A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?
Interval | Function |
(-∞, ∞) | x3 – 3x2 + 3x + 3 |
Interval | Function |
[2, ∞) | 2x3 – 3x2 – 12x + 6 |
Interval | Function |
(-∞, 1/3] | 3x2 – 2x + 1 |
Interval | Function |
(-∞, 1/3] | 3x2 – 2x + 1 |
A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to
–f(x)
f(x)
f(a) + f(a – x)
f(a) + f(a – x)
Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is
a function
reflexive
not symmetric
not symmetric
The range of the function 7-xPx-3 is
{1, 2, 3}
{1, 2, 3, 4, 5}
{1, 2, 3, 4}
{1, 2, 3, 4}
Let f and g be differentiable functions satisfying g′(a) = 2, g(a) = b and fog = I (identity function). Then f ′(b) is equal to
1/2
2
2/3
2/3
Let S = {t ∈ R: f(x) = |x-π|.(e|x| - 1) sin |x| is not differentiable at t}. Then the set S is equal to
{0,π}
{0}
{π}
Let S = { x ∈ R : x ≥ 0 and Then S:
Contains exactly four elements
Is an empty set
Contains exactly one element
Contains exactly two elements
On the set R of real numbers we define xPy if and only if . Then, the relation P is
reflexive but not symmetric
symmetric but not reflexive
transitive but not reflexive
reflexive and symmetric but not transitive
D.
reflexive and symmetric but not transitive
For every real number x,
Hence, P is reflective.
Now, let (x, y) P
Hence, P is symmetric.
Again, (- 1, 0) P and P
But, (- 1, 2) P
as (- 1, 2) = - 2 < 0
P is not transitive.
On R, the relation p be defined by 'xy holds if and only if x- y is zero or irrational'. Then,
s reflexive and transitive but not symmetric
s reflexive and symmetric but not transitive
s symmetric and transitive but not reflexive
is equivalence relation