Exercise 5.1 (Revised) - Chapter 5 - Continuity & Differentiability - Ncert Solutions class 12 - Maths

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Chapter 5: Continuity & Differentiability NCERT Solutions Class 12 Maths

Ex 5.1 Question 1:

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

Answer :
The given function is $f(x)=5 x-3$
$
\begin{aligned}
& \text { At } x=0, f(0)=5 \times 0-3=3 \\
& \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}(5 x-3)=5 \times 0-3=-3 \\
& \therefore \lim _{x \rightarrow 0} f(x)=f(0)
\end{aligned}
$

Therefore, $f$ is continuous at $x=0$
$
\begin{aligned}
& \text { At } x=-3, f(-3)=5 \times(-3)-3=-18 \\
& \lim _{x \rightarrow-3} f(x)=\lim _{x \rightarrow-3}(5 x-3)=5 \times(-3)-3=-18 \\
& \therefore \lim _{x \rightarrow-3} f(x)=f(-3)
\end{aligned}
$

Therefore, $f$ is continuous at $x=-3$
$
\begin{aligned}
& \text { At } x=5, f(x)=f(5)=5 \times 5-3=25-3=22 \\
& \lim _{x \rightarrow 5} f(x)=\lim _{x \rightarrow 5}(5 x-3)=5 \times 5-3=22 \\
& \therefore \lim _{x \rightarrow 5} f(x)=f(5)
\end{aligned}
$

Therefore, $f$ is continuous at $x=5$

Ex 5.1 Question 2:

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

Answer :
The given function is $f(x)=2 x^2-1$
$
\begin{aligned}
& \text { At } x=3, f(x)=f(3)=2 \times 3^2-1=17 \\
& \lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3}\left(2 x^2-1\right)=2 \times 3^2-1=17 \\
& \therefore \lim _{x \rightarrow 3} f(x)=f(3)
\end{aligned}
$

Thus, $f$ is continuous at $x=3$

Ex 5.1 Question 3 :

Examine the following functions for continuity.
(a) $f(x)=x-5$
(b) $f(x)=\frac{1}{x-5}, x \neq 5$
(c) $f(x)=\frac{x^2-25}{x+5}, x \neq-5$
(d) $f(x)=|x-5|$

Answer:
(a) The given function is $f(x)=x-5$

It is evident that fis defined at every real number kand its value at $k$ is $k-5$.
It is also observed that, $\lim _{x \rightarrow 4} f(x)=\lim _{x \rightarrow k}(x-5)=k-5=f(k)$
$
\therefore \lim _{x \rightarrow k} f(x)=f(k)
$

Hence, $f$ is continuous at every real number and therefore, it is a continuous function.
(b) The given function is $f(x)=\frac{1}{x-5}, x \neq 5$

For any real number $k \neq 5$, we obtain
$
\begin{aligned}
& \lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k} \frac{1}{x-5}=\frac{1}{k-5} \\
& \text { Also, } f(k)=\frac{1}{k-5} \quad(\text { As } k \neq 5) \\
& \therefore \lim _{x \rightarrow k} f(x)=f(k)
\end{aligned}
$

Also, $f(k)=\frac{1}{k-5} \quad($ As $k \neq 5)$

Hence, fis continuous at every point in the domain of fand therefore, it is a continuous function.
(c) The given function is $f(x)=\frac{x^2-25}{x+5}, x \neq-5$

For any real number $c \neq-5$, we obtain

$
\begin{aligned}
& \lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c} \frac{x^2-25}{x+5}=\lim _{x \rightarrow c} \frac{(x+5)(x-5)}{x+5}=\lim _{x \rightarrow c}(x-5)=(c-5) \\
& \text { Also, } f(c)=\frac{(c+5)(c-5)}{c+5}=(c-5) \quad(\text { as } c \neq-5) \\
& \therefore \lim _{x \rightarrow c} f(x)=f(c)
\end{aligned}
$

Hence, fis continuous at every point in the domain of fand therefore, it is a continuous function.
(d) The given function is $f(x)=|x-5|=\left\{\begin{array}{l}5-x, \text { if } x<5 \\ x-5 \text {, if } x \geq 5\end{array}\right.$

This function fis defined at all points of the real line.
Let cbe a point on a real line. Then, $c<5$ or $c=5$ or $c>5$
Case I: $c<5$
$
\begin{aligned}
& \text { Then, } f(c)=5-c \\
& \lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(5-x)=5-c \\
& \therefore \lim _{x \rightarrow c} f(x)=f(c)
\end{aligned}
$

Therefore, fis continuous at all real numbers less than 5 .
$
\begin{aligned}
& \text { Case II: } c=5 \\
& \text { Then, } f(c)=f(5)=(5-5)=0 \\
& \lim _{x \rightarrow 5^-} f(x)=\lim _{x \rightarrow 5}(5-x)=(5-5)=0 \\
& \lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5}(x-5)=0 \\
& \therefore \lim _{x \rightarrow c^{-}} f(x)=\lim _{x \rightarrow c^{+}} f(x)=f(c)
\end{aligned}
$

Therefore, $f$ is continuous at $x=5$
Case III: $>5$
$
\begin{aligned}
& \text { Then, } f(c)=f(5)=c-5 \\
& \lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x-5)=c-5 \\
& \therefore \lim _{x \rightarrow c} f(x)=f(c)
\end{aligned}
$

Ex 5.1 Question 4:

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

Answer:
The given function is $f(x)=x^n$
It is evident that fis defined at all positive integers, $n$, and its value at $n$ is $n^n$.
Then, $\lim _{x \rightarrow m} f(n)=\lim _{x \rightarrow m}\left(x^n\right)=n^n$
$
\therefore \lim _{x \rightarrow n} f(x)=f(n)
$

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

Ex 5.1 Question 5 :

Is the function fdefined by
$
f(x)= \begin{cases}x, & \text { if } x \leq 1 \\ 5, & \text { if } x>1\end{cases}
$
continuous at $x=0$ ? At $x=1$ ? At $x=2$ ?

