Question

The value of \(c\) satisfied by the Rolle's theorem for the function \(f(x) = x^2(1-x)^2\), \(x\in [0,1]\) is...

• A. \(0\)
• B. \(1\)
• C. \(\frac{1}{2}\)
• D. \(-1\)

Hint:

Rolle's theorem gives a point c where f'(c) = 0.

Answer 1

The Correct Option is C

Step 1: Symmetry Argument:
$f(x)=\big(x(1-x)\big)^2$ is symmetric about $x=\tfrac12$, because $f(1-x)=f(x)$. A symmetric function with equal endpoint values has a turning point at the middle.

Step 2: Confirm:
Let $g=x(1-x)$, so $f=g^2$ and $f'=2gg'$. $g'=1-2x=0$ at $x=\tfrac12$ (and $g=0$ at the ends).

Step 3: Answer:
The interior point is $c=\tfrac12$. Option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C) } \frac{1}{2}} \]