Question

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

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

Answer 1

The Correct Option is D

\(\begin{array}{l} f\left(x\right)=3^{\left(x^2-2\right)^3+4}, x\in R \end{array}\)
\(\begin{array}{l} f\left(x\right)=81.3^{\left(x^2-2\right)^3} \end{array}\)
\(\begin{array}{l} f’\left(x\right)=81.3^{\left(x^2-2\right)^3}\text{ln}2.3\left(x^2-2\right)2x\end{array}\)
\(\begin{array}{l} =\left(486\ \text{ln}2\right)\left(3^{\left(x^2-2\right)^3}\left(x^2-2\right)x\right)\end{array}\)

⇒ x = 0 is the local minima.
f′′(x) = (486 ln2)
\(\begin{array}{l} \begin{pmatrix}3^{\left(x^2-2\right)^3}\cdot\left(x^2-2\right) \\\left(5x^2-2+6x^2\text{ln}3\left(x^2-2\right)\right)\end{pmatrix}\end{array}\)
\(\begin{array}{l} f”\left(x\right)=0~~~~~x=\sqrt{2}\end{array}\)
\(\begin{array}{l} f”\left(\sqrt{2}^+\right)>0\end{array}\)
\(\begin{array}{l} \Rightarrow x=\sqrt{2}\end{array}\)
is point of inflection
\(\begin{array}{l} f”\left(x\right)>0~\forall~x>\sqrt{2}\end{array}\)
⇒ f′(x) is increasing for \(x > \sqrt2\)