Question

Let $f:[-2,2] \rightarrow R$ 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.
$\Rightarrow f(x)=\sqrt{2} [\because f(\sqrt{2})=\sqrt{2}]$
$\therefore f(\sqrt{2}-1)=\sqrt{2}$