Let a = max_{x in R} {8^{2 sin 3x} cdot 4^{4 | vedclass

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(B) Given $\alpha = \max \{8^{2 \sin 3x} \cdot 4^{4 \cos 3x}\}$ and $\beta = \min \{8^{2 \sin 3x} \cdot 4^{4 \cos 3x}\}$.We can rewrite the expression

Vedclass · Accepted Answer

$42$

Vedclass · Answer

(B) Given $\alpha = \max \{8^{2 \sin 3x} \cdot 4^{4 \cos 3x}\}$ and $\beta = \min \{8^{2 \sin 3x} \cdot 4^{4 \cos 3x}\}$.We can rewrite the expression as $2^{6 \sin 3x} \cdot 2^{8 \cos 3x} = 2^{6 \sin 3x + 8 \cos 3x}$.The range of $6 \sin 3x + 8 \cos 3x$ is $[-\sqrt{6^2 + 8^2}, \sqrt{6^2 + 8^2}] = [-10, 10]$.Thus,$\alpha = 2^{10}$ and $\beta = 2^{-10}$.Then $\alpha^{1/5} = (2^{10})^{1/5} = 2^2 = 4$ and $\beta^{1/5} = (2^{-10})^{1/5} = 2^{-2} = 1/4$.The roots of $8x^2 + bx + c = 0$ are $4$ and $1/4$.Sum of roots: $4 + 1/4 = 17/4 = -b/8 \Rightarrow b = -34$.Product of roots: $4 \cdot 1/4 = 1 = c/8 \Rightarrow c = 8$.Therefore,$c - b = 8 - (-34) = 42$.

Let $a = \max_{x \in R} \{8^{2 \sin 3x} \cdot 4^{4 \cos 3x}\}$ and $\beta = \min_{x \in R} \{8^{2 \sin 3x} \cdot 4^{4 \cos 3x}\}$. 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:

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