Let α =undersetx∈ R max 8^2 sin3x· 4^4 cos3x and β =undersetx∈ R min 8^2 sin3x· 4^4 cos3x. If 8x^2+bx+c=0 is a quadratic equation whose roots are α…

$$\alpha =\mathrm{max}{{8}^{2\mathrm{sin}3x}\cdot {4}^{4\mathrm{cos}3x}}$$α=max82sin3x⋅44cos3x

$$=\mathrm{max}{{2}^{6\mathrm{sin}3x}\cdot {2}^{8\mathrm{cos}3x}}$$=max26sin3x⋅28cos3x

$$=\mathrm{max}{{2}^{6\mathrm{sin}3x+8\mathrm{cos}3x}}$$=max26sin3x+8cos3x

and $\beta =\mathrm{min}{{8}^{2\mathrm{sin}3x}\cdot {4}^{4\mathrm{cos}3x}}=\mathrm{min}{{2}^{6\mathrm{sin}3x+8\mathrm{cos}3x}}$β=min82sin3x⋅44cos3x=min26sin3x+8cos3x

Now range of $6\mathrm{sin}3x+8\mathrm{cos}3x$6sin3x+8cos3x

$$=[-\sqrt{{6}^{2}+{8}^{2}},+\sqrt{{6}^{2}+{8}^{2}}]=[-10,10]$$=[−62+82​,+62+82​]=[−10,10]

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

So, ${\alpha }^{1/5}={2}^{2}=4$α1/5=22=4

$$\Rightarrow {\beta }^{1/5}={2}^{-2}=\frac{1}{4}$$⇒β1/5=2−2=41​

The quadratic $8{x}^{2}+bx+c=0,$8x2+bx+c=0,

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

$\therefore c-b=8\times [$∴c−b=8×[ (product of roots$)+($)+(sum of roots$)]$)]

$$=8\times [4\times \frac{1}{4}+4+\frac{1}{4}]=8\times [\frac{21}{4}]=42$$=8×[4×41​+4+41​]=8×[421​]=42