1
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
2
JEE Main 2020 (Online) 6th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
The negation of the Boolean expression p $$ \vee $$ (~p $$ \wedge $$ q) is equivalent to :
A
$$p \wedge \sim q$$
B
$$ \sim $$$$p \vee \sim q$$
C
$$ \sim p \wedge q$$
D
$$ \sim p \wedge \sim q$$
3
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
$$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$
4
JEE Main 2020 (Online) 6th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
Let L1 be a tangent to the parabola y2 = 4(x + 1)
and L2 be a tangent to the parabola y2 = 8(x + 2)
such that L1 and L2 intersect at right angles. Then L1 and L2 meet on the straight line :
A
x + 3 = 0
B
x + 2y = 0
C
x + 2 = 0
D
2x + 1 = 0
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