Question

Let $f:R \to R,g:R \to R{\text{ and }}h:R \to R$ be a differential function such that $f\left( x \right) = {x^3} + 3x + 2,g\left( {f\left( x \right)} \right) = x{\text{ and }}h\left( {g\left( {g\left( x \right)} \right)} \right) = x{\text{ for all }}x \in R$ . Then which of the following options is correct:
(A) $g{'}\left( 2 \right) = \dfrac{1}{{15}}$
(B) $h{'}\left( 1 \right) = 666$
(C) $h\left( 0 \right) = 16$
(D) $h\left( {g\left( 3 \right)} \right) = 36$

Answer 1

Use the given information to evaluate the options $g{'}\left( 2 \right)$ , $h{'}\left( 1 \right)$ , $h\left( 0 \right)$ and $h\left( {g\left( 3 \right)} \right)$. For option (A), differentiate $g\left( {f\left( x \right)} \right) = x$ using the chain rule and then find for what value of $x$ does $f\left( x \right) = 2$. For $h{'}\left( 1 \right)$, differentiate $h\left( {g\left( {g\left( x \right)} \right)} \right) = x$ using the chain rule and substitute the value of each term separately. Compare all the values using options to determine the correct answer.

Complete step by step answer:
We were given with the function $f$ , defined as $f\left( x \right) = {x^3} + 3x + 2$ , and function $g$ where $g\left( {f\left( x \right)} \right) = x$ and another function $h$ which has property $h\left( {g\left( {g\left( x \right)} \right)} \right) = x$ . With this given information we need to check which of the given options is correct.
Let’s first consider option (A).
Differentiating relation $g\left( {f\left( x \right)} \right) = x$ with respect to $x$ using the chain rule, i.e. $\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f{'}\left( {g\left( x \right)} \right)g{'}\left( x \right)$ , we get: $g{'}\left( {f\left( x \right)} \right) \times f{'}\left( x \right) = \dfrac{{d\left( x \right)}}{{dx}} = 1$
Therefore, we get the equation $g{'}\left( {f\left( x \right)} \right) \times f{'}\left( x \right) = 1$ ……………. (1)
And $f{'}\left( x \right)$ will be the first derivative of $f\left( x \right)$ , i.e. $f{'}\left( x \right) = \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {{x^3} + 3x + 2} \right) = 3{x^2} + 3$
For $x = 0$ , $f\left( 0 \right) = {0^3} + 3 \times 0 + 2 = 2$
Now, let’s put $x = 0$in relation (1), this will give:
$\Rightarrow g{'}\left( {f\left( 0 \right)} \right) \times f{'}\left( 0 \right) = 1$
Putting the value of $f{'}\left( x \right)$ and $f\left( 0 \right)$ , we get
$\Rightarrow g{'}\left( 2 \right) \times \left( {3 \times 0 + 3} \right) = 1 \Rightarrow g{'}\left( 2 \right) = \dfrac{1}{3}$
Let's put $x = 3$ in $f\left( x \right)$, we get: $f\left( 3 \right) = {3^3} + 3 \times 3 + 2 = 38$
So, we use this in $g\left( {f\left( x \right)} \right) = x$as follows:
$\Rightarrow g\left( {f\left( 3 \right)} \right) = 3 \Rightarrow g\left( {38} \right) = 3$
Now, we can use $x = 38$ in $h\left( {g\left( {g\left( x \right)} \right)} \right) = x$ and we will get:
$\Rightarrow h\left( {g\left( {g\left( {38} \right)} \right)} \right) = 38 \Rightarrow h\left( {g\left( 3 \right)} \right) = 38$
For $x = 2$ in $f\left( x \right) = {x^3} + 3x + 2$, we get: $f\left( 2 \right) = {2^3} + 3 \times 2 + 2 = 8 + 6 + 2 = 16$
Now for $x = 2$ in $g\left( {f\left( x \right)} \right) = x$ , we get:
$\Rightarrow g\left( {f\left( 2 \right)} \right) = 2 \Rightarrow g\left( {16} \right) = 2$
Similarly, for $x = 0$ in $g\left( {f\left( x \right)} \right) = x$ , we get:
$\Rightarrow g\left( {f\left( 0 \right)} \right) = 0 \Rightarrow g\left( 2 \right) = 0$
