Question

$g\left(x\right)=f\left(x^2\right)+f(1-x^2)$ and $f^{\prime \prime }\left(x\right)<0$ for $0<x<1$ then

• A. $g\left(x\right)$ decreases in $(0,\frac{1}{\sqrt{2}})$
• B. $g\left(x\right)$ increases in $(\frac{1}{\sqrt{2}},1)$
• C. $g\left(x\right)$ decreases in $(\frac{1}{\sqrt{2}},1)$
• D. $g\left(x\right)$ increases in $(0,\frac{1}{\sqrt{2}})$

Answer 1

Answer: (B), (C)

$g\left(x\right)$ increases in $(\frac{1}{\sqrt{2}},1)$
$g\left(x\right)$ decreases in $(\frac{1}{\sqrt{2}},1)$

$g^{\prime }\left(x\right)=2x\{f^{\prime }\left(x^2\right)-f^{\prime }(1-x^2)\}$
Since f'(x) is decreasing
hence for $x^2<1-x^2$
$\Rightarrow x<\frac{1}{\sqrt{2}},g^{\prime }\left(x\right)>0$
and for $x^2>1-x^2\Rightarrow x>\frac{1}{\sqrt{2}},g^{\prime }\left(x\right)<0$
$\Rightarrow g\left(x\right)$ is decreasing in $(\frac{1}{\sqrt{2}},1)$ and increasing in $(0,\frac{1}{\sqrt{2}})$