tan ^{-1} x \end{aligned} $

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

Example 11 If $y=\sin ^{-1} \bigg \lbrace x \sqrt{1-x}-\sqrt{x} \sqrt{1-x^{2}} \bigg \rbrace $ and $0<x<1$, then find $\frac{d y}{d x}$.

Solution We have $y=\sin ^{-1} \lbrace x \sqrt{1-x}-\sqrt{x} \sqrt{1-x^{2}} \rbrace $, where $ 0<x<1$.

Put

$ x=\sin A \text{ and } \sqrt{x}=\sin B $

Therefore, $y=\sin ^{-1} \bigg \lbrace \sin A \sqrt{1-\sin ^{2} B}-\sin B \sqrt{1-\sin ^{2} A} \bigg \rbrace $

$ \begin{aligned} & =\sin ^{-1} \lbrace \sin A \cos B-\sin B \cos A \rbrace \\ & =\sin ^{-1}\lbrace \sin (A-B)\rbrace =A-B \end{aligned} $

Thus

$ y=\sin ^{-1} x-\sin ^{-1} \sqrt{x} $

Differentiating w.r.t. $x$, we get

$ \begin{gathered} \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-\sqrt{(x)}^{2}}} \cdot \frac{d}{d x}(\sqrt{x}) \\ =\frac{1}{\sqrt{1-x^{2}}}-\frac{1}{2 \sqrt{x} \sqrt{1-x}} \end{gathered} $

Example 12 If $x=a \sec ^{3} \theta$ and $y=a \tan ^{3} \theta$, find $\frac{d y}{d x}$ at $\theta=\frac{\pi}{3}$.

Solution We have $x=a \sec ^{3} \theta$ and $y=a \tan ^{3} \theta$.

Differentiating w.r.t. $\theta$, we get

$ \frac{d x}{d \theta}=3 a \sec ^{2} \theta \frac{d}{d \theta}(\sec \theta)=3 a \sec ^{3} \theta \tan \theta $

and $\quad \frac{d y}{d \theta}=3 a \tan ^{2} \theta \frac{d}{d \theta}(\tan \theta)=3 a \tan ^{2} \theta \sec ^{2} \theta$.

Thus $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 a \tan ^{2} \theta \sec ^{2} \theta}{3 a \sec ^{3} \theta \tan \theta}=\frac{\tan \theta}{\sec \theta}=\sin \theta$.

Hence, $ ( \frac{d y}{d x})_{at \theta=\frac{\pi}{3}}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$.

Example 13 If $x^{y}=e^{x-y}$, prove that $\frac{d y}{d x}=\frac{\log x}{(1+\log x)^{2}}$.

Solution We have $x^{y}=e^{x-y}$. Taking logarithm on both sides, we get

$ y \log x=x-y $

$ \Rightarrow y(1+\log x)=x$

$\text{ i.e. } y=\frac{x}{1+\log x} $

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

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

Example 14 If $y=\tan x+\sec x$, prove that $\frac{d^{2} y}{d x^{2}}=\frac{\cos x}{(1-\sin x)^{2}}$.

Solution We have $y=\tan x+\sec x$. Differentiating w.r.t. $x$, we get

$ \begin{gathered} \frac{d y}{d x}=\sec ^{2} x+\sec x \tan x \\ =\frac{1}{\cos ^{2} x}+\frac{\sin x}{\cos ^{2} x}=\frac{1+\sin x}{\cos ^{2} x}=\frac{1+\sin x}{(1+\sin x)(1-\sin x)} . \end{gathered} $

thus $\frac{d y}{d x}=\frac{1}{1-\sin x}$.

Now, differentiating again w.r.t. $x$, we get

$ \frac{d^{2} y}{d x^{2}}=\frac{-(-\cos x)}{(1-\sin x)^{2}}=\frac{\cos x}{(1-\sin x)^{2}} $

Example 15 If $f(x)=|\cos x|$, find $f^{\prime} \bigg ( \frac{3 \pi}{4} \bigg ) $.

Solution When $\frac{\pi}{2}<x<\pi, \cos x<0$ so that $|\cos x|=-\cos x$, i.e., $f(x)=-\cos x$

$\Rightarrow f^{\prime}(x)=\sin x$.

Hence, $f^{\prime} (\frac{3 \pi}{4})=\sin (\frac{3 \pi}{4})=\frac{1}{\sqrt{2}}$

Example 16 If $f(x)=|\cos x-\sin x|$, find $f^{\prime} (\frac{\pi}{6})$.

