Given the problem:
If $f (x) = a x + b$ and $f^{- 1} (x) = b x + a$ with a and b real, find the appropriate matches based on the given solution.
Match the items in Column I with the appropriate expressions or properties in Column II based on the given solution.
$| x |,$ represents absolute value of x
Match the Following Based on the Given Problem and Solution

Column I

Column II

1 $a^{2} + b$ A
0
2 $| a + b |$ B
1
3 $| a |$ C
2
4 $| \frac{a}{b} |$ D
-1

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1-A,2-C,3-B,4-B

1-A,2-C,3-D,4-A

1-A,2-C,3-A,4-B

1-A,2-B,3-B,4-C

answer is D.

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Detailed Solution

Given:
$\begin{array}{l} f (x) = a x + b \\ f^{- 1} (x) = b x + a \end{array}$
To find $f (f^{- 1} (x)) = x$ :
$f (f^{- 1} (x)) = f (b x + a) = a (b x + a) + b = a b x + a^{2} + b$
This must equal x:
$a b x + a^{2} + b = x$
For this to hold true for all x:
$\begin{array}{l} a b = 1 \\ a^{2} + b = 0 \end{array}$
From $a^{2} + b = 0$ :
$b = - a^{2}$
Substitute b into $a b = 1$ :
$\begin{array}{l} a (- a^{2}) = 1 \\ - a^{3} = 1 \\ a^{3} = - 1 \\ a = - 1 \end{array}$
Substitute $a = - 1$ into : $b = - a^{2}$
$\begin{array}{l} b = - {(- 1)}^{2} \\ b = - 1 \end{array}$
Thus, $a + b = - 1 - 1 = - 2$ .
Therefore, the value of $a + b$ is $- 2$ .