Question

If $\alpha$ is such a minimum value for which the inverse of $f(x)=x^{2}+3 x-3$ exists in $[\alpha, \infty)$ and $g$ is the inverse of the $f$ then ar $x=\alpha+\frac{5}{2}, \frac{d g}{d x}=$

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

Answer: (D)

We have, $f(x)=x^{2}+3 x-3$
$\Rightarrow y=x^{2}+3 x-3 $
$\Rightarrow y=x^{2}+3 x-3+\frac{9}{4}-\frac{9}{4} $
$\Rightarrow y=\left(x+\frac{3}{2}\right)^{2}-\frac{21}{4} $
$\Rightarrow y+\frac{21}{4}=\left(x+\frac{3}{2}\right)^{2}$
$\Rightarrow x+\frac{3}{2}=\sqrt{y+\frac{21}{4}} $
$\Rightarrow x=\sqrt{y+\frac{21}{4}}-\frac{3}{2}$
$\Rightarrow f^{-1}(y)=\sqrt{y+\frac{21}{4}}-\frac{3}{2} $
$\Rightarrow f^{-1}(x)=\sqrt{x+\frac{21}{4}}-\frac{3}{2}=g(x) $ (given)
Minimum value of $f^{-1}(x)$ is $\frac{-3}{2}$
$\therefore \alpha=-\frac{3}{2}$
$\therefore x=\alpha+\frac{5}{2}=-\frac{3}{2}+\frac{5}{2}=\frac{2}{2}=1$
Now, $g(x)=\sqrt{x+\frac{21}{4}}-\frac{3}{2}$
$\Rightarrow g'(x)=\frac{1}{\sqrt{[2]} x+\frac{21}{4}}$
at $x=1, g'(x)=\frac{1}{2 \sqrt{1+\frac{21}{4}}}=\frac{1}{5}$