Solution When $0<x<\frac{\pi}{4}, \cos x>\sin x$, so that $\cos x-\sin x>0$, i.e.,

$f(x)=\cos x-\sin x$

$\Rightarrow f^{\prime}(x)=-\sin x-\cos x$

Hence $f^{\prime} \quad \bigg (\frac{\pi}{6} \bigg )=-\sin \frac{\pi}{6}-\cos \frac{\pi}{6}=-\frac{1}{2}(1+\sqrt{3})$.

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

Solution Consider $f(x)=\sin 2 x$ in $[0, \frac{\pi}{2}]$. Note that:

(i) The function $f$ is continuous in $ \bigg [0, \frac{\pi}{2} \bigg]$, as $f$ is a sine function, which is always continuous.

(ii) $\quad f^{\prime}(x)=2 \cos 2 x$, exists in $ \bigg(0, \frac{\pi}{2} \bigg )$, hence $f$ is derivable in $0,\bigg (\frac{\pi}{2} \bigg )$.

(iii) $\quad f(0)=\sin 0=0$ and $f \bigg (\frac{\pi}{2} \bigg )=\sin \pi=0 \Rightarrow f(0)=f \bigg(\frac{\pi}{2} \bigg )$.

Conditions of Rolle’s theorem are satisfied. Hence there exists at least one $c \in \bigg (0, \frac{\pi}{2} \bigg )$ such that $f^{\prime}(c)=0$. Thus

$ 2 \cos 2 c=0 \quad \Rightarrow \quad 2 c=\frac{\pi}{2} \quad \Rightarrow \quad c=\frac{\pi}{4} . $

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

Solution (i) Function $f$ is continuous in $[3,5]$ as product of polynomial functions is a polynomial, which is continuous.

(ii) $f^{\prime}(x)=3 x^{2}-36 x+99$ exists in $(3,5)$ and hence derivable in $(3,5)$.

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

$ \begin{aligned} & f^{\prime}(c)=\frac{f(5)-f(3)}{5-3} \\ & \Rightarrow 3 c^{2}-36 c+99=\frac{8-0}{2}=4 \\ & \Rightarrow c=6 \pm \sqrt{\frac{13}{3}} . \end{aligned} $

Hence $c=6-\sqrt{\frac{13}{3}}$ (since other value is not permissible).

Long Answer (L.A.)

Example 19 If $f(x)=\frac{\sqrt{2} \cos x-1}{\cot x-1}, x \neq \frac{\pi}{4}$

find the value of $f \big(\frac{\pi}{4} \big) $ so that $f(x)$ becomes continuous at $x=\frac{\pi}{4}$.

Solution Given, $f(x)=\frac{\sqrt{2} \cos x-1}{\cot x-1}, x \neq \frac{\pi}{4}$

Therefore, $\quad \lim _{x \to \frac{\pi}{4}} f(x)=\lim _{x \to (\frac{\pi}{4})} \frac{\sqrt{2} \cos x-1}{\cot x-1}$

$ \begin{aligned} & =\lim _{x \to \frac{\pi}{4}} \frac{(\sqrt{2} \cos x-1) \sin x}{\cos x-\sin x} \\ & =\lim _{x \to \frac{\pi}{4}} \frac{(\sqrt{2} \cos x-1)}{(\sqrt{2} \cos x+1)} \cdot \frac{(\sqrt{2} \cos x+1)}{(\cos x-\sin x)} \cdot \frac{(\cos x+\sin x)}{(\cos x+\sin x)} \cdot \sin x \end{aligned} $

$ \begin{aligned} & =\lim _{x \to \frac{\pi}{4}} \frac{2 \cos ^{2} x-1}{\cos ^{2} x-\sin ^{2} x} \cdot \frac{\cos x+\sin x}{\sqrt{2} \cos x+1} \cdot(\sin x) \\ & =\lim _{x \to \frac{\pi}{4}} \frac{\cos 2 x}{\cos 2 x} \cdot \Bigg ( \frac{\cos x+\sin x}{\sqrt{2} \cos x+1} \Bigg ) \cdot(\sin x) \\ & =\lim _{x \to \frac{\pi}{4}} \frac{(\cos x+\sin x)}{\sqrt{2} \cos x+1} \sin x \\ & =\frac{\frac{1}{\sqrt{2}} (\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})}{\sqrt{2} \cdot \frac{1}{\sqrt{2}}+1}=\frac{1}{2} \end{aligned} $

Thus, $\quad \lim _{x \to \frac{\pi}{4}} f(x)=\frac{1}{2}$

If we define $f (\frac{\pi}{4})=\frac{1}{2}$, then $f(x)$ will become continuous at $x=\frac{\pi}{4}$. Hence for $f$ to be continuous at $x=\frac{\pi}{4}, f (\frac{\pi}{4})=\frac{1}{2}$.

