Let α = max x ∈ R 8 2 sin 3 x · 4 4 cos 3 x and β = min x ∈ R 8 2 sin 3 x ·…

Let α=maxx∈R82sin3x·44cos3x and β=minx∈R82sin3x·44cos3x. If 8x2+bx+c=0 is a quadratic equation whose roots are α1/5 and β1/5, then the value of c-b is equal to :
  1. 42
  2. 47
  3. 43
  4. 50

Solution

$\alpha = \max \{8^{2\sin^3 x} \cdot 4^{4\cos^3 x}\}$ $= \max \{2^{6\sin^3 x} \cdot 2^{8\cos^3 x}\}$ $= \max \{2^{6\sin^3 x + 8\cos^3 x}\}$ and $\beta = \min \{8^{2\sin^3 x} \cdot 4^{4\cos^3 x}\} = \min \{2^{6\sin^3 x + 8\cos^3 x}\}$ Now range of $6\sin^3 x + 8\cos^3 x$ $= [-\sqrt{6^2 + 8^2}, +\sqrt{6^2 + 8^2}] = [-10, 10]$ $\alpha = 2^{10}$ and $\beta = 2^{-10}$ So, $\alpha^{1/5} = 2^2 = 4$ $\Rightarrow \beta^{1/5} = 2^{-2} = \frac{1}{4}$ The quadratic $8x^2 + bx + c = 0$, sum of roots = $\frac{-b}{8}$ and product of roots = $\frac{c}{8}$ $\Rightarrow c - b = 8 \times$ (product of roots + sum of roots) $= 8 \times \left[4 \times \frac{1}{4} + 4 + \frac{1}{4}\right] = 8 \times \left[\frac{21}{4}\right] = 42$

Asked in: JEE Main 2021 (27 Jul Shift 2)

Practice more Quadratic Equation questions on Aicharya