Q: Is the function $f$ defined by $f(x)=\left\{\begin{array}{l}x, \text { if } x \leq 1 \\ 5, \text { if } x>1\end{array}\right.$ continuous at $x=0$ ? At $x=1$ ? At $x=2$ ?

The given function is $f(x)=\left\{\begin{array}{l}x, \text { if } x \leq 1 \\ 5, \text { if } x>1\end{array}\right.$ At $x=0$, It is evident that $f$ is defined at 0 and its value at 0 is 0 . Then, $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}(x)=0$ $\therefore \lim _{x \rightarrow 0} f(x)=f(0)$ Therefore, $f$ is continuous at $x=0$. At $x=1$, It is evident that $f$ is defined at 1 and its value at 1 is 1 . The left hand limit of $f$ at $x=1$ is, $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x)=1$ The right hand limit of $f$ at $x=1$ is, $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(5)=5$ $\therefore \lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x)$ Therefore, $f$ is not continuous at $x=1$. At $x=2$, It is evident that $f$ is defined at 2 and its value at 2 is 5 . $\lim _{x \rightarrow 2} f(x)=\lim _{x \rightarrow 2}(5)=5$ $\therefore \lim _{x \rightarrow 1} f(x)=f(2)$ Therefore, $f$ is continuous at $x=2$.

Q: Find all points of discontinuity of $f$, where $f$ is defined by $f(x)=\left\{\begin{array}{l}2 x+3, \text { if } x \leq 2 \\ 2 x-3, \text { if } x>2 .\end{array}\right.$

The given function is $f(x)=\left\{\begin{array}{l}2 x+3, \text { if } x \leq 2 \\ 2 x-3, \text { if } x>2\end{array}\right.$ It is evident that the given function $f$ is defined at all the points of the real line. Let $c$ be a point on the real line. Then, three cases arise. $c 2$ $c=2$ Case I: $c 2$ Then, $f(c)=2 c-3$ $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2 x-3)=2 c-3$ $\therefore \lim _{x \rightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x>2$ Case III: $c=2$ Then, the left hand limit of $f$ at $x=2$ is, $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}(2 x+3)=2(2)+3=7$ The right hand limit of $f$ at $x=2$ is, $$ \lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x-3)=2(2)-3=1 $$ It is observed that the left and right hand limit of $f$ at $x=2$ do not coincide. Therefore, $f$ is not continuous at $x=2$. Hence, $x=2$ is the only point of discontinuity of $f$.

Q: Find all points of discontinuity of $f$, where $f$ is defined by $$ f(x)=\left\{\begin{array}{l} |x|+3, \text { if } x \leq-3 \\ -2 x, \text { if }-3<x<3 \\ 6 x+2, \text { if } x \geq 3 \end{array}\right. $$

The given function is $f(x)=\left\{\begin{array}{l}|x|+3, \text { if } x \leq-3 \\ -2 x, \text { if }-3<x<3 \\ 6 x+2, \text { if } x \geq 3\end{array}\right.$ The given function $f$ is defined at all the points of the real line. Let $c$ be a point on the real line. Case I: If $c<-3$, then $f(c)=-c+3$ $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-x+3)=-c+3$ $\therefore \lim _{x \rightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x<-3$. Case II: If $c=-3$, then $f(-3)=-(-3)+3=6$ $\lim _{x \rightarrow-3^{-}} f(x)=\lim _{x \rightarrow-3^{-}}(-x+3)=-(-3)+3=6$ $\lim _{x \rightarrow-3^{+}} f(x)=\lim _{x \rightarrow-3^{+}}(-2 x)=-2(-3)=6$ $\therefore \lim _{x \rightarrow-3} f(x)=f(-3)$ Therefore, $f$ is continuous at $x=-3$. \section*{Case III:} If $-3<c<3$, then $f(c)=-2 c$ $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-2 x)=-2 c$ $\therefore \lim _{x \rightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous in $(-3,3)$. \section*{Case IV:} If $c=3$, then the left hand limit of $f$ at $x=3$ is, $\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(-2 x)=-2(3)=-6$ The right hand limit of $f$ at $x=3$ is, $\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}(6 x+2)=6(3)+2=20$ It is observed that the left and right hand limit of $f$ at $x=3$ do not coincide. Therefore, $f$ is not continuous at $x=3$. \section*{Case V:} If $c>3$, then $f(c)=6 c+2$ $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(6 x+2)=6 c+2$ $\therefore \lim _{x \rightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x>3$. Hence, $x=3$ is the only point of discontinuity of $f$.

Q: Find all points of discontinuity of $f$, where $f$ is defined by $\left\{f(x)=\left\{\begin{array}{l}\frac{x}{|x|}, \text { if } x<0 \\ -1, \text { if } x \geq 0 .\end{array}\right.\right.$

The given function is $f(x)=\left\{\begin{array}{l}\frac{x}{|x|}, \text { if } x<0 \\ -1, \text { if } x \geq 0\end{array}\right.$ It is known that $x<0 \Rightarrow|x|=-x$ Therefore, the given function can be rewritten as $$ \begin{aligned} & f(x)=\left\{\begin{array}{l} \frac{x}{|x|}=\frac{x}{-x}=-1, \text { if } x<0 \\ -1, \text { if } x \geq 0 \end{array}\right. \\ & \Rightarrow f(x)=-1 \forall x \in R \end{aligned} $$ Let $c$ be any real number. Then, $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-1)=-1$ Also, $f(c)=-1=\lim _{x \rightarrow c} f(x)$ Therefore, the given function is a continuous function. Hence, the given function has no point of discontinuity.

Q: Find all points of discontinuity of $f$, where $f$ is defined by $f(x)=\left\{\begin{array}{l}x+1, \text { if } x \geq 1 \\ x^{2}+1, \text { if } x<1\end{array}\right.$

The given function is $f(x)=\left\{\begin{array}{l}x+1, \text { if } x \geq 1 \\ x^{2}+1, \text { if } x<1\end{array}\right.$ The given function $f$ is defined at all the points of the real line. Let $c$ be a point on the real line. \section*{Case I:} If $c<1$, then $f(c)=c^{2}+1$ $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{2}+1\right)=c^{2}+1$ $\therefore \lim _{v \rightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x<1$. Case II: If $c=1$, then $f(c)=f(1)=1+1=2$ The left hand limit of $f$ at $x=1$ is, $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(x^{2}+1\right)=1^{2}+1=2$ The right hand limit of $f$ at $x=1$ is, $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x+1)=1+1=2$ $\therefore \lim _{x \rightarrow 1} f(x)=f(1)$ Therefore, $f$ is continuous at $x=1$. Case III: If $c>1$, then $f(c)=c+1$ $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x+1)=c+1$ $\therefore \lim _{x \rightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x>1$. Hence, the given function $f$ has no point of discontinuity.

Question 1

Prove that the function $f(x)=5 x-3$ is continuous at $x=0, x=-3$ and at $x=5$.

Accepted Answer

The given function is $f(x)=5 x-3$

At $x=0, f(0)=5(0)-3=-3$

$\lim _{x ightarrow 0} f(x)=\lim _{x ightarrow 0}(5 x-3)=5(0)-3=-3$

$	herefore \lim _{x ightarrow 0} f(x)=f(0)$

Therefore, $f$ is continous at $x=0$.

At $x=-3, f(-3)=5(-3)-3=-18$

$\lim _{x ightarrow-3} f(x)=\lim _{x ightarrow-3}(5 x-3)=5(-3)-3=-18$

$	herefore \lim _{x ightarrow-3} f(x)=f(-3)$

Therefore, $f$ is continous at $x=-3$.

At $x=5, f(5)=5(5)-3=22$

$\lim _{x ightarrow 5} f(x)=\lim _{x ightarrow 5}(5 x-3)=5(5)-3=22$

$	herefore \lim _{x ightarrow 5} f(x)=f(5)$

Therefore, $f$ is continous at $x=5$.

Question 2

Examine the continuity of the function $f(x)=2 x^{2}-1$ at $x=3$.

