If `f:[1,10]->[1,10]` is a non-decreasing function and `g:[1,10]->[1,10]` is a non-increasing function. Let `h(x)=f(g(x))` with `h(1)=1,` then `h(2)`

Options

A. lies in (1,2)
B. is more than two
C. is equal to one
D. is not defined

Answer

To solve the problem step by step, we will analyze the functions and their properties based on the information given.

### Step 1: Understand the properties of the functions
- We are given that \( f: [1,10] \to [1,10] \) is a **non-decreasing function**. This means that for any \( x_1 < x_2 \), \( f(x_1) \leq f(x_2) \).
- We are also given that \( g: [1,10] \to [1,10] \) is a **non-increasing function**. This means that for any \( x_1 < x_2 \), \( g(x_1) \geq g(x_2) \).

### Step 2: Analyze the function \( h(x) \)
- We define \( h(x) = f(g(x)) \).
- We know that \( h(1) = 1 \). This means \( h(1) = f(g(1)) = 1 \).

### Step 3: Find \( h(2) \)
- We need to find \( h(2) = f(g(2)) \).
- Since \( 2 > 1 \) and \( g \) is a non-increasing function, we have:
\[ g(2) \leq g(1) \]
- Since \( h(1) = 1 \) implies \( f(g(1)) = 1 \), we can conclude that \( g(1) \) must be such that \( f(g(1)) = 1 \).

### Step 4: Use the properties of \( f \)
- Since \( f \) is a non-decreasing function, if \( g(2) \leq g(1) \), then:
\[ f(g(2)) \leq f(g(1)) \]
- Therefore, we have:
\[ h(2) = f(g(2)) \leq f(g(1)) = 1 \]

### Step 5: Conclusion
- Since \( h(2) \leq 1 \) and \( h(1) = 1 \), we can conclude that:
\[ h(2) \leq 1 \]
- The only possible value for \( h(2) \) that satisfies this condition, given that \( h(x) \) must also be in the range [1, 10], is:
\[ h(2) = 1 \]

Thus, the final answer is:
\[ h(2) = 1 \]