Example 20 Show that the function $f$ given by f(x)= $\begin{cases} \frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}, & if x \neq 0 \\ 0, & if x=0 \end{cases}$

is discontinuous at $x=0$.

Solution The left hand limit of $f$ at $x=0$ is given by

$ \lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{-}} \frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}=\frac{0-1}{0+1}=-1 $

Similarly, $\quad \lim _{x \to 0^{+}} f(x)=\lim _{x \to 0^{+}} \frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}$

$ =\lim _{x \to 0^{+}} \frac{1-\frac{1}{e^{\frac{1}{x}}}}{1+\frac{1}{e^{\frac{1}{x}}}}=\lim _{x \to 0^{+}} \frac{1-e^{\frac{-1}{x}}}{1+e^{\frac{-1}{x}}}=\frac{1-0}{1+0}=1 $

Thus $\lim _{x \to 0^{-}} f(x) \neq \lim f(x)$, therefore, $\lim _{x \to 0^{+}} f(x)$ does not exist. Hence $f$ is discontinuous at $x=0$.

Example 21 Let f(x)= $\begin{cases} \frac{1-\cos 4 x}{x^{2}} &\text{, if } x<0 \\ a & ,if x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4} & \text{, if } x>0 \end{cases} $

For what value of $a, f$ is continuous at $x=0$ ?

Solution Here $f(0)=a$ Left hand limit of $f$ at 0 is

$ \begin{aligned} & \lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{-}} \frac{1-\cos 4 x}{x^{2}}=\lim _{x \to 0^{-}} \frac{2 \sin ^{2} 2 x}{x^{2}} \\ & =\lim _{2 x \to 0^{-}} 8 \Bigg (\frac{\sin 2 x}{2 x}\Bigg)^2 =8(1)^{2}=8 . \end{aligned} $

and right hand limit of $f$ at 0 is

$ \begin{aligned} & \lim _{x \to 0^{+}} f(x)=\lim _{x \to 0^{+}} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4} \\ = & \lim _{x \to 0^{+}} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{(\sqrt{16+\sqrt{x}}+4)(\sqrt{16+\sqrt{x}}-4)} \end{aligned} $

$ = \lim _{x \to 0^{+}} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16}=\lim _{x \to 0^{+}}(\sqrt{16+\sqrt{x}}+4)=8 $

Thus, $\lim _{x \to 0^{+}} f(x)=\lim _{x \to 0^{-}} f(x)=8$. Hence $f$ is continuous at $x=0$ only if $a=8$.

Example 22 Examine the differentiability of the function $f$ defined by

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

Solution The only doubtful points for differentiability of $f(x)$ are $x=-2$ and $x=0$. Differentiability at $x=-2$.

Now L $f^{\prime}(-2)=\lim _{h \to 0^{-}} \frac{f(-2+h)-f(-2)}{h}$

$ =\lim _{h \to 0^{-}} \frac{2(-2+h)+3-(-2+1)}{h}=\lim _{h \to 0^{-}} \frac{2 h}{h}=\lim _{h \to 0^{-}} 2=2 . $

and $R f^{\prime}(-2)=\lim _{h \to 0^{+}} \frac{f(-2+h)-f(-2)}{h}$

$ \begin{aligned} & =\lim _{h \to 0^{-}} \frac{-2+h+1-(-2+1)}{h} \\ & =\lim _{h \to 0^{-}} \frac{h-1-(-1)}{h}=\lim _{h \to 0^{+}} \frac{h}{h}=1 \end{aligned} $

Thus $R f^{\prime}(-2) \neq L f^{\prime}(-2)$. Therefore $f$ is not differentiable at $x=-2$. Similarly, for differentiability at $x=0$, we have

$ \begin{aligned} L(f^{\prime}(0). & =\lim _{h \to 0^{-}} \frac{f(0+h)-f(0)}{h} \\ & =\lim _{h \to 0^{-}} \frac{0+h+1-(0+2)}{h} \\ & =\lim _{h \to 0^{-}} \frac{h-1}{h}=\lim _{h \to 0^{-}} \bigg (1-\frac{1}{h}\bigg) \end{aligned} $

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

Example 23 Differentiate $\tan ^{-1} ( \frac{\sqrt{1-x^{2}}}{x})$ with respect to $\cos ^{-1}(2 x \sqrt{1-x^{2}})$, where $x \in ( \frac{1}{\sqrt{2}}, 1 ) $.

