1
JEE Main 2019 (Online) 11th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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)
2
JEE Main 2019 (Online) 11th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If  $$\left| {\matrix{ {a - b - c} & {2a} & {2a} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|$$

      = (a + b + c) (x + a + b + c)2, x $$ \ne $$ 0,

then x is equal to :
A
–2(a + b + c)
B
2(a + b + c)
C
abc
D
–(a + b + c)
3
JEE Main 2019 (Online) 11th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If   $$\int {{{x + 1} \over {\sqrt {2x - 1} }}} \,dx$$ = f(x) $$\sqrt {2x - 1} $$ + C, where C is a constant of integration, then f(x) is equal to :
A
$${2 \over 3}$$ (x $$-$$ 4)
B
$${1 \over 3}$$ (x + 4)
C
$${1 \over 3}$$ (x + 1)
D
$${2 \over 3}$$ (x + 2)
4
JEE Main 2019 (Online) 11th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
$$\mathop {\lim }\limits_{x \to 0} {{x\cot \left( {4x} \right)} \over {{{\sin }^2}x{{\cot }^2}\left( {2x} \right)}}$$ is equal to :
A
0
B
4
C
1
D
2

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