Practicing Success
Let $f(x) = x^3+3 x^2-33 x-33$ for x > 0 and g be its inverse such that kg'(2) = 1, then the value of k is |
-36 -42 12 none of these |
none of these |
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, $f(x)=x^3+3 x^2-33 x-33$ $\Rightarrow f'(x)=3 x^2+6 x-33$ $\Rightarrow f'(5)=75+30-33=72 \Rightarrow f'(g(2))=72 \Rightarrow k=72$ |