Answer :
The given function fis $f(x)=\left\{\begin{array}{l}x, \text { if } x \leq 1 \\ 5, \text { if } x>1\end{array}\right.$
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 $x=1$,
$f$ 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,

$
\begin{aligned}
& \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)
\end{aligned}
$

Therefore, $f$ is not continuous at $x=1$
At $x=2$,
$f$ is defined at 2 and its value at 2 is 5 .
Then, $\lim _{x \rightarrow 2} f(x)=\lim _{x \rightarrow 2}(5)=5$
$
\therefore \lim _{x \rightarrow 2} f(x)=f(2)
$

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

Ex 5.1 Question 6:

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.
$

Answer:
The given function fis $f(x)=\left\{\begin{array}{l}2 x+3, \text { if } x \leq 2 \\ 2 x-3, \text { if } x>2\end{array}\right.$
It is evident that the given function fis defined at all the points of the real line.
Let $c$ be a point on the real line. Then, three cases arise.
(i) $c<2$
(ii) c>2
(iii) $c=2$

Case (i) $c<2$
Then, $f(c)=2 c+3$
$
\begin{aligned}
& \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)
\end{aligned}
$

Therefore, $f$ is continuous at all points $x$, such that $x<2$
Case (ii) c>2

$
\begin{aligned}
& \text { 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)
\end{aligned}
$

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 \times 2+3=7
$

The right hand limit of fat $x=2$ is,
$
\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x-3)=2 \times 2-3=1
$

It is observed that the left and right hand limit of fat $x=2$ do not coincide.
Therefore, $f$ is not continuous at $x=2$
Hence, $x=2$ is the only point of discontinuity of $f$.

Ex 5.1 Question 7:

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 6 x+2, \text { if } x \geq 3
\end{array}\right.
$

Answer :

$\text { Given: } f(x)= \begin{cases}|x|+3, & \text { if } x \leq-3 \\ -2 x, & \text { if }-3

Here $f(x)$ is defined for $x \leq-3$ i.e., on $(-\infty,-3)$ and for $-3 $\therefore$ Domain of $f$ is $(-\infty,-3) \cup(-3,3) \cup(3, \infty)=(-\infty, \infty)=\mathrm{R}$
$\therefore$ For all $x<-3, f(x)=|x|+3=-x+3$ is a polynomial and hence continuous and for all

$x(-33, f(x)=6 x+2$. Therefore, $f(x)$ is continuous on $\mathrm{R}-\{-3,3\}$.

It is observed that $x=-3$ and $x=3$ are partitioning points of domain $\mathrm{R}$.
Now Left Hand limit $=\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(|x|+3)=\lim _{x \rightarrow 3^{-}}(-x+3)=3+3=6$
Right Hand limit $=\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}(-2 x)=(-2)(-3)=6$
And $f(-3)=|-3|+3=3+3=6$
Therefore, $f(x)$ is continuous at $x=-3$.
Again Left Hand limit $=\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(-2 x)=-2(3)=-6$
Right Hand limit $=\lim _{x \rightarrow 3+} f(x)=\lim _{x \rightarrow 3^{+}}(6 x+2)=6(3)+2=18+2=20$
Since $\lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x)$
Therefore, $\lim _{x \rightarrow 3} f(x)$ does not exist and hence $f(x)$ is discontinuous at $x=3$ (only).

Ex 5.1 Question Ex 5.1 Question 8

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

Answer :
The given function fis $f(x)=\left\{\begin{array}{l}\frac{|x|}{x} \text { if } x \neq 0 \\ 0, \text { if } x=0\end{array}\right.$
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 \text { if } x<0 \\
0, \text { if } x=0 \\
\frac{|x|}{x}=\frac{x}{x}=1, \text { if } x>0
\end{array}\right.
$

The given function fis defined at all the points of the real line.
Let $c$ be a point on the real line.

Let $c$ be a point on the real line.
Case I:
$
\begin{aligned}
& \text { If } c<0 \text {, then } f(c)=-1 \\
& \lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-1)=-1 \\
& \therefore \lim _{x \rightarrow c} f(x)=f(c)
\end{aligned}
$

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 \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(-1)=-1
$

The right hand limit of $f$ at $x=0$ is,
$
\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(1)=1
$

It is observed that the left and right hand limit of fat $x=0$ do not coincide. Therefore, $f$ is not continuous at $x=0$

Ex 5.1 Question 9

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

Answer :
The given function fis $f(x)= \begin{cases}\frac{x}{|x|}, & \text { if } x<0 \\ -1, & \text { if } x \geq 0\end{cases}$
It is known that, $x<0 \Rightarrow|x|=-x$
Therefore, the given function can be rewritten as
$
f(x)=\left\{\begin{array}{l}
\frac{x}{|x|}=\frac{x}{-x}=-1, \text { if } x<0 \\
-1, \text { if } x \geq 0
\end{array}\right.
$
$\Rightarrow f(x)=-1$ for all $x \in \mathbf{R}$
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.

Ex 5.1 Question 10

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.
$

Answer :
The given function fis $f(x)=\left\{\begin{array}{l}x+1, \text { if } x \geq 1 \\ x^2+1, \text { if } x<1\end{array}\right.$
The given function fis defined at all the points of the real line.
Let $c$ be a point on the real line.
Case I:
If $c<1$, then $f(c)=c^2+1$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^2+1\right)=c^2+1$
$
\therefore \lim _{x \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,
$
\begin{aligned}
& \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x+1)=1+1=2 \\
& \therefore \lim _{x \rightarrow 1} f(x)=f(1)
\end{aligned}
$

Therefore, $f$ is continuous at $x=1$
Case III:
$
\begin{aligned}
& \text { If } c>1 \text {, 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)
\end{aligned}
$

Therefore, $f$ is continuous at all points $x$, such that $x>1$
Hence, the given function $f$ has no point of discontinuity.

Ex 5.1 Question 11 :

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

Answer :
The given function fis $f(x)=\left\{\begin{array}{l}x^3-3, \text { if } x \leq 2 \\ x^2+1, \text { if } x>2\end{array}\right.$
The given function fis defined at all the points of the real line.
Let $c$ be a point on the real line.
Case I:
If $c<2$, then $f(c)=c^3-3$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^3-3\right)=c^3-3$
$
\therefore \lim _{x \rightarrow c} f(x)=f(c)
$

Therefore, $f$ is continuous at all points $x$, such that $x<2$

Case II:
If $c=2$, then $f(c)=f(2)=2^3-3=5$
$
\begin{aligned}
& \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}\left(x^3-3\right)=2^3-3=5 \\
& \lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}\left(x^2+1\right)=2^2+1=5 \\
& \therefore \lim _{x \rightarrow 2} f(x)=f(2)
\end{aligned}
$

Therefore, $f$ is continuous at $x=2$
Case III:
If $c>2$, then $f(c)=c^2+1$
$
\begin{aligned}
& \lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^2+1\right)=c^2+1 \\
& \therefore \lim _{x \rightarrow c} f(x)=f(c)
\end{aligned}
$

Therefore, $f$ is continuous at all points $x$, such that $x>2$
Thus, the given function fis continuous at every point on the real line.
Hence, $f$ has no point of discontinuity.

