Question

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

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

Answer 1

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

f" (x) > 0
$\Rightarrow$ f' is increasing function
To find : where g is necessary Increasing function.
g is increasing $\Rightarrow g^{\prime }>0$
$\Rightarrow \frac{1}{4}{f}^{\prime }(2x^2-1)\left(4x\right)+\frac{1}{2}{f}^{\prime }(1-x^2)(-2x)>0\Rightarrow x\left\{f^{\prime }\right(2x^2-1)-f^{\prime }(1-x^2\left)\right\}>0$
Case1: $x>0\to \left(1\right)f^{\prime }(2x^2-1)>f^{\prime }(1-x^2)$
$\Rightarrow 2x^2-1>1-x^2\Rightarrow x\in (-\infty ,\sqrt{\frac{2}{3}})\cup (\sqrt{\frac{2}{3}},\infty )\to \left(2\right)$
(1)$\cap$ (2)$\Rightarrow x\in \sqrt{\frac{2}{3}},\infty ....\left(3\right)$
Case II: $x<0\to \left(3\right)f^{\prime }(2x^2-1)<f^{\prime }(1-x^2)$
$\Rightarrow 2x^2-1<1-x^2\Rightarrow x\in (-\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}})\to \left(4\right)$
$\left(3\right)\cap \left(4\right)\Rightarrow \in (-\sqrt{\frac{2}{3},0})\to \left(6\right)\therefore gisincinx\in \left(5\right)\cup \left(6\right)\Rightarrow x\in (-\sqrt{\frac{2}{3}},0)\cup (\sqrt{\frac{2}{3}},\infty )$