Q50 / 160 · WBJEE 2015 · Mathematics

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
    $f(\sqrt{2}-1)=\sqrt{2}-1$
  • C
    $f(\sqrt{2}-1)=\sqrt{2}+1$
  • Correct answer
    D
    $f(\sqrt{2}-1)=\sqrt{2}$

Solution

If a function $f(x)$ assumes only irrational values which is also continuous, then $f(x)$ must be constant function. $\Rightarrow \quad f(x)=\sqrt{2}$ $$ [\because f(\sqrt{2})=\sqrt{2} \mid $$ $\therefore \quad f(\sqrt{2}-1)=\sqrt{2}$

Hence, the correct answer is option: 'D'.

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