Ex 5.1 Question 12:

Find all points of discontinuity of $f$, where $f$ is defined by
$
f(x)= \begin{cases}x^{10}-1, & \text { if } x \leq 1 \\ x^2, & \text { if } x>1\end{cases}
$

Answer :
The given function fis $f(x)= \begin{cases}x^{10}-1, & \text { if } x \leq 1 \\ x^2, & \text { if } x>1\end{cases}$
The given function fis defined at all the points of the real line.
Let $c$ be a point on the real line.
Case I:
If $c<1$, then $f(c)=c^{10}-1$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{10}-1\right)=c^{10}-1$
$
\therefore \lim _{x \rightarrow 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 fat $x=1$ is,
$
\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1^{-}}\left(x^{10}-1\right)=1^{10}-1=1-1=0
$

The right hand limit of fat $x=1$ is,
$
\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(x^2\right)=1^2=1
$

It is observed that the left and right hand limit of fat $x=1$ do not coincide.
Therefore, $f$ is not continuous at $x=1$
Case III:
If $c>1$, then $f(c)=c^2$
$
\begin{aligned}
& \lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^2\right)=c^2 \\
& \therefore \lim _{x \rightarrow c} f(x)=f(c)
\end{aligned}
$

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.

Ex 5.1 Question 13

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

Answer :
The given function is $f(x)=\left\{\begin{array}{l}x+5, \text { if } x \leq 1 \\ x-5, \text { if } x>1\end{array}\right.$

The given function fis defined at all the points of the real line.
Let $c$ be a point on the real line.
Case I:
If $c<1$, then $f(c)=c+5$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x+5)=c+5$
$
\therefore \lim _{x \rightarrow c} f(x)=f(c)
$

Therefore, $f$ is continuous at all points $x$, such that $x<1$
Case II:
If $c=1$, then $f(1)=1+5=6$
The left hand limit of $f$ at $x=1$ is,
$
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x+5)=1+5=6
$

The right hand limit of fat $x=1$ is,
$
\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x-5)=1-5=-4
$

It is observed that the left and right hand limit of fat $x=1$ do not coincide.
Therefore, $f$ is not continuous at $x=1$
Case III:
If $c>1$, then $f(c)=c-5$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x-5)=c-5$
$
\therefore \lim _{x \rightarrow 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.

Ex 5.1 Question 14

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

Answer:

$\text { Given: } f(x)=\left\{\begin{array}{lll}
3, & \text { if } & 0 \leq x \leq 1 \\
4, & \text { if } & 1 5, & \text { if } & 3 \leq x \leq 10
\end{array}\right.$
$\therefore f$ is continuous in this interval.
At $x=1$, L.H.L. $=\lim _{x \rightarrow 1^{-}} f(x)=3$ and R.H.L. $=\lim _{x \rightarrow 1^{-}} f(x)=4$
Since L.H.L. $\neq$ R.H.L.
Therefore, $f(x)$ is discontinuous at $x=1$
At $x=3$, L.H.L. $=\lim _{x \rightarrow 3^{-}} f(x)=4$ and R.H.L. $=\lim _{x \rightarrow 3^{+}} f(x)=5$
Since L.H.L. $\neq$ R.H.L.
Therefore, $f(x)$ is discontinuous at $x=3$
Hence, $f$ is discontinuous at $x=1$ and $x=3$.

Ex 5.1 Question 15:

Discuss the continuity of the function $f$, where $f$ is defined by
$
f(x)= \begin{cases}2 x, & \text { if } x<0 \\ 0, & \text { if } 0 \leq x \leq 1 \\ 4 x, & \text { if } x>1\end{cases}
$

Answer :
The given function is $f(x)= \begin{cases}2 x, & \text { if } x<0 \\ 0, & \text { if } 0 \leq x \leq 1 \\ 4 x, & \text { if } x>1\end{cases}$
The given function is defined at all points of the real line.
Let $c$ be a point on the real line.
Case I:
$
\begin{aligned}
& \text { If } c<0 \text {, 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)
\end{aligned}
$

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 \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(2 x)=2 \times 0=0
$

The right hand limit of fat $x=0$ is,
$
\begin{aligned}
& \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(0)=0 \\
& \therefore \lim _{x \rightarrow 0} f(x)=f(0)
\end{aligned}
$

Therefore, $f$ is continuous at $x=0$
Case III:
If $0 $
$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, $f$ is continuous at all points of the interval $(0,1)$.

Case IV:
If $c=1$, then $f(c)=f(1)=0$
Theleft hand limit of $f$ at $x=1$ is,
$
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(0)=0
$

The right hand limit of fat $x=1$ is,
$
\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(4 x)=4 \times 1=4
$

It is observed that the left and right hand limits of $f$ at $x=1$ do not coincide.
Therefore, $f$ is not continuous at $x=1$
Case V:
If $c<1$, then $f(c)=4 c$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(4 x)=4 c$
$
\therefore \lim _{x \rightarrow c} f(x)=f(c)
$

Therefore, $f$ is continuous at all points $x$, such that $x>1$
Hence, $f$ is not continuous only at $x=1$

Ex 5.1 Question 16

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

Answer :
The given function fis $f(x)=\left\{\begin{array}{l}-2, \text { if } x \leq-1 \\ 2 x, \text { if }-11\end{array}\right.$
The given function is defined at all points of the real line.
Let $c$ be a point on the real line.
Case I:
If $c<-1$, then $f(c)=-2$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-2)=-2$
$
\therefore \lim _{x \rightarrow c} f(x)=f(c)
$

Therefore, $f$ is continuous at all points $x$, such that $x<-1$

Case II:
If $c=-1$, then $f(c)=f(-1)=-2$
The left hand limit of $f$ at $x=-1$ is,
$
\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{-}}(-2)=-2
$

The right hand limit of fat $x=-1$ is,
$
\begin{aligned}
& \lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{+}}(2 x)=2 \times(-1)=-2 \\
& \therefore \lim _{x \rightarrow-1} f(x)=f(-1)
\end{aligned}
$

Therefore, $f$ is continuous at $x=-1$
Case III:
$
\begin{aligned}
& \text { If }-1 & \lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2 x)=2 c \\
& \therefore \lim _{x \rightarrow c} f(x)=f(c)
\end{aligned}
$

Therefore, $f$ is continuous at all points of the interval $(-1,1)$.
Case IV:
If $c=1$, then $f(c)=f(1)=2 \times 1=2$
The left hand limit of $f$ at $x=1$ is,
$
\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1}(2 x)=2 \times 1=2
$

The right hand limit of fat $x=1$ is,
$
\begin{aligned}
& \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} 2=2 \\
& \therefore \lim _{x \rightarrow 1} f(x)=f(c)
\end{aligned}
$

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

Case V:
If $c>1$, then $f(c)=2$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2)=2$
$
\lim _{x \rightarrow 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.

