#### Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 10

The correct option is (d)

Hint:

Use the given formula:

(i) A function f(x) is said to be continuous at a point x = a of its domain, if

$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$

Given:

$f(x)=\left\{\begin{array}{l} \frac{x-4}{|x-4|}+a \\ a+b \\ \frac{x-4}{|x-4|}+b \end{array}\right.$    \begin{aligned} , \quad &x<4 \\ , \quad &x=4 \\ , \quad &x>4 \end{aligned}

and f(x) is continuous at x = 4

Solution:

Using RHL at x = 4

\begin{aligned} &=\lim _{x \rightarrow 4^{+}} f(x) \\ &=\lim _{h \rightarrow 0} f(4+h) \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0}\left(\frac{4+h-4}{|4+h-4|}+b\right) \\ &=\lim _{h \rightarrow 0}\left(\frac{h}{|h|}+b\right) \\ &=1+b \end{aligned}

Using LHL

\begin{aligned} &\lim _{x \rightarrow 4^{-}} f(x)=\lim _{h \rightarrow 0} f(4-h) \\ &=\lim _{h \rightarrow 0}\left(\frac{4-h-4}{|4-h-4|}+a\right) \\ &=\lim _{h \rightarrow \infty}\left(\frac{-h}{|-h|}+a\right) \\ &=a-1 \end{aligned}

f(x) will be continuous at x = 4, if

\begin{aligned} &\lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4^{-}} f(x)=f(4) \\ &a-1=b+1=a+b \\ &b=-1 \text { and } a=1 \end{aligned}

So, the correct option is (d)