Let f(x) = x^3+3 x^2-33 x-33 for x > 0 and g be its inverse such that…

We have,

$f(x) = x^3+3 x^2-33 x-33, x > 0$

Since g is the inverse of f. Therefore,

fog(x) = x  for all x

$\Rightarrow \frac{d}{d x}\{f(g(x))\}=1$ for all x

$\Rightarrow f'(g(x)) . g'(x)=1$ for all x

$\Rightarrow f'(g(2)) . g'(2)=1$

$\Rightarrow f'(g(2))=k$                       [∵ kg'(2) = 1 (Given)]

Now, let

$g(2)=x$

$\Rightarrow f(g(2))=f(x)$

$\Rightarrow fog(2)=f(x)$

$\Rightarrow 2=x^3+3 x^2-33 x-33$

$\Rightarrow x^3+3 x^2-33 x-35=0$

$\Rightarrow (x+1)\left(x^2+2 x-35\right)=0$

$\Rightarrow (x+1)(x+7)(x-5)=0$

$\Rightarrow x=-1,-7,5 \Rightarrow x=5$

∴   $g(2)=5$              [∵ x > 0]

Also,

$f(x)=x^3+3 x^2-33 x-33$

$\Rightarrow f'(x)=3 x^2+6 x-33$

$\Rightarrow f'(5)=75+30-33=72 \Rightarrow f'(g(2))=72 \Rightarrow k=72$