A function $$s(x)$$ is defined as $$g(x)=14f2x2$$−$$1+12f1$$−$$x2$$ and f′(x) is an increasing function. Then $$g(x)$$ is increasing in the interval

Correct option is 2−23,0∪23,∞

Questions

A function s(x) is defined as $g (x) = \frac{1}{4} f (2 x^{2} - 1) + \frac{1}{2} f (1 - x^{2}) and f^{'} (x)$ is an increasing function. Then g(x) is increasing in the interval

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(-1, 1)

−23,0∪23,∞

−23,23

none of these

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detailed solution

Correct option is B

g′(x)=xf′2x2−1−xf′1−x2=xf2x2−1−f1−x2 ⇒ g′(x)>0If x>0,2x2−1>1−x2 (as f ' is an increasing function)or 3x2>2 or x∈−∞,−23∪23,∞or x∈23,∞If x<0,2x2−1<1−x2or 3x2<2 or x∈−23,23 or x∈−23,0

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A function s(x) is defined as g(x)=14f2x2−1+12f1−x2 and f′(x) is an increasing function. Then g(x) is increasing in the interval

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A function s(x) is defined as $g (x) = \frac{1}{4} f (2 x^{2} - 1) + \frac{1}{2} f (1 - x^{2}) and f^{'} (x)$ is an increasing function. Then g(x) is increasing in the interval