To determine the remainder when [tex]\( f(x) \)[/tex] is divided by [tex]\( (x - 1)(x - 2) \)[/tex], we need to solve for the constants in an expression of the form [tex]\( ax + b \)[/tex] since the degree of [tex]\( (x - 1)(x - 2) \)[/tex] is 2, which means the remainder will be a first-degree polynomial.
Given:
- When [tex]\( f(x) \)[/tex] is divided by [tex]\( (x - 1) \)[/tex], the remainder is 5. Therefore, [tex]\( f(1) = 5 \)[/tex].
- When [tex]\( f(x) \)[/tex] is divided by [tex]\( (x - 2) \)[/tex], the remainder is 7. Therefore, [tex]\( f(2) = 7 \)[/tex].
We consider the remainder in the form [tex]\( ax + b \)[/tex].
1. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( ax + b \)[/tex]:
[tex]\[
a(1) + b = 5 \implies a + b = 5
\][/tex]
2. Substitute [tex]\( x = 2 \)[/tex] into [tex]\( ax + b \)[/tex]:
[tex]\[
a(2) + b = 7 \implies 2a + b = 7
\][/tex]
Now we have a system of linear equations:
[tex]\[
\begin{cases}
a + b = 5 \\
2a + b = 7
\end{cases}
\][/tex]
To solve this system, we subtract the first equation from the second:
[tex]\[
(2a + b) - (a + b) = 7 - 5
\][/tex]
[tex]\[
2a + b - a - b = 2
\][/tex]
[tex]\[
a = 2
\][/tex]
Substituting [tex]\( a = 2 \)[/tex] back into the first equation:
[tex]\[
2 + b = 5
\][/tex]
[tex]\[
b = 3
\][/tex]
Thus, the remainder when [tex]\( f(x) \)[/tex] is divided by [tex]\( (x - 1)(x - 2) \)[/tex] is:
[tex]\[
2x + 3
\][/tex]
So, the correct answer is:
(a) [tex]\( 2x + 3 \)[/tex]