Solution Let $u=\tan ^{-1} ( \frac{\sqrt{1-x^{2}}}{x}) $ and $v=\cos ^{-1}(2 x \sqrt{1-x^{2}})$.

We want to find $\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$

Now $u=\tan ^{-1} ( \frac{\sqrt{1-x^{2}}}{x} ) $. Put $x=\sin \theta . \quad ( \frac{\pi}{4}<\theta<\frac{\pi}{2} )$.

Then $u=\tan ^{-1} ( \frac{\sqrt{1-\sin ^{2} \theta}}{\sin \theta})=\tan ^{-1}(\cot \theta)$

$=\tan ^{-1} \lbrace \tan ( \frac{\pi}{2}-\theta ) \rbrace =\frac{\pi}{2}-\theta=\frac{\pi}{2}-\sin ^{-1} x$

Hence

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

Now

$ \begin{aligned} v & =\cos ^{-1}(2 x \sqrt{1-x^{2}}) \\ & =\frac{\pi}{2}-\sin ^{-1}(2 x \sqrt{1-x^{2}}) \\ & =\frac{\pi}{2}-\sin ^{-1}(2 \sin \theta \sqrt{1-\sin ^{2} \theta})=\frac{\pi}{2}-\sin ^{-1}(\sin 2 \theta) \\ & =\frac{\pi}{2}-\sin ^{-1} \lbrace \sin (\pi-2 \theta) \rbrace \quad[\text{ since } \frac{\pi}{2}<2 \theta<\pi] \end{aligned} $

$\frac{\pi}{2}-(\pi-2 \theta)=\frac{-\pi}{2}+2 \theta $

$\Rightarrow v=\frac{-\pi}{2}+2 \sin ^{-1} x $

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

$\text{ Hence } \frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{-1}{\sqrt{1-x^{2}}}}{\frac{2}{\sqrt{1-x^{2}}}}=\frac{-1}{2}$ .

Objective Type Questions

Choose the correct answer from the given four options in each of the Examples 24 to 35.

Example 24 The function f(x)= $\begin{cases} \frac{\sin x}{x}+\cos x & \text{, if } x \neq 0 \\ k \quad & \text{, if } x=0\end{cases} $

is continuous at $x=0$, then the value of $k$ is

(A) 3

(B) 2

(C) 1

(D) 1.5

Solution (B) is the Correct answer.

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

(A) 4

(B) -2

(C) 1

(D) $ 1.5$

Solution (D) is the correct answer. The greatest integer function $[x]$ is discontinuous at all integral values of $x$. Thus $D$ is the correct answer.

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

(A) 1

(B) 2

(C) 3

(D) none of these

Solution (D) is the correct answer. As $x-[x]=0$, when $x$ is an integer so $f(x)$ is discontinuous for all $x \in \mathbf{Z}$.

Example 27 The function given by $f(x)=\tan x$ is discontinuous on the set

(A) $\quad \lbrace n \pi: n \in Z \rbrace $

(B) $\quad \lbrace 2 n \pi: n \in Z \rbrace $

(C) $\quad \lbrace 2 n+1) \frac{\pi}{2}: n \in Z \rbrace $

(D) $\quad \lbrace \frac{n \pi}{2}: n \in \mathbb{Z} \rbrace $

Solution C is the correct answer.

Example 28 Let $f(x)=|\cos x|$. Then,

(A) $\quad f$ is everywhere differentiable.

(B) $\quad f$ is everywhere continuous but not differentiable at $n=n \pi, n \in Z$.

(C) $\quad f$ is everywhere continuous but not differentiable at $x=(2 n+1) \frac{\pi}{2}$, $n \in \mathbf{Z}$.

(D) none of these.

Solution $C$ is the correct answer.

Example 29 The function $f(x)=|x|+|x-1|$ is

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

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

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

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

Solution Correct answer is A.

Example 30 The value of $k$ which makes the function defined by

f(x)= $\begin{cases} \sin \frac{1}{x}, & \text{ if } x \neq 0 \\ k, & \text{ if } x=0\end{cases} $, continuous at $x=0$ is

(A) 8

(B) 1

(C) $-1$

(D) none of these

Solution (D) is the correct answer. Indeed $\lim _{x \to 0} \sin \frac{1}{x}$ does not exist.

Example 31 The set of points where the functions $f$ given by $f(x)=|x-3| \cos x$ is differentiable is

(A) $\mathbf{R}$

(B) $ \mathbf{R}-\lbrace 3 \rbrace$

(C) $(0, \infty)$

(D) none of these

Solution B is the correct answer.

