JEE Main 2020 — Application Of Derivatives Question with Solution
From: JEE Main 2020 (Online) 7th January Evening Slot
Question
The value of c in the Lagrange's mean value theorem for the function
ƒ(x) = x3 - 4x2 + 8x + 11, when x [0, 1] is:
ƒ(x) = x3 - 4x2 + 8x + 11, when x [0, 1] is:
Choose an option
Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
ƒ(x) = x3
- 4x2
+ 8x + 11
f(0) = 11
f(1) = 16
Using LMVT
f'(c) =
3c2 – 8c + 8 =
3c2 – 8c + 3 = 0
c =
c = as c [0, 1]
f(0) = 11
f(1) = 16
Using LMVT
f'(c) =
3c2 – 8c + 8 =
3c2 – 8c + 3 = 0
c =
c = as c [0, 1]
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