1
JEE Main 2020 (Online) 3rd September Morning Slot
+4
-1
If $$\alpha$$ and $$\beta$$ are the roots of the equation
x2 + px + 2 = 0 and $${1 \over \alpha }$$ and $${1 \over \beta }$$ are the
roots of the equation 2x2 + 2qx + 1 = 0, then
$$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right)$$ is equal to :
A
$${9 \over 4}\left( {9 - {q^2}} \right)$$
B
$${9 \over 4}\left( {9 + {q^2}} \right)$$
C
$${9 \over 4}\left( {9 - {p^2}} \right)$$
D
$${9 \over 4}\left( {9 + {p^2}} \right)$$
2
JEE Main 2020 (Online) 2nd September Evening Slot
+4
-1
Let f(x) be a quadratic polynomial such that
f(–1) + f(2) = 0. If one of the roots of f(x) = 0
is 3, then its other root lies in :
A
(–3, –1)
B
(1, 3)
C
(–1, 0)
D
(0, 1)
3
JEE Main 2020 (Online) 2nd September Morning Slot
+4
-1
Let $$\alpha$$ and $$\beta$$ be the roots of the equation
5x2 + 6x – 2 = 0. If Sn = $$\alpha$$n + $$\beta$$n, n = 1, 2, 3...., then :
A
5S6 + 6S5 = 2S4
B
5S6 + 6S5 + 2S4 = 0
C
6S6 + 5S5 + 2S4 = 0
D
6S6 + 5S5 = 2S4
4
JEE Main 2020 (Online) 9th January Evening Slot
+4
-1
Let a, b $$\in$$ R, a $$\ne$$ 0 be such that the equation, ax2 – 2bx + 5 = 0 has a repeated root $$\alpha$$, which is also a root of the equation, x2 – 2bx – 10 = 0. If $$\beta$$ is the other root of this equation, then $$\alpha$$2 + $$\beta$$2 is equal to :
A
28
B
24
C
26
D
25
JEE Main Subjects
Physics
Mechanics
Electricity
Optics
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
Mathematics
Algebra
Trigonometry
Coordinate Geometry
Calculus
EXAM MAP
Joint Entrance Examination