JEE Mains · Maths · STD 12 - 5. continuity and differentiation
The value of \(c\) in the Lagrange's mean value theorem for the function \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}-4 \mathrm{x}^{2}+8 \mathrm{x}+11\) when \(\mathrm{x} \in[0,1]\) is
- A \(\frac{2}{3}\)
- B \(\frac{\sqrt{7}-2}{3}\)
- C \(\frac{4-\sqrt{5}}{3}\)
- D \(\frac{4-\sqrt{7}}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{4-\sqrt{7}}{3}\)
Step-by-step Solution
Detailed explanation
\(f(0)=11\) \(f(1)=16\) \(\frac{\mathrm{f}(1)-\mathrm{f}(0)}{1-0}=3 \mathrm{c}^{2}-8 \mathrm{c}+8\) \(\Rightarrow 3 c^{2}-8 c+8=5\) \(\Rightarrow 3 c^{2}-8 c+3=0\) \(c \in[0,1] \Rightarrow c=\frac{4-\sqrt{7}}{3}\)
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