1
JEE Main 2019 (Online) 12th January Evening Slot
+4
-1
If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, y) on it as $${{{x^2} - 2y} \over x}$$, then the curve also passes through the point :
A
(–1, 2)
B
$$\left( { - \sqrt 2 ,1} \right)$$
C
$$\left( { \sqrt 3 ,0} \right)$$
D
(3, 0)
2
JEE Main 2019 (Online) 12th January Morning Slot
+4
-1
Let y = y(x) be the solution of the differential equation, x$${{dy} \over {dx}}$$ + y = x loge x, (x > 1). If 2y(2) = loge 4 $$-$$ 1, then y(e) is equal to :
A
$$- {e \over 2}$$
B
$$- {{{e^2}} \over 2}$$
C
$${{{e^2}} \over 4}$$
D
$${e \over 4}$$
3
JEE Main 2019 (Online) 11th January Evening Slot
+4
-1
The solution of the differential equation,

$${{dy} \over {dx}}$$ = (x – y)2, when y(1) = 1, is :
A
$$-$$ loge $$\left| {{{1 + x - y} \over {1 - x + y}}} \right|$$ = x + y $$-$$ 2
B
loge $$\left| {{{2 - x} \over {2 - y}}} \right|$$ = x $$-$$ y
C
loge $$\left| {{{2 - y} \over {2 - x}}} \right|$$ = 2(y $$-$$ 1)
D
$$-$$ loge $$\left| {{{1 - x + y} \over {1 + x - y}}} \right|$$ = 2(x $$-$$ 1)
4
JEE Main 2019 (Online) 11th January Morning Slot
+4
-1
If  xloge(logex) $$-$$ x2 + y2 = 4(y > 0), then $${{dy} \over {dx}}$$ at x = e is equal to :
A
$${{\left( {1 + 2e} \right)} \over {2\sqrt {4 + {e^2}} }}$$
B
$${{\left( {1 + 2e} \right)} \over {\sqrt {4 + {e^2}} }}$$
C
$${{\left( {2e - 1} \right)} \over {2\sqrt {4 + {e^2}} }}$$
D
$${e \over {\sqrt {4 + {e^2}} }}$$
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