QJEEAdv16Ma38f1: JEE Advanced 2016 Mathematics - Class 12 JEEAdv16Ma38f1 Solution: Let and be differentiable functions such that and for all . Then

Option (A): It is given that $f: \mathbb{R} \rightarrow \mathbb{R}, g: \mathbb{R} \rightarrow \mathbb{R}$ and $h: \mathbb{R} \rightarrow \mathbb{R}$ are differentiable functions.

Now, $\quad f(x)=x^3+3 x+2$

Differentiating w.r.t. to $x$ , we get

$f^{\prime}(x)=3 x^2+3$

Also, $g(f(x))=x$

Now,

$\begin{aligned}& g^{\prime}(f(x)) \cdot f^{\prime}(x)=1 \\\\& f(x)=2 \Rightarrow x^3+3 x+2=2 \\\\& \Rightarrow x^3+3 x=0 \Rightarrow x\left(x^2+3\right)=0 \\\\& \Rightarrow x=0\end{aligned}$ Now,

$g^{\prime}(f(0))=\frac{1}{f^{\prime}(0)} \Rightarrow g^{\prime}(2)=\frac{1}{3}$

Hence, option (A) is incorrect.

Option (B): For all $x \in \mathbb{R}$ :

$\begin{aligned}& h(g(g(f(x)))=x \\\\& h(g(g(x)))=x \\\\& \text { Now, } \quad x \rightarrow f(x) \Rightarrow h(g(g(f(x)))=f(x) \\\\& \Rightarrow \quad h(g(x))=f(x) \\\\& \Rightarrow \quad h^{\prime}(g(x)) \cdot g^{\prime}(x)=f^{\prime}(x)=3 x^2+3~~...(1)\end{aligned}$

For all $x \in \mathbb{R}: g(f(x))=x$

Now, $\quad x=1 \Rightarrow g(f(1))=1 \Rightarrow g(6)=1 \quad[\because f(1)=6]$

Substituting $x=6$ in Eq. (1), we get

$h^{\prime}(g(6)) \cdot g^{\prime}(6)=3\left(6^2\right)+3=111$

Therefore,

$h^{\prime}(1)=\frac{111}{g^{\prime}(6)} \quad\left(g^{\prime}(6)=\frac{1}{f^{\prime}(1)}\right)$

That is, $h^{\prime}(1)=111 \cdot f^{\prime}(x)=111 \times(3+3)=666$

Hence, option (B) is correct.

Option (C): $h(g(g(x)))=x$ .

For $g(g(x))=0$ , we have

$\begin{array}{ll} & g(x)=g^{-1}(0)=2 \\\\\Rightarrow & x=g^{-1}(2)=f(2)=16 \\\\\Rightarrow & h(0)=16\end{array}$

Hence, option (C) is correct.

Option (D): Here, $g(g(x))=g(3)$ which implies that

$g(x)=3 \Rightarrow x=g^{-1}(3)=f(3)=38$

Hence, option (D) is incorrect.