Question

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

• 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)

$f(x)=81\cdot 3^{\left(x^2-2^3\right)}$
$f^{\prime }(x)=81\cdot 3^{(x^2-2^3})\ln 3\cdot 3(x^2-2^2\cdot 2x$
$=(81\ne 6)3^{(x^2-2^3}x\left(x^2\right)-2^2\ln 3$
$x=6$ is point of local min
$f^{\prime }(x)=\underset{k}{\underbrace{(486\cdot \ln 3}}\underset{g(x)}{\underbrace{3{)}^{(x^2-2^3})(x^2-2^2}})$
$g^{\prime }(x)=3^{(x^2-2^3}(x^2-2^2+x\cdot 3^{(x^2-2^3}\cdot 4x\cdot \left(x^2-2\right)$
$+x\cdot \left(x^2\right)2^2)\cdot 3^{(x^2-2^3}\ln 3\cdot 3\left(x^2\right)2^2\cdot 2x$
$=3^{(x^2-2^3}(x)-2\left[x^2-2+4x^2+6x^2\ln 3\left(x^2\right)-2^3\right]$
$g^{\prime }(x)=3^{(x^2-2^3}\left(x^2-2[5)x^2-2+6x^2\ln 3x^2-2^3\right]$
$f^{\prime \prime }(x)=k\cdot g^{\prime }(x)$
$f^{\prime \prime }(\sqrt{2})=0,f^{\prime \prime }(\sqrt{2}+)>0,f^{\prime \prime }(\sqrt{2}-)<0$
$x^{-}=\sqrt{2}\text{ is point of inflection }$
$f^{\prime \prime }(x)>0\text{ for }x>\sqrt{2}\text{ so }{f}^{\prime }(x)\text{ is increasing }$