1
JEE Main 2020 (Online) 4th September Morning Slot
+4
-1
If 1+(1–22.1)+(1–42.3)+(1-62.5)+......+(1-202.19)= $$\alpha$$ - 220$$\beta$$,
then an ordered pair $$\left( {\alpha ,\beta } \right)$$ is equal to:
A
(11, 103)
B
(10, 103)
C
(10, 97)
D
(11, 97)
2
JEE Main 2020 (Online) 4th September Morning Slot
+4
-1
Let x0 be the point of Local maxima of $$f(x) = \overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)$$, where
$$\overrightarrow a = x\widehat i - 2\widehat j + 3\widehat k$$, $$\overrightarrow b = - 2\widehat i + x\widehat j - \widehat k$$, $$\overrightarrow c = 7\widehat i - 2\widehat j + x\widehat k$$. Then the value of
$$\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a$$ at x = x0 is :
A
14
B
-30
C
-4
D
-22
3
JEE Main 2020 (Online) 4th September Morning Slot
+4
-1
The integral $$\int {{{\left( {{x \over {x\sin x + \cos x}}} \right)}^2}dx}$$ is equal to
(where C is a constant of integration):
A
$$\sec x - {{x\tan x} \over {x\sin x + \cos x}} + C$$
B
$$\sec x + {{x\tan x} \over {x\sin x + \cos x}} + C$$
C
$$\tan x - {{x\sec x} \over {x\sin x + \cos x}} + C$$
D
$$\tan x + {{x\sec x} \over {x\sin x + \cos x}} + C$$
4
JEE Main 2020 (Online) 4th September Morning Slot
+4
-1
A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If $$\angle BAC = {90^o}$$ and area$$\left( {\Delta ABC} \right) = 5\sqrt 5$$ s units, then the abscissa of the vertex C is :
A
$$1 + 2\sqrt 5$$
B
$$2\sqrt 5 - 1$$
C
$$1 + \sqrt 5$$
D
$$2 + \sqrt 5$$
EXAM MAP
Joint Entrance Examination