Question

Let $g\left(x\right)=\frac{1}{4}f(2x^2-1)+\frac{1}{2}f(1-x^2)\forall xϵR$, where $f^{\prime \prime }\left(x\right)>0\forall xϵR$. g(x) is necessarily increasing in the interval

• A. $(-\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}})$
• B. $(-\sqrt{\frac{2}{3}},0)\cup (\sqrt{\frac{2}{3}},\infty )$
• C. (-1, 1)
• D. None of these

Answer 1

The correct option is B $(-\sqrt{\frac{2}{3}},0)\cup (\sqrt{\frac{2}{3}},\infty )$

$g\left(x\right)=\frac{1}{4}f(2x^2-1)+\frac{1}{2}f(1-x^2)$
$g^{\prime }\left(x\right)=\frac{1}{4}{f}^{\prime }(2x^2-1).4x-\frac{1}{2}{f}^{\prime }(1-x^2).2x=x\left(f^{\prime }\right(2x^2-1)-f^{\prime }(1-x^2\left)\right)$
for $g\left(x\right)$ to be increasing function
$g^{\prime }\left(x\right)=x\left(f^{\prime }\right(2x^2-1)-f^{\prime }(1-x^2\left)\right)>0$
also $f^{\prime }\left(x\right)>0\Rightarrow f\left(x\right)$ is increasing function
so if $x>0$
$\left(f^{\prime }\right(2x^2-1)-f^{\prime }(1-x^2\left)\right)>0\Rightarrow \left(f^{\prime }\right(2x^2-1)>f^{\prime }(1-x^2\left)\right)\Rightarrow 2x^2-1>1-x^2$
$\Rightarrow x^2>\frac{2}{3}\Rightarrow xϵ(\sqrt{\frac{2}{3}},\infty )$ (i)
and if $x<0$
$f^{\prime }(2x^2-1)-f^{\prime }(1-x^2)<0\Rightarrow f^{\prime }(2x^2-1)<f^{\prime }(1-x^2)$
$\Rightarrow 1-x^2>2x^2-1\Rightarrow 3x^2<2\Rightarrow x\in (-\sqrt{\frac{2}{3}},0)$ (ii) since $x<0$ in this case.
So the final answer is Union of (i) and (ii) $(-\sqrt{\frac{2}{3}},0)U(\sqrt{\frac{2}{3}},\infty )$