Question

Assertion :Let the function $f\left(x\right)=x^2-x+1\forall x\ge \frac{1}{2}$ and $g\left(x\right)=\frac{1}{2}+\sqrt{x-\frac{3}{4}}$, then $f\left(x\right)=g\left(x\right)$ has two solutions. 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. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

$f\left(x\right)=x^2-x+1$

Substitute $f\left(x\right)=y$

$y=x^2-x+1$

$\Rightarrow x^2-x+1-y=0$

$\Rightarrow x=\frac{1\pm \sqrt{4y-3}}{2}$

$\Rightarrow x=\frac{1}{2}\pm \sqrt{y-\frac{3}{4}}$

Since, $x\ge \frac{1}{2}$

$\Rightarrow x=\frac{1}{2}+\sqrt{y-\frac{3}{4}}$

$\Rightarrow f^{-1}\left(y\right)=\frac{1}{2}+\sqrt{y-\frac{3}{4}}$

$\Rightarrow f^{-1}\left(x\right)=\frac{1}{2}+\sqrt{x-\frac{3}{4}}$ ....(1)
Given, $g\left(x\right)=\frac{1}{2}+\sqrt{x-\frac{3}{4}}$ .....(2)

From (1) and (2), we have

$g\left(x\right)=f^{-1}\left(x\right)$

Therefore, $f\left(x\right)$ and $g\left(x\right)$ are inverses of each other.

Hence, reason is true.
Since ,$f\left(x\right)=g\left(x\right)$

$f\left(x\right)=f^{-1}\left(x\right)$

$\Rightarrow f\left(x\right)=x$

$\Rightarrow x^2-x+1=x$

$\Rightarrow (x-1{)}^2=0$

$\Rightarrow x=1$

i.e there is only one solution.

Hence, assertion is incorrect.