1
JEE Main 2020 (Online) 6th September Evening Slot
+4
-1
For a suitably chosen real constant a, let a

function, $$f:R - \left\{ { - a} \right\} \to R$$ be defined by

$$f(x) = {{a - x} \over {a + x}}$$. Further suppose that for any real number $$x \ne - a$$ and $$f(x) \ne - a$$,

(fof)(x) = x. Then $$f\left( { - {1 \over 2}} \right)$$ is equal to :
A
$${1 \over 3}$$
B
–3
C
$$- {1 \over 3}$$
D
3
2
JEE Main 2020 (Online) 6th September Morning Slot
+4
-1
If f(x + y) = f(x)f(y) and $$\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2$$ , x, y $$\in$$ N, where N is the set of all natural number, then the value of $${{f\left( 4 \right)} \over {f\left( 2 \right)}}$$ is :
A
$${2 \over 3}$$
B
$${1 \over 9}$$
C
$${1 \over 3}$$
D
$${4 \over 9}$$
3
JEE Main 2020 (Online) 5th September Morning Slot
+4
-1
A survey shows that 73% of the persons working in an office like coffee, whereas 65% like tea. If x denotes the percentage of them, who like both coffee and tea, then x cannot be :
A
63
B
36
C
54
D
38
4
JEE Main 2020 (Online) 5th September Morning Slot
+4
-1
If the minimum and the maximum values of the function $$f:\left[ {{\pi \over 4},{\pi \over 2}} \right] \to R$$, defined by
$$f\left( \theta \right) = \left| {\matrix{ { - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \cr { - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \cr {12} & {10} & { - 2} \cr } } \right|$$ are m and M respectively, then the ordered pair (m,M) is equal to :
A
$$\left( {0,2\sqrt 2 } \right)$$
B
(-4, 0)
C
(-4, 4)
D
(0, 4)
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