Accepted Answer

The given function is $f(x)=2 x^{2}-1$

At $x=3, f(3)=2(3)^{2}-1=17$

$\lim _{x ightarrow 3} f(x)=\lim _{x ightarrow 3}\left(2 x^{2}-1ight)=2\left(3^{2}ight)-1=17$

$	herefore \lim _{x ightarrow 3} f(x)=f(3)$

Therefore, $f$ is continous at $x=3$.

Question 3

Examine the following functions for continuity.

(i)

$$

f(x)=x-5

$$

(ii) $f(x)=\frac{1}{x-5}, x 
eq 5$

(iii) $f(x)=\frac{x^{2}-25}{x+5}, x 
eq-5$

(iv) $f(x)=|x-5|, x 
eq 5$

Accepted Answer

(i) The given function is $f(x)=x-5$ It is evident that $f$ is defined at every real number $k$ and its value at $k$ is $k-5$. It is also observed that $$ \begin{aligned} & \lim _{x ightarrow k} f(x)=\lim _{x ightarrow k}(x-5)=k-5=f(k) \ & herefore \lim _{x ightarrow k} f(x)=f(k) \end{aligned} $$ Hence, $f$ is continuous at every real number and therefore, it is a continuous function. (ii) The given function is $f(x)=\frac{1}{x-5}, x eq 5$ For any real number $k eq 5$, we obtain $$ \lim _{x ightarrow k} f(x)=\lim _{x ightarrow k} \frac{1}{x-5}=\frac{1}{k-5} $$ Also, $$ \begin{aligned} & f(k)=\frac{1}{k-5} \quad( ext { As } k eq 5) \ & herefore \lim _{x ightarrow k} f(x)=f(k) \end{aligned} $$ Hence, $f$ is continuous at every point in the domain of $f$ and therefore, it is a continuous function. (iii) The given function is $f(x)=\frac{x^{2}-25}{x+5}, x eq-5$ For any real number $c eq-5$, we obtain $$ \lim _{x ightarrow c} f(x)=\lim _{x ightarrow c} \frac{x^{2}-25}{x+5}=\lim _{x ightarrow c} \frac{(x+5)(x-5)}{x+5}=\lim _{x ightarrow c}(x-5)=(c-5) $$ Also, $$ \begin{aligned} & f(c)=\frac{(c+5)(c-5)}{c+5}=(c-5) \ & herefore \lim _{x ightarrow c} f(x)=f(c) \end{aligned} $$ Hence, $f$ is continuous at every point in the domain of $f$ and therefore, it is a continuous function. (iv) The given function is $f(x)=|x-5|=\left\{\begin{array}{l}5-x, ext { if } x<5 \ x-5, ext { if } x \geq 5\end{array} ight\}$ This function $f$ is defined at all points of the real line. Let c be a point on a real line. Then, $c<5, c=5$ or $c>5$ Case I: $c<5$ Then, $f(c)=5-c$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(5-x)=5-c$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all real numbers less than 5 . Case II: $c=5$ Then, $f(c)=f(5)=(5-5)=0$ $\lim _{x ightarrow 5^{-}} f(x)=\lim _{x ightarrow 5}(5-x)=(5-5)=0$ $\lim _{x ightarrow 5^{+}} f(x)=\lim _{x ightarrow 5}(x-5)=0$ $ herefore \lim _{x ightarrow c^{-}} f(x)=\lim _{x ightarrow c^{+}} f(x)=f(c)$ Therefore, $f$ is continuous at $x=5$ Case III: $c>5$ Then, $f(c)=f(5)=c-5$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(x-5)=c-5$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all real numbers greater than 5 . Hence, $f$ is continuous at every real number and therefore, it is a continuous function.

Question 4

Prove that the function $f(x)=x^{n}$ is continuous at $x=n$, where $n$ is a positive integer.

Accepted Answer

The given function is $f(x)=x^{n}$

It is observed that $f$ is defined at all positive integers, $n$, and its value at $n$ is $n^{n}$.

Then,

$\lim _{x ightarrow n} f(n)=\lim _{x ightarrow n}\left(x^{n}ight)=x^{n}$

$	herefore \lim _{x ightarrow n} f(x)=f(n)$

Therefore, $f$ is continuous at $n$, where $n$ is a positive integer.

Question 5

Is the function $f$ defined by $f(x)=\left\{\begin{array}{l}x, 	ext { if } x \leq 1 \ 5, 	ext { if } x>1\end{array}ight.$ continuous at $x=0$ ? At $x=1$ ? At $x=2$ ?

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}x, 	ext { if } x \leq 1 \ 5, 	ext { if } x>1\end{array}ight.$

At $x=0$,

It is evident that $f$ is defined at 0 and its value at 0 is 0 .

Then,

$\lim _{x ightarrow 0} f(x)=\lim _{x ightarrow 0}(x)=0$

$	herefore \lim _{x ightarrow 0} f(x)=f(0)$

Therefore, $f$ is continuous at $x=0$.

At $x=1$,

It is evident that $f$ is defined at 1 and its value at 1 is 1 .

The left hand limit of $f$ at $x=1$ is,

$\lim _{x ightarrow 1^{-}} f(x)=\lim _{x ightarrow 1^{-}}(x)=1$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x ightarrow 1^{+}} f(x)=\lim _{x ightarrow 1^{+}}(5)=5$

$	herefore \lim _{x ightarrow 1^{-}} f(x) 
eq \lim _{x ightarrow 1^{+}} f(x)$

Therefore, $f$ is not continuous at $x=1$.

At $x=2$,

It is evident that $f$ is defined at 2 and its value at 2 is 5 .

$\lim _{x ightarrow 2} f(x)=\lim _{x ightarrow 2}(5)=5$

$	herefore \lim _{x ightarrow 1} f(x)=f(2)$

Therefore, $f$ is continuous at $x=2$.

Question 6

Find all points of discontinuity of $f$, where $f$ is defined by $f(x)=\left\{\begin{array}{l}2 x+3, 	ext { if } x \leq 2 \ 2 x-3, 	ext { if } x>2 .\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}2 x+3, ext { if } x \leq 2 \ 2 x-3, ext { if } x>2\end{array} ight.$ It is evident that the given function $f$ is defined at all the points of the real line. Let $c$ be a point on the real line. Then, three cases arise. $c<2$ $c>2$ $c=2$ Case I: $c<2$ $f(c)=2 c+3$ Then, $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(2 x+3)=2 c+3$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x<2$. Case II: $c>2$ Then, $f(c)=2 c-3$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(2 x-3)=2 c-3$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x>2$ Case III: $c=2$ Then, the left hand limit of $f$ at $x=2$ is, $\lim _{x ightarrow 2^{-}} f(x)=\lim _{x ightarrow 2^{-}}(2 x+3)=2(2)+3=7$ The right hand limit of $f$ at $x=2$ is, $$ \lim _{x ightarrow 2^{+}} f(x)=\lim _{x ightarrow 2^{+}}(2 x-3)=2(2)-3=1 $$ It is observed that the left and right hand limit of $f$ at $x=2$ do not coincide. Therefore, $f$ is not continuous at $x=2$. Hence, $x=2$ is the only point of discontinuity of $f$.

Question 7

Find all points of discontinuity of $f$, where $f$ is defined by

$$

f(x)=\left\{\begin{array}{l}

|x|+3, 	ext { if } x \leq-3 \

-2 x, 	ext { if }-3<x<3 \

6 x+2, 	ext { if } x \geq 3

\end{array}ight.

$$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}|x|+3, ext { if } x \leq-3 \ -2 x, ext { if }-33$, then $f(c)=6 c+2$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(6 x+2)=6 c+2$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x>3$. Hence, $x=3$ is the only point of discontinuity of $f$.

