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Q.

If $g (x) = \frac{1}{4} f (2 x^{2} - 1) + \frac{1}{2} f (1 - x^{2})$ where $f^{''} (x) > 0 & x < 0$ then g(x) is increasing on

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a

$(- \sqrt{\frac{2}{3}}, 0)$

b

$(- 1, - \sqrt{\frac{2}{3}})$

c

$(- \infty, 0)$

d

$(- 2, - \sqrt{\frac{2}{3}})$

answer is A.

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Detailed Solution

$\begin{array}{l} g^{1} (x) = \frac{1}{4} f^{1} (2 x^{2} - 1) 4 x + \frac{1}{2} f^{1} (1 - x^{2}) (- 2 x) \\ = x (f^{1} (2 x^{2} - 1) - f^{1} (1 - x^{2})) > 0 \\ f^{1} (2 x^{2} - 1) - f^{1} (1 - x^{2}) < 0 \\ f^{1} (2 x^{2} - 1) < f^{1} (1 - x^{2}) \\ 2 x^{2} - 1 < 1 - x^{2} \\ 3 x^{2} - 2 < 0 \\ (x + \sqrt{\frac{2}{3}}) (x - \sqrt{\frac{2}{3}}) < 0 \end{array}$

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