Question

Let \( f : \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \rightarrow \mathbb{R} \) be a differentiable function such that \( f(0) = \frac{1}{2} \). If the \( \lim_{x \to 0} \frac{\int_{0}^{x} f(t) \, dt}{e^{x^2} - 1} = \alpha \), then \( 8\alpha^2 \) is equal to:

• A. 16
• B. 2
• C. 1
• D. 4

Answer 1

The Correct Option is B

To determine the value of \(\alpha\), we must evaluate the limit:

\[\lim_{x \to 0} \frac{\int_{0}^{x} f(t) \, dt}{e^{x^2} - 1} = \alpha\]

Given that \(f(0) = \frac{1}{2}\) and \(f(x)\) is differentiable, L'Hôpital's Rule is applied because both the numerator and the denominator approach 0 as \(x \to 0\).

The derivatives of the numerator and the denominator with respect to \(x\) are as follows:

• The derivative of the numerator \(\int_{0}^{x} f(t) \, dt\) is \(f(x)\) by the Fundamental Theorem of Calculus.
• The derivative of the denominator \(\left(e^{x^2} - 1\right)\) is \(2xe^{x^2}\).

Applying L'Hôpital's Rule yields:

\[\lim_{x \to 0} \frac{f(x)}{2x e^{x^2}} = \alpha\]

Substituting \(f(0) = \frac{1}{2}\), we examine the limit at \(x = 0\):

\[\lim_{x \to 0} \frac{f(x)}{2x e^{x^2}} = \lim_{x \to 0} \frac{\frac{1}{2} + \frac{d}{dx}(f(x) - \frac{1}{2})}{2x e^{x^2}}\]

As higher-order terms in \(x\) vanish more rapidly than \(x\) itself, the limit simplifies to:

\[\lim_{x \to 0} \frac{\frac{1}{2}}{2x} = \frac{1}{4}\]

Therefore, \(\alpha = \frac{1}{4}\).

Next, we compute \(8\alpha^{2}\):

\[8 \alpha^2 = 8 \left(\frac{1}{4}\right)^2 = 8 \times \frac{1}{16} = \frac{1}{2} \times 2 = 2\]

The final result is 2.