Question 8

Find all points of discontinuity of $f$, where $f$ is defined by $f(x)=\left\{\begin{array}{l}\frac{|x|}{x} 	ext {, if } x 
eq 0 \ 0 	ext {, if } x=0\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}\frac{|x|}{x}, ext { if } x eq 0 \ 0, ext { if } x=0\end{array} ight.$ It is known that, $x<0 \Rightarrow|x|=-x$ and $x>0 \Rightarrow|x|=x$ Therefore, the given function can be rewritten as $f(x)=\left\{\begin{array}{l}\frac{|x|}{x}=\frac{-x}{x}=-1, ext { if } x<0 \ 0, ext { if } x=0 \ \frac{|x|}{x}=\frac{x}{x}=1, ext { if } x>0\end{array} ight.$ The given function $f$ is defined at all the points of the real line. Let $c$ be a point on the real line. Case I: If $c<0$, then $f(c)=-1$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(-1)=-1$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x<0$. Case II: If $c=0$, then the left hand limit of $f$ at $x=0$ is, $\lim _{x ightarrow 0^{-}} f(x)=\lim _{x ightarrow 0^{-}}(-1)=-1$ The right hand limit of $f$ at $x=0$ is, $\lim _{x ightarrow 0^{+}} f(x)=\lim _{x ightarrow 0^{+}}(1)=1$ It is observed that the left and right hand limit of $f$ at $x=0$ do not coincide. Therefore, $f$ is not continuous at $x=0$. \section*{Case III:} If $c>0$, then $f(c)=1$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(1)=1$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x>0$. Hence, $x=0$ is the only point of discontinuity of $f$.

Question 9

Find all points of discontinuity of $f$, where $f$ is defined by $\left\{f(x)=\left\{\begin{array}{l}\frac{x}{|x|}, 	ext { if } x<0 \ -1, 	ext { if } x \geq 0 .\end{array}ight.ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}\frac{x}{|x|}, 	ext { if } x<0 \ -1, 	ext { if } x \geq 0\end{array}ight.$

It is known that $x<0 \Rightarrow|x|=-x$

Therefore, the given function can be rewritten as

$$

\begin{aligned}

& f(x)=\left\{\begin{array}{l}

\frac{x}{|x|}=\frac{x}{-x}=-1, 	ext { if } x<0 \

-1, 	ext { if } x \geq 0

\end{array}ight. \

& \Rightarrow f(x)=-1 \forall x \in R

\end{aligned}

$$

Let $c$ be any real number.

Then, $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(-1)=-1$

Also, $f(c)=-1=\lim _{x ightarrow c} f(x)$

Therefore, the given function is a continuous function.

Hence, the given function has no point of discontinuity.

Question 10

Find all points of discontinuity of $f$, where $f$ is defined by $f(x)=\left\{\begin{array}{l}x+1, 	ext { if } x \geq 1 \ x^{2}+1, 	ext { if } x<1\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}x+1, ext { if } x \geq 1 \ x^{2}+1, ext { if } x<1\end{array} ight.$ The given function $f$ is defined at all the points of the real line. Let $c$ be a point on the real line. \section*{Case I:} If $c<1$, then $f(c)=c^{2}+1$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}\left(x^{2}+1 ight)=c^{2}+1$ $ herefore \lim _{v ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x<1$. Case II: If $c=1$, then $f(c)=f(1)=1+1=2$ The left hand limit of $f$ at $x=1$ is, $\lim _{x ightarrow 1^{-}} f(x)=\lim _{x ightarrow 1^{-}}\left(x^{2}+1 ight)=1^{2}+1=2$ The right hand limit of $f$ at $x=1$ is, $\lim _{x ightarrow 1^{+}} f(x)=\lim _{x ightarrow 1^{+}}(x+1)=1+1=2$ $ herefore \lim _{x ightarrow 1} f(x)=f(1)$ Therefore, $f$ is continuous at $x=1$. Case III: If $c>1$, then $f(c)=c+1$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(x+1)=c+1$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x>1$. Hence, the given function $f$ has no point of discontinuity.

Question 11

Find all points of discontinuity of $f$, where $f$ is defined by $f(x)=\left\{\begin{array}{l}x^{3}-3, 	ext { if } x \leq 2 \ x^{2}+1, 	ext { if } x>2\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}x^{3}-3, ext { if } x \leq 2 \ x^{2}+1, ext { if } x>2\end{array} ight.$ The given function $f$ is defined at all the points of the real line. Let $c$ be a point on the real line. Case I: If $c<2$, then $f(c)=c^{3}-3$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}\left(x^{3}-3 ight)=c^{3}-3$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x<2$. Case II: If $c=2$, then $f(c)=f(2)=2^{3}-3=5$ $\lim _{x ightarrow 2^{-}} f(x)=\lim _{x ightarrow 2^{-}}\left(x^{3}-3 ight)=2^{3}-3=5$ $\lim _{x ightarrow 2^{+}} f(x)=\lim _{x ightarrow 2^{+}}\left(x^{2}+1 ight)=2^{2}+1=5$ $ herefore \lim _{x ightarrow 2} f(x)=f(2)$ Therefore, $f$ is continuous at $x=2$. Case III: If $c>2$, then $f(c)=c^{2}+1$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}\left(x^{2}+1 ight)=c^{2}+1$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x>2$. Thus, the given function $f$ is continuous at every point on the real line. Hence, $f$ has no point of discontinuity.

Question 12

For what value of $\lambda$ is the function defined by $f(x)=\left\{\begin{array}{l}\lambda\left(x^{2}-2 xight), 	ext { if } x \leq 0 \ 4 x+1, 	ext { if } x>0 \quad 	ext { is continous at }\end{array}ight. x=0$ ? What about continuity at $x=1$ ?

Accepted Answer

The given function is

$$

f(x)=\left\{\begin{array}{l}

\lambda\left(x^{2}-2 xight), 	ext { if } x \leq 0 \

4 x+1, 	ext { if } x>0

\end{array}ight.

$$

If $f$ is continuous at $x=0$, then

$\lim _{x ightarrow 0^{-}} f(x)=\lim _{x ightarrow 0^{+}} f(x)=f(0)$

$\Rightarrow \lim _{x ightarrow 0^{-}} \lambda\left(x^{2}-2 xight)=\lim _{x ightarrow 0^{+}}(4 x+1)=\lambda\left(0^{2}-2 	imes 0ight)$

$\Rightarrow \lambda\left(0^{2}-2 	imes 0ight)=4(0)+1=0$

$\Rightarrow 0=1=0 \quad$ [which is not possible]

Therefore, there is no value of $\lambda$ for which $f$ is continuous at $x=0$.

At $x=1$

$f(1)=4 x+1=4(1)+1=5$

$\lim _{x ightarrow 1}(4 x+1)=4(1)+1=5$

$	herefore \lim _{x ightarrow 1} f(x)=f(1)$

Therefore, for any values of $\lambda, f$ is continuous at $x=1$.

Question 13

Find all points of discontinuity of $f$, where $f$ is defined by $f(x)=\left\{\begin{array}{l}x^{10}-1, 	ext { if } x \leq 1 \ x^{2}, 	ext { if } x>1\end{array}ight.$.

Accepted Answer

The given function is $$ f(x)=\left\{\begin{array}{l} x^{10}-1, ext { if } x \leq 1 \ x^{2}, ext { if } x>1 \end{array} ight. $$ The given function $f$ is defined at all the points of the real line. Let $c$ be a point on the real line. Case I: If $c<1$, then $f(c)=c^{10}-1$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}\left(x^{10}-1 ight)=c^{10}-1$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x<1$. Case II: If $c=1$, then the left hand limit of $f$ at $x=1$ is, $\lim _{x ightarrow 1^{-}} f(x)=\lim _{x ightarrow 1^{-}}\left(x^{10}-1 ight)=1^{10}-1=1-1=0$ The right hand limit of $f$ at $x=1$ is, $\lim _{x ightarrow 1^{+}} f(x)=\lim _{x ightarrow 1^{+}}\left(x^{2} ight)=1^{2}=1$ It is observed that the left and right hand limit of $f$ at $x=1$ do not coincide. Therefore, $f$ is not continuous at $x=1$. Case III: If $c>1$, then $f(c)=c^{2}$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}\left(x^{2} ight)=c^{2}$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x>1$. Thus from the above observation, it can be concluded that $x=1$ is the only point of discontinuity of $f$.

