#### Need solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 19

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

\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \\ &\text { (ii) } \lim _{x \rightarrow a}\left\{\frac{f(x)}{g(x)}\right\}=\frac{1}{m}, \end{aligned}

${\text { Where, }} \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m$

Given:

$f(x)=\frac{2-(256-7 x)^{\frac{1}{8}}}{(5 x+32)^{\frac{1}{5}}-2} \text { is continuous everywhere }$

Solution:

$f(x)=\frac{2-(256-7 x)^{\frac{1}{8}}}{(5 x+32)^{\frac{1}{5}}-2}$

Now,

$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2-(256-7 x)^{\frac{1}{8}}}{(5 x+32)^{\frac{1}{5}}-2}$

Using factorization method

$\Rightarrow \lim _{x \rightarrow 0} \frac{(256)^{1 / 8}-(256-7 x)^{1 / 8}}{(5 x+32)^{1 / 5}-(32)^{1 / 5}}=x(0)$

$\Rightarrow \lim _{x \rightarrow 0} \frac{x}{x}\left\{\frac{(256)^{1 / 8}-(256-7 x)^{1 / 8}}{(5 x+32)^{1 / 5}-(32)^{1 / 5}}\right\}=f(0)$

$\Rightarrow \lim _{x \rightarrow 0} \frac{\frac{(256)^{1 / 8}-(256-7 x)^{1 / 8}}{x}}{\frac{(5 x+32)^{1 / 5}-(32)^{1 / 5}}{5}}=f(0)$

$\Rightarrow+\frac{7}{5} \lim _{x \rightarrow 0} \frac{\frac{(256-7 x)^{1 / 8}-(256)^{1 / 8}}{-7 x}}{\frac{(5 x+32)^{1 / 5}-(32)^{1 / 5}}{5 x}}=f(0)$

$\Rightarrow \frac{7}{5} \lim _{x \rightarrow 0} \frac{\frac{(256-7 x)^{1 / 8}-(256)^{1 / 8}}{256-7 x-256}}{\frac{(5 x+32)^{1 / 5-(32)^{1 / 5}}}{32+5 x-32}}=f(0)$

$\Rightarrow \frac{7 \frac{1}{8}(256)^{1 / 8-1}}{5} \frac{1}{\frac{1}{5}(32)^{1 / 5-1}}=f(0) \quad\left[\therefore \lim _{x \rightarrow a} \frac{x^{n}-a^{4}}{x-a}=n a^{n-1}\right]$

\begin{aligned} &\Rightarrow \frac{7}{5} \times \frac{1}{8} \times \frac{5}{1} \frac{(256)^{-7 / 8}}{(32)^{-4 / 5}}=f(0) \\ &\Rightarrow \frac{7}{8} \frac{(32)^{\frac{4}{5}}}{(256)^{-\frac{1}{8}}}=f(0) \end{aligned}

\begin{aligned} \Rightarrow f(0) &=\frac{7}{8} \times \frac{\left.72^{5}\right)^{\frac{4}{5}}}{\left(2^{8}\right)^{\frac{7}{8}}} \\ &=\frac{7}{2^{3}} \times \frac{2^{4}}{2^{7}} \\ &=\frac{7}{2^{6}}=\frac{7}{64} \end{aligned}

So the correct-option is (d)