Test: Functions

The correct option is B

Since the function is defined from 

XY.So the domain will be the set X which is {1,2,3,4}. Codomain is  what may possibly come out of a function. So codomain is the whole set Y which is {1,2,3,...,20}. Range is The range is the set of possible output values which will be the values in the set of Y which have a pre-image. That is, the value of squares. So the range is {1,4,,9,16}. 

A relation f from a set A to a set B is said to be function if every element of set A has one and only one image in set B. In other words, a function f is a relation such that no two pairs in the relation has the same first element. The notation f : X →Y means that f is a function from X to Y.

4 - x² = 0
=> (2+x)(2-x) = 0
x = +-2
Now select a test point, let it be x = 0. Then y = (4 - (0)²)^1/2 = 2, so the function is defined on [-2,2]
Thus, the graph of y = (4 - x²)^1/2 is a semicircle with radius 2 and domain is [-2,2]

X = {1, 2, 3, 4}
Y = {1, 2, …, 10}
f: X→ Y be defined by
f(x) = 2x + 1
f (1) = 3
f (2) = 5
f (3) = 7
f (4) = 9
Range(f) = {3, 5, 7, 9}

 f(x) = x2 – x + 1; g(x) = 7x – 3
(f+g)(x) = (x2 - x + 1 + 7x - 3)
=(x² - x + 7x + 1 - 3)
= x² + 6x - 2
(f+g)(3) = x² + 6x - 2
= (3)² + 6(3) - 2
= 9 + 18 - 2
= 25

A function is a relation in which no two ordered pairs have the same first element. A function associates each element in its domain with one and only one element in its range. R={(1,2), (1,4)(3,1),(5,1)} is not  a function because all the first elements are not different.

Option A has value b & c for 2.
Option C has value b & c for 1.
Option D has value a & c for 1.

Example: y = x². 
If x = 2, y = 4. 
If x = -2, y = 4. So there you have a many-to-one relation and it is DEFINITELY a function.
Now, a one-to-many relation is NOT a function.

 f(x) = 3x/2x-8
denominator can't be zero so
2x - 8 not equal to 0
x should not equal to 4
we have domain R - {4}