Question

If $f\left(x\right)=4x^3-6x^2\cos 2a+3x\cdot \sin 2a\cdot \sin 6a+\sqrt{\log (2a-a^2)}$, then show that $f^{\prime }\left(\frac{1}{2}\right)>0$

Answer 1

Given, $f\left(x\right)=4x^3-6x^2\cos 2a+3x\sin 2a\cdot \sin 6a+\sqrt{\log (2a-a^2)}$
For $f\left(x\right)$ to exist, $\log (2a-a^2)\ge 0\Rightarrow (2a-a^2)\ge e^0$
ie, $2a-a^2\ge 1$ or $a^2-2a+1\le 0$
$\Rightarrow (a-1{)}^2\le 0$
Which is only possible, if $(a-1{)}^2=0$ i.e., $a=1$
$\therefore f\left(x\right)=4x^3-6x^2\cos 2+3x\sin 2\cdot \sin 6$
$\Rightarrow f^{\prime }\left(x\right)=12x^2-12x\cos 2+3\sin 2\sin 6$
$\Rightarrow f^{\prime }\left(\frac{1}{2}\right)=3-6\cos 2+3\sin 2\sin 6=3(1+\sin 2\sin 6)-6\cos 2$
$\Rightarrow f^{\prime }\left(\frac{1}{2}\right)>0$