Question

Let $f:R\to R,g:R\to R$ and $h:R\to R$ be differentiable functions such that $f\left(x\right)=x^3+3x+2,g\left(f\right(x\left)\right)=x$ 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. $h\left(g\right(3\left)\right)=36$

Answer 1

Answer: (B), (C)

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

$f\left(x\right)=x^3+3x+2$
$f^{\prime }\left(x\right)=3x^2+3$
Now, we have $g\left(f\right(x\left)\right)=x\Rightarrow g^{\prime }\left(f\right(x\left)\right).f^{\prime }\left(x\right)=1$
We need to find $g^{\prime }\left(2\right)\Rightarrow f\left(x\right)=2\Rightarrow x=0$
$\therefore g^{\prime }\left(2\right).f^{\prime }\left(0\right)=1\Rightarrow g^{\prime }\left(2\right)=\frac{1}{3}$

Now, we have $h\left(g\right(g\left(x\right)\left)\right)=x$
Replacing $x$ by $f\left(x\right)$ in the above expression:
$\Rightarrow h\left(g\right(g\left(f\right(x\left)\right)\left)\right)=f\left(x\right)\Rightarrow h\left(g\right(x\left)\right)=f\left(x\right)...\left[1\right]$
Replacing $x$ by $f\left(x\right)$ in the above expression:
$\Rightarrow h\left(g\right(f\left(x\right)\left)\right)=f\left(f\right(x\left)\right)\Rightarrow h\left(x\right)=f\left(f\right(x\left)\right)\Rightarrow h^{\prime }\left(x\right)=f^{\prime }\left(f\right(x\left)\right)\cdot f^{\prime }\left(x\right)$

$h\left(0\right)=f\left(f\right(0\left)\right)=f\left(2\right)=16$
$h^{\prime }\left(1\right)=f^{\prime }\left(f\right(1\left)\right)\cdot f^{\prime }\left(1\right)=f^{\prime }\left(6\right)\times 6h^{\prime }\left(1\right)=111\times 6=666$

$h\left(g\right(3\left)\right)=f\left(3\right)\left(\text{From}\right[1\left]\right)\Rightarrow h\left(g\right(3\left)\right)=3^3+3\times 3+2=38$