If f(x) is a quadratic polynomial in x such that
`6int_0^1 f(x)dx-{f(0)+4f((1)/(2))}=kf(1)`, then k=`

Options

A. -1
B. 0
C. 1
D. 2

Answer

To solve the problem step-by-step, we will analyze the given information and derive the required value of \( k \).

### Step 1: Define the quadratic polynomial
Let \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants.

### Step 2: Calculate \( f(0) \), \( f\left(\frac{1}{2}\right) \), and \( f(1) \)
- \( f(0) = c \)
- \( f\left(\frac{1}{2}\right) = a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + c = \frac{a}{4} + \frac{b}{2} + c \)
- \( f(1) = a(1)^2 + b(1) + c = a + b + c \)

### Step 3: Set up the integral
We need to calculate the integral \( \int_0^1 f(x) \, dx \):
\[ \int_0^1 f(x) \, dx = \int_0^1 (ax^2 + bx + c) \, dx \]
Using the power rule for integration, we have:
\[ \int_0^1 ax^2 \, dx = \left[ \frac{a}{3} x^3 \right]_0^1 = \frac{a}{3} \]
\[ \int_0^1 bx \, dx = \left[ \frac{b}{2} x^2 \right]_0^1 = \frac{b}{2} \]
\[ \int_0^1 c \, dx = [cx]_0^1 = c \]
Thus,
\[ \int_0^1 f(x) \, dx = \frac{a}{3} + \frac{b}{2} + c \]

### Step 4: Substitute into the given equation
The equation given in the problem is:
\[ 6 \int_0^1 f(x) \, dx - (f(0) + 4f\left(\frac{1}{2}\right)) = k f(1) \]
Substituting the values we calculated:
\[ 6\left(\frac{a}{3} + \frac{b}{2} + c\right) - \left(c + 4\left(\frac{a}{4} + \frac{b}{2} + c\right)\right) = k(a + b + c) \]
This simplifies to:
\[ 6\left(\frac{a}{3} + \frac{b}{2} + c\right) - \left(c + a + 2b + 4c\right) = k(a + b + c) \]

### Step 5: Simplify the left-hand side
Calculating the left-hand side:
\[ 6\left(\frac{a}{3}\right) + 6\left(\frac{b}{2}\right) + 6c - (c + a + 2b + 4c) \]
This becomes:
\[ 2a + 3b + 6c - (5c + a + 2b) = 2a + 3b + 6c - 5c - a - 2b \]
Combining like terms:
\[ (2a - a) + (3b - 2b) + (6c - 5c) = a + b + c \]

### Step 6: Set the equation equal to the right-hand side
Now we have:
\[ a + b + c = k(a + b + c) \]
Assuming \( a + b + c \neq 0 \), we can divide both sides by \( a + b + c \):
\[ 1 = k \]
Thus, we find that:
\[ k = 1 \]

### Final Answer
The value of \( k \) is \( 1 \).