Answer is: option3

Solution:

From (1), substitute \( g(x) = f(x) - n \) into (2):

\[ \int_0^4 f(x)\,dx - \int_4^6 [f(x) - n]\,dx = 1 \]

\[ \int_0^4 f(x)\,dx - \left( \int_4^6 f(x)\,dx - \int_4^6 n\,dx \right) = 1 \]

\[ \int_0^4 f(x)\,dx - \int_4^6 f(x)\,dx + \int_4^6 n\,dx = 1 \]

Since \( n \) is constant and \( \int_4^6 n\,dx = 2n \), we get:

\[ \int_0^4 f(x)\,dx - \int_4^6 f(x)\,dx + 2n = 1 \]

Now use (3): \( \int_4^6 f(x)\,dx = 5n - 1 \)

\[ \int_0^4 f(x)\,dx - (5n - 1) + 2n = 1 \]

\[ \int_0^4 f(x)\,dx = 1 + (5n - 1) - 2n = 3n \]

Use substitution to evaluate \( \int_0^2 f(2x)\,dx \)

Let \( u = 2x \Rightarrow dx = \frac{1}{2}du \)

When \( x = 0, u = 0 \); when \( x = 2, u = 4 \)

So \[ \int_0^2 f(2x)\,dx = \int_0^4 f(u) \cdot \frac{1}{2} \,du = \frac{1}{2} \int_0^4 f(u)\,du \]

But \( \int_0^4 f(u)\,du = \int_0^4 f(x)\,dx = 3n \)

Therefore,

\[ \int_0^2 f(2x)\,dx = \frac{1}{2} \cdot 3n = \frac{3n}{2} \]

So, comparing to \( kn \), we have:

\[ kn = \frac{3n}{2} \Rightarrow k = \frac{3}{2} \]

Correct Answer: C. \( \frac{3}{2} \)