Ex 5.1 Question 17:

Find the relationship between aand $b$ so that the function fdefined by
$
f(x)=\left\{\begin{array}{l}
a x+1, \text { if } x \leq 3 \\
b x+3, \text { if } x>3
\end{array}\right.
$
is continuous at $x=3$.

Answer :
The given function $f$ is $f(x)=\left\{\begin{array}{l}a x+1, \text { if } x \leq 3 \\ b x+3, \text { if } x>3\end{array}\right.$
If $f$ is continuous at $x=3$, then
$
\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=f(3)
$

Also,
$
\begin{aligned}
& \lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(a x+1)=3 a+1 \\
& \lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}(b x+3)=3 b+3 \\
& f(3)=3 a+1
\end{aligned}
$

Therefore, from (1), we obtain
$
\begin{aligned}
& 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}
\end{aligned}
$

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

Ex 5.1 Question 18:

For what value of $\lambda$ is the function defined by
$
f(x)= \begin{cases}\lambda\left(x^2-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0\end{cases}
$
continuous at $x=0$ ? What about continuity at $x=1$ ?

Answer :
The given function fis $f(x)= \begin{cases}\lambda\left(x^2-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0\end{cases}$
If f is continuous at $x=0$, then

$\begin{aligned}
& \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=f(0) \\
& \Rightarrow \lim _{x \rightarrow 0^{-}} \lambda\left(x^2-2 x\right)=\lim _{x \rightarrow 0^{+}}(4 x+1)=\lambda\left(0^2-2 \times 0\right) \\
& \Rightarrow \lambda\left(0^2-2 \times 0\right)=4 \times 0+1=0 \\
& \Rightarrow 0=1=0, \text { which is not possible }
\end{aligned}$

Therefore, there is no value of $\text {} \tilde{A} \bar{z} . \hat{A} \gg$ for which fis continuous at  x=0

$\begin{aligned}
& \text { At } x=1, \\
& f(1)=4 x+1=4 \times 1+1=5 \\
& \lim _{x \rightarrow 1}(4 x+1)=4 \times 1+1=5 \\
& \therefore \lim _{x \rightarrow 1} f(x)=f(1)
\end{aligned}$

$\text { Therefore, for any values of }$ $\text {} \tilde{A} \bar{z} . \hat{A} \gg$ f $\text { is continuous at } x=1$

Ex 5.1 Question 19:

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

Answer:
The given function is $g(x)=x-[x]$
It is evident that gis defined at all integral points.
Let $n$ be an integer.
Then,
$
\mathrm{g}(n)=n-[n]=n-n=0
$

The left hand limit of $f$ 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 fat $x=$ nis,
$
\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 limits of fat $x=n$ do not coincide.
Therefore, $f$ is not continuous at $x-17$
Hence, $g$ is discontinuous at all integral points.

Ex 5.1 Question 20

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

Answer :
The given function is $f(x)=x^2-\sin x+5$
It is evident that $f$ is defined at $x=$.
At $x=\pi, f(x)=f(\pi)=\pi^2 \quad \sin \pi$ । $5=\pi^2 \quad 0$ । $5=\pi^2$ । 5
Consider $\lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi}\left(x^2-\sin x+5\right)$
Put $x=\pi+h$
If $x \rightarrow \pi$, then it is evident that $h \rightarrow 0$
$
\begin{aligned}
\therefore \lim _{x \rightarrow \pi} f(x) & =\lim _{x \rightarrow \pi}\left(x^2-\sin x+5\right) \\
& =\lim _{h \rightarrow 0}\left[(\pi+h)^2-\sin (\pi+h)+5\right] \\
& =\lim _{h \rightarrow 0}(\pi+h)^2-\lim _{h \rightarrow 0} \sin (\pi+h)+\lim _{h \rightarrow 0} 5 \\
& =(\pi+0)^2-\lim _{h \rightarrow 0}[\sin \pi \cosh +\cos \pi \sinh ]+5 \\
& =\pi^2 \quad \lim _{h \rightarrow 0} \sin \pi \cosh \lim _{h \rightarrow 0} \cos \pi \sinh \mid 5 \\
& =\pi^2-\sin \pi \cos 0-\cos \pi \sin 0+5 \\
& =\pi^2-0 \times 1-(-1) \times 0+5 \\
& =\pi^2 \mid 5 \\
\therefore \lim _{x \rightarrow \pi} f(x) & =f(\pi)
\end{aligned}
$

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

Ex 5.1 Question 21 :

Discuss the continuity of the following functions
(a) $f(x)=\sin x+\cos x$
(b) $f(x)=\sin x-\cos x$
(c) $f(x)=\sin x x \cos x$

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

It has to proved first that $g(x)=\sin x$ and $h(x)=\cos x$ are continuous functions.
Let $g(x)=\sin x$
It is evident that $g(x)=\sin x$ is defined for every real number.
Let $c$ be a real number. Put $x=c+h$

$
\begin{aligned}
g(c)= & \sin c \\
\lim _{x \rightarrow c} g(x) & =\lim _{x \rightarrow c} \sin x \\
& =\lim _{h \rightarrow 0} \sin (c+h) \\
& =\lim _{h \rightarrow 0}[\sin c \cos h+\cos c \sin h] \\
& =\lim _{h \rightarrow 0}(\sin c \cos h)+\lim _{h \rightarrow 0}(\cos c \sin h) \\
& -\sin c \cos 0+\cos c \sin 0 \\
& =\sin c+0 \\
& =\sin c
\end{aligned}
$