Example 32 Differential coefficient of sec $(\tan ^{-1} x)$ w.r.t. $x$ is

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

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

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

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

Solution (A) is the correct answer.

Example 33 If $u= \sin ^{-1} ( \frac{2 x}{1+x^{2}})$ and $v=\tan ^{-1} (\frac{2 x}{1-x^{2}})$, then $\frac{d u}{d v}$ is

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

(B) $x$

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

(D) 1

Solution (D) is the correct answer.

Example 34 The value of $c$ in Rolle’s Theorem for the function $f(x)=e^{x} \sin x$, $x \in [0, \pi]$ is

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

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

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

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

Solution (D) is the correct answer.

Example 35 The value of $c$ in Mean value theorem for the function $f(x)=x(x-2)$, $x \in[1,2]$ is

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

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

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

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

Solution (A) is the correct answer.

Example 36 Match the following

$ \hspace{20 mm}$ COLUMN-I $\hspace{60 mm}$ COLUMN-II

(A) If a function f(x)= $ \begin{cases} & \frac{\sin 3 x}{x}, \text{ if } x \neq 0 \\ & \frac{k}{2}, \text{ if } x=0\end{cases} \hspace{40 mm} (a) |x| $ $ \newline $ is continuous at x=0, then k is equal to

(B) Every continuous function is differentiable $\hspace{35 mm}$ (b) True

(C) An example of a function which is continuous $\hspace{34 mm} $ (c) 6 $ \newline $ everywhere but not differentiable at exactly one point

(D) The identity function i.e. $f(x)=x \forall x \in R \newline $ is a continuous function $ \hspace{67 mm } $ (d) False

Solution $A \to c, B \to d, \quad C \to a, D \to b$

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

Example 37 The number of points at which the function $f(x)=\frac{1}{\log |x|}$ is discontinuous is________________

Solution The given function is discontinuous at $x=0, \pm 1$ and hence the number of points of discontinuity is 3 .

Example 38 If f(x)=$\begin{cases}a x+1 \text{ if } x \geq 1 \\ x+2 \text{ if } x<1\end{cases}$ is continuous, then $a$ should be equal to__________

Solution $a=2$

Example 39 The derivative of $\log_{10} x$ w.r.t. $x$ is_______________

Solution $(\log _{10} e) \frac{1}{x}$.

Example 40 If $y=\sec ^{-1} \bigg (\frac{\sqrt{x}+1}{\sqrt{x}-1} \bigg ) +\sin ^{-1} \bigg (\frac{\sqrt{x}-1}{\sqrt{x}+1} \bigg )$, then $\frac{d y}{d x}$ is equal to_____________

Solution 0 .

Example 41 The deriative of $\sin x$ w.r.t. $\cos x$ is______________

Solution $-\cot x$

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

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

Solution True .

Example 43 $y=|x-1|$ is a continuous function.

Solution True .

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

Solution False .

Example 45 $ |\sin x|$ is a differentiable function for every value of $x$.

Solution False.

Example 46 $\cos |x|$ is differentiable everywhere.

Solution True.

5.3 EXERCISE

Short Answer (S.A.)

1. Examine the continuity of the function

$ f(x)=x^{3}+2 x^{2}-1 \text{ 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{cases} 3 x+5, & \text{ if } x \geq 2 \\ x^{2}, & \text{ if } x<2\end{cases} \quad $ at $x=2$

3. f(x)=$ \begin{cases}\frac{1-\cos 2 x}{x^{2}} & if x \neq 0 \\ 5 & \text{ if } x=0 \end{cases}\quad $ at $x=0$

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

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

6. f(x)=$\begin{cases} |x| \cos \frac{1}{x}, & \text{ if } x \neq 0 \\ 0, & \text{ if } x=0\end{cases}\quad $ at $x=0$

7. $f(x)=\begin{cases} |x-a| \sin \frac{1}{x-a}, & \text{ if } x \neq 0 \\ 0, & \text{ if } x=a\end{cases}\quad $ at $x=a$

8. f(x)=$ \begin{cases}\frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}, & if x \neq 0 ,\\ 0 & if x=0 \end{cases} \quad $ $\text{ at } x=0 $

9. $f(x)=\begin{cases} & \frac{x^{2}}{2}, & \text{ if } 0 \leq x \leq 1 \\ & 2 x^{2}-3 x+\frac{3}{2}, & \text{ if } 1<x \leq 2\end{cases} \quad $ at $x=1$

10. $f(x)=|x|+|x-1|$ 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{cases} 3 x-8, & \text{ if } x \leq 5 \\ 2 k, & \text{ if } x>5\end{cases} $ at $x=5$

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

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

14. $f(x)=\begin{cases} \frac{1-\cos k x}{x \sin x} & \text{ if } x \neq 0 \\ \frac{1}{2} & \text{,if } x=0\end{cases} \quad $ at $x=0$

15. Prove that the function $f$ defined by

f(x)= $ \begin{cases} \frac{x}{|x|+2 x^{2}}, & x \neq 0 \\ k \quad, & x=0 \end{cases} $

remains discontinuous at $x=0$, regardless the choice of $k$.

