If $x=1$ is a critical point of the function $f(x)=\left(3 x^{2}+a x-2-a\right) e^{x}$, then

(A) $x=1$ is a local minima and $x=-\frac{2}{3}$ is a local maxima of $f$.

(B) $x=1$ and $x=\frac{-2}{3}$ are local minima of $f$.

(C) $x=1$ is a local maxima and $x=-\frac{2}{3}$ is a local minima of $f$.

(D) $x=1$ and $x=-\frac{2}{3}$ are local maxima of $f$.

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Application of Derivatives

x = 1

is a critical point of the function

f (x) = (3 x^{2} + a x - 2 - a) e^{x}

, then

(A) $x = 1$ is a local minima and $x = - \frac{2}{3}$ is a local maxima of $f$ .✓ Correct
(B) $x = 1$ and $x = \frac{- 2}{3}$ are local minima of $f$ .
(C) $x = 1$ is a local maxima and $x = - \frac{2}{3}$ is a local minima of $f$ .
(D) $x = 1$ and $x = - \frac{2}{3}$ are local maxima of $f$ .

Answer: (A)

Here

f^{'} (x) = e^{x} (3 x^{2} + (a + 6) x - 2)

. Since

x = 1

is a critical point,

f^{'} (1) = 3 + (a + 6) - 2 = 0

gives

a = - 7

, so

f^{'} (x) = e^{x} (3 x + 2) (x - 1)

. The sign of

f^{'}

changes from positive to negative at

x = - \frac{2}{3}

, giving a local maximum there, and from negative to positive at

x = 1

, giving a local minimum.

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option