1
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If $${{3 + i\sin \theta } \over {4 - i\cos \theta }}$$, $$\theta $$ $$ \in $$ [0, 2$$\theta $$], is a real number, then an argument of
sin$$\theta $$ + icos$$\theta $$ is :
A
$$\pi - {\tan ^{ - 1}}\left( {{3 \over 4}} \right)$$
B
$$ - {\tan ^{ - 1}}\left( {{3 \over 4}} \right)$$
C
$${\tan ^{ - 1}}\left( {{4 \over 3}} \right)$$
D
$$\pi - {\tan ^{ - 1}}\left( {{4 \over 3}} \right)$$
2
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If $$\theta $$1 and $$\theta $$2 be respectively the smallest and the largest values of $$\theta $$ in (0, 2$$\pi $$) - {$$\pi $$} which satisfy the equation,
2cot2$$\theta $$ - $${5 \over {\sin \theta }}$$ + 4 = 0, then
$$\int\limits_{{\theta _1}}^{{\theta _2}} {{{\cos }^2}3\theta d\theta } $$ is equal to :
A
$${\pi \over 9}$$
B
$${{2\pi } \over 3}$$
C
$${{\pi } \over 3}$$
D
$${\pi \over 3} + {1 \over 6}$$
3
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let y = y(x) be a function of x satisfying

$$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} $$ where k is a constant and

$$y\left( {{1 \over 2}} \right) = - {1 \over 4}$$. Then $${{dy} \over {dx}}$$ at x = $${1 \over 2}$$, is equal to :
A
$${2 \over {\sqrt 5 }}$$
B
$$ - {{\sqrt 5 } \over 2}$$
C
$${{\sqrt 5 } \over 2}$$
D
$$ - {{\sqrt 5 } \over 4}$$
4
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let y = y(x) be the solution curve of the differential equation,

$$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$$, satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is :
A
2 + e
B
-e
C
2
D
2 - e
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