Let `f(x)=(e^(x)-1)^(2n)/(sin^n (x//a)(log (1+(x//a)))^n` for `x ne 0` . If `f(0)=16^n` and f is a continuous function, then the value of a is

Options

A. 16
B. 2
C. 8
D. 4

Answer

To solve the problem, we need to find the value of \( a \) such that the function

\[ f(x) = \frac{(e^x - 1)^{2n}}{\sin^n\left(\frac{x}{a}\right) \left(\log\left(1 + \frac{x}{a}\right)\right)^n} \]

is continuous at \( x = 0 \) and \( f(0) = 16^n \).

### Step 1: Find \( f(0) \)

Since \( f(0) = 16^n \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 equals \( 16^n \).

### Step 2: Evaluate the limit as \( x \to 0 \)

We need to compute:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{(e^x - 1)^{2n}}{\sin^n\left(\frac{x}{a}\right) \left(\log\left(1 + \frac{x}{a}\right)\right)^n} \]

### Step 3: Use Taylor expansions

Using Taylor series expansions:

- \( e^x - 1 \approx x \) as \( x \to 0 \)
- \( \sin\left(\frac{x}{a}\right) \approx \frac{x}{a} \) as \( x \to 0 \)
- \( \log\left(1 + \frac{x}{a}\right) \approx \frac{x}{a} \) as \( x \to 0 \)

### Step 4: Substitute the approximations

Substituting these approximations into the limit gives:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{(x)^{2n}}{\left(\frac{x}{a}\right)^n \left(\frac{x}{a}\right)^n} \]

This simplifies to:

\[ = \lim_{x \to 0} \frac{x^{2n}}{\frac{x^n}{a^n} \cdot \frac{x^n}{a^n}} = \lim_{x \to 0} \frac{x^{2n}}{\frac{x^{2n}}{a^{2n}}} = \lim_{x \to 0} a^{2n} = a^{2n} \]

### Step 5: Set the limit equal to \( f(0) \)

Since \( f(0) = 16^n \), we have:

\[ a^{2n} = 16^n \]

### Step 6: Solve for \( a \)

Taking the \( n \)-th root on both sides gives:

\[ a^2 = 16 \implies a = 4 \quad (\text{since } a \text{ must be positive}) \]

### Conclusion

Thus, the value of \( a \) is

\[ \boxed{4} \]