#### Need solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 3

The correct option is (a) and (d)

Hint:

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

\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)\\ &\text { (ii) } \lim _{h \rightarrow 0}\left\{\frac{f(a+h)-f(a)}{h}\right\}=f^{\prime}\left(a^{+}\right) \text {for right hand derivative }\\ &\lim _{h \rightarrow 0}\left\{\frac{f(a-h)-f(a)}{-h}\right\}=f^{\prime}\left(a^{-}\right) \text {for left hand derivative } \end{aligned}

Given:

$f(x)=\left|\log _{10} x\right|$

Solution:

$f(x)=\left|\log _{10} x\right|$

$=\left|\frac{\log x}{\log _{e} 10}\right|=\left|\log _{\mathrm{e}} x \log _{10} e\right|$

Calculate right hand derivative

Using limit at x = 1

Apply limit at x = 1

\begin{aligned} &=\lim _{h \rightarrow 0}\left\{\frac{f(1+h)-f(1)}{h}\right\} \\ &=\lim _{h \rightarrow 0}\left\{\frac{\left|\log _{10} e \log (1+h)\right|-\left|\log _{10} e \log 1\right|}{h}\right\} \\ &=\lim _{h \rightarrow 0} \frac{\log _{10} e|\log (1+h)|}{h} \end{aligned}

Therefore applying L-Hospital rule

\begin{aligned} &=\log _{10} e \times 1 \\ &=f^{\prime}\left(1^{+}\right) \end{aligned}

For left hand limit

$=\lim _{h \rightarrow 0}\left\{\frac{f(1-h)-f(1)}{-h}\right\}$

\begin{aligned} &=\lim _{h \rightarrow 0}\left\{\frac{\left|\log _{10} e \log (1-h)\right|-\left|\log _{10} e \log 1\right|}{-h}\right\} \\ &=\lim _{h \rightarrow 0} \frac{-\log _{10} e|\log (1-h)|}{h} \\ &-\log _{10} e=f^{\prime}\left(1^{-}\right) \end{aligned}

\begin{aligned} &\text { The function } f(x)=\left|\log _{10} x\right| \text { is continuous and in the right hand derivative }\\ &f^{\prime}\left(1^{-}\right)=-\log _{10} e \end{aligned}

So the correct option is (a) and (d)