CUET · MATHS · PYQ PAPER 2023
The value of \(c\) for which \(f(x)=x(x-3)^2, 0 \leq x \leq 3\), satisfies Rolle's theorem is:
- A \(0\)
- B 1
- C 2
- D 3
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
\(f(0)=0(0-3)^2 = 0\) \(f(3)=3(3-3)^2 = 0\) \(f'(x) = \frac{d}{dx}(x^3-6x^2+9x) = 3x^2-12x+9\) \(f'(c)=0 \implies 3c^2-12c+9=0 \implies c^2-4c+3=0\) \((c-1)(c-3)=0 \implies c=1\) or \(c=3\) Since \(c \in (0,3)\), \(c=1\)
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