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Question: Answered & Verified by Expert
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
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Options:
  • A $f(0)=0$
  • B $f(\sqrt{2}-1)=\sqrt{2}-1$
  • C $f(\sqrt{2}-1)=\sqrt{2}+1$
  • D $f(\sqrt{2}-1)=\sqrt{2}$
Solution:
2485 Upvotes Verified Answer
The correct answer is: $f(\sqrt{2}-1)=\sqrt{2}$
If a function $f(x)$ assumes only irrational values which is also continuous, then $f(x)$ must be constant function.
$\Rightarrow \quad f(x)=\sqrt{2}$
$$
[\because f(\sqrt{2})=\sqrt{2} \mid
$$
$\therefore \quad f(\sqrt{2}-1)=\sqrt{2}$

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