Solve this following
Question:

Let $\alpha=\max _{x \in \mathbf{R}}\left\{8^{2 \sin 3 x} \cdot 4^{4 \cos 3 x}\right\}$ and

$\beta=\min _{x \in \mathbb{R}}\left\{8^{2 \sin 3 x} \cdot 4^{4 \cos 3 x}\right\} .$ If $8 x^{2}+b x+c=0$ is a

quadratic equation whose roots are $\alpha^{1 / 5}$ and $\beta^{1 / 5}$,

then the value of $\mathrm{c}-\mathrm{b}$ is equal to :

1. 42

2. 47

3. 43

4. 50

Correct Option: 1

Solution:

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

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

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

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

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

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

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

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

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

quadratic $8 x^{2}+b x+c=0, c-b=$

$8 \times[$ (product of roots $]+($ sum of roots $)$

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