If x = 1 is a critical point of the function f ( x ) = (3 x ^ 2 + ax - 2 - a) e ^ x , then

Correct answer: D. x = 1 is a local minima and x = - (2)/(3) are local maxima of f f ( x ) = (3 x ^ 2 + ax - 2 - a) e ^ x f ' ( x ) = (3 x ^ 2 + ax - 2 - a) e ^ x + e ^ x ( 6x + a ) = e ^ x (3 x ^ 2 + ( a + 6 ) x - 2) ∵ x = 1 is a critical point ∴ f ^ ' ( 1 ) = 0 ∴ 3 + a + 6 - 2 = 0 a = - 7 ∴ f ^ ' ( x ) = e ^ x (3 x ^ 2 - x - 2) = e ^ x (3 x ^ 2 - 3x + 2x - 2) = e ^ x ( 3x + 2 ) ( x - 1 ) ∴ maxima at x = (- 2)/(3) ∴ minima at x = 1

JEE Main 20205 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 5 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2020 · 5 September · Evening Shift · Mathematics · Application of Derivatives

If x = 1 is a critical point of the function f ( x ) = (3 x ^ 2 +...

x = 1

is a critical point of the function $f ( x ) = \left(3{x}^{2} + ax - 2 - a\right){e}^{x}, then

Options

A $x = 1$ and $x = - \dfrac{2}{3}$ are local minima of $f$
B $x = 1$ and $x = - \dfrac{2}{3}$ is a local maxima of $f$
C $x = 1$ is a local maxima and $x = - \dfrac{2}{2}$ is a local minima of $f$
D $x = 1$ is a local minima and $x = - \dfrac{2}{3}$ are local maxima of $f$

Correct answer

D. $x = 1$ is a local minima and $x = - \dfrac{2}{3}$ are local maxima of $f$

Step-by-step solution

$f ( x ) = \left(3{x}^{2} + ax - 2 - a\right){e}^{x}$
$f ' ( x ) = \left(3{x}^{2} + ax - 2 - a\right){e}^{x} + {e}^{x} ( 6x + a ) = {e}^{x}\left(3{x}^{2} + ( a + 6 ) x - 2\right)$

$∵ x = 1$ is a critical point $∴ {f}^{'} ( 1 ) = 0$

$∴ 3 + a + 6 - 2 = 0$
$a = - 7$

$∴ {f}^{'} ( x ) = {e}^{x}\left(3{x}^{2} - x - 2\right) = {e}^{x}\left(3{x}^{2} - 3x + 2x - 2\right) = {e}^{x} ( 3x + 2 ) ( x - 1 )$

$∴$ maxima at $x = \dfrac{- 2}{3} ∴$ minima at $x = 1$

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