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

$ f(x)=\begin{cases} \frac{x-4}{|x-4|}+a & \text{, if } x<4 \\ a+b & \text{, if } x=4 \\ \frac{x-4}{|x-4|}+b & \text{, if } x>4 \end{cases} $

is a continuous function at $x=4$.

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

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

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

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

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

21. $f(x)=\begin{cases} x^{2} \sin \frac{1}{x} & \text{, if } x \neq 0 \\ 0 & \text{, if } x=0\end{cases} \quad $ at $x=0$.

22. $f(x)=\begin{cases} 1+x & \text{, if } x \leq 2 \\ 5-x & \text{, if } x>2\end{cases} \quad $ at $x=2$.

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

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

Differentiate each of the following w.r.t. $x$ (Exercises 25 to 43) :

25. $2^{\cos ^{2} x}$

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

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

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

29. $\sin \sqrt{x}+\cos ^{2} \sqrt{x}$

30. $\sin ^{n}(a x^{2}+b x+c)$

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

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

33. $\sin ^{-1} (\frac{1}{\sqrt{x+1}})$

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

35. $\sin ^{m} x \cdot \cos ^{n} x$

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

37. $\cos ^{-1} ( \frac{\sin x+\cos x}{\sqrt{2}}), \frac{- \pi}{4}<x<\frac{\pi}{4} \quad$

38. $\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}},-\frac{\pi}{4}<x<\frac{\pi}{4}$

39. $\tan ^{-1}(\sec x+\tan x),-\frac{\pi}{2}<x<\frac{\pi}{2}$

40. $\tan ^{-1} \frac{a \cos x-b \sin x}{b \cos x+a \sin x},-\frac{\pi}{2}<x<\frac{\pi}{2}$ and $\frac{a}{b} \tan x>-1$

41. $\sec ^{-1} \frac{1}{4 x^{3}-3 x}, 0<x<\frac{1}{\sqrt{2}} \quad$

42. $\tan ^{-1} \frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}, \frac{-1}{\sqrt{3}}<\frac{x}{a}<\frac{1}{\sqrt{3}}$

43. $\tan ^{-1} \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}},-1<x \triangleleft, x \neq 0$

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

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

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

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

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

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

49. If $x=e^{\cos 2 t}$ and $y=e^{\sin 2 t}$, prove that $\frac{d y}{d x}=\frac{-y \log x}{x \log y}$.

50. If $x=a \sin 2 t(1+\cos 2 t)$ and $y=b \cos 2 t(1-\cos 2 t)$, show that $ (\frac{d y}{d x})_{\text{ at } t=\frac{\pi}{4}} = \frac{b}{a} $

51. If $x=3 \sin t-\sin 3 t, y=3 \cos t-\cos 3 t$, find $\frac{d y}{d x}$ at $t=\frac{\pi}{3}$.

52. Differentiate $\frac{x}{\sin x}$ w.r.t. $\sin x$.

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

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

54. $\quad \sin (x y)+\frac{x}{y}=x^{2}-y$

55. $\quad \sec (x+y)=x y$

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

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

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

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

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

61. If $y=(\cos x)^{(\cos x)^{(\cos x) \ldots \infty}}$, show that $\frac{d y}{d x}=\frac{y^{2} \tan x}{y \log \cos x-1}$.

62. If $x \sin (a+y)+\sin a \cos (a+y)=0$, prove that $\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$.

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

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

Verify the Rolle’s theorem for each of the functions in Exercises 65 to 69.

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

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

67. $f(x)=\log (x^{2}+2)-\log 3$ in $[-1,1]$.

68. $f(x)=x(x+3) e^{-x / 2}$ in $[-3,0]$.

69. $f(x)=\sqrt{4-x^{2}}$ in $[-2,2]$.

70. Discuss the applicability of Rolle’s theorem on the function given by

$ f(x)=\begin{cases} & x^{2}+1, \text{ if } 0 \leq x \leq 1 \\ & 3-x, \text{ if } 1 \leq x \leq 2 \end{cases} . $

71. Find the points on the curve $y=(\cos x-1)$ in $[0,2 \pi]$, where the tangent is parallel to $x$-axis.

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 $x$-axis.

