JEE Main 8 Apr 2019 Paper 2 Solved Paper

Section: Mathematics

© examsiri.com

Question : 73 of 90

Marks: +1, -0

Let f:R⇒R be a differentiable function Satisfying f′(3)+f′(2)=0. Then

lim

x⇒0

(

1+f(3+x)−f(3)

1+f(2−x)−f(2)

)

1

x

is equal to

[8 Apr 2019 Shift 2]

1
e
e–1
e2

Go to Question:

Prev Question

Next Question