Differentiation and its Application: If x=1 is a critical point of the function f(x)= (3x^2+ax-2-a )e^x ,… | JEE Main 2020

JEE Main 2020 (Online) 5th September Evening ShiftQuestion 1

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

Differentiation and its Application

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

Solution

Concept. $f'(1)=0$ fixes $a$; then factorise $f'$. Formulas used. (1) $f'(x)=e^x\left(3x^2+(a+6)x-2\right)$Step 1 - Find $a$. By (1), $f'(1)=e(3+a+6-2)=0$, so $a=-7$. Step 2 - Factorise. $f'(x)=e^x(3x^2-x-2)=e^x(3x+2)(x-1)$. Step 3 - Signs. $+,-,+$: maximum at $x=-\frac23$, minimum at $x=1$. Answer: (D)

Differentiation and its Application - JEE Mathematics question (JEE Main 2020 (Online) 5th September Evening Shift)

More from Differentiation and its Application