Mathematics - Application of Derivative Question with Solution | TestHub

Given that $f^{\prime \prime }(x)>0$, it implies that $f^{\prime }$ is an increasing function. To find where $g$ is increasing, we need $g^{\prime }>0$.

$$\frac{1}{4}\cdot f^{\prime }(2x^2-1)(4x)+\frac{1}{2}P(1-x^2)(-2x)>0$$

$$x\left\{f^{\prime }(2x^2-1)-f^{\prime }(1-x^2)\right\}>0$$

Case 1: $x>0\to (1)$

$f^{\prime }(2x^2-1)>f^{\prime }(1-x^2)$

Since $f^{\prime }$ is an increasing function,

$$2x^2-1>1-x^2$$

$$3x^2>2$$

$$x^2>\frac{2}{3}$$

$$x\in \left(-\infty ,-\sqrt{\frac{2}{3}}\right)\cup \left(\sqrt{\frac{2}{3}},\infty \right)\to (2)$$

Combining (1) and (2) gives $x\in \left(\sqrt{\frac{2}{3}},\infty \right)$.

Case II: $x<0\to (3)$

$f^{\prime }(2x^2-1)<f^{\prime }(1-x^2)$

Since $f^{\prime }$ is an increasing function,

$$2x^2-1<1-x^2$$

$$3x^2<2$$

$$x^2<\frac{2}{3}$$

$$x\in \left(-\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}}\right)\to (4)$$

Combining (3) and (4) gives $x\in \left(-\sqrt{\frac{2}{3}},0\right)$.

Therefore, $g$ is increasing when $x\in \left(-\sqrt{\frac{2}{3}},0\right)\cup \left(\sqrt{\frac{2}{3}},\infty \right)$.