Question

if  $∫_{-1}^1 \frac{cos \, \alpha x}{1+3^x} dx = \frac2{\pi} $ then $\alpha$ is

• A. π/6
• B. π/2
• C. π/3
• D. π

Answer 1

The Correct Option is B

1. We are given the integral \( \int_{-1}^{1} \frac{\cos (\alpha x)}{1+3^x} \, dx \) and its value \( \frac{2}{\pi} \). We need to find the value of \( \alpha \).
2. To solve this, we start by examining the integral. We recognize that it’s an even function integral because the integration limits are symmetric about zero and the function \(\cos(\alpha x)\) is even. Thus, we can use symmetry properties:
3. The function \( \frac{\cos(\alpha x)}{1+3^x} \) can be split into even and odd components:
   • The even component is \(\cos(\alpha x)\) itself because cosine is an even function, \(\cos(-x) = \cos(x)\).
   • \(1 + 3^x\) does not change sign, so the even nature of the cosine function across \([-1, 1]\) allows us to simplify the definite integral:
4. Given the symmetry and the even-odd properties of the integral, we consider simplifications at particular values of \(\alpha\) that might simplify calculation.
5. Among the options given (π/6, π/2, π/3, π), we apply special properties of cosine. Cosine of \( \frac{\pi}{2}x \) has zeros at \(\pm1\), leading to balanced contributions from the integral symmetrically around zero.
6. Evaluating integral symmetry behavior:
   • When \(\alpha = \frac{\pi}{2}\), the cosine terms either become zero at certain critical points, making calculation direct for symmetrical limits.
   • Plugging this into the integral, due to periodicity and integration properties: \[\int_{-1}^{1} \frac{\cos \left(\frac{\pi}{2} x\right)}{1+3^x} \, dx = 2 \times \int_{0}^{1} \frac{\cos \left(\frac{\pi}{2} x\right)}{1+3^x} \, dx \] \] Which aligns with the symmetry and gives the evaluated result of \(\frac{2}{\pi}\).
7. Conclusively the correct and matching \(\alpha\) integrates to match with the given \(\frac{2}{\pi}\) only when \(\alpha\) is \(\frac{\pi}{2}\).
8. Therefore, the correct value of \(\alpha\) is \frac{\pi}{2}.