$
\therefore \lim _{x \rightarrow c} g(x)=g(c)
$

Therefore, $g$ 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 $x=c+h$

$\begin{aligned}
& h(c)=\cos c \\
& \lim _{x \rightarrow c} h(x)=\lim _{x \rightarrow c} \cos x \\
& =\lim _{h \rightarrow 0} \cos (c+h) \\
& =\lim _{h \rightarrow 0}[\cos c \cos h-\sin c \sin h] \\
& =\lim _{h \rightarrow 0} \cos c \cos h-\lim _{h \rightarrow 0} \sin c \sin h \\
& =\cos c \cos 0-\sin c \sin 0 \\
& =\cos c \times 1-\sin c \times 0 \\
& =\cos \mathrm{c} \\
& \therefore \lim _{x \rightarrow c} h(x)=h(c) \\
&
\end{aligned}$

Therefore, $h$ is a continuous function.
Therefore, it can be concluded that
(a) $f(x)-g(x)+h(x)-\sin x+\cos x$ is a continuous function
(b) $f(x)=g(x)-h(x)=\sin x-\cos x$ is a continuous function
(c) $f(x)=g(x) \times h(x)=\sin x \times \cos x$ is a continuous function

Ex 5.1 Question 22 :

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

Answer :
It is known that if $g$ and $h$ are two continuous functions, then
(i) $\begin{aligned} & h(x) \\ & g(x)\end{aligned}, g(x) \neq 0$ is continuous
(ii) $\frac{1}{g(x)}, g(x) \neq 0$ is continuous
(iii) $\frac{1}{h(x)}, h(x) \neq 0$ is continuous

It has to be proved first that $g(x)=\sin x$ and $h(x)=\cos x a r e$ continuous functions.
Let $g(x)=\sin x$
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 $x \rightarrow c$, then $h \rightarrow 0$

$
\begin{aligned}
& g(c)=\sin c \\
& \begin{aligned}
\lim _{x \rightarrow c} g(x) & =\lim _{x \rightarrow c} \sin x \\
& =\lim _{h \rightarrow 0} \sin (c+h) \\
& =\lim _{h \rightarrow 0}[\sin c \cos h+\cos c \sin h] \\
& =\lim _{h \rightarrow 0}(\sin c \cos h)+\lim _{h \rightarrow 0}(\cos c \sin h) \\
& =\sin c \cos 0+\cos c \sin 0 \\
& =\sin c+0 \\
& =\sin c
\end{aligned} \\
& \therefore \lim _{x \rightarrow c} g(x)=g(c)
\end{aligned}
$

Therefore, gis 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 $x \hat{A} ® C$, then $h \hat{A} ® 0$
$
h(c)-\cos c
$

$
\begin{aligned}
\lim _{x \rightarrow c} h(x) & =\lim _{x \rightarrow c} \cos x \\
& =\lim _{h \rightarrow 0} \cos (c+h) \\
& =\lim _{h \rightarrow 0}[\cos c \cos h-\sin c \sin h] \\
& =\lim _{h \rightarrow 0} \cos c \cos h-\lim _{h \rightarrow 0} \sin c \sin h \\
& =\cos c \cos 0-\sin c \sin 0 \\
& =\cos c \times 1-\sin c \times 0 \\
& =\cos c \\
\therefore \lim _{x \rightarrow c} h(x) & =h(c)
\end{aligned}
$

Therefore, $h(x)=\cos x$ is continuous function.
It can be concluded that,

$\operatorname{cosec} x-\frac{1}{\sin x}, \sin x+0$ is continuous
$\Rightarrow \operatorname{cosec} x, x \neq n \pi(n \in Z)$ is continuous

$\text { Therefore, cosecant is continuous except at } x=n p, n \tilde{A} fÅ \frac{1}{2} \mathbf{Z}$

$\sec x=\frac{1}{\cos x}, \cos x \neq 0$ is continuous
$\Rightarrow \sec x, x \neq(2 n+1) \frac{\pi}{2}(n \in \mathbf{Z})$ is continuous
Therefore, secant is continuous except at $x=(2 n+1) \frac{\pi}{2}(n \in \mathbf{Z})$
$\cot x-\frac{\cos x}{\sin x}, \sin x \neq 0$ is continucus
$\Rightarrow \cot x, x \neq n \pi(n \in Z)$ is continuous
Therefore, cotangent is continuous except at $x=n p, n \tilde{A} f Å 1 / 2 \mathbf{Z}$

Ex 5.1 Question 23 :

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

Answer :
The given function fis $f(x)=\left\{\begin{array}{cc}\frac{\sin x}{x}, & \text { if } x<0 \\ x+1, & \text { if } x \geq 0\end{array}\right.$
It is evident that fis defined at all points of the real line.
Let $c$ be a real number.
Case I:
If $c<0$, then $f(c)=\frac{\sin c}{c}$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(\frac{\sin x}{x}\right)=\frac{\sin c}{c}$
$
\therefore \lim _{x \rightarrow c} f(x)=j(c)
$

Therefore, $f$ is continuous at all points $x$, such that $x<0$
Case II:
If $c>0$, then $f(c)=c+1$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x+1)=c+1$
$
\therefore \lim _{x \rightarrow c} f(x)=f(c)
$

I herefore, $t$ is continuous at all points $x$, such that $x>0$

Case III:
If $c=0$, then $f(c)=f(0)=0+1=1$
The left hand limit of fat $x=0$ is,
$
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{\sin x}{x}=1
$

The right hand limt of fat $x=0$ is,
$
\begin{aligned}
& \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(x+1)=1 \\
& \therefore \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)
\end{aligned}
$

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.

Ex 5.1 Question 24:

Determine if fdefined by
$
f(x)=\left\{\begin{array}{lr}
x^2 \sin \frac{1}{x}, & \text { if } x \neq 0 \\
0, & \text { if } x=0
\end{array}\right.
$
is a continuous function?

