Important Key points

Example 1.6
Let X = {1,2, 3, 4} and Y = {2, 4,6, 8,10} and R = {(1,2),(2,4),(3,6),(4,8)}. Show that R is a function and find its domain, co-domain and range?

Solution :
To prove R is a function, all elements in X have only one image in Y.
From R = {(1,2),(2,4),(3,6),(4,8)}, we see that every element of X that are 1,2,3,4 related with 2,4,6,8 that are in Y.
Therefore R is a function.
Domain X = {1,2,3,4}; Co-domain Y = {2,4,6,8,10}; Range of f = {2,4,6,8}.

Example 1.7
A relation f :X →Y is defined by f (x) =$x^2$ −2 where, X = {−2,−1, 0, 3} and Y = R(Real Numbers).
(i) List the elements of f
(ii) Is f a function?

Solution :
f (x) =$x^2$ −2 where, X = {−2,−1, 0, 3}
(i) $\hspace{0.70cm}$f (-2) = $(−2)^2$ – 2 = 4 - 2 = 2 ;
$\hspace{1cm}$f (-1) = $(−1)^2$ – 2 = 1 - 2 = −1
$\hspace{1cm}$f (0) = $(0)^2$ – 2 = 0 - 2 = −2 ;
$\hspace{1cm}$f (3) = $(3)^2$ – 2 = 9 - 2 =7
∴ f = {(−2,2),(−1,−1),(0,−2),(3,7)}
(ii) We note that each element in the domain of f has a unique image. Therefore, f is a function.

Example 1.8
If X = {–5,1,3,4} and Y = {a,b,c}, then which of the following relations are functions from X to Y ?
(i) R1 = {(–5,a), (1,a), (3,b)}
(ii) R2 = {(–5,b), (1,b), (3,a),(4,c)}
(iii) R3 = {(–5,a), (1,a), (3,b),(4,c),(1,b)}

Solution :
(i) To prove $R_1$ is a function, all elements in X have only one image in Y.
$R_1$ = {(–5,a), (1,a), (3,b)}
We may represent the relation $R_1$ in an arrow diagram
*
$R_1$ is not a function as 4 $\in$ X does not have an image in Y.
(ii) To prove $R_2$ is a function, all elements in X have only one image in Y.
$R_2$ = {(–5,b), (1,b), (3,a),(4,c)}
Arrow diagram of $R_2$ is shown in Fig.
*
$R_2$ is a function as each element of X has an unique image in Y.
(iii) To prove $R_3$ is a function, all elements in have only one image in Y.
$R_3$ = {(–5,a), (1,a), (3,b),(4,c),(1,b)}
Representing $R_3$ in an arrow diagram (Fig.1.15(c)).
$R_3$ is not a function as 1 $\in$ X has two images a$\in$Y and b$\in$Y .
Note that the image of an element should always be unique.

Example 1.9
Given f (x) = $2x – x^2$ ,
find (i)f (1) (ii) f (x+1) (iii) f (x) + f (1)

Solution :
(i) x = 1, we get
$\hspace{0.5cm}$f (1) = 2(1) – $(1)^2$ =2 – 1 =1
(ii) x = x+1, we get
$\hspace{0.5cm}$f(x+1) = 2(x +1) – $(x +1)^2$ =2x +2 – ($x^2$ +2x +1) = −$x^2$ +1
(iii) f (x) + f (1) = (2x – $x^2$)+1 = −$x^2$ +2x +1

Exercise 1.3

1. Let f = {(x, y)|x, y ∈ N and y = 2x} be a relation on N. Find the domain, co-domain and range. Is this relation a function?

Solution :
F = {(x, y)|x, y ∈ N and y = 2x}
x = {1, 2, 3,…}
y = {2$\times$1, 2$\times$2, 2$\times$3, 2$\times$5, 2$\times$5 …}
R = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10),…}
Domain of R = {1, 2, 3, 4,…}(Note : First coordinates of R)
Co-domain = {1, 2, 3…..}
Range of R = {2, 4, 6, 8, 10,…} (Note : Second coordinates of R)
Yes, Since every element of Domain have a unique image in Co-domain. Thus this relation is a function.

2. Let X = {3, 4, 6, 8}. Determine whether the relation R = {(x, f(x))|x ∈ X, f(x) = $x^2$ + 1} is a function from X to N ?

Solution :
x = {3,4, 6, 8}
R = ((x, f(x))|x ∈ X, f(x) = X2 + 1}
f(x) = $x^2$ + 1
f(3) = $3^2$ + 1 = 10
f(4) = $4^2$ + 1 = 17
f(6) = $6^2$ + 1 = 37
f(8) = $8^2$ + 1 = 65
R = {(3, 10), (4, 17), (6, 37), (8, 65)}
Yes, Since Each element in X have a unique image in N(Co-domain). Thus R is a function from X to N.

3. Given the function
f : x → $x^2$ – 5x + 6, evaluate
(i) f(-1)
(ii) f(2 a)
(iii) f(2)
(iv) f(x – 1)

Solution :
Take f(x) = $x^2$ – 5x + 6
(i) f (-1) = $(-1)^2$ – 5 (-1) + 6
$\hspace{1.2cm}$= 1 + 5 + 6
$\hspace{1.2cm}$= 12
(ii) f (2a) = $(2a)^2$ – 5 (2a) + 6
$\hspace{1.2cm}$= 4$a^2$ – 10a + 6
(iii) f(2) = $2^2$ – 5(2) + 6
$\hspace{1.2cm}$= 4 – 10 + 6
$\hspace{1.2cm}$= 0
(iv) f(x – 1) = $(x – 1)^2$ – 5 (x – 1) + 6
$\hspace{1.5cm}$= $x^2$ – 2x + 1 – 5x + 5 + 6
$\hspace{1.5cm}$= $x^2$ – 7x + 12

4. A graph representing the function f(x) is given in figure it is clear that f(9) = 2.
(i) Find the following values of the function
(a) f(0)
(b) f(7)
(c) f(2)
(d) f(10)
(ii) For what value of x is f (x) = 1?
(iii) Describe the following
(i) Domain
(ii) Range.
(iv) What is the image of 6 under f?

