Question
A function $$g\left( x \right)$$ is defined as $$g\left( x \right) = \frac{1}{4}f\left( {2{x^2} - 1} \right) + \frac{1}{2}f\left( {1 - {x^2}} \right)$$ and $$f'\left( x \right)$$ is an increasing function. Then $$g\left( x \right)$$ is increasing in the interval :
A.
$$\left( { - 1,\,1} \right)$$
B.
$$\left( { - \sqrt {\frac{2}{3}} ,\,0} \right) \cup \left( {\sqrt {\frac{2}{3}} ,\,\infty } \right)$$
C.
$$\left( { - \sqrt {\frac{2}{3}} ,\,\sqrt {\frac{2}{3}} } \right)$$
D.
none of these
Answer :
$$\left( { - \sqrt {\frac{2}{3}} ,\,0} \right) \cup \left( {\sqrt {\frac{2}{3}} ,\,\infty } \right)$$
Solution :
$$\eqalign{
& g'\left( x \right) = xf'\left( {2{x^2} - 1} \right) - xf'\left( {1 - {x^2}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = x\left( {f'\left( {2{x^2} - 1} \right) - f'\left( {1 - {x^2}} \right)} \right) \cr
& g'\left( x \right) > 0 \cr} $$
$${\text{If }}x > 0,\,2{x^2} - 1 > 1 - {x^2}$$ (as $$f'$$ is an increasing function)
$$\eqalign{
& {\text{or }}3{x^2} > 2 \cr
& {\text{or }}x\, \in \left( { - \infty ,\, - \sqrt {\frac{2}{3}} } \right) \cup \left( {\sqrt {\frac{2}{3}} ,\,\infty } \right) \cr
& {\text{or }}x\, \in \left( {\sqrt {\frac{2}{3}} ,\,\infty } \right) \cr
& {\text{If }}x < 0,\,2{x^2} - 1 < 1 - {x^2} \cr
& {\text{or }}3{x^2} < 2 \cr
& {\text{or }}x\, \in \left( { - \sqrt {\frac{2}{3}} ,\,\sqrt {\frac{2}{3}} } \right) \cr
& {\text{or }}x\, \in \left( { - \sqrt {\frac{2}{3}} ,\,0} \right) \cr} $$