Question

Suppose f : (−1, 1) → \(\R\) is an infinitely differentiable function such that the series \(\sum\limits_{j=0}^{\infin}a_j\frac{x^j}{j^!}\) converges to f(x) for each x ∈ (−1, 1), where,
\(a_j=\int\limits_{0}^{\pi/2}\theta^j\cos^j(\tan\theta)d\theta+\int\limits^{\pi}_{\pi/2}(\theta-\pi)^2\cos^j(\tan\theta)d\theta\)
for j ≥ 0. Then

• A. f(x) = 0 for all x ∈ (−1, 1)
• B. f is a non-constant even function on (−1, 1)
• C. f is a non-constant odd function on (−1, 1)
• D. f is NEITHER an odd function NOR an even function on (−1, 1)

Answer 1

The Correct Option is B

To solve this problem, we need to determine the nature of the function \( f(x) \) defined by the given series. The steps are as follows:

1. Understanding the series: \(\sum\limits_{j=0}^{\infin} a_j \frac{x^j}{j^!}\) converges to \( f(x) \). The terms \( a_j \) are given by: \(a_j = \int\limits_{0}^{\pi/2}\theta^j\cos^j(\tan\theta)d\theta + \int\limits^{\pi}_{\pi/2}(\theta-\pi)^2\cos^j(\tan\theta)d\theta\).
2. Analyzing the symmetry:
   • The first integral, \(\int\limits_{0}^{\pi/2}\theta^j\cos^j(\tan\theta)d\theta\), involves powers of \(\theta\) and \(\cos^j(\tan\theta)\). Due to the dependency on \(\theta\) symmetrically distributed around \(\pi/4\), this part can exhibit even or odd properties, dependent on \( j \).
   • The second integral, \(\int\limits^{\pi}_{\pi/2}(\theta-\pi)^2\cos^j(\tan\theta)d\theta\), is symmetric concerning \(\pi\). Here, (\theta-\pi)^2 is always non-negative and always positive within the considered interval, implying an even symmetry.
3. Convergence and Behavior:
   • The term \(\cos^j(\tan\theta)\) affects how fast the terms decay as \( j \) grows. For this function to be infinitely differentiable and converge on \( (-1, 1) \), the powers suggest even values maintain balance across negative and positive inputs.
4. Conclusion: The function \( f(x) \) determined by the series is even because:
   • Each term constructed \( a_j \) integrates over symmetric limits and functions with respect to their even properties.
   • Even functions satisfy \( f(x) = f(-x) \) which corresponds to the balancing of the power series terms.
   Finally, since the series converges to \( f(x) \) and \( f(x) \) across an interval of symmetric properties, \( f(x) \) must be a non-constant even function on \((-1, 1)\).

Therefore, the correct option is: f is a non-constant even function on \((-1, 1)\).