Question

Let $f:R\to R,g:R\to R$ and $h:R\to R$ be differentiable function such that $f\left(x\right)=x^3+3x+2,g\left(f\left(x\right)\right)=x$ and $h\left(g\right(g\left(x\right)\left)\right)=x$ for all $x\in R$. Then

• A. $g^{\prime }\left(2\right)=\frac{1}{15}$
• B. $h^{\prime }\left(1\right)=666$
• C. $h\left(0\right)=16$
• D. $h\left(g\left(3\right)\right)=36$

Answer 1

Answer: (B), (C)

The correct options are
B $h^{\prime }\left(1\right)=666$
C $h\left(0\right)=16$

Here,

$f\left(x\right)=x^3+3x+2$

and, $g\left(f\right(x\left)\right)=x$
and, $f^{\prime }\left(x\right)=3x^2+3$

$g\left(f\right(x\left)\right)=x$

Differentiating both sides, we get

$g^{\prime }\left(f\right(x\left)\right)f^{\prime }\left(x\right)=1$

$\Rightarrow g^{\prime }\left(f\right(x\left)\right)=\frac{1}{f^{\prime }\left(x\right)}$

$\because f\left(x\right)=2$ at $x=0$

$\therefore g^{\prime }\left(f\right(0\left)\right)=g^{\prime }\left(2\right)=\frac{1}{f^{\prime }\left(0\right)}$

$\Rightarrow g^{\prime }\left(0\right)=\frac{1}{3}$

Given:

$h\left(g\right(g\left(x\right)\left)\right)=x$

$\Rightarrow h\left(g\right(g\left(f\right(x\left)\right)\left)\right)=f\left(x\right)$

$\Rightarrow h\left(g\right(x\left)\right)=f\left(x\right)$

$\Rightarrow h\left(g\right(f\left(x\right)\left)\right)=f\left(f\right(x\left)\right)$

$\Rightarrow h\left(x\right)=f\left(f\right(x\left)\right)$

At $x=0$,

$h\left(0\right)=f\left(f\right(0\left)\right)$

$\Rightarrow h\left(0\right)=f\left(2\right)$

$\Rightarrow h\left(0\right)=16$

Now,

$h\left(x\right)=f\left(f\right(x\left)\right)$

Differentiating both sides,

$h^{\prime }\left(x\right)=f^{\prime }\left(f\right(x\left)\right)f^{\prime }\left(x\right)$

At $x=1$, $h^{\prime }\left(1\right)=f^{\prime }\left(f\right(1\left)\right)f^{\prime }\left(1\right)$

$\Rightarrow h^{\prime }\left(1\right)=f^{\prime }\left(6\right)\cdot 6$

$\Rightarrow h^{\prime }\left(1\right)=3\times 37\times 6=666$

Now,

$h\left(g\right(x\left)\right)=f\left(x\right)$

$\Rightarrow h\left(g\right(3\left)\right)=f\left(3\right)$

$\Rightarrow h\left(g\right(3\left)\right)=38$