Let `n in N`, if the value of c prescribed in Roole's theorem for the function `f(x)=2x(x-3)^(n)` on `[0, 3]" is "(3)/(4)`, then n is equat to

Options

A. 1
B. 3
C. 5
D. 7

Answer

To solve the problem, we will follow the steps outlined in the video transcript and apply Rolle's Theorem to find the value of \( n \).

### Step-by-Step Solution:

1. **Understand the Function and Interval**:
We are given the function \( f(x) = 2x(x - 3)^n \) and the interval \([0, 3]\). We need to apply Rolle's theorem, which requires the function to be continuous on the closed interval and differentiable on the open interval.

2. **Check Continuity and Differentiability**:
The function \( f(x) \) is a polynomial and is continuous and differentiable everywhere, including the interval \([0, 3]\).

3. **Apply Rolle's Theorem**:
According to Rolle's theorem, there exists at least one \( c \) in the interval \((0, 3)\) such that:
\[ f'(c) = \frac{f(3) - f(0)}{3 - 0} \]

4. **Calculate \( f(3) \) and \( f(0) \)**:
- \( f(3) = 2 \cdot 3 \cdot (3 - 3)^n = 0 \) (since \( (3 - 3)^n = 0 \))
- \( f(0) = 2 \cdot 0 \cdot (0 - 3)^n = 0 \) (since \( 2 \cdot 0 = 0 \))

Thus, we have:
\[ f(3) - f(0) = 0 - 0 = 0 \]

5. **Set Up the Equation**:
From Rolle's theorem, we have:
\[ f'(c) = \frac{0}{3} = 0 \]
Therefore, we need to find \( c \) such that \( f'(c) = 0 \).

6. **Differentiate \( f(x) \)**:
Using the product rule:
\[ f'(x) = 2(x - 3)^n + 2x \cdot n(x - 3)^{n-1} \cdot 1 \]
Simplifying this gives:
\[ f'(x) = 2(x - 3)^{n-1} \left[ (x - 3) + nx \right] \]

7. **Set \( f'(c) = 0 \)**:
For \( f'(c) = 0 \), we need:
\[ 2(c - 3)^{n-1} \left[ (c - 3) + nc \right] = 0 \]
This implies either \( (c - 3)^{n-1} = 0 \) or \( (c - 3) + nc = 0 \).

Since \( c \) is in the interval \((0, 3)\), \( (c - 3)^{n-1} = 0 \) does not hold. Thus, we solve:
\[ c - 3 + nc = 0 \implies c(n + 1) = 3 \implies c = \frac{3}{n + 1} \]

8. **Set \( c = \frac{3}{4} \)**:
Given that \( c = \frac{3}{4} \):
\[ \frac{3}{n + 1} = \frac{3}{4} \]

9. **Solve for \( n \)**:
Cross-multiplying gives:
\[ 3 \cdot 4 = 3(n + 1) \implies 12 = 3n + 3 \implies 3n = 9 \implies n = 3 \]

### Final Answer:
Thus, the value of \( n \) is \( 3 \).

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