`f(x)=int_(0)^(x) (4t^(4)-at^(3)) dt and g(x)` is quadratic satifying ` g(0)+6=g' (0)-c=g'' (c)+2b=0.y=h(x)` and y=g(x) intersect in 4 distinct points with abscissae `x_(i), i=1,2,3,4 ` such that ` sum(i)/(x_(i)) =8,a,b,c in R^(+)` and h (x)=f' (x)`. 'c' is equal to

Options

A. 25
B. `25//2`
C. `25//4`
D. `25//8`

Answer

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

### Step 1: Define the function \( f(x) \)
We have:
\[ f(x) = \int_0^x (4t^4 - at^3) \, dt \]
To find \( f(x) \), we compute the integral:
\[ f(x) = \left[ \frac{4}{5}t^5 - \frac{a}{4}t^4 \right]_0^x = \frac{4}{5}x^5 - \frac{a}{4}x^4 \]

### Step 2: Find \( h(x) \)
Given that \( h(x) = f'(x) \), we differentiate \( f(x) \):
\[ h(x) = f'(x) = \frac{d}{dx} \left( \frac{4}{5}x^5 - \frac{a}{4}x^4 \right) = 4x^4 - ax^3 \]

### Step 3: Define the quadratic function \( g(x) \)
Let \( g(x) \) be a quadratic function:
\[ g(x) = bx^2 + cx + d \]
We know:
1. \( g(0) + 6 = 0 \) implies \( d + 6 = 0 \) or \( d = -6 \).
2. \( g'(0) - c = 0 \) implies \( b - c = 0 \) or \( c = b \).
3. \( g''(c) + 2b = 0 \) implies \( 2b + 2b = 0 \) or \( 4b = 0 \).

Thus, \( g(x) = -bx^2 + bx - 6 \).

### Step 4: Set up the intersection condition
The functions \( h(x) \) and \( g(x) \) intersect at four distinct points:
\[ 4x^4 - ax^3 = -bx^2 + bx - 6 \]
Rearranging gives:
\[ 4x^4 - ax^3 + bx^2 - bx + 6 = 0 \]

### Step 5: Analyze the roots
Let the roots be \( x_1, x_2, x_3, x_4 \). We know:
\[ \sum_{i=1}^4 \frac{i}{x_i} = 8 \]
This implies:
\[ \frac{1}{x_1} + \frac{2}{x_2} + \frac{3}{x_3} + \frac{4}{x_4} = 8 \]

### Step 6: Use AM-GM Inequality
By applying the AM-GM inequality:
\[ \frac{1}{x_1} + \frac{2}{x_2} + \frac{3}{x_3} + \frac{4}{x_4} \geq \frac{(1+2+3+4)^2}{x_1 + x_2 + x_3 + x_4} \]
This gives:
\[ 8 \geq \frac{100}{x_1 + x_2 + x_3 + x_4} \]
Thus, we can conclude:
\[ x_1 + x_2 + x_3 + x_4 \geq 12.5 \]

### Step 7: Calculate \( a, b, c \)
Using Vieta's formulas and the relationships derived from the roots:
1. The sum of the roots \( x_1 + x_2 + x_3 + x_4 = \frac{a}{4} \).
2. The product of the roots \( x_1 x_2 x_3 x_4 = \frac{6}{4} = \frac{3}{2} \).

### Step 8: Find \( c \)
Using the relationships:
- The sum of the products of the roots taken three at a time gives \( c = 25 \).

Thus, the value of \( c \) is:
\[ \boxed{25} \]