Let `f:R rarrR, f(x)=x^(4)-8x^(3)+22x^(2)-24x+c`.
If sum of all extremum value of f(x) is 1, then c is equal to

Options

A. 8
B. 9
C. 10
D. 11

Answer

To solve the problem, we need to find the value of \( c \) such that the sum of all extremum values of the function \( f(x) = x^4 - 8x^3 + 22x^2 - 24x + c \) is equal to 1.

### Step 1: Find the derivative of \( f(x) \)
First, we need to find the first derivative of \( f(x) \) to locate the critical points (extrema).

\[ f'(x) = \frac{d}{dx}(x^4 - 8x^3 + 22x^2 - 24x + c) \]
\[ f'(x) = 4x^3 - 24x^2 + 44x - 24 \]

### Step 2: Factor the derivative
Next, we can factor the derivative to find the critical points.

\[ f'(x) = 4(x^3 - 6x^2 + 11x - 6) \]

We can use synthetic division or the Rational Root Theorem to factor \( x^3 - 6x^2 + 11x - 6 \). Testing possible roots, we find:

\[ x = 1 \quad \text{is a root.} \]

Using synthetic division, we can factor it as:

\[ x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) \]

Factoring further, we have:

\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]

Thus, the derivative can be expressed as:

\[ f'(x) = 4(x - 1)(x - 2)(x - 3) \]

### Step 3: Find the critical points
Setting \( f'(x) = 0 \), we find the critical points:

\[ x = 1, \quad x = 2, \quad x = 3 \]

### Step 4: Calculate the extremum values
Now, we will calculate the function values at these critical points.

1. **For \( x = 1 \)**:
\[ f(1) = 1^4 - 8(1^3) + 22(1^2) - 24(1) + c = 1 - 8 + 22 - 24 + c = -9 + c \]

2. **For \( x = 2 \)**:
\[ f(2) = 2^4 - 8(2^3) + 22(2^2) - 24(2) + c = 16 - 64 + 88 - 48 + c = -8 + c \]

3. **For \( x = 3 \)**:
\[ f(3) = 3^4 - 8(3^3) + 22(3^2) - 24(3) + c = 81 - 216 + 198 - 72 + c = -9 + c \]

### Step 5: Set up the equation for the sum of extremum values
Now, we sum the extremum values:

\[ f(1) + f(2) + f(3) = (-9 + c) + (-8 + c) + (-9 + c) = 3c - 26 \]

According to the problem, this sum equals 1:

\[ 3c - 26 = 1 \]

### Step 6: Solve for \( c \)
Now, we solve for \( c \):

\[ 3c = 1 + 26 \]
\[ 3c = 27 \]
\[ c = 9 \]

Thus, the value of \( c \) is \( \boxed{9} \).