Question 14

Is the function defined by $f(x)=\left\{\begin{array}{l}x+5, 	ext { if } x \leq 1 \ x-5, 	ext { if } x>1 	ext { a continous function? }\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}x+5, ext { if } x \leq 1 \ x-5, ext { if } x>1\end{array} ight.$ The given function $f$ is defined at all the points of the real line. Let $c$ be a point on the real line. \section*{Case I:} If $c<1$, then $f(c)=c+5$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(x+5)=c+5$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x<1$. \section*{Case II:} If $c=1$, then $f(1)=1+5=6$ The left hand limit of $f$ at $x=1$ is, $\lim _{x ightarrow 1^{-}} f(x)=\lim _{x ightarrow 1^{-}}(x+5)=1+5=6$ The right hand limit of $f$ at $x=1$ is, $\lim _{x ightarrow 1^{+}} f(x)=\lim _{x ightarrow 1^{+}}(x-5)=1-5=-4$ It is observed that the left and right hand limit of $f$ at $x=1$ do not coincide. Therefore, $f$ is not continuous at $x=1$. \section*{Case III:} If $c>1$, then $f(c)=c-5$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(x-5)=c-5$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x>1$. From the above observation it can be concluded that, $x=1$ is the only point of discontinuity of $f$.

Question 15

Discuss the continuity of the function $f$, where $f$ is defined by $f(x)=\left\{\begin{array}{l}3, 	ext { if } 0 \leq x \leq 1 \ 4, 	ext { if } 1<x<3 \ 5, 	ext { if } 3 \leq x \leq 10\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}3, 	ext { if } 0 \leq x \leq 1 \ 4, 	ext { if } 1<x<3 \ 5, 	ext { if } 3 \leq x \leq 10\end{array}ight.$

The given function $f$ is defined at all the points of the interval $[0,10]$.

Let $c$ be a point in the interval $[0,10]$.

\section*{Case I:}

If $0 \leq c<1$, then $f(c)=3$

$\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(3)=3$

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

Therefore, $f$ is continuous in the interval $[0,1)$.

Case II:

If $c=1$, then $f(3)=3$

The left hand limit of $f$ at $x=1$ is,

$\lim _{x ightarrow 1^{-}} f(x)=\lim _{x ightarrow 1^{-}}(3)=3$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x ightarrow 1^{+}} f(x)=\lim _{x ightarrow 1^{+}}(4)=4$

It is observed that the left and right hand limit of $f$ at $x=1$ do not coincide.

Therefore, $f$ is not continuous at $x=1$.

\section*{Case III:}

If $1<c<3$, then $f(c)=4$

$\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(4)=4$

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

Therefore, $f$ is continuous at in the interval $(1,3)$.

Case IV:

If $c=3$, then $f(c)=5$

The left hand limit of $f$ at $x=3$ is,

$\lim _{x ightarrow 3^{-}} f(x)=\lim _{x ightarrow 3^{-}}(4)=4$

The right hand limit of $f$ at $x=3$ is,

$$

\lim _{x ightarrow 3^{+}} f(x)=\lim _{x ightarrow 3^{+}}(5)=5

$$

It is observed that the left and right hand limit of $f$ at $x=3$ do not coincide.

Therefore, $f$ is discontinuous at $x=3$.

\section*{Case V:}

If $3<c \leq 10$, then $f(c)=5$

$\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(5)=5$

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

Therefore, $f$ is continuous at all points of the interval $(3,10]$.

Hence, $f$ is discontinuous at $x=1$ and $x=3$.

Question 16

Discuss the continuity of the function $f$, where $f$ is defined by $f(x)=\left\{\begin{array}{l}2 x, 	ext { if } x<0 \ 0, 	ext { if } 0 \leq x \leq 1 \ 4 x, 	ext { if } x>1\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}2 x, ext { if } x<0 \ 0, ext { if } 0 \leq x \leq 1 \ 4 x, ext { if } x>1\end{array} ight.$ The given function $f$ is defined at all the points of the real line. Let $c$ be a point on the real line. \section*{Case I:} If $c<0$, then $f(c)=2 c$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(2 x)=2 c$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x<0$. Case II: If $c=0$, then $f(c)=f(0)=0$ The left hand limit of $f$ at $x=0$ is, $\lim _{x ightarrow 0^{-}} f(x)=\lim _{x ightarrow 0^{-}}(2 x)=2(0)=0$ The right hand limit of $f$ at $x=0$ is, $\lim _{x ightarrow 0^{+}} f(x)=\lim _{x ightarrow 0^{+}}(0)=0$ $ herefore \lim _{x ightarrow 0} f(x)=f(0)$ Therefore, $f$ is continuous at $x=0$ \section*{Case III:} If $01$. Hence, $f$ is not continuous only at $x=1$.

Question 17

Discuss the continuity of the function $f$, where $f$ is defined by $f(x)=\left\{\begin{array}{l}-2, 	ext { if } x \leq-1 \ 2 x, 	ext { if }-1<x \leq 1 \ 2, 	ext { if } x>1\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}-2, ext { if } x \leq-1 \ 2 x, ext { if }-11\end{array} ight.$ The given function $f$ is defined at all the points. Let $c$ be a point on the real line. \section*{Case I:} If $c<-1$, then $f(c)=-2$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(-2)=-2$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x<-1$. Case II: If $c=-1$, then $f(c)=f(-1)=-2$ The left hand limit of $f$ at $x=-1$ is, $\lim _{x ightarrow-1^{-}} f(x)=\lim _{x ightarrow-1^{-}}(-2)=-2$ The right hand limit of $f$ at $x=-1$ is, $\lim _{x ightarrow-1^{+}} f(x)=\lim _{x ightarrow-1^{+}}(2 x)=2(-1)=-2$ $ herefore \lim _{x ightarrow-1} f(x)=f(-1)$ Therefore, $f$ is continuous at $x=-1$ \section*{Case III:} If $-11$, then $f(c)=2$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(2)=2$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x>1$. Thus, from the above observations, it can be concluded that $f$ is continuous at all points of the real line.

Question 18

Find the relationship between $a$ and $b$ so that the function $f$ defined by $f(x)= \begin{cases}a x+1, & 	ext { if } x \leq 3 \ b x+3, & 	ext { if } x>3\end{cases}$ is continous at $x=3$.

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}a x+1, 	ext { if } x \leq 3 \ b x+3, 	ext { if } x>3\end{array}ight.$

For $f$ to be continuous at $x=3$, then

$$

\begin{equation*}

\lim _{x ightarrow 3^{-}} f(x)=\lim _{x ightarrow 3^{+}} f(x)=f(3) 	ag{1}

\end{equation*}

$$

Also,

$\lim _{x ightarrow 3^{-}} f(x)=\lim _{x ightarrow 3^{-}}(a x+1)=3 a+1$

$\lim _{x ightarrow 3^{+}} f(x)=\lim _{x ightarrow 3^{+}}(b x+3)=3 b+3$

$f(3)=3 a+1$

Therefore, from (1), we obtain

$3 a+1=3 b+3=3 a+1$

$\Rightarrow 3 a+1=3 b+3$

$\Rightarrow 3 a=3 b+2$

$\Rightarrow a=b+\frac{2}{3}$

Therefore, the required relationship is given by, $a=b+\frac{2}{3}$.

Question 19

Show that the function defined by $g(x)=x-[x]$ is discontinuous at all integral point. Here $[x]$ denotes the greatest integer less than or equal to $x$.

Accepted Answer

The given function is $g(x)=x-[x]$

It is evident that $g$ is defined at all integral points.

Let $n$ be an integer.

Then,

$g(n)=n-[n]=n-n=0$

The left hand limit of $g$ at $x=n$ is,

$\lim _{x \rightarrow n^{-}} g(x)=\lim _{x \rightarrow n^{-}}(x-[x])=\lim _{x \rightarrow n^{-}}(x)-\lim _{x \rightarrow n^{-}}[x]=n-(n-1)=1$

The right hand limit of $g$ at $x=n$ is,

$\lim _{x \rightarrow n^{+}} g(x)=\lim _{x \rightarrow n^{+}}(x-[x])=\lim _{x \rightarrow n^{+}}(x)-\lim _{x \rightarrow n^{+}}[x]=n-n=0$

It is observed that the left and right hand limit of $g$ at $x=n$ do not coincide.

Therefore, $g$ is not continuous at $x=n$.

