Let A B and be two non-empty sets. Then, a function f from set A to Bis a rule which associates elements of set A to elements of B such that all elements of set A are associated to elements of set B in unique way.
If f associates x $\in$ A to y $\in$ B, then we say that y is the image of the element x and denote it by f(x) and write as y = f(x). The element x is called the pre-image or inverse image in B.
A function is denoted by f : A → B or A $\overset{f}{\rightarrow}$ B.
Note
- There may exist some elements in set B which are not the images of any element in set A.
- To each and every independent element in Athere corresponds one and only one image in B.
- Every function is a relation but every relation may or may not be a function.
- The number of functions from a finite set A into finite set B = $[n(B)]^{[n(A)]}$
Difference between a Relation and a Function
Geometrically, if we draw a vertical parallel line (VPL) i.e., any line which is parallel to y-axis (x = a), then if this line intersects the graph of the expression in more than one point, then the expression is a relation else if it intersects at only one point, the expression is a function, e.g.,
Domain, Codomain and Range of a Function
Let f : A → B, then A is known as domain of f while B is known as codomain of f.
Also, set f(A) = {f(x) : x $\in$ A} is known as range of f.
Clearly, f(A) $\subseteq$ B
e.g., Let A = {1, –1, 2, –2}, B = {1, 4, 9}
f : A $\overset{x^2}{\rightarrow}$ B i.e f(x) = $x^2$
Also, codomain of function = {1, 4, 9}.
Sample Problem 1
Find the domain and range of the
function f(x) = $\sqrt{9-x^2}$. [NCERT]
(a) [– 3, 3], [0, 3]
(b) [-3, 3), [0, 3]
(c) (– 3, 3), (0, 3)
(d) None of these
Answer: (a)
Given, f(x) = $\sqrt{9-x^2}$
For function to be defined,
9 - $x^2$ $\geqslant$ 0
⇒ $x^2$ - 9 $\leqslant$ 0 (*)
⇒ {x - (x - 3)} (x - 3) $\leqslant$ 0
⇒ -3 $\leqslant$ x $\leqslant$ 3
{if a < b and (x - a) (x - b) $\leqslant$ ⇒ a $\leqslant$ x $\leqslant$ b}
⇒ Domain = [-3, 3]
or Domain = {x: x $\in$ R and -3 $\leqslant$ x $\leqslant$ 3}
Let f(x) = y, then
⇒ y = $\sqrt{9-x^2}$
On squaring both sides, we get
$y^2$ = $\sqrt{9-x^2}$
⇒ $x^2$ = $\sqrt{9-y^2}$ (**)
On putting the value of $x^2$ in Eq. (*), we get
⇒ 9 - $y^2$ - 9 $\leqslant$ 0
⇒ -$y^2$ $\leqslant$ 0
⇒ $y^2$ $\geqslant$ 0
- $\infty$ < y < $\infty$
But y cannot be negative f(x) = $\sqrt{9-x^2}$ cannot contain negative values.
So, y $\geqslant$ 0
Also, from Eq. (**), x = $\sqrt{9 - y^2}$
⇒ 9 - $y^2$ $\geqslant$ 0
⇒ $y^2$ - 9 $\leqslant$ 0
-3 $\leqslant$ 0 y $\leqslant$ 3
⇒ Range = [0, 3] or {y:y $\in$ R, 0 $\leqslant$ y $\leqslant$ 3}
Sample Problem 2
Let f = {(1, 1), (2, 3), (0, -1), (-1, -3)} be
a function from Z to Z defined by f(x) = ax + b for some integers
a and b. Find, a, b.
(a) 1, 2
(b) 2, –1
(c) –1, –2
(d) None of these
Answer: (b)
Given, f = {(1, 1), (2, 3), (0, -1), (-1, -3)}
and f(x) = ax + b or y = ax + b
at x = 1, y = 1
⇒ 1 = a + b (*)
at, x = 2, y = 3
⇒ 3 = 2a + b (**)
On subtracting Eq. (*) from Eq. (**), we get a = 2
On putting the value of a in Eq. (*), we get
b = 1 - 2 ⇒ b = -1
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