QJEEM22Maa1bb: JEE Main 2022 Mathematics - Class 12 JEEM22Maa1bb Solution: If for , the function is continuous at , then :

$$f(x)=\frac{\sqrt[7]{p(729+x)}-3}{\sqrt[3]{729+qx}-9}$$

for continuity at $x = 0$ , $\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$

Now, $\therefore$ $$\underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}\frac{\sqrt[7]{p(729+x)}-3}{\sqrt[3]{729+qx}-9}$$

$\Rightarrow p = 3$ (To make indeterminant form)

So, $\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {{{{\left( {{3^7} + 3x} \right)}^{{1 \over 7}}} - 3} \over {{{\left( {729 + qx} \right)}^{{1 \over 3}}} - 9}}$

$= \mathop {\lim }\limits_{x \to 0} {{3\left[ {{{\left( {1 + {x \over {{3^6}}}} \right)}^{{1 \over 7}}} - 1} \right]} \over {9\left[ {{{\left( {1 + {q \over {729}}x} \right)}^{{1 \over 3}}} - 1} \right]}} = {1 \over 3}\,.\,{{{1 \over 7}\,.\,{1 \over {{3^6}}}} \over {{1 \over 3}\,.\,{q \over {729}}}}$

$\therefore$ $f(0) = {1 \over {7q}}$

$\therefore$ Option (B) is correct.