Hence, $g$ is discontinuous at all integral points.

Question 20

Is the function defined by $f(x)=x^{2}-\sin x+5$ continuous at $x=\pi$ ?

Accepted Answer

The given function is $f(x)=x^{2}-\sin x+5$

It is evident that $f$ is defined at $x=\pi$.

At $x=\pi, f(x)=f(\pi)=\pi^{2}-\sin \pi+5=\pi^{2}-0+5=\pi^{2}+5$

Consider $\lim _{x ightarrow \pi} f(x)=\lim _{x ightarrow \pi}\left(x^{2}-\sin x+5ight)$

Put $x=\pi+h$, it is evident that if $x ightarrow \pi$, then $h ightarrow 0$

$$

\begin{aligned}

herefore \lim _{x ightarrow \pi} f(x) & =\lim _{x ightarrow \pi}\left(x^{2}-\sin xight)+5 \

& =\lim _{h ightarrow 0}\left[(\pi+h)^{2}-\sin (\pi+h)+5ight] \

& =\lim _{h ightarrow 0}(\pi+h)^{2}-\lim _{h ightarrow 0} \sin (\pi+h)+\lim _{h ightarrow 0} 5 \

& =(\pi+0)^{2}-\lim _{h ightarrow 0}[\sin \pi \cos h+\cos \pi \sin h]+5 \

& =\pi^{2}-\lim _{h ightarrow 0} \sin \pi \cos h-\lim _{h ightarrow 0} \cos \pi \sin h+5 \

& =\pi^{2}-\sin \pi \cos 0-\cos \pi \sin 0+5 \

& =\pi^{2}-0(1)-(-1) 0+5 \

& =\pi^{2}+5 \

& =f(\pi)

\end{aligned}

$$

Therefore, the given function $f$ is continuous at $x=\pi$.

Question 21

Discuss the continuity of the following functions.

(i) $f(x)=\sin x+\cos x$

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

(iii) $f(x)=\sin x 	imes \cos x$

Accepted Answer

It is known that if $g$ and $h$ are two continuous functions, then $g+h, g-h$ and $g, h$ are also continuous.

Let $g(x)=\sin x$ and $h(x)=\cos x$ are continuous functions.

It is evident that $g(x)=\sin x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+h$

If $x ightarrow c$, then $h ightarrow 0$

$$

\begin{aligned}

& g(c)=\sin c \

& \begin{aligned}

\lim _{x ightarrow c} g(x) & =\lim _{x ightarrow c} \sin x \

& =\lim _{h ightarrow 0} \sin (c+h) \

& =\lim _{h ightarrow 0}[\sin c \cos h+\cos c \sin h] \

& =\lim _{h ightarrow 0}(\sin c \cos h)+\lim _{h ightarrow 0}(\cos c \sin h) \

& =\sin c \cos 0+\cos c \sin 0 \

& =\sin c(1)+\cos c(0) \

& =\sin c

\end{aligned} \

& 	herefore \lim _{x ightarrow c} g(x)=g(c)

\end{aligned}

$$

Therefore, $g(x)=\sin x$ is a continuous function.

Let $h(x)=\cos x$

It is evident that $h(x)=\cos x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+h$

If $x ightarrow c$, then $h ightarrow 0$

$h(c)=\cos c$

$$

\begin{aligned}

\lim _{x ightarrow c} h(x) & =\lim _{x ightarrow c} \cos x \

& =\lim _{h ightarrow 0} \cos (c+h) \

& =\lim _{h ightarrow 0}[\cos c \cos h-\sin c \sin h] \

& =\lim _{h ightarrow 0}(\cos c \cos h)-\lim _{h ightarrow 0}(\sin c \sin h) \

& =\cos c \cos 0-\sin c \sin 0 \

& =\cos c(1)-\sin c(0) \

& =\cos c

\end{aligned}

$$

$	herefore \lim _{x ightarrow c} h(x)=h(c)$

Therefore, $h(x)=\cos x$ is a continuous function.

Therefore, it can be concluded that,

(i) $f(x)=g(x)+h(x)=\sin x+\cos x$ is a continuous function.

(ii) $f(x)=g(x)-h(x)=\sin x-\cos x$ is a continuous function.

(iii) $f(x)=g(x) 	imes h(x)=\sin x 	imes \cos x$ is a continuous function.

Question 22

Discuss the continuity of the cosine, cosecant, secant, and cotangent functions.

Accepted Answer

It is known that if $g$ and $h$ are two continuous functions, then

(i) $\frac{h(x)}{g(x)}, g(x) 
eq 0$ is continuous.

(ii) $\frac{1}{g(x)}, g(x) 
eq 0$ is continuous.

(iii) $\frac{1}{h(x)}, h(x) 
eq 0$ is continuous.

Let $g(x)=\sin x$ and $h(x)=\cos x$ are continuous functions.

It is evident that $g(x)=\sin x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+h$

If $x ightarrow c$, then $h ightarrow 0$

$$

\begin{aligned}

g(c) & =\sin c \

\lim _{x ightarrow c} g(x) & =\lim _{x ightarrow c} \sin x \

& =\lim _{h ightarrow 0} \sin (c+h) \

& =\lim _{h ightarrow 0}[\sin c \cos h+\cos c \sin h] \

& =\lim _{h ightarrow 0}(\sin c \cos h)+\lim _{h ightarrow 0}(\cos c \sin h) \

& =\sin c \cos 0+\cos c \sin 0 \

& =\sin c(1)+\cos c(0) \

& =\sin c

\end{aligned}

$$

$	herefore \lim _{x ightarrow c} g(x)=g(c)$

Therefore, $g(x)=\sin x$ is a continuous function.

Let $h(x)=\cos x$

It is evident that $h(x)=\cos x$ is defined for every real number.

Let $c$ be a real number. Put $x=c+h$

If $x ightarrow c$, then $h ightarrow 0$

$$

\begin{aligned}

h(c) & =\cos c \

\lim _{x ightarrow c} h(x) & =\lim _{x ightarrow c} \cos x \

& =\lim _{h ightarrow 0} \cos (c+h) \

& =\lim _{h ightarrow 0}[\cos c \cos h-\sin c \sin h] \

& =\lim _{h ightarrow 0}(\cos c \cos h)-\lim _{h ightarrow 0}(\sin c \sin h) \

& =\cos c \cos 0-\sin c \sin 0 \

& =\cos c(1)-\sin c(0) \

& =\cos c

\end{aligned}

$$

$	herefore \lim _{x ightarrow c} h(x)=h(c)$

Therefore, $h(x)=\cos x$ is a continuous function.

Therefore, it can be concluded that, $\operatorname{cosec} x=\frac{1}{\sin x}, \sin x 
eq 0$ is continuous.

$\Rightarrow \operatorname{cosec} x, x 
eq n \pi(n \in Z)$ is continuous.

Therefore, cosecant is continuous except at $x=n \pi(n \in Z)$

$\sec x=\frac{1}{\cos x}, \cos x 
eq 0$ is continuous.

$\Rightarrow \sec x, x 
eq(2 n+1) \frac{\pi}{2}(n \in Z)$ is continuous.

Therefore, secant is continuous except at $x=(2 n+1) \frac{\pi}{2}(n \in Z)$

$\cot x=\frac{\cos x}{\sin x}, \sin x 
eq 0$ is continuous.

$\Rightarrow \cot x, x 
eq n \pi(n \in Z)$ is continuous.

Therefore, cotangent is continuous except at $x=n \pi(n \in Z)$.

