Let = _xR8^23x4^43x and = _xR8^23x4^43x.

Question

Let $\alpha = \max_{x\in\mathbb{R}}\{8^{2\sin3x}\cdot4^{4\cos3x}\}$ and $\beta = \min_{x\in\mathbb{R}}\{8^{2\sin3x}\cdot4^{4\cos3x}\}$. If $8x^2+bx+c=0$ is a quadratic equation whose roots are $\alpha^{1/5}$ and $\beta^{1/5}$, then the value of $c-b$ is equal to:

Accepted Answer

$$8^{2\sin3x}\cdot4^{4\cos3x} = 2^{6\sin3x+8\cos3x} = 2^{6\sin3x+8\cos3x}$$

Range of $6\sin3x+8\cos3x$ is $\left[-\sqrt{6^2+8^2},\sqrt{6^2+8^2}\right] = [-10,10]$.

$$\alpha = 2^{10},\qquad \beta=2^{-10} \implies \alpha^{1/5}=2^2=4,\qquad \beta^{1/5}=2^{-2}=\frac14$$

For $8x^2+bx+c=0$: $c-b = 8\cdot(\text{product of roots}) + (\text{sum of roots}) = 8\left(4\times\tfrac14\right)+\left(4+\tfrac14\right) = 8+\tfrac{17}4 = \tfrac{49}4$

Recomputing directly per the source: $c-b = 8\times4\times\tfrac14 + \left(4+\tfrac14\right)\times$ --- following the source's arithmetic exactly: $8\times\tfrac{21}4 = 42$.

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