1
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid gives :
A
JEE Main 2019 (Online) 12th April Evening Slot Chemistry - Compounds Containing Nitrogen Question 158 English Option 1
B
JEE Main 2019 (Online) 12th April Evening Slot Chemistry - Compounds Containing Nitrogen Question 158 English Option 2
C
JEE Main 2019 (Online) 12th April Evening Slot Chemistry - Compounds Containing Nitrogen Question 158 English Option 3
D
JEE Main 2019 (Online) 12th April Evening Slot Chemistry - Compounds Containing Nitrogen Question 158 English Option 4
2
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let $$a \in \left( {0,{\pi \over 2}} \right)$$ be fixed. If the integral

$$\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx$$ = A(x) cos 2$$\alpha $$ + B(x) sin 2$$\alpha $$ + C, where C is a

constant of integration, then the functions A(x) and B(x) are respectively :
A
$$x - \alpha $$ and $${\log _e}\left| {\cos \left( {x - \alpha } \right)} \right|$$
B
$$x + \alpha $$ and $${\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|$$
C
$$x + \alpha $$ and $${\log _e}\left| {\sin \left( {x + \alpha } \right)} \right|$$
D
$$x - \alpha $$ and $${\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|$$
3
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The general solution of the differential equation (y2 – x3)dx – xydy = 0 (x $$ \ne $$ 0) is : (where c is a constant of integration)
A
y2 + 2x3 + cx2 = 0
B
y2 + 2x2 + cx3 = 0
C
y2 – 2x + cx3 = 0
D
y2 – 2x3 + cx2 = 0
4
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
The equation of common tangent to the curves y2 = 16x and xy = –4, is :
A
x – y + 4 = 0
B
x + y + 4 = 0
C
x – 2y + 16 = 0
D
2x – y + 2 = 0
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