Question 23

Find the points of discontinuity of $f$, where $\quad f(x)=\left\{\begin{array}{l}\frac{\sin x}{x}, 	ext { if } x<0 \ x+1, 	ext { if } x \geq 0\end{array}ight.$

Accepted Answer

The given function is $f(x)= \begin{cases}\frac{\sin x}{x}, & ext { if } x<0 \ x+1, & ext { if } x \geq 0\end{cases}$ The given function $f$ is defined at all the points of the real line. Let $c$ be a point on the real line. Case I: If $c<0$, then $f(c)=\frac{\sin c}{c}$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}\left(\frac{\sin x}{x} ight)=\frac{\sin c}{c}$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x<0$. Case II: If $c>0$, then $f(c)=c+1$ $\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(x+1)=c+1$ $ herefore \lim _{x ightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x>0$. \section*{Case III:} If $c=0$, then $f(c)=f(0)=0+1=1$ The left hand limit of $f$ at $x=0$ is, $\lim _{x ightarrow 0^{-}} f(x)=\lim _{x ightarrow 0^{-}}\left(\frac{\sin x}{x} ight)=1$ The right hand limit of $f$ at $x=0$ is, $\lim _{x ightarrow 0^{+}} f(x)=\lim _{x ightarrow 0^{+}}(x+1)=1$ $ herefore \lim _{x ightarrow 0^{-}} f(x)=\lim _{x ightarrow 0^{+}} f(x)=f(0)$ Therefore, $f$ is continuous at $x=0$ From the above observations, it can be concluded that $f$ is continuous at all points of the real line. Thus, $f$ has no point of discontinuity.

Question 24

Determine if $f$ defined by $f(x)=\left\{\begin{array}{l}x^{2} \sin \frac{1}{x}, 	ext { if } x 
eq 0 \ 0, 	ext { if } x=0 \quad 	ext { is a continuous function? }\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}x^{2} \sin \frac{1}{x}, 	ext { if } x 
eq 0 \ 0, 	ext { if } x=0\end{array}ight.$

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c 
eq 0$, then $f(c)=c^{2} \sin \frac{1}{c}$

$\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}\left(x^{2} \sin \frac{1}{x}ight)=\left(\lim _{x ightarrow c} x^{2}ight)\left(\lim _{x ightarrow c} \sin \frac{1}{x}ight)=c^{2} \sin \frac{1}{c}$

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

Therefore, $f$ is continuous at all points $x$, such that $x 
eq 0$.

Case II:

If $c=0$, then $f(0)=0$

$\lim _{x ightarrow 0^{-}} f(x)=\lim _{x ightarrow 0^{-}}\left(x^{2} \sin \frac{1}{x}ight)=\lim _{x ightarrow 0}\left(x^{2} \sin \frac{1}{x}ight)$

It is known that, $-1 \leq \sin \frac{1}{x} \leq 1, x 
eq 0$

$\Rightarrow-x^{2} \leq x^{2} \sin \frac{1}{x} \leq x^{2}$

$\Rightarrow \lim _{x ightarrow 0}\left(-x^{2}ight) \leq \lim _{x ightarrow 0}\left(x^{2} \sin \frac{1}{x}ight) \leq \lim _{x ightarrow 0} x^{2}$

$\Rightarrow 0 \leq \lim _{x ightarrow 0}\left(x^{2} \sin \frac{1}{x}ight) \leq 0$

$\Rightarrow \lim _{x ightarrow 0}\left(x^{2} \sin \frac{1}{x}ight)=0$

$	herefore \lim _{x ightarrow 0^{-}} f(x)=0$

Similarly,

$\lim _{x ightarrow 0^{+}} f(x)=\lim _{x ightarrow 0^{+}}\left(x^{2} \sin \frac{1}{x}ight)=\lim _{x ightarrow 0}\left(x^{2} \sin \frac{1}{x}ight)=0$

$	herefore \lim _{x ightarrow 0^{-}} f(x)=f(0)=\lim _{x ightarrow 0^{+}} f(x)$

Therefore, $f$ is continuous at $x=0$.

From the above observations, it can be concluded that $f$ is continuous at every point of the real line.

Thus, $f$ is a continuous function.

Question 25

Examine the continuity of $f$, where $f$ is defined by $f(x)=\left\{\begin{array}{l}\sin x-\cos x, 	ext { if } x 
eq 0 \ -1, 	ext { if } x=0\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}\sin x-\cos x, 	ext { if } x 
eq 0 \ -1, 	ext { if } x=0\end{array}ight.$

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

\section*{Case I:}

If $c 
eq 0$, then $f(c)=\sin c-\cos c$

$\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(\sin x-\cos x)=\sin c-\cos c$

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

Therefore, $f$ is continuous at all points $x$, such that $x 
eq 0$.

\section*{Case II:}

If $c=0$, then $f(0)=-1$

$\lim _{x ightarrow 0^{-}} f(x)=\lim _{x ightarrow 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1$

$\lim _{x ightarrow 0^{+}} f(x)=\lim _{x ightarrow 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1$

$\lim _{x ightarrow 0^{-}} f(x)=\lim _{x ightarrow 0^{+}} f(x)=f(0)$

Therefore, $f$ is continuous at $x=0$.

From the above observations, it can be concluded that $f$ is continuous at every point of the real line.

Thus, $f$ is a continuous function.

Question 26

Find the values of $k$ so that the function $f$ is continuous at the indicated point $f(x)=\left\{\begin{array}{l}\frac{k \cos x}{\pi-2 x}, 	ext { if } x 
eq \frac{\pi}{2} \ 3, 	ext { if } x=\frac{\pi}{2} \quad 	ext { at } \quad x=\frac{\pi}{2}\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}\frac{k \cos x}{\pi-2 x}, 	ext { if } x 
eq \frac{\pi}{2} \ 3, 	ext { if } x=\frac{\pi}{2}\end{array}ight.$

The given function $f$ is continuous at $x=\frac{\pi}{2}$, if $f$ is defined at $x=\frac{\pi}{2}$ and if the value of the $f$ at $x=\frac{\pi}{2}$ equals the limit of $f$ at $x=\frac{\pi}{2}$.

It is evident that $f$ is defined at $x=\frac{\pi}{2}$ and $f\left(\frac{\pi}{2}ight)=3$

$\lim _{x ightarrow \frac{\pi}{2}} f(x)=\lim _{x ightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}$

Put $x=\frac{\pi}{2}+h$

Then $x ightarrow \frac{\pi}{2} \Rightarrow h ightarrow 0$

$	herefore \lim _{x ightarrow \frac{\pi}{2}} f(x)=\lim _{x ightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{h ightarrow 0} \frac{k \cos \left(\frac{\pi}{2}+hight)}{\pi-2\left(\frac{\pi}{2}+hight)}$

$$

=k \lim _{h ightarrow 0} \frac{-\sin h}{-2 h}=\frac{k}{2} \lim _{h ightarrow 0} \frac{\sin h}{h}=\frac{k}{2} \cdot 1=\frac{k}{2}

$$

$	herefore \lim _{x ightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}ight)$

$\Rightarrow \frac{k}{2}=3$

$\Rightarrow k=6$

Therefore, the value of $k=6$.

Question 27

Find the values of $k$ so that the function $f$ is continuous at the indicated point. $f(x)=\left\{\begin{array}{l}k x^{2}, 	ext { if } x \leq 2 \ 3, 	ext { if } x>2\end{array}ight.$ at $x=2$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}k x^{2}, 	ext { if } x \leq 2 \ 3, 	ext { if } x>2\end{array}ight.$

The given function $f$ is continuous at $x=2$, if $f$ is defined at $x=2$ and if the value of the $f$ at $x=2$ equals the limit of $f$ at $x=2$.

It is evident that $f$ is defined at $x=2$ and $f(2)=k(2)^{2}=4 k$

$\lim _{x ightarrow 2^{-}} f(x)=\lim _{x ightarrow 2^{+}} f(x)=f(2)$

$\Rightarrow \lim _{x ightarrow 2^{-}}\left(k x^{2}ight)=\lim _{x ightarrow 2^{+}}(3)=4 k$

$\Rightarrow k 	imes 2^{2}=3=4 k$

$\Rightarrow 4 k=3$

$\Rightarrow k=\frac{3}{4}$

Therefore, the value of $k=\frac{3}{4}$.

Question 28

Find the values of $k$ so that the function $f$ is continuous at the indicated point $f(x)=\left\{\begin{array}{l}k x+1, 	ext { if } x \leq \pi \ \cos x, 	ext { if } x>\pi 	ext { at } x=\pi\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}k x+1, 	ext { if } x \leq \pi \ \cos x, 	ext { if } x>\pi\end{array}ight.$

The given function $f$ is continuous at $x=\pi$, if $f$ is defined at $x=\pi$ and if the value of the $f$ at $x=\pi$ equals the limit of $f$ at $x=\pi$.

