Discussion Forum

Let  $f:R\rightarrow R,g:R\rightarrow R$ and  $h:R\rightarrow R$ be differentiable functions such that  $f(x)=x^{3}+3x+2$, $g(f(x))=x$   and  $h(g(g(x)))=x$ for all  $x \epsilon R$   , Then

Answer:

Explanation:

As  , g(f(x))=x

Thus, g(x) is inverse of f(x)

$\Rightarrow$    g(f(x))=x

$\Rightarrow$     $g'(f(x)).f'(x)=1$

$\therefore$     $g'(f(x))=\frac{1}{f'(x)}$             ............(i)

 [ where, $f'(x)=3x^{2}+3$ ]

when    f(x)=2, then

 $x^{3}+3x+2=2$

$\Rightarrow$    x=0

 i.e, when x=0 , then f(x)=2

$\therefore$    $g'(f(x))=\frac{1}{3x^{2}+3}$  at (0,2)

$\Rightarrow$   $g'(2)=\frac{1}{3}$

$\therefore$   Option  (a) is incorrect.

Now,   $h(g(g(x)))=x$

 $\Rightarrow$    $h(g(g(f(x)))=f(x)$

 $\Rightarrow$   $h(g(x))=f(x)$            ......(ii)

As   $g(f(x))=x$ 

 $\therefore$     $h(g(3))=f(3)=3^{3}+3(3)+2=38$

 $\therefore$   Option (d) is incorrect.

 From Eq.(ii).

h(g(x))=f(x)

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

$\Rightarrow$    $h(x)=f(f(x))$    .......(iii)

[using g(f(x))=x]

$\Rightarrow$   $h'(x)=f'(f(x)).f'(x)$         .........(iv)

Putting x=1, we get

$h'(1)=f'(f(1)).f'(1)$

$=(3\times 36+3)\times 6$

=  $111\times 6=666$

  $\therefore$    Option (b) is correct.

  Putting x=0 in E.q. (iii) , we get

   $h(0)=f(f(0))=f(2)$

   =8+6+2=16

 $\therefore$  Option (c) is correct.