Let `f,g` and `h` be real-valued functions defined on the interval `[0,1]` by `f(x)=e^(x^2)+e^(-x^2)` , `g(x)=x e^(x^2)+e^(-x^2)` and `h(x)=x^2 e^(x^2)+e^(-x^2)`. if `a,b` and `c` denote respectively, the absolute maximum of `f,g` and `h` on `[0,1]` then

Options

A. `a=bandcneb`
B. `a=c and a ne b`
C. `a ne b c ne b`
D. `a=b=c`

Answer

To solve the problem, we need to find the absolute maximum values of the functions \( f(x) \), \( g(x) \), and \( h(x) \) on the interval \([0, 1]\) and determine the relationship between these maximum values.

### Step 1: Define the Functions
The functions are given as:
- \( f(x) = e^{x^2} + e^{-x^2} \)
- \( g(x) = x e^{x^2} + e^{-x^2} \)
- \( h(x) = x^2 e^{x^2} + e^{-x^2} \)

### Step 2: Evaluate the Functions at the Endpoints
We will first evaluate each function at the endpoints \( x = 0 \) and \( x = 1 \).

1. **For \( f(x) \)**:
- \( f(0) = e^{0^2} + e^{-0^2} = 1 + 1 = 2 \)
- \( f(1) = e^{1^2} + e^{-1^2} = e + \frac{1}{e} \)

2. **For \( g(x) \)**:
- \( g(0) = 0 \cdot e^{0^2} + e^{-0^2} = 0 + 1 = 1 \)
- \( g(1) = 1 \cdot e^{1^2} + e^{-1^2} = e + \frac{1}{e} \)

3. **For \( h(x) \)**:
- \( h(0) = 0^2 \cdot e^{0^2} + e^{-0^2} = 0 + 1 = 1 \)
- \( h(1) = 1^2 \cdot e^{1^2} + e^{-1^2} = e + \frac{1}{e} \)

### Step 3: Compare the Values
Now we compare the values we calculated:
- \( f(0) = 2 \), \( f(1) = e + \frac{1}{e} \)
- \( g(0) = 1 \), \( g(1) = e + \frac{1}{e} \)
- \( h(0) = 1 \), \( h(1) = e + \frac{1}{e} \)

### Step 4: Determine the Maximum Values
From the evaluations:
- At \( x = 0 \), \( f(0) = 2 \) is the maximum.
- At \( x = 1 \), \( g(1) = e + \frac{1}{e} \) and \( h(1) = e + \frac{1}{e} \) are equal.

### Step 5: Conclusion
Thus, we have:
- \( a = f(1) = e + \frac{1}{e} \)
- \( b = g(1) = e + \frac{1}{e} \)
- \( c = h(1) = e + \frac{1}{e} \)

This implies:
- \( a = b = c \)

### Final Answer
The correct relationship is:
- \( a = b = c \)