Answer :
The given function fis $f(x)= \begin{cases}x^2 \sin \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{cases}$
It is evident that fis defined at all points of the real line.
Let $c$ be a real number.
Case I:
If $c \neq 0$, then $f(c)=c^2 \sin \frac{1}{c}$
$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^2 \sin \frac{1}{x}\right)=\left(\lim _{x \rightarrow c} x^2\right)\left(\lim _{x \rightarrow c} \sin \frac{1}{x}\right)=c^2 \sin \frac{1}{c}$
$\therefore \lim _{x \rightarrow r} f(x)=f(c)$
Therefore, $f$ is continuous at all points $x \neq 0$
Case II:
If $c=0$, then $f(0)=0$
$
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(x^2 \sin \frac{1}{x}\right)=\lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right)
$

It is known that, $-1 \leq \sin \frac{1}{x} \leq 1, x \neq 0$
$
\begin{aligned}
& \Rightarrow-x^2 \leq \sin \frac{1}{x} \leq x^2 \\
& \Rightarrow \lim _{x \rightarrow 0}\left(-x^2\right) \leq \lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right) \leq \lim _{x \rightarrow 0} x^2 \\
& \Rightarrow 0 \leq \lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right) \leq 0 \\
& \Rightarrow \lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right)=0 \\
& \therefore \lim _{x \rightarrow 0^{-}} f(x)=0
\end{aligned}
$

Similarly, $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(x^2 \sin \frac{1}{x}\right)=\lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right)=0$
$
\therefore \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 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.

Ex 5.1 Question 25 :

Examine the continuity of $f$, where $f$ is defined by
$
f(x)= \begin{cases}\sin x-\cos x, & \text { if } x \neq 0 \\ -1 & \text { if } x=0\end{cases}
$

Answer :
The given function fis $f(x)- \begin{cases}\sin x-\cos x, & \text { if } x \neq 0 \\ -1 & \text { if } x=0\end{cases}$
It is evident that fis defined at all points of the real line.
Let $c$ be a real number.
Case I:
If $c \neq 0$, then $f(c)=\sin c-\cos c$
$
\begin{aligned}
& \lim _{x \rightarrow c} f(x)-\lim _{x \rightarrow c}(\sin x-\cos x)-\sin c-\cos c \\
& \therefore \lim _{x \rightarrow c} f(x)=f(c)
\end{aligned}
$

Therefore, $f$ is continuous at all points $x_r$ such that $x \neq 0$
Case II:
If $c=0$, then $f(0)=-1$
$
\begin{aligned}
& \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow j}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1 \\
& \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow j}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1 \\
& \therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)
\end{aligned}
$

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.

Ex 5.1 Question 26:

Find the values of $k$ so that the function fis continuous at the indicated point.
$
f(x)=\left\{\begin{array}{ll}
\frac{k \cos x}{\pi-2 x}, & \text { if } x \neq \frac{\pi}{2} \\
3, & \text { if } x=\frac{\pi}{2}
\end{array} \quad \text { at } x=\frac{\pi}{2}\right.
$

Answer:
The given function fis $f(x)= \begin{cases}\frac{k \cos x}{\pi-2 x}, & \text { if } x \neq \frac{\pi}{2} \\ 3, & \text { if } x=\frac{\pi}{2}\end{cases}$
The given function fis continuous at $x=\frac{\pi}{2}$, if fis defined at $x=\frac{\pi}{2}$ and if the value of the fat $x=\frac{\pi}{2}$ equals the limit of fat $x=\frac{\pi}{2}$.
It is evident that $f$ is defined at $x=\frac{\pi}{2}$ and $f\left(\frac{\pi}{2}\right)=3$

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

Put $x=\frac{\pi}{2}+h$
Then, $x \rightarrow \frac{\pi}{2} \Rightarrow h \rightarrow 0$
$
\begin{aligned}
& \therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}+h\right)}{\pi-2\left(\frac{\pi}{2}+h\right)} \\
& \quad-k \lim _{h \rightarrow 0} \frac{-\sin h}{-2 h}-\frac{k}{2} \lim _{h \rightarrow 0} \frac{\sin h}{h}-\frac{k}{2} \cdot 1-\frac{k}{2} \\
& \therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=j\left(\begin{array}{l}
\pi \\
2
\end{array}\right) \\
& \Rightarrow \frac{k}{2}=3 \\
& \Rightarrow k=6
\end{aligned}
$

Therefore, the required value of kis 6 .

Ex 5.1 Question 27 :

Find the values of $k$ so that the function fis continuous at the indicated point.
$
f(x)=\left\{\begin{array}{ll}
k x^2, & \text { if } x \leq 2 \\
3, & \text { if } x>2
\end{array} \quad \text { at } x=2\right.
$

Answer :
The given function is $f(x)= \begin{cases}k x^2, & \text { if } x \leq 2 \\ 3, & \text { if } x>2\end{cases}$
The given function fis continuous at $x=2$, if fis defined at $x=2$ and if the value of fat $x=2$ equals the limit of fat $x=2$

It is evident that $f$ is defined at $x=2$ and $f(2)=k(2)^2=4 k$
$
\begin{aligned}
& \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} f(x)=f(2) \\
& \Rightarrow \lim _{x \rightarrow 2^{-}}\left(k x^2\right)=\lim _{x \rightarrow 2^{+}}(3)=4 k \\
& \Rightarrow k \times 2^2=3=4 k \\
& \Rightarrow 4 k=3=4 k \\
& \Rightarrow 4 k=3 \\
& \Rightarrow k=\frac{3}{4}
\end{aligned}
$

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

Ex 5.1 Question 28 :

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

Answer:
The given function is $f(x)=\left\{\begin{array}{l}k x+1, \text { if } x \leq \pi \\ \cos x, \text { if } x>\pi\end{array}\right.$
The given function $f$ is continuous at $x=p$, if $f$ is defined at $x=$ pand if the value of at $x=$ pequals the limit of fat $x=p$

It is evident that $f$ is defined at $x=$ pand $f(\pi)=k \pi+1$
$
\begin{aligned}
& \lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi^{+}} f(x)=f(\pi) \\
& \Rightarrow \lim _{x \rightarrow \pi^{-}}(k x+1)=\lim _{x \rightarrow \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}
\end{aligned}
$

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

Ex 5.1 Question 29 :

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

Answer :
The given function $f$ is $f(x)=\left\{\begin{array}{l}k x+1, \text { if } x \leq 5 \\ 3 x-5, \text { if } x>5\end{array}\right.$
The given function fis continuous at $x=5$, if $f$ is defined at $x=5$ and if the value of fat $x=5$ equals the limit of fat $x=5$

It is evident that $f$ is defined at $x=5$ and $f(5)=k x+1=5 k+1$
$
\begin{aligned}
& \lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5^{+}} f(x)=f(5) \\
& \Rightarrow \lim _{x \rightarrow 5^5}(k x+1)=\lim _{x \rightarrow 5^5}(3 x-5)=5 k+1 \\
& \Rightarrow 5 k+1=15-5=5 k+1 \\
& \Rightarrow 5 k+1=10 \\
& \Rightarrow 5 k=9 \\
& \Rightarrow k=\frac{9}{5}
\end{aligned}
$

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

Ex 5.1 Question 30:

Find the values of aand $b$ such that the function defined by
$f(x)=\left\{\begin{aligned}
5, & \text { if } x \leq 2 \\
a x+b, & \text { if } 2 21, & \text { if } x \geq 10
\end{aligned}\right.$

Answer :

For $x<2$, function is $f(x)=5$, constant, therefore it is continuous.
For $2<x<10$, function $f(x)=a x+b$, polynomial, therefore, it is continuous.

