1
JEE Main 2020 (Online) 6th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The area (in sq. units) of the region
A = {(x, y) : |x| + |y| $$ \le $$ 1, 2y2 $$ \ge $$ |x|}
A
$${1 \over 6}$$
B
$${5 \over 6}$$
C
$${1 \over 3}$$
D
$${7 \over 6}$$
2
JEE Main 2020 (Online) 6th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The general solution of the differential equation

$$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} $$ + xy$${{dy} \over {dx}}$$ = 0 is :

(where C is a constant of integration)
A
$$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$$
B
$$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$$
C
$$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$
D
$$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$
3
JEE Main 2020 (Online) 6th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The region represented by
{z = x + iy $$ \in $$ C : |z| – Re(z) $$ \le $$ 1} is also given by the
inequality : {z = x + iy $$ \in $$ C : |z| – Re(z) $$ \le $$ 1}
A
y2 $$ \le $$ $$2\left( {x + {1 \over 2}} \right)$$
B
y2 $$ \le $$ $${x + {1 \over 2}}$$
C
y2 $$ \ge $$ 2(x + 1)
D
y2 $$ \ge $$ x + 1
4
JEE Main 2020 (Online) 6th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let a , b, c , d and p be any non zero distinct real numbers such that
(a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :
A
a, c, p are in G.P.
B
a, b, c, d are in G.P.
C
a, b, c, d are in A.P.
D
a, c, p are in A.P.
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