Let `f(x)="max"(1+s in x ,1,1-cosx),x in [0,2pi],a n dg(x)=max{1,|x-1|}, x in Rdot` Then `g(f(0))=1` (b) `g(f(1))=1` `f(f(1))=1` (d) `f(g(0))= 1 + sin1`

Options

A. `g(f(0))=1`
B. `g(f(1))=1`
C. `f(f(1))=1`
D. `f(g(0))=1+sin 1`

Explanation

To solve the problem step by step, we will evaluate the functions \( f(x) \) and \( g(x) \) as defined in the question.

### Step 1: Define the Functions
The functions are defined as follows:
- \( f(x) = \max(1 + \sin x, 1, 1 - \cos x) \) for \( x \in [0, 2\pi] \)
- \( g(x) = \max(1, |x - 1|) \) for \( x \in \mathbb{R} \)

### Step 2: Calculate \( f(0) \)
Substituting \( x = 0 \) into \( f(x) \):
\[ f(0) = \max(1 + \sin(0), 1, 1 - \cos(0)) = \max(1 + 0, 1, 1 - 1) = \max(1, 1, 0) = 1 \]

### Step 3: Calculate \( g(f(0)) \)
Since \( f(0) = 1 \):
\[ g(f(0)) = g(1) = \max(1, |1 - 1|) = \max(1, 0) = 1 \]

### Step 4: Calculate \( f(1) \)
Substituting \( x = 1 \) into \( f(x) \):
\[ f(1) = \max(1 + \sin(1), 1, 1 - \cos(1)) \]
Calculating the values:
- \( 1 + \sin(1) \) (approximately 1.8415)
- \( 1 \)
- \( 1 - \cos(1) \) (approximately 0.4597)

Thus,
\[ f(1) = \max(1.8415, 1, 0.4597) = 1 + \sin(1) \]

### Step 5: Calculate \( g(f(1)) \)
Now substituting \( f(1) = 1 + \sin(1) \):
\[ g(f(1)) = g(1 + \sin(1)) = \max(1, |(1 + \sin(1)) - 1|) = \max(1, \sin(1)) \]
Since \( \sin(1) \) is approximately 0.8415, we have:
\[ g(f(1)) = \max(1, 0.8415) = 1 \]

### Step 6: Calculate \( f(f(1)) \)
Since \( f(1) = 1 + \sin(1) \):
\[ f(f(1)) = f(1 + \sin(1)) = \max(1 + \sin(1 + \sin(1)), 1, 1 - \cos(1 + \sin(1))) \]
However, we can see that \( f(1 + \sin(1)) \) will be greater than 1, and thus will not equal 1.

### Step 7: Calculate \( f(g(0)) \)
First, calculate \( g(0) \):
\[ g(0) = \max(1, |0 - 1|) = \max(1, 1) = 1 \]
Then, substituting into \( f \):
\[ f(g(0)) = f(1) = 1 + \sin(1) \]

### Summary of Results
- \( g(f(0)) = 1 \) (True)
- \( g(f(1)) = 1 \) (True)
- \( f(f(1)) \neq 1 \) (False)
- \( f(g(0)) = 1 + \sin(1) \) (True)

### Final Answers
- (a) True: \( g(f(0)) = 1 \)
- (b) True: \( g(f(1)) = 1 \)
- (c) False: \( f(f(1)) \neq 1 \)
- (d) True: \( f(g(0)) = 1 + \sin(1) \)