JEE Main 2020 — Application Of Derivatives Question with Solution
From: JEE Main 2020 (Online) 5th September Evening Slot
Question
If x = 1 is a critical point of the function
f(x) = (3x2 + ax – 2 – a)ex , then :
f(x) = (3x2 + ax – 2 – a)ex , then :
Choose an option
Show full solutionCorrect option: D
Correct answer
Dx = 1 is a local minima and x = is a local
maxima of f.
Step-by-step explanation
f(x) = (3x2
+ ax – 2 – a)ex
f'(x) = ex(6x + a) + (3x2 + ax – 2 – a)ex
= ex(3x2 + x(6 + a) – 2)
f '(x) = 0 at x = 1
3 + (6 + a) – 2 = 0
a = -7
f'(x) = ex(3x2 – x – 2)
= ex(x – 1) (3x + 2)
x = 1 is a local minima and x = is a local maxima of f.
f'(x) = ex(6x + a) + (3x2 + ax – 2 – a)ex
= ex(3x2 + x(6 + a) – 2)
f '(x) = 0 at x = 1
3 + (6 + a) – 2 = 0
a = -7
f'(x) = ex(3x2 – x – 2)
= ex(x – 1) (3x + 2)
x = 1 is a local minima and x = is a local maxima of f.
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