Mathematics - Quadratic Equation Question with Solution | TestHub

$\alpha =\max \left\{8^{2\sin 3x}·4^{4\cos 3x}\right\}$

$=\max \left\{2^{6\sin 3x}·2^{8\cos 3x}\right\}$

$=\max \left\{2^{6\sin 3x+8\cos 3x}\right\}$

and $\beta =\min \left\{8^{2\sin 3x}·4^{4\cos 3x}\right\}=\min \left\{2^{6\sin 3x+8\cos 3x}\right\}$

Now range of $6\sin 3x+8\cos 3x$

$=\left[-\sqrt{6^2+8^2},+\sqrt{6^2+8^2}\right]=\left[-10,10\right]$

$\alpha =2^{10} \& \beta =2^{-10}$

So, ${\alpha }^{1/5}=2^2=4$

$\Rightarrow {\beta }^{1/5}=2^{-2}=\frac{1}{4}$

The quadratic $8x^2+bx+c=0,$

sum of roots $=\frac{-b}{8}$ and product of roots $=\frac{c}{8}$

$\therefore c-b=8\times [$ (product of roots$)+($sum of roots$\left)\right]$

$=8\times \left[4\times \frac{1}{4}+4+\frac{1}{4}\right]=8\times \left[\frac{21}{4}\right]=42$