Question

Let $f:[−2,2]→R$ be a continuous function such that $f(x)$ assumes only irrational values. If $f(\sqrt{2})=\sqrt{2}$, then

• A. $f(0)=0$
• B. $f\left(\sqrt{2}−1\right)=\sqrt{2}−1$
• C. $f\left(\sqrt{2}−1\right)=\sqrt{2}+1$
• D. $f\left(\sqrt{2}−1\right)=\sqrt{2}$

Answer 1

Answer: (D)

If a function $f(x)$ assumes only irrational values which is also continuous, then $f(x)$ must be a constant function.
$⇒f(x)=\sqrt{2}[âˆ\mu f(\sqrt{2})=\sqrt{2}]$
$∴f(\sqrt{2}−1)=\sqrt{2}$