If g(x)=\frac{1}{6} f(3 x^{2}-1)+\frac{1}{2} f(1-x^{2}), \forallx \in\R, where f"(x)>0, \\forallx \\in\R. Then g(x) is increasing in the interval ______

Correct Answer is : (−1√2,0)∪(1√2,∞)

0, \\forallx \\in\R. Then g(x) is increasing in the interval ______ Options: ({-1}/{\\sqrt{2}, 0) \\cup({1}/{\\sqrt{2}, \infty) ({-1}/{\\sqrt{2}, {1}/{\\sqrt{2}) (-1,0) \\cup(1,2) (-\infty, {-1}/{\\sqrt{2}) \\cup({1}/{\\sqrt{2}, \infty) Solution: g(x)=\frac{1}{6} f(3 x^{2}-1)+\frac{1}{2} f(1-x^{2}) g'(x)=\frac{1}{6} f'(3 x^{2}-1)(6 x) +\frac{1}{2} f'(1-x)^{2}(-2 x) =x(f'(3 x^{2} 1)+(-1) f'(1-x^{2})) g(x) is increasing, when g'(x)>0 Case 1 f'(3 x^{2}-1)-f'(1-x^{2})>0 and x > 0 ...(a) \Rightarrowf'(3x^2-1) > f'(1-x^2) ...(i) Given that, f"(x) > 0 \Rightarrowf(x) is increasing. \Rightarrow f'(x)>0 From Eq. (i), 3x^2-1 > 1-x^2 4x^2> 2\Rightarrowx^2>1/2 \Rightarrowx 1/\sqrt2 \Rightarrowx\in(-\infty,-1/\sqrt2)\cup(1/\sqrt2,\infty)...(b) From (a) and (b) \Rightarrowx\in (1/\sqrt2,\infty) Case 2 When x" >