1
JEE Main 2020 (Online) 8th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
The length of the perpendicular from the origin, on the normal to the curve,
x2 + 2xy – 3y2 = 0 at the point (2,2) is
A
$$\sqrt 2 $$
B
$$4\sqrt 2 $$
C
2
D
$$2\sqrt 2 $$
2
JEE Main 2020 (Online) 8th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If $$I = \int\limits_1^2 {{{dx} \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}} $$, then :
A
$${1 \over 16} < {I^2} < {1 \over 9}$$
B
$${1 \over 8} < {I^2} < {1 \over 4}$$
C
$${1 \over 9} < {I^2} < {1 \over 8}$$
D
$${1 \over 6} < {I^2} < {1 \over 2}$$
3
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
4
JEE Main 2020 (Online) 8th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
If a line, y = mx + c is a tangent to the circle, (x – 3)2 + y2 = 1 and it is perpendicular to a line L1, where L1 is the tangent to the circle, x2 + y2 = 1 at the point $$\left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$$, then :
A
c2 + 6c + 7 = 0
B
c2 - 7c + 6 = 0
C
c2 – 6c + 7 = 0
D
c2 + 7c + 6 = 0
JEE Main Papers
2023
2021
EXAM MAP
Medical
NEET
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
CBSE
Class 12