JEE MAIN - Mathematics (2020 - 5th September Evening Slot - No. 17)
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 :
x = 1 is a local maxima and x = $$ - {2 \over 3}$$ is a
local minima of f.
x = 1 and x = $$ - {2 \over 3}$$ are local maxima of f.
x = 1 and x = $$ - {2 \over 3}$$ are local minima of f.
x = 1 is a local minima and x = $$ - {2 \over 3}$$ is a local
maxima of f.
Explanation
f(x) = (3x2
+ ax – 2 – a)ex
$$ \therefore $$ f'(x) = ex(6x + a) + (3x2 + ax – 2 – a)ex
= ex(3x2 + x(6 + a) – 2)
f '(x) = 0 at x = 1
$$ \Rightarrow $$ 3 + (6 + a) – 2 = 0
$$ \Rightarrow $$ a = -7
$$ \therefore $$ f'(x) = ex(3x2 – x – 2)
= ex(x – 1) (3x + 2)_5th_September_Evening_Slot_en_17_1.png)
x = 1 is a local minima and x = $$ - {2 \over 3}$$ is a local maxima of f.
$$ \therefore $$ f'(x) = ex(6x + a) + (3x2 + ax – 2 – a)ex
= ex(3x2 + x(6 + a) – 2)
f '(x) = 0 at x = 1
$$ \Rightarrow $$ 3 + (6 + a) – 2 = 0
$$ \Rightarrow $$ a = -7
$$ \therefore $$ f'(x) = ex(3x2 – x – 2)
= ex(x – 1) (3x + 2)
x = 1 is a local minima and x = $$ - {2 \over 3}$$ is a local maxima of f.
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