(a) Showing \( f \) is even:

Solution:

1. A function is even if \( f(-x) = f(x) \) for all \( x \) in its domain.

2. Calculate \( f(-x) \):

\[ f(-x) = \frac{\sin^2(k(-x))}{(-x)^2} = \frac{\sin^2(-kx)}{x^2} \]

3. Using \( \sin(-\theta) = -\sin(\theta) \):

\[ f(-x) = \frac{(-\sin(kx))^2}{x^2} = \frac{\sin^2(kx)}{x^2} = f(x) \]

4. Since \( f(-x) = f(x) \), \( f \) is even.

[2 marks]

(b) Finding \( k \):

Solution:

1. Evaluate the limit:

\[ \lim_{x \to 0} \frac{\sin^2(kx)}{x^2} \]

2. Using the standard limit \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \):

\[ = \lim_{x \to 0} \left(\frac{\sin(kx)}{kx}\right)^2 \cdot k^2 = k^2 \]

3. Set equal to given limit:

\[ k^2 = 16 \]

4. Solve for positive \( k \):

\[ k = 4 \]

Alternative method using L’Hôpital’s rule:

1. First application:

\[ \lim_{x \to 0} \frac{2k\sin(kx)\cos(kx)}{2x} = \lim_{x \to 0} \frac{k\sin(2kx)}{2x} \]

2. Second application:

\[ = \lim_{x \to 0} \frac{2k^2\cos(2kx)}{2} = k^2 \]

3. Therefore \( k^2 = 16 \) ⇒ \( k = 4 \)

[6 marks]