TS EAMCET · Maths · Application of Derivatives
The value \(C\) of the Lagrange's mean value theorem for the function \(f(x)=x(x-1)(x-2)\) in the interval \(\left[0, \frac{1}{2}\right]\) is
- A \(1-\frac{\sqrt{7}}{2 \sqrt{3}}\)
- B \(1-\frac{\sqrt{7}}{\sqrt{3}}\)
- C \(\frac{1}{3}\)
- D \(\frac{1}{6}\)
Answer & Solution
Correct Answer
(A) \(1-\frac{\sqrt{7}}{2 \sqrt{3}}\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} f(x) & =x(x-1)(x-2) \\ & =x\left(x^2-3 x+2\right)=x^3-3 x^2+2 x \\ \therefore \quad f^{\prime}(x) & =3 x^2-6 x+2 \\ \Rightarrow \quad f^{\prime}(c) & =3 c^2-6 c+2 \end{aligned} \] According to LMVT…
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