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: Check the Conditions:
\(f(x)=x^2(1-x)^2\) is a polynomial, so it is continuous and differentiable. Also \(f(0)=0=f(1)\). Rolle\'s theorem applies.

Step 2: Differentiate:
\[ f'(x)=2x(1-x)^2-2x^2(1-x)=2x(1-x)\left[(1-x)-x\right]=2x(1-x)(1-2x) \]

Step 3: Solve f'(c) = 0:
The roots are \(x=0,\ x=1,\ x=\tfrac12\). The point \(c\) must lie in the open interval \((0,1)\), so \(c=\tfrac12\).

Step 4: Check the Options:
Options 0 and 1 are the endpoints, which are not in the open interval. Option \(-1\) lies outside \([0,1]\). So (C) is correct.

Final Answer:
\(c=\dfrac12\), option (C). \[ \boxed{\text{(C) } \frac{1}{2}} \]