MHT CET · Maths · Application of Derivatives
In the 5. Mean Value Theorem, \(\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}\), if \(a=0, \mathrm{~b}=\frac{1}{2}\) and \(\mathrm{f}(x)=x(x-1)(x-2)\), then the value of c is
- A \(1-\frac{\sqrt{15}}{6}\)
- B \(1-\frac{\sqrt{13}}{6}\)
- C \(1-\frac{\sqrt{21}}{6}\)
- D \(1+\frac{\sqrt{21}}{6}\)
Answer & Solution
Correct Answer
(C) \(1-\frac{\sqrt{21}}{6}\)
Step-by-step Solution
Detailed explanation
\(f(x) = x^3 - 3x^2 + 2x\) \(f'(x) = 3x^2 - 6x + 2\)
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