Question

If \( \alpha = 3 + 4 + 8 + 9 + 13 + \dots \) up to 40 terms and \( (\tan \beta)^{\frac{\alpha}{1020}} \) is the root of the equation \( x^2 - x - 2 = 0 \), then the value of \( \sin^2 \beta + 3\cos^2 \beta \) is:

• A. 1
• B. \(\frac{2}{3}\)
• C. \(\frac{5}{3}\)
• D. \(\frac{1}{3}\)

Answer 1

The Correct Option is C

Step 1: Understanding the Series Structure:
The series $3, 4, 8, 9, 13, 14, \ldots$ is formed by interleaving two arithmetic progressions (APs): \[ \text{Series 1 (odd positions)}: 3, 8, 13, \ldots \quad (a = 3,\, d = 5) \] \[ \text{Series 2 (even positions)}: 4, 9, 14, \ldots \quad (a = 4,\, d = 5) \] Since there are 40 terms total, each AP contributes 20 terms.
Step 2: Computing $\alpha$:
Sum of Series 1 (20 terms, $a=3$, $d=5$): \[ S_1 = \frac{20}{2}\left[2(3) + 19(5)\right] = 10\left[6 + 95\right] = 10 \times 101 = 1010 \] Sum of Series 2 (20 terms, $a=4$, $d=5$): \[ S_2 = \frac{20}{2}\left[2(4) + 19(5)\right] = 10\left[8 + 95\right] = 10 \times 103 = 1030 \] Therefore: \[ \alpha = S_1 + S_2 = 1010 + 1030 = 2040 \] Step 3: Finding $\tan^2\beta$:
\[ (\tan\beta)^{\alpha/1020} = (\tan\beta)^{2040/1020} = (\tan\beta)^2 = \tan^2\beta \] The roots of $x^2 - x - 2 = 0$ are found by factoring: \[ x^2 - x - 2 = (x-2)(x+1) = 0 \implies x = 2 \text{ or } x = -1 \] Since $\tan^2\beta \geq 0$, we must have $\tan^2\beta = 2$.
Step 4: Computing $\sin^2\beta + 3\cos^2\beta$:
We use the identity $\sin^2\beta + \cos^2\beta = 1$ and $\tan^2\beta = \dfrac{\sin^2\beta}{\cos^2\beta} = 2$.
From $\tan^2\beta = 2$: $\sin^2\beta = 2\cos^2\beta$.
Substituting into $\sin^2\beta + \cos^2\beta = 1$: \[ 2\cos^2\beta + \cos^2\beta = 1 \implies \cos^2\beta = \frac{1}{3} \] Therefore $\sin^2\beta = \dfrac{2}{3}$, and: \[ \sin^2\beta + 3\cos^2\beta = \frac{2}{3} + 3\cdot\frac{1}{3} = \frac{2}{3} + 1 = \frac{5}{3} \] Step 5: Final Answer:
\[ \sin^2\beta + 3\cos^2\beta = \frac{5}{3} \] The answer is Option (3).