AP EAMCET · Maths · Application of Derivatives
If \(f(x)=(x-1)(x-2)(x-3)\) for \(x \in[0,4]\), then the value of \(c \in(0,4)\) satisfying Lagrange's mean value theorem, is
- A \(3 \pm \frac{\sqrt{2}}{3}\)
- B \(2 \pm \frac{2 \sqrt{3}}{3}\)
- C \(2 \pm \frac{\sqrt{3}}{2}\)
- D \(3 \pm \frac{\sqrt{3}}{3}\)
Answer & Solution
Correct Answer
(B) \(2 \pm \frac{2 \sqrt{3}}{3}\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} & f(x)=(x-1)(x-2)(x-3) \\ & f(x)=x^3-6 x^2+11 x-6 \end{aligned} \] Given function is algebraic function so. it is continuous and difference in \([0,4]\) Now,…
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