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Question

The function $$f(x)=\sqrt{1+\dfrac{4}{x}}-\sqrt{1-x}$$ attains its minimum value at $$x$$ equal to $$($$0 < x \le 1$$)

A

$$2\sqrt{2}-2$$

B

$$\dfrac{1}{2\sqrt{2}}$$

C

$$\sqrt{2}-1$$

D

$$\dfrac{1}{\sqrt{2}}$$

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Solution

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Correct option is C. $$\dfrac{1}{\sqrt{2}}$$

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