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. $g^{\prime }(2)=\frac{1}{3}$
• C. $h^{\prime }(1)=666$
• D. $h(0)=16$

Answer 1

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

Given $f(x)=x^3+3x+2,g(f(x))=x$ &
$h(g(g(x)))=x$
$\therefore f^{\prime }(x)=3x^2+3$
$\therefore f^{\prime }(0)=3$
and $f(x)=x^3+3x+2$
$\therefore g(0)=2$
Again $g(f(x))=x$
$\therefore g^{\prime }(f(x))f^{\prime }(x)=1$
$g^{\prime }(f(x))=\frac{1}{f^{\prime }(x)}.....(\ast )$
$\therefore$ For $g^{\prime }(2)$, we need to solve $f(x)=2$
$\Rightarrow x=0$
$\Rightarrow g^{\prime }(2)=g(f(0))=\frac{1}{f^{\prime }(0)}=\frac{1}{3}$
Also, $h(g(g(x)))=x$, on differentiating w. r to $x$ we get
$\therefore h^{\prime }\left(g\left(g\left(x\right)\right)\right).g^{\prime }\left(g\left(x\right)\right).g^{\prime }\left(x\right)=1$
$\Rightarrow h^{\prime }\left(g\left(g\left(x\right)\right)\right)=\frac{1}{g^{\prime }\left(g\left(x\right)\right)g^{\prime }\left(x\right)}$
Now for $h^{\prime }(1)$ we need to solve $g(g(x))=1$
$\Rightarrow g(x)=g^{\prime }(1)$ 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, $h^{\prime }(g(g(x)))=\frac{1}{g^{\prime }(g(x))g^{\prime }(x)}=\frac{1}{g^{\prime }(6)g^{\prime }(236)}$
$=\frac{1}{\frac{1}{(f^{\prime }(x){)}_{x=1}}\times \frac{1}{(f^{\prime }(x){)}_{x=6}}}$ (using $\ast$)
$\therefore h^{\prime }(1)=\frac{1}{\frac{1}{6}\times \frac{1}{111}}=666$
$($As $f(x)=x^3+3x+2,f^{\prime }(x)=3x^2+3$,
$f^{\prime }(1)$ & $f^{\prime }(6)$ 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
$x^3+3x-4=0$ & $x^3+3x-234=0$
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 $(x^3+3x+2{)}_{x=4}$
$=64+12+2=78$