If `f(x)=x+sinx ,g(x)=e^(-x),u=sqrt(c+1)-sqrt(c)` `v=sqrt(c)` `-sqrt(c-1),(c >1),` then `fog(u)gof(v)` (d) `fog(u)

Options

A. `fog(u)ltfog(v)`
B. `gof(u)ltgof(v)`
C. `gof(u)gtgof(v)`
D. `fog(u)ltfog(v)`

Explanation

To solve the problem, we need to analyze the functions \( f(x) = x + \sin x \) and \( g(x) = e^{-x} \) along with the expressions for \( u \) and \( v \).

### Step 1: Understanding the Functions
1. **Function \( f(x) \)**:
- The function \( f(x) = x + \sin x \) is an increasing function because its derivative \( f'(x) = 1 + \cos x \) is always positive (since \( \cos x \) oscillates between -1 and 1, \( f'(x) \) is always greater than or equal to 0).

2. **Function \( g(x) \)**:
- The function \( g(x) = e^{-x} \) is a decreasing function because its derivative \( g'(x) = -e^{-x} \) is always negative.

### Step 2: Analyzing \( u \) and \( v \)
Given:
- \( u = \sqrt{c+1} - \sqrt{c} \)
- \( v = \sqrt{c} - \sqrt{c-1} \)

Since \( c > 1 \), we can analyze \( u \) and \( v \):
- **For \( u \)**:
- As \( c \) increases, \( \sqrt{c+1} \) increases faster than \( \sqrt{c} \), thus \( u \) is positive.

- **For \( v \)**:
- \( v = \sqrt{c} - \sqrt{c-1} \) is also positive since \( \sqrt{c} > \sqrt{c-1} \) for \( c > 1 \).

### Step 3: Comparing \( u \) and \( v \)
To compare \( u \) and \( v \):
- We can rationalize both expressions:
- \( u = \frac{(\sqrt{c+1} - \sqrt{c})(\sqrt{c+1} + \sqrt{c})}{\sqrt{c+1} + \sqrt{c}} = \frac{1}{\sqrt{c+1} + \sqrt{c}} \)
- \( v = \frac{(\sqrt{c} - \sqrt{c-1})(\sqrt{c} + \sqrt{c-1})}{\sqrt{c} + \sqrt{c-1}} = \frac{1}{\sqrt{c} + \sqrt{c-1}} \)

Since \( \sqrt{c+1} > \sqrt{c} \) and \( \sqrt{c} > \sqrt{c-1} \), we can conclude that \( v > u \).

### Step 4: Applying the Functions
Now we apply the functions:
1. **For \( f(u) \) and \( f(v) \)**:
- Since \( f \) is increasing, \( f(u) < f(v) \).

2. **For \( g(f(u)) \) and \( g(f(v)) \)**:
- Since \( g \) is decreasing, \( g(f(u)) > g(f(v)) \).

### Conclusion
From the analysis:
- \( f(u) < f(v) \) implies \( g(f(u)) > g(f(v)) \).

Thus, we have:
- \( g(f(u)) < g(f(v)) \) is **false**.
- \( g(f(u)) > g(f(v)) \) is **true**.
- \( f(g(u)) < f(g(v)) \) is **true**.

### Final Answer
The correct inequalities are:
- \( g(f(u)) < g(f(v)) \)
- \( g(f(u)) > g(f(v)) \)