It is evident that $f$ is defined at $x=\pi$ and $f(\pi)=k \pi+1$

$\lim _{x ightarrow \pi^{-}} f(x)=\lim _{x ightarrow \pi^{+}} f(x)=f(\pi)$

$\Rightarrow \lim _{x ightarrow \pi^{-}}(k x+1)=\lim _{x ightarrow \pi^{+}}(\cos x)=k \pi+1$

$\Rightarrow k \pi+1=\cos \pi=k \pi+1$

$\Rightarrow k \pi+1=-1=k \pi+1$

$\Rightarrow k=-\frac{2}{\pi}$

Therefore, the value of $k=-\frac{2}{\pi}$.

Question 29

Find the values of $k$ so that the function $f$ is continuous at the indicated point $f(x)=\left\{\begin{array}{l}k x+1, 	ext { if } x \leq 5 \ 3 x-5, 	ext { if } x>5 	ext { at } x=5 .\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}k x+1, 	ext { if } x \leq 5 \ 3 x-5, 	ext { if } x>5\end{array}ight.$

The given function $f$ is continuous at $x=5$, if $f$ is defined at $x=5$ and if the value of the $f$ at $x=5$ equals the limit of $f$ at $x=5$.

It is evident that $f$ is defined at $x=5$ and $f(5)=k x+1=5 k+1$

$\lim _{x ightarrow 5^{-}} f(x)=\lim _{x ightarrow 5^{+}} f(x)=f(5)$

$\Rightarrow \lim _{x ightarrow 5^{-}}(k x+1)=\lim _{x ightarrow 5^{+}}(3 x-5)=5 k+1$

$\Rightarrow 5 k+1=3(5)-5=5 k+1$

$\Rightarrow 5 k+1=15-5=5 k+1$

$\Rightarrow 5 k+1=10=5 k+1$

$\Rightarrow 5 k+1=10$

$\Rightarrow 5 k=9$

$\Rightarrow k=\frac{9}{5}$

Therefore, the value of $k=\frac{9}{5}$.

Question 30

Find the values of $a \& b$ such that the function defined by $f(x)=\left\{\begin{array}{l}5, 	ext { if } x \leq 2 \ a x+b, 	ext { if } 2<x<10 \ 21, 	ext { if } x \geq 10, 	ext { is a }\end{array}ight.$ continuous function.

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}5, 	ext { if } x \leq 2 \ a x+b, 	ext { if } 2<x<10 \ 21, 	ext { if } x \geq 10\end{array}ight.$

It is evident that $f$ is defined at all points of the real line.

If $f$ is a continuous function, then $f$ is continuous at all real numbers.

In particular, $f$ is continuous at $x=2$ and $x=10$

Since $f$ is continuous at $x=2$, we obtain

$\lim _{x ightarrow 2^{-}} f(x)=\lim _{x ightarrow 2^{+}} f(x)=f(2)$

$\Rightarrow \lim _{x ightarrow 2^{-}}(5)=\lim _{x ightarrow 2^{+}}(a x+b)=5$

$\Rightarrow 5=2 a+b=5$

$$

\begin{equation*}

\Rightarrow 2 a+b=5 	ag{1}

\end{equation*}

$$

Since $f$ is continuous at $x=10$, we obtain

$\lim _{x ightarrow 10^{-}} f(x)=\lim _{x ightarrow 10^{+}} f(x)=f(10)$

$\Rightarrow \lim _{x ightarrow 10^{-}}(a x+b)=\lim _{x ightarrow 10^{+}}(21)=21$

$\Rightarrow 10 a+b=21=21$

$$

\begin{equation*}

\Rightarrow 10 a+b=21 	ag{2}

\end{equation*}

$$

On subtracting equation (1) from equation (2), we obtain

$8 a=16$

$\Rightarrow a=2$

By putting $a=2$ in equation (1), we obtain

$2(2)+b=5$

$\Rightarrow 4+b=5$

$\Rightarrow b=1$

Therefore, the values of $a$ and $b$ for which $f$ is a continuous function are 2 and 1 respectively.

Question 31

Show that the function defined by $f(x)=\cos \left(x^{2}\right)$ is a continuous function.

Accepted Answer

The given function is $f(x)=\cos \left(x^{2}ight)$.

This function $f$ is defined for every real number and $f$ can be written as the composition of two functions as,

$f=g o h$, where $g(x)=\cos x$ and $h(x)=x^{2}$

$\left[\because(g o h)(x)=g(h(x))=g\left(x^{2}ight)=\cos \left(x^{2}ight)=f(x)ight]$

It has to be proved first that $g(x)=\cos x$ and $h(x)=x^{2}$ are continuous functions.

It is evident that $g$ is defined for every real number.

Let $c$ be a real number.

Let $g(c)=\cos c$. Put $x=c+h$

If $x ightarrow c$, then $h ightarrow 0$

$$

\begin{aligned}

\lim _{x ightarrow c} g(x) & =\lim _{x ightarrow c} \cos x \

& =\lim _{h ightarrow 0} \cos (c+h) \

& =\lim _{h ightarrow 0}[\cos c \cos h-\sin c \sin h] \

& =\lim _{h ightarrow 0}(\cos c \cos h)-\lim _{h ightarrow 0}(\sin c \sin h) \

& =\cos c \cos 0-\sin c \sin 0 \

& =\cos c(1)-\sin c(0) \

& =\cos c

\end{aligned}

$$

$	herefore \lim _{x ightarrow c} g(x)=g(c)$

Therefore, $g(x)=\cos x$ is a continuous function.

Let $h(x)=x^{2}$

It is evident that $h$ is defined for every real number.

Let $k$ be a real number, then $h(k)=k^{2}$

$\lim _{x ightarrow k} h(x)=\lim _{x ightarrow k} x^{2}=k^{2}$

$	herefore \lim _{x ightarrow k} h(x)=h(k)$

Therefore, $h$ is a continuous function.

It is known that for real valued functions $g$ and $h$, such that $(g o h)$ is defined at $c$, if $g$ is continuous at $c$ and if $f$ is continuous at $g(c)$, then $(f o g)$ is continuous at $c$.

Therefore, $f(x)=(g o h)(x)=\cos \left(x^{2}ight)$ is a continuous function.

Question 32

Show that the function defined by $f(x)=|\cos x|$ is a continuous function.

Accepted Answer

The given function is $f(x)=|\cos x|$. This function $f$ is defined for every real number and $f$ can be written as the composition of two functions as, $f=g o h$, where $g(x)=|x|$ and $h(x)=\cos x$ $\lceil\because(g o h)(x)=g(h(x))=g(\cos x)=|\cos x|=f(x) ceil$ It has to be proved first that $g(x)=|x|$ and $h(x)=\cos x$ are continuous functions. $g(x)=|x|$ can be written as $g(x)=\left\{\begin{array}{l}-x, ext { if } x<0 \ x, ext { if } x \geq 0\end{array} ight.$ It is evident that $g$ is defined for every real number. Let $c$ be a real number. Case I: If $c<0$, then $g(c)=-c$ $\lim _{x ightarrow c} g(x)=\lim _{x ightarrow c}(-x)=-c$ $ herefore \lim _{x ightarrow c} g(x)=g(c)$ Therefore, $g$ is continuous at all points $x$, such that $x<0$. \section*{Case II:} If $c>0$, then $g(c)=c$ $\lim _{x ightarrow c} g(x)=\lim _{x ightarrow c}(x)=c$ $ herefore \lim _{x ightarrow c} g(x)=g(c)$ Therefore, $g$ is continuous at all points $x$, such that $x>0$. Case III: If $c=0$, then $g(c)=g(0)=0$ $\lim _{x ightarrow 0^{-}} g(x)=\lim _{x ightarrow 0^{-}}(-x)=0$ $\lim _{x ightarrow 0^{+}} g(x)=\lim _{x ightarrow 0^{+}}(x)=0$ $ herefore \lim _{x ightarrow 0^{-}} g(x)=\lim _{x ightarrow 0^{+}}(x)=g(0)$ Therefore, $g$ is continuous at all $x=0$. From the above three observations, it can be concluded that $g$ is continuous at all points. Let $h(x)=\cos x$ It is evident that $h(x)=\cos x$ is defined for every real number. Let $c$ be a real number. Put $x=c+h$ If $x ightarrow c$, then $h ightarrow 0$ $$ \begin{aligned} h(c) & =\cos c \ \lim _{x ightarrow c} h(x) & =\lim _{x ightarrow c} \cos x \ & =\lim _{h ightarrow 0} \cos (c+h) \ & =\lim _{h ightarrow 0}[\cos c \cos h-\sin c \sin h] \ & =\lim _{h ightarrow 0}(\cos c \cos h)-\lim _{h ightarrow 0}(\sin c \sin h) \ & =\cos c \cos 0-\sin c \sin 0 \ & =\cos c(1)-\sin c(0) \ & =\cos c \ herefore \lim _{x ightarrow c} h(x) & =h(c) \end{aligned} $$ Therefore, $h(x)=\cos x$ is a continuous function. It is known that for real valued functions $g$ and $h$, such that $(g o h)$ is defined at $c$, if $g$ is continuous at $c$ and if $f$ is continuous at $g(c)$, then $(f o g)$ is continuous at $c$. Therefore, $f(x)=(g o h)(x)=g(h(x))=g(\cos x)=|\cos x|$ is a continuous function.