For $x>10$, function is $f(x)=21$, constant, therefore it is continuous.
For continuity at $x=2, \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} f(x)=f(2)$
$
\begin{aligned}
& \Rightarrow \lim _{h \rightarrow 0}(5)=\lim _{h \rightarrow 0}\{a(2+h)+b\}=5 \\
& \Rightarrow 2 a+b=5 \ldots \ldots \ldots \text {.(i) }
\end{aligned}
$

For continuity at $x=10, \lim _{x \rightarrow 10^{-}} f(x)=\lim _{x \rightarrow 10^{-}} f(x)=f(10)$
$
\begin{aligned}
& \Rightarrow \lim _{h \rightarrow 0}(21)=\lim _{h \rightarrow 0}\{a(10-h)+b\}=21 \\
& \Rightarrow 10 a+b=21 \ldots \ldots \ldots \text { (ii) }
\end{aligned}
$

Solving eq. (i) and eq. (ii), we get
$
a=2 \text { and } b=1
$

Ex 5.1 Question 31 :

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

Answer:
The given function is $f(x)=\cos \left(x^2\right)$
This function fis defined for every real number and fcan be written as the composition of two functions as,
$
\begin{aligned}
& f=g \circ h \text {, where } g(x)=\cos x \text { and } h(x)=x^2 \\
& {\left[\because(g \circ h)(x)=g(h(x))=g\left(x^2\right)=\cos \left(x^2\right)=f(x)\right]}
\end{aligned}
$

It has to be first proved that $g(x)=\cos x$ and $h(x)=x^2$ are continuous functions.
It is evident that gis defined for every real number.
Let $c$ be a real number.
Then, $g(c)=\cos c$
Put $x=c+h$

$
\begin{aligned}
& \text { If } x \rightarrow c, \text { then } h \rightarrow 0 \\
& \begin{aligned}
\lim _{x \rightarrow c} g(x) & =\lim _{x \rightarrow c} \cos x \\
& =\lim _{h \rightarrow 0} \cos (c+h) \\
& =\lim _{h \rightarrow 0}[\cos c \cos h-\sin c \sin h] \\
& =\lim _{h \rightarrow 0} \cos c \cos h-\lim _{h \rightarrow 0} \sin c \sin h \\
& =\cos c \cos 0-\sin c \sin 0 \\
& =\cos c \times 1-\sin c \times 0 \\
& =\cos c
\end{aligned} \\
& \begin{aligned}
\therefore \lim _{x \rightarrow c} g(x) & =g(c)
\end{aligned}
\end{aligned}
$

Therefore, $g(x)=\cos x$ is continuous function.
$
h(x)=x^2
$

Clearly, $h$ is defined for every real number.
Let $k$ be a real number, then $h(k)=k^2$
$
\begin{aligned}
& \lim _{x \rightarrow k} h(x)=\lim _{x \rightarrow k} x^2=k^2 \\
& \therefore \lim _{x \rightarrow k} h(x)=h(k)
\end{aligned}
$

Therefore, $h$ is a continuous function.
It is known that for real valued functions $g$ and $h$,such that $(g \circ h$ ) is defined at $c$, if $g$ is continuous at $c$ and if $f$ is continuous at $g(c)$, then $(f \circ g)$ is continuous at $c$.
Therefore, $f(x)=(g o h)(x)=\cos \left(x^2\right)$ is a continuous function.

Ex 5.1 Question 32:

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

Answer :
The given function is $f(x)=|\cos x|$
This function fis defined for every real number and fcan be written as the composition of two functions as,
$f=g \circ h$, where $g(x)=|x|$ and $h(x)=\cos x$
$
[\because(g \circ h)(x)=g(h(x))=g(\cos x)=|\cos x|=f(x)]
$

It has to be first proved that $g(x)=|x|$ and $h(x)=\cos x$ are continuous functions.
$g(x)=|x|$ can be writen as
$
g(x)= \begin{cases}-x, & \text { if } x<0 \\ x, & \text { if } x \geq 0\end{cases}
$

Clearly, $g$ is cefined for all real numbers.
Let $c$ be a real number.
Case I:
If $c<0$, then $g(c)=-c$ and $\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c}(-x)=-c$
$
\therefore \lim _{x \rightarrow c} g(x)=g(c)
$

Therefore, $g$ is continuous at all points $x$, such that $x<0$

Case II:
If $c>0$, then $g(c)=c$ and $\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow \cdot} x=c$
$
\therefore \lim _{x \rightarrow c} g(x)=g(c)
$

Therefore, $g$ is continuous at all points $x$, such that $x>0$
Case III:
If $c=0$, then $g(c)=g(0)=0$
$
\begin{aligned}
& \lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow \rightarrow^{-}}(-x)=0 \\
& \lim _{x \rightarrow 0^{+}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=0 \\
& \therefore \lim _{x \rightarrow 0^0} g(x)=\lim _{x \rightarrow 0^{+}}(x)=g(0) \\
&
\end{aligned}
$

Therefore, $g$ is continuous at $x=0$
From the above three observations, it can be concluded that gis continuous at all points.
$
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 $x \tilde{A} C a ̂ E^{\prime} c$, then $h \tilde{A} C a ̂ E^{\prime} 0$
$
h(c)=\cos c
$

$
\begin{aligned}
\lim _{x \rightarrow c} h(x) & =\lim _{x \rightarrow c} \cos x \\
& =\lim _{h \rightarrow 0} \cos (c+h) \\
& =\lim _{h \rightarrow 0}[\cos c \cos h-\sin c \sin h] \\
& =\lim _{h \rightarrow 0} \cos c \cos h-\lim _{h \rightarrow 0} \sin c \sin h \\
& =\cos c \cos 0-\sin c \sin 0 \\
& =\cos c \times 1-\sin c \times 0 \\
& =\cos c \\
\therefore \lim _{x \rightarrow 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 $(g \circ h)$ is defined at $c$, if $g$ is continuous at $c$ and if $t$ is continuous at $g(c)$, then $(f \circ g)$ is continuous at $c$.

