Question

The function \( e^{\cos x} \) is Taylor expanded about \( x = 0 \). The coefficient of \( x^2 \) is:

• A. \( -\frac{1}{2} \)
• B. \( -\frac{e}{2} \)
• C. \( \frac{e}{2} \)
• D. Zero

Hint:

When expanding exponentials involving trigonometric functions, first expand the trigonometric term, then exponentiate and collect powers systematically.

Answer 1

The Correct Option is B

Step 1: Expand \( \cos x \) in a Taylor series.
We know that \( \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \). Therefore,
\[ e^{\cos x} = e^{1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots}. \]
Step 2: Separate the constant term.
Let \( e^{\cos x} = e \cdot e^{-\frac{x^2}{2!} + \frac{x^4}{4!} - \cdots} \).
Step 3: Expand the exponential term \( e^{-\frac{x^2}{2!} + \cdots} \).
Using the exponential expansion \( e^y = 1 + y + \frac{y^2}{2!} + \cdots \), we get:
\[ e^{-\frac{x^2}{2} + \cdots} = 1 - \frac{x^2}{2} + \cdots \]
Step 4: Multiply by \( e \).
\[ e^{\cos x} = e(1 - \frac{x^2}{2} + \cdots) \]
Thus, the coefficient of \( x^2 \) is \( e \times (-\frac{1}{2}) = -\frac{e}{2}. \)
Step 5: Final Answer.
The coefficient of \( x^2 \) is \( -\frac{e}{2} \).