Question 33

Show that the function defined by $f(x)=|\sin x|$ is a continuous function.

Accepted Answer

The given function is $f(x)=|\sin x|$. This function $f$ is defined for every real number and $f$ can be written as the composition of two functions as, $f=g o h$, where $g(x)=|x|$ and $h(x)=\sin x$ $\lceil\because(g o h)(x)=g(h(x))=g(\sin x)=|\sin x|=f(x) ceil$ It has to be proved first that $g(x)=|x|$ and $h(x)=\sin x$ are continuous functions. $g(x)=|x|$ can be written as $g(x)=\left\{\begin{array}{l}-x, ext { if } x<0 \ x, ext { if } x \geq 0\end{array} ight.$ It is evident that $g$ is defined for every real number. Let $c$ be a real number. \section*{Case I:} If $c<0$, then $g(c)=-c$ $\lim _{x ightarrow c} g(x)=\lim _{x ightarrow c}(-x)=-c$ $ herefore \lim _{x ightarrow c} g(x)=g(c)$ Therefore, $g$ is continuous at all points $x$, such that $x<0$. \section*{Case II:} If $c>0$, then $g(c)=c$ $\lim _{x ightarrow c} g(x)=\lim _{x ightarrow c}(x)=c$ $ herefore \lim _{x ightarrow c} g(x)=g(c)$ Therefore, $g$ is continuous at all points $x$, such that $x>0$. \section*{Case III:} If $c=0$, then $g(c)=g(0)=0$ $\lim _{x ightarrow 0^{-}} g(x)=\lim _{x ightarrow 0^{-}}(-x)=0$ $\lim _{x ightarrow 0^{+}} g(x)=\lim _{x ightarrow 0^{+}}(x)=0$ $ herefore \lim _{x ightarrow 0^{-}} g(x)=\lim _{x ightarrow 0^{+}}(x)=g(0)$ Therefore, $g$ is continuous at all $x=0$. From the above three observations, it can be concluded that $g$ is continuous at all points. Let $h(x)=\sin x$ It is evident that $h(x)=\sin x$ is defined for every real number. Let $c$ be a real number. Put $x=c+k$ If $x ightarrow c$, then $k ightarrow 0$ $$ \begin{aligned} h(c) & =\sin c \ \lim _{x ightarrow c} h(x) & =\lim _{x ightarrow c} \sin x \ & =\lim _{k ightarrow 0} \sin (c+k) \ & =\lim _{k ightarrow 0}[\sin c \cos k+\cos c \sin k] \ & =\lim _{k ightarrow 0}(\sin c \cos k)+\lim _{k ightarrow 0}(\cos c \sin k) \ & =\sin c \cos 0+\cos c \sin 0 \ & =\sin c(1)+\cos c(0) \ & =\sin c \end{aligned} $$ $ herefore \lim _{x ightarrow c} h(x)=h(c)$ Therefore, $h(x)=\sin x$ is a continuous function. It is known that for real valued functions $g$ and $h$, such that $(g o h)$ is defined at $c$, if $g$ is continuous at $c$ and if $f$ is continuous at $g(c)$, then $(f o g)$ is continuous at $c$. Therefore, $f(x)=(g \circ h)(x)=g(h(x))=g(\sin x)=|\sin x|$ is a continuous function.

Question 34

Find all the points of discontinuity of $f$ defined by $f(x)=|x|-|x+1|$.

Accepted Answer

The given function is $f(x)=|x|-|x+1|$. The two functions, $g$ and $h$ are defined as $g(x)=|x|$ and $h(x)=|x+1|$. Then, $f=g-h$ The continuity of $g$ and $h$ are examined first. $g(x)=|x|$ can be written as $g(x)=\left\{\begin{array}{l}-x, ext { if } x<0 \ x, ext { if } x \geq 0\end{array} ight.$ It is evident that $g$ is defined for every real number. Let $c$ be a real number. \section*{Case I:} If $c<0$, then $g(c)=-c$ $\lim _{x ightarrow c} g(x)=\lim _{x ightarrow c}(-x)=-c$ $ herefore \lim _{x ightarrow c} g(x)=g(c)$ Therefore, $g$ is continuous at all points $x$, such that $x<0$. \section*{Case II:} If $c>0$, then $g(c)=c$ $\lim _{x ightarrow c} g(x)=\lim _{x ightarrow c}(x)=c$ $ herefore \lim _{x ightarrow c} g(x)=g(c)$ Therefore, $g$ is continuous at all points $x$, such that $x>0$. Case III: If $c=0$, then $g(c)=g(0)=0$ $\lim _{x ightarrow 0^{-}} g(x)=\lim _{x ightarrow 0^{-}}(-x)=0$ $\lim _{x ightarrow 0^{+}} g(x)=\lim _{x ightarrow 0^{+}}(x)=0$ $ herefore \lim _{x ightarrow 0^{-}} g(x)=\lim _{x ightarrow 0^{+}}(x)=g(0)$ Therefore, $g$ is continuous at all $x=0$. From the above three observations, it can be concluded that $g$ is continuous at all points. $h(x)=|x+1|$ can be written as $h(x)=\left\{\begin{array}{l}-(x+1), ext { if } x<-1 \ x+1, ext { if } x \geq-1\end{array} ight.$ It is evident that $h$ is defined for every real number. Let $c$ be a real number. Case I: If $c<-1$, then $h(c)=-(c+1)$ $\lim _{x ightarrow c} h(x)=\lim _{x ightarrow c}[-(x+1)]=-(c+1)$ $ herefore \lim _{x ightarrow c} h(x)=h(c)$ Therefore, $h$ is continuous at all points $x$, such that $x<-1$. Case II: If $c>-1$, then $h(c)=c+1$ $\lim _{x ightarrow c} h(x)=\lim _{x ightarrow c}(x+1)=c+1$ $ herefore \lim _{x ightarrow c} h(x)=h(c)$ Therefore, $h$ is continuous at all points $x$, such that $x>-1$. Case III: If $c=-1$, then $h(c)=h(-1)=-1+1=0$ $\lim _{x ightarrow-1^{-}} h(x)=\lim _{x ightarrow-1^{-}}[-(x+1)]=-(-1+1)=0$ $\lim _{x ightarrow-1^{+}} h(x)=\lim _{x ightarrow-1^{+}}(x+1)=(-1+1)=0$ $ herefore \lim _{x ightarrow-1^{-}} h(x)=\lim _{x ightarrow-1^{+}} h(x)=h(-1)$ Therefore, $h$ is continuous at $x=-1$. From the above three observations, it can be concluded that $h$ is continuous at all points. It concludes that $g$ and $h$ are continuous functions. Therefore, $f=g-h$ is also a continuous function. Therefore, $f$ has no point of discontinuity. \section*{EXERCISE 5.2}