JEE MAIN - Mathematics (2022 - 29th July Morning Shift - No. 16)

Let $$f(x)=3^{\left(x^{2}-2\right)^{3}+4}, x \in \mathrm{R}$$. Then which of the following statements are true?

$$\mathrm{P}: x=0$$ is a point of local minima of $$f$$

$$\mathrm{Q}: x=\sqrt{2}$$ is a point of inflection of $$f$$

$$R: f^{\prime}$$ is increasing for $$x>\sqrt{2}$$

Only P and Q

Only P and R

Only Q and R

All P, Q and R

$$f(x) = {3^{{{({x^2} - 2)}^3} + 4}},\,x \in R$$

$$f(x) = {81.3^{{{({x^2} - 2)}^3}}}$$

$$f'(x) = {81.3^{{{({x^2} - 2)}^3}}}\ln 2.3({x^2} - 2)2x$$

$$ = (486\ln 2)\left( {{3^{{{({x^2} - 2)}^3}}}({x^2} - 2)x} \right)$$

JEE Main 2022 (Online) 29th July Morning Shift Mathematics - Application of Derivatives Question 54 English Explanation

$$ \Rightarrow x = 0$$ is the local minima.

$$f''(x) = (486\ln 2)\left( {{3^{{{({x^2} - 2)}^3}}}\,.\,({x^2} - 2)(5{x^2} - 2 + 6{x^2}\ln 3({x^2} - 2))} \right)$$

$$f''(x) = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \sqrt 2 $$

$$f''\left( {{{\sqrt 2 }^ + }} \right) > 0$$

$$f''\left( {{{\sqrt 2 }^ - }} \right) < 0$$

$$ \Rightarrow x = \sqrt 2 $$ is point of inflection

$$f''(x) > 0\,\forall x > \sqrt 2 $$

$$ \Rightarrow f'(x)$$ is increasing for $$x > \sqrt 2 $$