Solution :
(i)From the graph
(a) f(0) = 9
(b) f(7) = 6
(c) f(2) = 6
(d) f(10) = 0
(ii) At x = 9.5, f(x) = 1
(iii) Domain = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
$\hspace{1.0cm}$= {x |0 < x < 10, x ∈ R}
Range = {x|0 < x < 9, x ∈ R}
$\hspace{1.0cm}$= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(iv) The image of 6 under f is 5.

5. Let f(x) = 2x + 5. If x ≠ 0 then find $\Large \frac{f(x+2)-f(2}{x}$.

Solution :
Given f(x) = 2x + 5, x ≠ 0.
To find f(x+2):
f(x) = 2x + 5 ;
f(x+2) = 2(x+2)+ 5 ; (Replaced 'x' by 'x+2' )
$\hspace{1cm}$= 2x +4 +5
$\hspace{1cm}$=2x + 9
To find f(2) :
f(x) = 2x + 5 ;
f(2)= 2(2) + 5
$\hspace{0.5cm}$=4+5
$\hspace{0.5cm}$=9
To find $\Large \frac{f(x+2)-f(2}{x}$ = $\Large \frac{2x+9-9}{x}$
$\hspace{1.1in}$= $\Large \frac{2x}{x}$ = 2

6. A function f is defined by f(x) = 2x – 3
(i) find $\large\frac{f(0)+f(1)}{2}$
(ii) find x such that f(x) = 0.
(iii) find x such that f(x) = x.
(iv) find x such that f(x) = f(1 – x).

Solution :
Given f(x) = 2x – 3
(i) find $\large\frac{f(0)+f(1)}{2}$
f(0) = 2(0) – 3 = -3
f(1) = 2(1) – 3 = -1
$\large\frac{f(0)+f(1)}{2}$ = $\large\frac{-3-1}{2}$ = $\large\frac{-4}{2}$ = 2

(ii) f(x) = 0
⇒ 2x – 3 = 0
2x = 3
x = $\frac{3}{2}$

(iii) f(x) = x
⇒ 2x – 3 = x ⇒ 2x – x = 3
x = 3

(iv) f(x) = f(1 – x)
2x – 3 = 2(1 – x) – 3
2x – 3 = 2x – 2x – 3
2x + 2x = 2 – 3 + 3
4x = 2
x = $\frac{2}{4}$
x = $\frac{1}{2}$

7. An open box is to be made from a square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown in figure. Express the volume V of the box as a function of x.

Solution :
Volume of the box = Volume of the cuboid
= l × b × h cu. units
Here l = 24 – 2x, b = 24 – 2x, h = x
∴$\hspace{0.2cm}$ V = (24 – 2x) (24 – 2x) × (x)
$\hspace{0.5cm}$= (576 – 48x – 48x + 4$x^2$)x
$\hspace{0.5cm}$= 4$x^3$ – 96$x^2$ + 576x

8. A function f is defined bv f(x) = 3 – 2x . Find x such that $f(x^2$) = $(f(x))^2$.

Solution :
f(x) = 3 – 2x
$f(x^2) = 3 – 2x^2$
$f(x)^2 = (3 – 2x)^2 = 9-12x+4x^2$
Given $f(x^2$) = $(f(x))^2$.
$\Rightarrow$ $3 – 2x^2$ = $9-12x+4x^2$
$6x^2-12x+6=0$
$x^2-2x+1=0$
$(x-1)(x-1)=0$
$x=1,1$

9. A plane is flying at a speed of 500 km per hour. Express the distance d travelled by the plane as function of time r in hours.

Solution :
Speed of the plane = 500 km/hr
Distance travelled in “t” hours = 500 × t (distance = speed × time) = 500 t

10. The data in the adjacent table depicts the length of a woman’s forehand and her corresponding height. Based on this data, a student finds a relationship between the height (y) and the forehand length(x) as y = ax + b, where a, b are constants.
(i) Check if this relation is a function.
(ii) Find a and b.
(iii) Find the height of a woman whose forehand length is 40 cm.
(iv) Find the length of forehand of a woman if her height is 53.3 inches.

Solution :
(i) Given y = ax + b …………. (1)
The ordered pairs are R = {(35, 56) (45, 65) (50, 69.5) (55, 74)}
∴ Hence this relation is a function.

(ii) Consider any two ordered pairs (35,56), (45, 65) Substituting in (1) and by solving (by elimination method) we get a = 0.9 b = 24.5
∴ y = 0.9x + 24.5 ...........(2)

(iii) Given x = 40 , y = ?
∴ (2) → y = 0.9 (40) + 24.5
⇒ y = 36 + 24.5
⇒ y = 60.5 inches

(iv) Given y = 53.3 inches, x = ?
substituting in eqn (2) → 53.3 = 0.9x + 24.5
⇒ 53.3 – 24.5 = 0.9x
⇒ 28.8 = 0.9x
⇒ x = 28.80.9 = 32 cm
∴ When y = 53.3 inches, x = 32 cm