Step 1: Understanding the Concept:
The problem has two parts: determining the linear function \( g(x) \) and evaluating an inverse trigonometric expression. Since \( g(x) \) is a linear function with a negative slope, it must map the lower bound of the domain to the upper bound of the codomain, and vice-versa.
Step 2: Key Formula or Approach:
1. For \( g(x) \): \( g(1) = 2 \) and \( g(3) = 0 \).
2. For the trig part: Range of \( |\sin x| + |\cos x| \) is \( [1, \sqrt{2}] \).
Step 3: Detailed Explanation:
Finding \( g(x) \):
Linear function \( g(x) = ax + b \). Since \( a<0 \), it is a decreasing function.
Onto mapping means:
\( g(1) = 2 \implies a + b = 2 \)
\( g(3) = 0 \implies 3a + b = 0 \)
Subtracting the equations: \( 2a = -2 \implies a = -1 \).
Then \( b = 3 \). Thus, \( g(x) = -x + 3 \).
Evaluating the trigonometric expression:
Let \( A = |\sin x| + |\cos x| \).
The domain of \( \cos^{-1}(u) \) and \( \sin^{-1}(u) \) is \( [-1, 1] \).
However, \( A \in [1, \sqrt{2}] \). The only value for which \( \cos^{-1} A \) and \( \sin^{-1} A \) are defined is \( A = 1 \).
(Note: If \( A>1 \), the expression is not defined in reals, but in multiple-choice math, we look for the valid boundary).
At \( A = 1 \) (which happens at \( x = 0, \pi/2, \dots \)):
The expression is \( \cot(\cos^{-1}(1) + \sin^{-1}(-1)) \).
\( \cos^{-1}(1) = 0 \).
\( \sin^{-1}(-1) = -\pi/2 \).
The expression becomes \( \cot(0 - \pi/2) = \cot(-\pi/2) = 0 \).
Now check options for \( g(x) = -x + 3 \):
(A) \( g(2) + g(3) = (-2+3) + (-3+3) = 1 + 0 = 1 \).
(B) \( g(2) = 1 \).
(C) \( g(3) = -3 + 3 = 0 \).
The result matches \( g(3) \).
Step 4: Final Answer:
The trigonometric expression evaluates to 0, which equals \( g(3) \).