The value of c in the Lagrange's mean value theorem for the function ƒ(x) = x3 - 4x2 + 8x + 11, when x $$ \in $$ [0, 1] is:
JEE · Math · previous-year question
- A.$${2 \over 3}$$
- B.$${{\sqrt 7 - 2} \over 3}$$
- C.$${{4 - \sqrt 5 } \over 3}$$
- D.$${{4 - \sqrt 7 } \over 3}$$correct
Answer
D. $${{4 - \sqrt 7 } \over 3}$$
Explanation
ƒ(x) = x3 - 4x2 + 8x + 11 f(0) = 11 f(1) = 16 Using LMVT f'(c) = $${{f\left( 1 \right) - f\left( 0 \right)} \over {1 - 0}}$$ $$ \Rightarrow $$ 3c2 – 8c + 8 = $${{16 - 11} \over {1 - 0}}$$ $$ \Rightarrow $$ 3c2 – 8c + 3 = 0 $$ \therefore $$ c = $${{8 \pm 2\sqrt 7 } \over 6}$$ $$ \therefore $$ c = $${{4 - \sqrt 7 } \over 3}$$ as c $$ \in $$ [0, 1]
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