1
JEE Main 2020 (Online) 8th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let S be the set of all functions ƒ : [0,1] $$ \to $$ R, which are continuous on [0,1] and differentiable on (0,1). Then for every ƒ in S, there exists a c $$ \in $$ (0,1), depending on ƒ, such that
A
$$\left| {f(c) - f(1)} \right| < \left| {f'(c)} \right|$$
B
$$\left| {f(c) + f(1)} \right| < \left( {1 + c} \right)\left| {f'(c)} \right|$$
C
$$\left| {f(c) - f(1)} \right| < \left( {1 - c} \right)\left| {f'(c)} \right|$$
D
None
2
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
$$\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}$$ is equal to
A
e
B
e2
C
$${1 \over {{e^2}}}$$
D
$${1 \over e}$$
3
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
$$\mathop {\lim }\limits_{x \to 0} {{x + 2\sin x} \over {\sqrt {{x^2} + 2\sin x + 1} - \sqrt {{{\sin }^2}x - x + 1} }}$$ is :
A
6
B
1
C
3
D
2
4
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x $$ \in $$ R. If f(x) attains maximum value at $$\alpha $$ and g(x) attains minimum value at $$\beta $$, then $$\mathop {\lim }\limits_{x \to -\alpha \beta } {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}}$$ is equal to :
A
$${1 \over 2}$$
B
$$-{1 \over 2}$$
C
$${3 \over 2}$$
D
$$-{3 \over 2}$$
JEE Main Subjects
EXAM MAP
Medical
NEETAIIMS
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
Defence
NDA
Staff Selection Commission
SSC CGL Tier I
CBSE
Class 12