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B
A local minima
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C
Neither a local maxima nor a local minima
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D
None of these
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Solution
The correct option is A A local maxima Let f(x)=x√1−x2 ⇒f′(x)=1−2x2√1−x2=0⇒x=±1√2 But as x > 0, we have x=1√2 Now, again f′′(x)=√1−x2(−4x)−(1−2x2)−x√1−x2(1−x2) =2x3−3x(1−x2)3/2 ⇒f"(1√2)=−ve. Then f(x)is maximum at x=1√2