Let $α = max x \in R {8^{2 sin 3 x} \cdot 4^{4 cos 3 x}}$ and $β = min x \in R {8^{2 sin 3 x} \cdot 4^{4 cos 3 x}} .$
If $8 x^{2} + b x + c = 0$ is a quadratic equation whose roots are $α^{1 / 5}$ and $β^{1 / 5}$ , then the value of $c - b$ is equal to

43

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42

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50

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47

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Solution

The correct option is B $42$
$α = max {2^{6 sin 3 x + 8 cos 3 x}} = 2^{10} β = min {2^{6 sin 3 x + 8 cos 3 x}} = 2^{- 10}$
$α^{1 / 5} = 4$ and $β^{1 / 5} = \frac{1}{4}$
Sum of roots $= \frac{17}{4}$ & Product of roots $= 1$
$\frac{- b}{8} = \frac{17}{4} \Rightarrow b = - 34$
and $\frac{c}{8} = 1 \Rightarrow c = 8$
Now, $c - b = 8 + 34 = 42$

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