So, let's use $x = 16$ in the given expression $h\left( {g\left( {g\left( x \right)} \right)} \right) = x$ and using the above-found values in it, this will result in:
$\Rightarrow h\left( {g\left( {g\left( {16} \right)} \right)} \right) = 16 \Rightarrow h\left( {g\left( 2 \right)} \right) = 16 \Rightarrow h\left( 0 \right) = 16$
Now, let's differentiate the equation $h\left( {g\left( {g\left( x \right)} \right)} \right) = x$using the chain rule
$\dfrac{d}{{dx}}\left( {h\left( {g\left( {g\left( x \right)} \right)} \right)} \right) = \dfrac{d}{{dx}}\left( x \right) \Rightarrow h{'}\left( {g\left( {g\left( x \right)} \right)} \right) \times g{'}\left( {g\left( x \right)} \right) \times g{'}\left( x \right) = 1$ ………. (2)
We need the value of $h{'}\left( 1 \right)$ , so for that, we need to make $h{'}\left( {g\left( {g\left( x \right)} \right)} \right) = h{'}\left( 1 \right)$ , i.e. $g\left( {g\left( x \right)} \right) = 1$
For $x = 1$ in $g\left( {f\left( x \right)} \right) = x$, we get: $g\left( {f\left( 1 \right)} \right) = 1 \Rightarrow g\left( {{1^3} + 3 \times 1 + 2} \right) = 1 \Rightarrow g\left( 6 \right) = 1$
Now for $x = 6$ in $g\left( {f\left( x \right)} \right) = x$, we get: $g\left( {f\left( 6 \right)} \right) = 6 \Rightarrow g\left( {{6^3} + 3 \times 6 + 2} \right) = 6 \Rightarrow g\left( {216 + 18 + 2} \right) = 6 \Rightarrow g\left( {236} \right) = 6$
So, for $x = 236$ the relation (2) will become:
$\Rightarrow h{'}\left( {g\left( {g\left( {236} \right)} \right)} \right) \times g{'}\left( {g\left( {236} \right)} \right) \times g{'}\left( {236} \right) = 1 \Rightarrow h{'}\left( {g\left( 6 \right)} \right) \times g{'}\left( 6 \right) \times g{'}\left( {236} \right) = 1 \Rightarrow h{'}\left( 1 \right) = \dfrac{1}{{g{'}\left( 6 \right) \times g{'}\left( {236} \right)}}$Therefore, we get: $h{'}\left( 1 \right) = \dfrac{1}{{g{'}\left( 6 \right) \times g{'}\left( {236} \right)}}$ ………. (3)
From relation (1), we can put $x = 6{\text{ and }}x = 236$
$\Rightarrow g{'}\left( {f\left( x \right)} \right) \times f{'}\left( x \right) = 1 \Rightarrow g{'}\left( {f\left( x \right)} \right) = \dfrac{1}{{f{'}\left( x \right)}}$
For $x = 1$ , we get: $g{'}\left( {f\left( 1 \right)} \right) = \dfrac{1}{{f{'}\left( 1 \right)}} \Rightarrow g{'}\left( {1 + 3 + 2} \right) = \dfrac{1}{{\left( {3 \times 1 + 3} \right)}} \Rightarrow g{'}\left( 6 \right) = \dfrac{1}{6}$
For $x = 6$ , we get: $g{'}\left( {f\left( 6 \right)} \right) = \dfrac{1}{{f{'}\left( 6 \right)}} \Rightarrow g{'}\left( {236} \right) = \dfrac{1}{{3 \times {6^2} + 3}} = \dfrac{1}{{108 + 3}} = \dfrac{1}{{111}}$
Now, let's substitute these values in the relation (3)
$\Rightarrow h{'}\left( 1 \right) = \dfrac{1}{{g{'}\left( 6 \right) \times g{'}\left( {236} \right)}} = \dfrac{1}{{\dfrac{1}{6} \times \dfrac{1}{{111}}}} = \dfrac{1}{{\dfrac{1}{{666}}}} = 666$
Hence, we get the results: $g{'}\left( 2 \right) = \dfrac{1}{3}$ , $h{'}\left( 1 \right) = 666$ , $h\left( 0 \right) = 16$ and $h\left( {g\left( 3 \right)} \right) = 38$

Therefore, only the option (B) and (C) are correct.

Note:
Analyse the given information and go step by step while proceeding through the solution. Notice that the use of the chain rule of differentiation is a crucial part of the solution to this problem. Be careful with the use of braces while solving to avoid any confusion. In questions like these, there{'}s no choice other than checking for each option one by one.