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Question: Answered & Verified by Expert
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 at $x=\alpha+\frac{5}{2}, \frac{d g}{d x}$
MathematicsFunctionsTS EAMCETTS EAMCET 2020 (10 Sep Shift 2)
Options:
  • A $\frac{1}{2}$
  • B $\frac{1}{3}$
  • C $\frac{1}{4}$
  • D $\frac{1}{5}$
Solution:
1116 Upvotes Verified Answer
The correct answer is: $\frac{1}{5}$
We have, $f(x)=x^2+3 x-3$
$\Rightarrow \quad y=x^2+3 x-3$
$\Rightarrow \quad y=x^2+3 x-3+\frac{9}{4}-\frac{9}{4}$
$\Rightarrow y=\left(x+\frac{3}{2}\right)^2-\frac{21}{4}$
$\Rightarrow \quad y+\frac{21}{4}=\left(x+\frac{3}{2}\right)^2$
$\Rightarrow \quad x+\frac{3}{2}=\sqrt{y+\frac{21}{4}} \Rightarrow x=\sqrt{y+\frac{21}{4}}-\frac{3}{2}$
$\Rightarrow f^{-1}(y)=\sqrt{y+\frac{21}{4}}-\frac{3}{2}$
$\Rightarrow f^{-1}(x)=\sqrt{x+\frac{21}{4}}-\frac{3}{2}=g(x)$ (given)
Minimum value of $f^{-1}(x)$ is $\frac{-3}{2}$
$\therefore \quad \alpha=-\frac{3}{2}$
$\therefore \quad x=\alpha+\frac{5}{2}=-\frac{3}{2}+\frac{5}{2}=\frac{2}{2}=1$
Now, $g(x)=\sqrt{x+\frac{21}{4}}-\frac{3}{2}$
$\Rightarrow \quad g^{\prime}(x)=\frac{1}{2 \sqrt{x+\frac{21}{4}}}$
at $\quad x=1, g^{\prime}(x)=\frac{1}{2 \sqrt{1+\frac{21}{4}}}=\frac{1}{5}$

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