Question

Assertion :Let $f\left(x\right)=(x+1{)}^2-1\forall x\ge -1$ and $g\left(x\right)=-1+\sqrt{x+1}$ then number of solutions of the equation $g\left(x\right)=f\left(x\right)$ is two Reason: $f\left(x\right)$ and $g\left(x\right)$ are inverse of each other.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$f\left(x\right)=y\therefore f^{-1}\left(y\right)=x$
$\Rightarrow {(x+1)}^2-1=y$

$\Rightarrow x=-1\pm \sqrt{1+y}$
$\Rightarrow x=-1+\sqrt{1+y}=g\left(y\right)\left(\text{say}\right)(\because x\ge -1)$

$g\left(y\right)=-1+\sqrt{1+y}$
$f^{-1}\left(x\right)=-1+\sqrt{1+x}=g\left(x\right)$ ($\therefore$ Reason is true )

$f\left(x\right)=f^{-1}\left(x\right)$
$\Rightarrow {(x+1)}^2-1=-1+\sqrt{1+x}$

$\Rightarrow {(x+1)}^2=\sqrt{1+x}$
$\Rightarrow \sqrt{(x+1)}({(x+1)}^{3/2}-1)=0$

$\Rightarrow x=-1$ and $x=0$
$\Rightarrow$ there exists two solutions

$\Rightarrow$ Assertion is true.