Question:
Mark (√) against the correct answer in the following:
Let $f: N \rightarrow X: f(x)=4 x^{2}+12 x+15 .$ Then, $f^{-1}(y)=?$
A. $\frac{1}{2}(\sqrt{\mathrm{y}-4}+3)$
B. $\frac{1}{2}(\sqrt{\mathrm{y}-6}-3)$
C. $\frac{1}{2}(\sqrt{\mathrm{y}-4}+5)$
D. None of these
Solution:
$f(x)=4 x^{2}+12 x+15$
$\Rightarrow y=4 x^{2}+12 x+15$
$\Rightarrow y=(2 x+3)^{2}+6$
$\Rightarrow \sqrt{(} y-6)=2 x+3$
$\Rightarrow \frac{1}{2}(\sqrt{y-6}-3)=x$
$f^{-1}(y)=\frac{1}{2}(\sqrt{y-6}-3)$