Key Concepts and Formulas

• Range of $a \sin \theta + b \cos \theta$: The range of $a \sin \theta + b \cos \theta$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
• Properties of Exponents: $(a^m)^n = a^{mn}$ and $a^m \cdot a^n = a^{m+n}$.
• Quadratic Equations - Sum and Product of Roots: For a quadratic equation $Ax^2 + Bx + C = 0$ with roots $r_1$ and $r_2$, $r_1 + r_2 = -\frac{B}{A}$ and $r_1 r_2 = \frac{C}{A}$.

Step-by-Step Solution

Step 1: Simplify the Given Expression

We are given the expression $8^{2\sin 3x} \cdot 4^{4\cos 3x}$. To simplify this, we express both terms with a common base of 2. This allows us to combine the exponents. Since $8 = 2^3$ and $4 = 2^2$, we can substitute these into the expression: $8^{2\sin 3x} \cdot 4^{4\cos 3x} = (2^3)^{2\sin 3x} \cdot (2^2)^{4\cos 3x}$ Using the power of a power rule, $(a^m)^n = a^{mn}$: $(2^3)^{2\sin 3x} \cdot (2^2)^{4\cos 3x} = 2^{6\sin 3x} \cdot 2^{8\cos 3x}$ Now, using the product of powers rule, $a^m \cdot a^n = a^{m+n}$: $2^{6\sin 3x} \cdot 2^{8\cos 3x} = 2^{6\sin 3x + 8\cos 3x}$ Let $E = 2^{6\sin 3x + 8\cos 3x}$.

Step 2: Find the Range of the Exponent

The exponent is of the form $a \sin \theta + b \cos \theta$, where $a=6$, $b=8$, and $\theta = 3x$. To find the maximum and minimum values of the entire expression, we need to find the range of the exponent. The range of $6\sin 3x + 8\cos 3x$ is given by $[-\sqrt{6^2 + 8^2}, \sqrt{6^2 + 8^2}]$. $[-\sqrt{6^2 + 8^2}, \sqrt{6^2 + 8^2}] = [-\sqrt{36 + 64}, \sqrt{36 + 64}] = [-\sqrt{100}, \sqrt{100}] = [-10, 10]$ So, the maximum value of the exponent is 10, and the minimum value is -10.

Step 3: Determine $\alpha$ and $\beta$

We have $E = 2^{\text{exponent}}$. Since the base of the exponential function, 2, is greater than 1, the maximum value of $E$ ($\alpha$) will occur when the exponent is at its maximum, and the minimum value of $E$ ($\beta$) will occur when the exponent is at its minimum. This is because exponential functions with a base greater than 1 are increasing.

Therefore: $\alpha = \mathop {\max }\limits_{x \in R} \{ {2^{6\sin 3x + 8\cos 3x}}\} = 2^{\text{maximum value of exponent}} = 2^{10}$ $\beta = \mathop {\min }\limits_{x \in R} \{ {2^{6\sin 3x + 8\cos 3x}}\} = 2^{\text{minimum value of exponent}} = 2^{-10}$

Step 4: Calculate $\alpha^{1/5}$ and $\beta^{1/5}$

We need to find the values of $\alpha^{1/5}$ and $\beta^{1/5}$. These are the roots of the quadratic. For $\alpha^{1/5}$: $\alpha^{1/5} = (2^{10})^{1/5}$ Using the exponent rule $(a^m)^n = a^{mn}$: $\alpha^{1/5} = 2^{10 \cdot (1/5)} = 2^{10/5} = 2^2 = 4$

For $\beta^{1/5}$: $\beta^{1/5} = (2^{-10})^{1/5}$ Using the exponent rule $(a^m)^n = a^{mn}$: $\beta^{1/5} = 2^{-10 \cdot (1/5)} = 2^{-10/5} = 2^{-2} = \frac{1}{2^2} = \frac{1}{4}$

Step 5: Formulate the Quadratic Equation and Find $b$ and $c$

We are given that $8x^2 + bx + c = 0$ is a quadratic equation whose roots are $\alpha^{1/5}$ and $\beta^{1/5}$. Let the roots be $r_1 = \alpha^{1/5} = 4$ and $r_2 = \beta^{1/5} = \frac{1}{4}$. The general form of a quadratic equation is $Ax^2 + Bx + C = 0$. In our case, $A=8$, $B=b$, and $C=c$.

Sum of roots: $r_1 + r_2 = -\frac{B}{A}$ $4 + \frac{1}{4} = -\frac{b}{8}$ $\frac{16+1}{4} = -\frac{b}{8}$ $\frac{17}{4} = -\frac{b}{8}$ Multiply both sides by 8: $17 \cdot \frac{8}{4} = -b$ $17 \cdot 2 = -b$ $34 = -b \implies b = -34$

Product of roots: $r_1 r_2 = \frac{C}{A}$ $4 \cdot \frac{1}{4} = \frac{c}{8}$ $1 = \frac{c}{8}$ Multiply both sides by 8: $c = 8$

Step 6: Calculate $c - b$

Finally, we need to find the value of $c - b$. $c - b = 8 - (-34)$ $c - b = 8 + 34$ $c - b = 42$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs, especially when dealing with the sum of roots formula ($-\frac{B}{A}$) and when calculating $c-b$.
• Base Conversion: Double-check that you have correctly converted to a common base before combining exponents.
• Range of Trigonometric Functions: Remember that the formula for the range of $a\sin\theta + b\cos\theta$ is $[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$.

Summary

We simplified the given expression using exponent rules and trigonometric identities. We found the range of the exponent, determined the maximum and minimum values of the expression, calculated the roots of the quadratic equation, and finally, found the value of $c - b$. The final answer is 42.

Final Answer

The final answer is \boxed{42}, which corresponds to option (A).