Verify mean value theorem for each of the functions given Exercises 73 to 76.

73. $f(x)=\frac{1}{4 x-1}$ in $[1,4]$.

74. $f(x)=x^{3}-2 x^{2}-x+3$ in $[0,1]$.

75. $f(x)=\sin x-\sin 2 x$ in $[0, \pi]$.

76. $f(x)=\sqrt{25-x^{2}}$ in $[1,5]$.

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 $(4,1)$.

78. Using mean value theorem, prove that there is a point on the curve $y=2 x^{2}-5 x+3$ between the points $A(1,0)$ and $B(2,1)$, where tangent is parallel to the chord $AB$. Also, find that point.

Long Answer (L.A.)

79. Find the values of $p$ and $q$ so that

$ f(x)=\begin{cases} & x^{2}+3 x+p, & \text{ if } x \leq 1 \\ & q x+2 \quad & \text{, if } x>1 \end{cases} $

is differentiable at $x=1$.

80. If $x^{m} \cdot y^{n}=(x+y)^{m+n}$, prove that

(i) $\frac{d y}{d x}=\frac{y}{x}$ and

(ii) $\frac{d^{2} y}{d x^{2}}=0$.

81. If $x=\sin t$ and $y=\sin p t$, prove that $(1-x^{2}) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+p^{2} y=0$.

82. Find $\frac{d y}{d x}$, if $y=x^{\tan x}+\sqrt{\frac{x^{2}+1}{2}}$.

Objective Type Questions

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) $\frac{g(x)}{f(x)}$

84. The function $f(x)=\frac{4-x^{2}}{4 x-x^{3}}$ is

(A) discontinuous at only one point

(B) discontinuous at exactly two points

(C) discontinuous at exactly three points

(D) none of these

85. The set of points where the function $f$ given by $f(x)=|2 x-1| \sin x$ is differentiable is

(A) $\mathbf{R}$

(B) $\mathbf{R}- \lbrace \frac{1}{2} \rbrace $

(C) $(0, \infty)$

(D) none of these

86. The function $f(x)=\cot x$ is discontinuous on the set

(A) {$ x=n \pi: n \in \mathbf{Z} $}

(B) {$ x=2 n \pi: n \in \mathbf{Z} $}

(C) $ \lbrace x=(2 n+1) \frac{\pi}{2} ; n \in \mathbf{Z} \rbrace $

(iv) $ \lbrace x=\frac{n \pi}{2} ; n \in \mathbf{Z} \rbrace $

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.

88. If $f(x)=x^{2} \sin \frac{1}{x}$, 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

89. If f(x)= $\begin{cases}m x+1 \text{, if } x \leq \frac{\pi}{2} \\ \sin x+n \text{, if } x>\frac{\pi}{2}\end{cases} \quad $, is continuous at $x=\frac{\pi}{2}$, then

(A) $m=1, n=0$

(B) $m=\frac{n \pi}{2}+1$

(C) $n=\frac{m \pi}{2}$

(D) $m=n=\frac{\pi}{2}$

90. Let $f(x)=|\sin x|$. Then

(A) $f$ is everywhere differentiable

(B) $f$ is everywhere continuous but not differentiable at $x=n \pi, n \in \mathbf{Z}$.

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

(D) none of these

91. If $y=\log ( \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) $\frac{-4 x^{3}}{1-x^{4}}$

92. If $y=\sqrt{\sin x+y}$, then $\frac{d y}{d x}$ is equal to

(A) $\frac{\cos x}{2 y-1}$

(B) $\frac{\cos x}{1-2 y}$

(C) $\frac{\sin x}{1-2 y}$

(D) $\frac{\sin x}{2 y-1}$

93. The derivative of $\cos ^{-1}(2 x^{2}-1)$ w.r.t. $\cos ^{-1} x$ is

(A) 2

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

(C) $\frac{2}{x}$

(D) $1-x^{2}$

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) $\frac{3}{4}$

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) $\frac{1}{3}$

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

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____________

98. Derivative of $x^{2}$ w.r.t. $x^{3}$ is____________

99. If $f(x)=|\cos x|$, then $f^{\prime} (\frac{\pi}{4})=$____________

100 . If $f(x)=|\cos x-\sin x|$, then $f^{\prime} (\frac{\pi}{3})=$____________

101. For the curve $\sqrt{x}+\sqrt{y}=1, \frac{d y}{d x}$ at $ (\frac{1}{4}, \frac{1}{4})$ is____________

State True or False for the statements in each of the Exercises 102 to 106.

