Consider f,g and h be three real valued functions defined on R. Let `f(x)={:{(-1", "xlt0),(0", "x=0"),(1", "xgto):}`"g(x)(1-x^(2))andh(x) "be such that" h''(x)=6x-4. Also, h(x) has local minimum value 5 at x=1
The area bounded by y=h(x),y=g(f(x))between x=0 and x=2 equals

Options

A. `23//2`
B. `20//3`
C. `32//3`
D. `40//3`

Answer

To solve the problem step by step, we will follow the given information and derive the necessary functions and area step by step.

### Step 1: Define the functions
We are given the following functions:
- \( f(x) = \begin{cases} -1 & \text{if } x < 0 \\ 0 & \text{if } x = 0 \\ 1 & \text{if } x > 0 \end{cases} \)

- \( g(x) = 1 - x^2 \)

- \( h''(x) = 6x - 4 \)

### Step 2: Find \( h'(x) \)
To find \( h(x) \), we first need to integrate \( h''(x) \):
\[ h'(x) = \int (6x - 4) \, dx = 3x^2 - 4x + C \]
where \( C \) is a constant.

### Step 3: Find \( h(x) \)
Now, we integrate \( h'(x) \):
\[ h(x) = \int (3x^2 - 4x + C) \, dx = x^3 - 2x^2 + Cx + D \]
where \( D \) is another constant.

### Step 4: Use the local minimum condition
We know that \( h(x) \) has a local minimum value of 5 at \( x = 1 \). This means:
1. \( h'(1) = 0 \)
2. \( h(1) = 5 \)

Using \( h'(1) = 0 \):
\[ h'(1) = 3(1)^2 - 4(1) + C = 0 \implies 3 - 4 + C = 0 \implies C = 1 \]

Now substituting \( C \) into \( h(x) \):
\[ h(x) = x^3 - 2x^2 + x + D \]

Using \( h(1) = 5 \):
\[ h(1) = (1)^3 - 2(1)^2 + (1) + D = 5 \implies 1 - 2 + 1 + D = 5 \implies D = 5 \]

Thus, we have:
\[ h(x) = x^3 - 2x^2 + x + 5 \]

### Step 5: Find \( g(f(x)) \)
Next, we need to find \( g(f(x)) \):
- For \( x < 0 \): \( f(x) = -1 \) so \( g(f(x)) = g(-1) = 1 - (-1)^2 = 0 \)
- For \( x = 0 \): \( f(x) = 0 \) so \( g(f(x)) = g(0) = 1 - 0^2 = 1 \)
- For \( x > 0 \): \( f(x) = 1 \) so \( g(f(x)) = g(1) = 1 - 1^2 = 0 \)

Thus, we can summarize:
\[ g(f(x)) = \begin{cases} 0 & \text{if } x < 0 \\ 1 & \text{if } x = 0 \\ 0 & \text{if } x > 0 \end{cases} \]

### Step 6: Calculate the area between \( h(x) \) and \( g(f(x)) \)
We need to find the area bounded by \( y = h(x) \) and \( y = g(f(x)) \) between \( x = 0 \) and \( x = 2 \).

Since \( g(f(x)) = 1 \) at \( x = 0 \) and \( 0 \) for \( x > 0 \), we will calculate the area from \( x = 0 \) to \( x = 2 \):
\[ \text{Area} = \int_0^2 (h(x) - g(f(x))) \, dx = \int_0^2 (h(x) - 0) \, dx = \int_0^2 h(x) \, dx \]

Substituting \( h(x) \):
\[ \text{Area} = \int_0^2 (x^3 - 2x^2 + x + 5) \, dx \]

### Step 7: Evaluate the integral
Calculating the integral:
\[ \int (x^3 - 2x^2 + x + 5) \, dx = \frac{x^4}{4} - \frac{2x^3}{3} + \frac{x^2}{2} + 5x \]

Now evaluate from \( 0 \) to \( 2 \):
\[ = \left[ \frac{2^4}{4} - \frac{2(2^3)}{3} + \frac{(2^2)}{2} + 5(2) \right] - \left[ 0 \right] \]
\[ = \left[ 4 - \frac{16}{3} + 2 + 10 \right] \]
\[ = 16 - \frac{16}{3} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3} \]

### Final Answer
The area bounded by \( y = h(x) \) and \( y = g(f(x)) \) between \( x = 0 \) and \( x = 2 \) is:
\[ \frac{32}{3} \]