Question

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

• A. 42
• B. 43
• C. 47
• D. 50

Hint:

Remember the range of the expression $a\sin\theta + b\cos\theta$ is $[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$. This is a very common pattern in finding the maximum and minimum values of trigonometric functions.

Answer 1

The Correct Option is A

\[ 8^{2\sin3x}\cdot4^{4\cos3x} = (2^3)^{2\sin3x}(2^2)^{4\cos3x} = 2^{6\sin3x+8\cos3x} \] Let \[ E = 6\sin3x + 8\cos3x \] For $a\sin\theta + b\cos\theta$, \[ -\sqrt{a^2+b^2} \le E \le \sqrt{a^2+b^2} \] \[ \sqrt{6^2+8^2} = 10 \] Hence, \[ \alpha = 2^{10}, \qquad \beta = 2^{-10} \] \[ \alpha^{1/5} = 4, \qquad \beta^{1/5} = \frac14 \] The quadratic equation is $8x^2+bx+c=0$. Sum of roots: \[ 4+\frac14=\frac{17}{4}=-\frac{b}{8} \Rightarrow b=-34 \] Product of roots: \[ 1=\frac{c}{8}\Rightarrow c=8 \] \[ c-b = 8-(-34)=42 \] \[ \boxed{42} \]