102. Rolle’s theorem is applicable for the function $f(x)=|x-1|$ in $[0,2]$.

103. If $f$ is continuous on its domain $D$, then $|f|$ is also continuous on $D$.

104. The composition of two continuous function is a continuous function.

105. Trigonometric and inverse - trigonometric functions are differentiable in their respective domain.

106. If $f . g$ is continuous at $x=a$, then $f$ and $g$ are separately continuous at $x=a$.

SOLUTIONS

1. Continuous at $x=1$

2. Discontinuous

3. Discontinuous

4. Continuous

5. Discontinuous

6. Continuous

7. Continuous

8. Discontinuous

9. Continuous

10. Continuous

11. $k=\frac{7}{2}$

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

13. $k=-1$

14. $k= \pm 1$

16. $a=1, b=-1$

17. Discontinuous at $x=-2$ and $x=\frac{-5}{2}$

18. Discontinuous at $x=1, \frac{1}{2}$ and 2

20. Not differentiable at $x=2$

21. Differentiable at $x=0$

22. Not differentiable at $x=2$

25. $-(\log 2) \cdot \sin 2 x \cdot 2^{\cos ^{2} x}$

26. $\frac{8^{x}}{x^{8}} \Big[\log 8-\frac{8}{x}\Big]$

27. $\frac{1}{\sqrt{x^{2}+a}}$

28. $\frac{5}{x \log (x^{5}) \log (\log x^{5})}$

29. $\frac{\cos \sqrt{x}}{2 \sqrt{x}}-\frac{\sin 2 \sqrt{x}}{2 \sqrt{x}}$

30. $n(2 a x+b) \sin ^{n-1}(a x^{2}+b x+c) \cos (a x^{2}+b x+c)$

31. $\frac{-1}{2 \sqrt{x+1}} \sin \Big(\tan \sqrt{x+1}\Big) \sec ^{2} \Big(\sqrt{x+1}\Big)$

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

33. $\frac{-1}{2 \sqrt{x}(x+1)}$

34. $(\sin x)^{\cos x} \Big[\frac{\cos ^{2} x}{\sin x}-\sin x \cdot \log \sin x\Big]$

35. $\sin ^{m x} x \cos ^{n} x(-n \tan x+m \cot x)$

36. $(x+1)(x+2)^{2}(x+3)^{3} \big[9 x^{2}+34 x+29\big]$

37. -1

38. $\frac{1}{2}$

39. $\frac{1}{2}$

40. -1

41. $\frac{-3}{\sqrt{1-x^{2}}}$

42. $\frac{3 a}{a^{2}+x^{2}}$

43. $\frac{-x}{\sqrt{1-x^{4}}}$

44. $\frac{t^{2}+1}{t^{2}-1}$

45. $e^{-2 \theta} \Big(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{3}+\theta^{2}+\theta-1}\Big)$

46. $\cot \theta$

47. 1

48. t

51. $-\frac{1}{\sqrt{3}}$

52. $\frac{\tan x-x}{\sin ^{2} x}$

53. $\frac{1}{2}$

54. $\frac{2 x y^{2}-y^{3} \cos (x y)-y}{x y^{2} \cos (x y)-x+y^{2}}$

55. $\frac{y-\sec (x+y) \tan (x+y)}{\sec (x+y) \tan (x+y)-x}$

56. $\frac{-x}{y}$

57. $\frac{y-4 x^{3}-4 x y^{2}}{4 y x^{2}+4 y^{3}-x}$

64. $-2 \sin y \cos ^{3} y$

70. Not applicable since $f$ is not differentiable at $x=1$

71. $(\pi,-2)$

72. $(2,-4)$

77. $\begin{pmatrix}\frac{7}{2}, & \frac{1}{4} \end{pmatrix}$

78. $\big(\frac{3}{2}, 0\big)$

79. $p=3, q=5$

82. $x^{\tan x} \Big(\sec ^{2} x \log x+\frac{\tan x}{x}\Big)+\frac{x}{\sqrt{2} \sqrt{x^{2}+1}} $

83. D

84. C

85. B

86. A

87. A

88. A

89. C

90. B

91. B

92. A

93. A

94. B

95. A

96. B

97. $|x|+|x-1|$

98. $\frac{2}{3 x}$

99. $\frac{-1}{\sqrt{2}}$

100. $\Big(\frac{\sqrt{3}+1}{2}\Big)$

101. -1

102. False

103. True

104. True

105. True

106. False