Let $f(x)=\left(2^x-1\right)\left(2^x-2\right)$ and $g(x)=2 \sin x+\cos 2 x$ in $[0, \pi]$

Question

Professor PC Thomas Classes · Answer

(A) $f^{\prime\prime}(x)>0\forall x\in R\Rightarrow f^{\prime}(x)$ is an increasing function.
Now $g(x)=-f(4-x)+f(2+x) \Rightarrow g^{\prime}(x)=f^{\prime}(4-x)+f^{\prime}(2+x)$
If $g^{\prime}(x)>0\Rightarrow f^{\prime}(2+x)>-f^{\prime}(4-x)$
(B) $f^{\prime}(x)=3(x-1)(x+1)\Rightarrow f^{\prime}(x)=0$ has roots $x=-1,1$
$f(x)=0$ will have exactly one real root if $f(-1)f(1)>0$
$(a+2)(a-2)>0 \Rightarrow a<-2$ or $a>2$
(C) $f^{\prime}(x)=-sin~x+a^{2}\ge0\forall x\in R \Rightarrow a^{2}\ge sin~x\forall x\in R$
(D) $f^{\prime}(x)=2e^{x}+ae^{-x}+2a+1=e^{-x}(2e^{2x}+(2a+1)e^{x}+a)=2e^{-x}(e^{x}+a)(e^{x}+\frac{1}{2})$
$f(x)$ increases for all $x$ if $f^{\prime}(x)\ge0\forall x\in R 	herefore e^{x}+a\ge0\forall x\in R\Rightarrow a\ge0$