Let $\alpha=\max _{x \in R}\left\{8^{2 \sin 3 x} \cdot 4^{4 \cos 3 x}\right\}$ and $\beta=\min _{x \in R}\left\{8^{2 \sin 3 x} \cdot 4^{4 \cos 3 x}\right\}$. 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 $c-b$ is equal to:

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Application of Derivatives

Let

α = max_{x \in R} {8^{2 s i n 3 x} \cdot 4^{4 c o s 3 x}}

and

β = min_{x \in R} {8^{2 s i n 3 x} \cdot 4^{4 c o s 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:

Answer: (A)

We have

α = max {2^{6 s i n 3 x} \cdot 2^{8 c o s 3 x}} = max {2^{6 s i n 3 x + 8 c o s 3 x}}

.

The maximum of

6 sin 3 x + 8 cos 3 x

36 + 64 = 10

, so

α = 2^{10}

.

Similarly, the minimum is

- 10

, so

β = 2^{- 10}

.

Thus

α^{1/5} = 4

and

β^{1/5} = \frac{1}{4}

.

Sum of roots:

4 + \frac{1}{4} = \frac{17}{4} = - \frac{b}{8}

, so

b = - 34

.

Product of roots:

4 \cdot \frac{1}{4} = 1 = \frac{c}{8}

, so

c = 8

.

Therefore,

c - b = 8 - (- 34) = 42

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option