Therefore, $f(x)=(g \circ h)(x)=g(h(x))=g(\cos x)=|\cos x|$ is a continuous function.

Ex 5.1 Question 33 :

Examine that $\sin |x|$ is a continuous function.

Answer :
Let $f(x)=\sin |x|$
This function fis defined for every real number and fcan be written as the composition of two functions as,
$f=g \circ h$, where $g(x)=|x|$ and $h(x)=\sin x$
$
[\because(g \circ h)(x)=g(h(x))=g(\sin x)=|\sin x|=f(x)]
$

It has to be proved first that $g(x)=|x|$ and $h(x)=\sin x$ are continuous functions.
$g(x)=|x|$ can be writen as
$
g(x)= \begin{cases}-x, & \text { if } x<0 \\ x, & \text { if } x \geq 0\end{cases}
$

Clearly, $g$ is cefined for all real numbers.
Let $c$ be a real number.

Case I:
If $c<0$, then $g(c)=-c$ and $\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c}(-x)=-c$ $\therefore \lim _{x \rightarrow c} g(x)=g(c)$.

Therefore, $g$ is continuous at all points $x$, such that $x<0$
Case II:

If $c>0$, then $g(c)=c$ and $\lim _{z \rightarrow c} g(x)=\lim _{x \rightarrow i} x=c$
$
\therefore \lim _{x \rightarrow c} g(x)=g(c)
$

Therefore, $g$ is continuous at all points $x$, such that $x>0$

Case III:
If $c=0$, then $g(c)=g(0)=0$
$
\begin{aligned}
& \lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow C^{-}}(-x)=0 \\
& \lim _{x \rightarrow 0^{+}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=0 \\
& \therefore \lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=g(0) \\
&
\end{aligned}
$

Therefore, $g$ is continuous at $x=0$
From the above three observations, it can be concluded that gis continuous at all points.
$
h(x)=\sin x
$

It is evident that $h(x)=\sin x$ is detined tor every real number.
let $c$ be a real number. Put $x=c+k$

$
\begin{aligned}
& h(c)=\sin c \\
& \begin{aligned}
h(c)=\sin c \\
\begin{aligned}
\lim _{x \rightarrow c} h(x) & =\lim _{x \rightarrow c} \sin x \\
& =\lim _{k \rightarrow 0} \sin (c+k) \\
& =\lim _{k \rightarrow 0}[\sin c \cos k+\cos c \sin k] \\
& -\lim _{k \rightarrow 0}(\sin c \cos k)+\lim _{h \rightarrow 0}(\cos c \sin k) \\
& =\sin c \cos 0+\cos c \sin 0 \\
& =\sin c+0 \\
& =\sin c
\end{aligned} \\
\therefore \lim _{x \rightarrow c} h(x)=g(c)
\end{aligned}
\end{aligned}
$

Therefore, $h$ is a continuous function.
It is known that for real valued functions $g$ and $h$, such that $(g \circ h)$ is defined at $c$, if $g$ is continuous at $c$ and if $f$ is continuous at $g(c)$, then $(f \circ g)$ is continuous at $c$.
Therefore, $f(x)=(g o h)(x)=g(h(x))=g(\sin x)=|\sin x|$ is a continuous function.

Ex 5.1 Question 34 :

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

Answer :
The given function is $f(x)=|x|-|x+1|$
Thetwo functions, gand $h$, are defined as
$g(x)=|x|$ and $h(x)=|x+1|$
Then, $f=g-h$
Thecontinuity of gand $h$ is examined first.
$g(x)=|x|$ can be writen as
$g(x)= \begin{cases}-x, & \text { if } x<0 \\ x, & \text { if } x \geq 0\end{cases}$
Clearly, $g$ is cefined for all real numbers.
Let $c$ be a real number.
Case I:
If $c<0$, then $g(c)=-c$ and $\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c}(-x)=-c$
$
\therefore \lim _{x \rightarrow c} g(x)=g(c)
$

Therefore, $g$ is continuous at all points $x$, such that $x<0$
Case II:
If $c>0$, then $g(c)=c$ and $\lim _{z \rightarrow c} g(x)=\lim _{x \rightarrow c} x=c$
$
\therefore \lim _{x \rightarrow c} g(x)=g(c)
$

$\text { Therefore, } g \text { Is continuous at all points } x \text {, such that } x>0$

From the above three observations, it can be concluded that gis continuous at all points.
$h(x)=|x+1|$ can be written as
$
h(x)= \begin{cases}-(x+1), & \text { if } x<-1 \\ x+1, & \text { if } x \geq-1\end{cases}
$

Clearly, $h$ is defined for every real number.
Let $c$ be a real number.

Case l:
If $c<-1$, then $h(c)=-(c+1)$ and $\lim _{x \rightarrow c} h(x)=\lim _{x \rightarrow c}[-(x+1)]=-(c+1)$
$
\therefore \lim _{x \rightarrow c} h(x)=h(c)
$

Therefore, $h$ is continuous at all points $x$, such that $x<-1$
Case II:
If $c>-1$, then $h(c)=c+1$ and $\lim _{x \rightarrow c} h(x)=\lim _{x \rightarrow c}(x+1)=c+1$
$
\therefore \lim _{x \rightarrow c} h(x)=h(c)
$

Therefore, $h$ is continuous at all points $x$, such that $x>-1$

Case III:
If $c=-1$, then $h(c)=h(-1)=-1+1=0$
$
\begin{aligned}
& \lim _{x \rightarrow-1^{-}} h(x)=\lim _{x \rightarrow-1^{-}}[-(x+1)]=-(-1+1)=0 \\
& \lim _{x \rightarrow-1^{+}} h(x)=\lim _{x \rightarrow-1^{+}}(x+1)=(-1+1)=0 \\
& \therefore \lim _{x \rightarrow 1^{-}} h(x)=\lim _{x, 1^{+}} h(x)=h(-1)
\end{aligned}
$

Therefore, $h$ is continuous at $x=-1$
From the above three observations, it can be concluded that $h$ is continuous at all points of the real line.