Question

The function f(x) = e^sinx is expanded as a Taylor series in x, around x = 0, in the form f(x) = \(\sum^{\infin}_{n=0}a_nx^n\). The value of a₀ + a₁ + a₂ is

• A. 0
• B. \(\frac{3}{2}\)
• C. \(\frac{5}{2}\)
• D. 5

Answer 1

The Correct Option is C

To solve this problem, we need to find the Taylor series expansion of the function \( f(x) = e^{\sin x} \) around \( x = 0 \) and then determine the sum \( a_0 + a_1 + a_2 \), where \( a_n \) are the coefficients of the series.

The Taylor series of a function \( f(x) \) around \( x = 0 \) (Maclaurin series) is given by:

\(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\)

First, we need to calculate the derivatives of \( f(x) = e^{\sin x} \) at \( x = 0 \).

1. Calculate \( f(0) \):
   Since \( \sin(0) = 0 \), we have:
   \(f(0) = e^0 = 1\)
   Thus, \( a_0 = 1 \).
2. Calculate \( f'(x) \):
   Using the chain rule:
   \(f'(x) = e^{\sin x} \cdot \cos x\)
   Evaluate at \( x = 0 \):
   \(f'(0) = e^0 \cdot \cos(0) = 1\)
   Thus, \( a_1 = \frac{1}{1!} = 1 \).
3. Calculate \( f''(x) \):
   Differentiate \( f'(x) \):
   \(f''(x) = e^{\sin x}(\cos^2 x - \sin x)\)
   Evaluate at \( x = 0 \):
   \(f''(0) = 1 \times (\cos^2(0) - \sin(0)) = 1 \times (1 - 0) = 1\)
   Thus, \( a_2 = \frac{1}{2!} = \frac{1}{2} \).

Therefore, the series coefficients are:

• \( a_0 = 1 \)
• \( a_1 = 1 \)
• \( a_2 = \frac{1}{2} \)

The sum \( a_0 + a_1 + a_2 \) is:

\(a_0 + a_1 + a_2 = 1 + 1 + \frac{1}{2} = \frac{5}{2}\)

Thus, the correct answer is \(\frac{5}{2}\).