Question

Let $f : R \to R , g : R \to R$ and $h : R \to R$ be differential functions such that $f(x ) = x^3 + 3x + 2, g(f (x)) = x$ & $h(g(g(x))) = x \,\forall\, x \in R$, then which of the following is true?

• A. $h(g(4)) = 78$
• B. $g'(2) = \frac{1}{3}$
• C. $h'(1) = 666$
• D. $h(0)= 16$

Answer 1

Answer: (A), (B), (C), (D)

Given $f(x) = x^3 + 3x + 2, g(f(x)) = x$ &
$h(g(g(x))) = x$
$\therefore f'(x) = 3x^2 + 3$
$\therefore f'(0) = 3$
and $f(x) = x^3 + 3x + 2$
$ \therefore g(0) = 2$
Again $g(f(x)) = x$
$\therefore g'(f(x))f'(x)= 1$
$g'(f(x)) =\frac{1}{f'(x)} \,\,.....(*)$
$\therefore $ For $g'(2)$, we need to solve $f(x) = 2$
$\Rightarrow x = 0$
$\Rightarrow g'(2) = g(f(0)) = \frac{1}{f'(0)} = \frac{1}{3}$
Also, $h(g(g(x))) = x$, on differentiating w. r to $x$ we get
$\therefore h'\left(g\left(g\left(x\right)\right)\right).g'\left(g\left(x\right)\right).g'\left(x\right) = 1 $
$ \Rightarrow h'\left(g\left(g\left(x\right)\right)\right) = \frac{1}{g'\left(g\left(x\right)\right)g'\left(x\right)}$
Now for $h'(1)$ we need to solve $g(g(x)) = 1$
$\Rightarrow g(x) = g'(1) $ but $g(s(x)) = x$
$\Rightarrow g^{-1} (x) = f(x)$
$\Rightarrow g^{-1} (1) = f(1), g^{-1}(6) = f(6) = 236$
$\Rightarrow g(x) = g^{-1} (1) = f(1) = 1^3 + 3(1) + 2 = 6$
Now, $h'(g(g(x))) = \frac{1}{g'(g(x))g'(x)} = \frac{1}{g'(6)g'(236)}$
$ = \frac{1}{\frac{1}{(f'(x))_{x = 1}} \times \frac{1}{(f'(x))_{x=6}}}$ (using $*$)
$\therefore h'(1) = \frac{1}{\frac{1}{6} \times \frac{1}{111}} = 666$
$($As $f(x) = x^3 + 3x + 2, f'(x) = 3x^2 + 3$,
$f'(1)$ & $f'(6)$ are $6$ & $ 111$ respectively)
where $x = 1, x = 6$ are solutions of the equations
$x^3 + 3x + 2 = 6$ & $x^3 + 3x + 2 = 236$
i.e., solutions of the equations
$x^3 + 3 x - 4 = 0$ & $x^3 + 3 x - 234 = 0$
For the value of $h(0)$ we need to solve $g(g(x)) = 0$
$\Rightarrow g(x) = g^{-1}(0) = f(0) = 2$ thus $x = g^{-1}(2) = f(2) = 16$
$\therefore h(0) = h(g(g(x)))$ (at $g(g(x)) = 0) = 16$
Again, $h(g(g(x))) = x$, replacing $x$ by $f(x)$
$\Rightarrow h(g(g(f(x)))) = f(x)$
$\Rightarrow h(g(x)) =f(x)$ using $g(f(x)) = x$
$\therefore h(g(4)) = f(4 ) =$ value of $(x^3 + 3x + 2)_{x = 4}$
$= 64 + 12 + 2 = 78$