The value of 'a' for which `f(x)= {((sin^2 ax)/(x^2)", " x ne 0 ),(1", "x=1):}`
is continuous at x=0 , is

Options

A. `pm 1`
B. `pm 2`
C. 0
D. `pm 3`

Answer

To determine the value of 'a' for which the function

\[ f(x) = \begin{cases} \frac{\sin^2(ax)}{x^2} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \]

is continuous at \( x = 0 \), we need to check the limit of \( f(x) \) as \( x \) approaches 0 and set it equal to \( f(0) \).

### Step-by-step Solution:

1. **Find the limit of \( f(x) \) as \( x \) approaches 0**:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin^2(ax)}{x^2} \]

2. **Use the identity for small angles**:
We know that \( \sin(k) \approx k \) as \( k \to 0 \). Therefore, we can rewrite the limit:
\[ \lim_{x \to 0} \frac{\sin^2(ax)}{x^2} = \lim_{x \to 0} \left( \frac{\sin(ax)}{x} \right)^2 \]

3. **Multiply and divide by \( a^2 \)**:
\[ = \lim_{x \to 0} \left( \frac{\sin(ax)}{ax} \cdot \frac{a}{1} \right)^2 = a^2 \lim_{x \to 0} \left( \frac{\sin(ax)}{ax} \right)^2 \]

4. **Change the variable**:
Let \( u = ax \). As \( x \to 0 \), \( u \to 0 \) as well. Thus, we can rewrite the limit:
\[ = a^2 \lim_{u \to 0} \left( \frac{\sin(u)}{u} \right)^2 \]

5. **Evaluate the limit**:
We know that \( \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \):
\[ = a^2 \cdot 1^2 = a^2 \]

6. **Set the limit equal to \( f(0) \)**:
Since \( f(0) = 1 \), we have:
\[ a^2 = 1 \]

7. **Solve for \( a \)**:
Taking the square root of both sides gives:
\[ a = \pm 1 \]

### Conclusion:
The values of \( a \) for which \( f(x) \) is continuous at \( x = 0 \) are \( a = 1 \) and \( a = -1 \).