#### Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 18

The correct option is (c)

Hint:

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:

$f(x)=\frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}},(x \neq 0)$

Solution:

Step 1: Understand that, if f(x) to be continuous at x = 0 then,

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

Therefore,

$=f(0)=\lim _{x \rightarrow 0} \frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}}$

$=f(0)=\lim _{x \rightarrow 0} \frac{(27-2 x)^{\frac{1}{3}}-(27)^{\frac{1}{3}}}{3\left[3-(243+5 x)^{\frac{1}{5}}\right]}$

$=\lim _{x \rightarrow 0} \frac{(27-2 x)^{1 / 3}-(27)^{4 / 3}}{3\left\{(243)^{15}-(243+5 x)^{1 / 5}\right\}}$

$=\frac{1}{3} \lim _{x \rightarrow 0} \frac{x\left\{(27-2 x)^{1 / 3}-(27)^{4 / 3}\right\}}{\left.x(243)^{1 / 5}-(243+5 x)^{1 / 5}\right\}}$

$=\frac{1}{3} \lim _{x \rightarrow 0} \frac{\frac{(27-2 x)^{1 / 3}-\left(2+1^{1 / 3}\right.}{x}}{\frac{(243)^{1 / 5}-(243+5 x)^{1 / 5}}{x}}$

$=\frac{2}{3} \lim _{x \rightarrow 0} \frac{\frac{(27-2 x)^{1 / 3}-(27)^{1 / 3}}{5}}{5 \frac{(243+5 x)^{1 / 5}-(243)^{1 / 5}}{5 x}}$

$=\frac{2}{15} \lim _{x \rightarrow 0} \frac{\frac{(27-2 x)^{1 / 3}-(27)^{1 / 3}}{27-2 x-27}}{\frac{(243+5 x)^{1 / 5}-(243)^{1 / 5}}{243+5 x-243}}$

$=f(0)=\frac{2}{15} \times \frac{\frac{1}{3} \times 27^{-\frac{2}{3}}}{\frac{1}{5} \times 243^{-\frac{4}{5}}}$

$=f(0)=\frac{2}{15} \times \frac{\frac{1}{3} \times \frac{1}{27^{\frac{2}{3}}}}{\frac{1}{5} \times \frac{1}{243^{\frac{4}{5}}}}$

\begin{aligned} &\begin{aligned} &=\frac{2}{9} \frac{(243)^{\frac{4}{5}}}{(27)^{\frac{2}{3}}} \\ =& \frac{2}{9} \frac{\left(3^{5}\right)^{4 / 5}}{\left(3^{3}\right)^{2 / 3}} \\ \therefore f(0) &=\frac{2}{9} \times \frac{3^{4}}{3^{2}} \\ &=2 . \end{aligned}\\ &\text { Therefore, } f(0)=2 \end{aligned}

Hence, the correct option is (c)

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