Mathematics - Application of Derivative Question with Solution | TestHub

$h(x)=f(g(x))$

$h^{{}^{\prime }}(x)=f^{{}^{\prime }}(g(x))g^{{}^{\prime }}(x)<0\forall x\in [0,\infty )$

$\because g^{{}^{\prime }}(x)<0\forall x\in [0,\infty )$ and $f^{{}^{\prime }}(g(x))>0\forall x\in [0,\infty )$

Also, $\mathrm{h}(0)=0$ and hence, $\mathrm{h}(\mathrm{x})<0\forall \mathrm{x}\in [0,\infty )$

$\mathrm{p}(\mathrm{x})=\mathrm{h}\left({\mathrm{x}}^3-2{\mathrm{x}}^2+2\mathrm{x}\right)-\mathrm{h}(4)$

$p^{{}^{\prime }}(x)=h^{{}^{\prime }}\left(x^3-2x^2+2x\right)\cdot \left(3x^2-4x+2\right)<0\forall x\in (0,2)$ $\because h^{{}^{\prime }}\left(x^3-2x^2+2x\right)<0\forall x\in (0,\infty )$ and $3x^2-4x+2>0\forall x\in R$

$\Rightarrow \mathrm{p}(\mathrm{x})$ is an decreasing function

$\Rightarrow \mathrm{p}(2)<\mathrm{p}(\mathrm{x})<\mathrm{p}(0)\forall \mathrm{x}\in (0,2)$

$\Rightarrow \mathrm{h}(4)-\mathrm{h}(4)<\mathrm{p}(\mathrm{x})<\mathrm{h}(0)-\mathrm{h}(4)$ $\Rightarrow 0<\mathrm{p}(\mathrm{x})<-\mathrm{h}(4)$