Question

Let $f(x)=3^{{\left(x^2-2\right)}^3+4},x\in R$. Then which of the following statements are true ? $P:x=0$ is a point of local minima of $f$ $Q:x=\sqrt{2}$ is a point of inflection of $f$ $R$ : $f$ ' is increasing for $x>\sqrt{2}$

• A. Only P and $Q$
• B. Only P and R
• C. Only Q and R
• D. All, P, Q and R

Answer 1

Answer: (D)

$f(x)=81\cdot 3^{{\left(x^2-2\right)}^3}$
$f^{\prime }(x)=81\cdot 3^{{\left(x^2-2\right)}^3}\cdot \ln 3\cdot 3{\left(x^2-2\right)}^2\cdot 2x$
$=(81\times 6)3^{{\left(x^2-2\right)}^3}\times {\left(x^2-2\right)}^2\ln 3$

$f^{\prime }(x)=\underset{k}{\underbrace{(486\cdot \ln 3)}}\underset{g(x)}{\underbrace{3^{{\left(x^2-2\right)}^3}x{\left(x^2-2\right)}^2}}$
$g^{\prime }(x)=3^{{\left(x^2-2\right)}^3}{\left(x^2-2\right)}^2+x\cdot 3^{{\left(x^2-2\right)}^3}\cdot 4x\cdot \left(x^2-2\right)$
$+x\cdot {\left(x^2-2\right)}^2\cdot 3^{{\left(x^2-2\right)}^3}\ln 3\cdot 3{\left(x^2-2\right)}^2\cdot 2x$
$=3^{{\left(x^2-2\right)}^3}\left(x^2-2\right)\left[x^2-2+4x^2+6x^2\ln 3{\left(x^2-2\right)}^3\right]$
$g^{\prime }(x)=3^{{\left(x^2-2\right)}^3}\left(x^2-2\right)\left[5x^2-2+6x^2\ln 3{\left(x^2-2\right)}^3\right]$
$f^{\prime \prime }(x)=k\cdot g^{\prime }(x)$
$f^{\prime \prime }(\sqrt{2})=0,f^{\prime \prime }(\sqrt{2}+>0,f^{\prime }(\sqrt{2})<0$
$x^{-}=\sqrt{2}\text{ is point of inflection }$
$f^{\prime \prime }(x)>0\text{ for }x>\sqrt{2}\text{ so }{f}^{\prime }(x)\text{ is increasing }$