#### Please solve RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 5 maths textbook solution

The correct option is (d)

Hint:

Use the given formula:

\begin{aligned} &\text { (i) If } \lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x) \text { then } f(x) \text { is discontinuous at } x=0 \text { . }\\ &\text { (ii)If } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=f(0) \text { then } f(x) \text { is continuous at } x=0 \end{aligned}

\begin{aligned} &\text { (iii) A function } f(x) \text { is said to be continuous at a point } x=a \text { of its domain, if }\\ &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \end{aligned}

Given:

$f(x)=\left\{\begin{array}{lr} \frac{\left|x^{2}-x\right|}{x^{2}-x} \\ 1 & \\ -1 & \end{array}\right.$    $\begin{array}{r} , \quad x \neq 0,1 \\ , \quad x=0 \\ , \quad x=1 \end{array}$

Solution:

Simplify the given function

$f(x)=\left\{\begin{array}{lr} \frac{\left|x^{2}-x\right|}{x^{2}-x} \\ 1 & \\ -1 & \end{array}\right.$    $\begin{array}{r} , \quad x \neq 0,1 \\ , \quad x=0 \\ , \quad x=1 \end{array}$

\begin{aligned} &f(x)=\left\{\begin{aligned} &1,\: x>1 \\ &1,\: x \leq 0 \\ &-1,\: 0 \leq x \leq 1 \end{aligned}\right. \\ &\text { Using RHL } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(h)=-1 \\ &\text { Using LHL } \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(h)=1 \end{aligned}

f(x) is discontinuous at x =0

Again, Using RHL,

\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=1\\ \end{aligned}

Using LHL,

\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=-1 \end{aligned}

f(x) is discontinuous at x = 1

Therefore,

f(x) is continuous for all except at x = 0 and x = 1

So, the correct option is (d)