Question

The function \(f(x) = x(x+3)e^{-(\frac{1}{2})x}\) satisfies all the conditions of Rolle's theorem in \([-3,0]\), then \(c =\)

• A. \(-3\)
• B. \(-2\)
• C. \(-1\)
• D. \(0\)

Hint:

In Rolle theorem the point c is where the derivative is zero inside the interval.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Rolle's theorem says that if \(f(a)=f(b)\), then some \(c\) in \((a,b)\) has \(f'(c)=0\). Here \(f(-3)=0\) and \(f(0)=0\).

Step 2: Differentiate
\[ f(x)=(x^2+3x)e^{-x/2} \]
\[ f'(x)=e^{-x/2}\left[(2x+3)-\frac{x^2+3x}{2}\right] \]

Step 3: Set to zero
The exponential is never zero, so
\[ 4x+6-x^2-3x=0\Rightarrow x^2-x-6=0 \]
\[ (x-3)(x+2)=0\Rightarrow x=3\ \text{or}\ x=-2 \]

Step 4: Choose c
Only \(x=-2\) lies inside \((-3,0)\). The root \(x=3\) is outside the interval. So \(c=-2\), option (B).

Final Answer:
Setting f'(x) = 0 gives x = -2 or 3, and only -2 lies in the interval, option (B). \[ \boxed{-2} \]