Let A = { 10, 11, 12, 13, 14 }; B = { 0, 1, 2, 3, 5 } and $f i : A \to B, i = 1, 2, 3$ . State the type of function for the following (give reason):
(i) $f_{1}$ = { (10, 1), (11, 2), (12, 3), (13, 5), (14, 3) }
(ii) $f_{2}$ = { (10, 1), (11, 1), (12, 1), (13, 1), (14, 1) }
(iii) $f_{3}$ = { (10, 0), (11, 1), (12, 2), (13, 3), (14, 5) }

Open in App

Solution

$(i) 12$ and $14$ have the same image hence $f_{1}$ is many-one,
also $d o m a i n \neq c o d o m a i n ∴ f$ is not onto.
$(i i)$ Every element has the same image $1$ so $f_{2}$ is a constant function. It is not onto.
$(i i i)$ Every element has a unique image and $d o m a i n = c o d o m a i n$ hence $f_{3}$ is one-one and onto.

Suggest Corrections

Similar questions

Q. Find the value of the given expression:

\frac{1}{4 - \sqrt{15}} - \frac{1}{\sqrt{15} - \sqrt{14}} + \frac{1}{\sqrt{14} - \sqrt{13}} - \frac{1}{\sqrt{13} - \sqrt{12}} + \frac{1}{\sqrt{12} - \sqrt{11}} - \frac{1}{\sqrt{11} - \sqrt{10}} + \frac{1}{\sqrt{10} - 3}

Prove that:
(i) $\frac{1}{3 + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + 1} = 1$
(ii) $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} = 2$

Q. Let

i^{2} = - 1

, then

(i^{10} - \frac{1}{i^{11}}) + (i^{11} - \frac{1}{i^{12}}) + (i^{12} - \frac{1}{i^{13}}) + (i^{13} - \frac{1}{i^{14}}) + (i^{14} + \frac{1}{i^{15}})

is equal to

Q. Which of the following relations are functions? Give reasons.

If it is a function determine its domain and range
(i)

{(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

(ii)

{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

(iii)

{(1, 3), (1, 5), (2, 5)}

∣ ∣ ∣ ∣ \begin{matrix} 10! & 11! & 12! 11! & 12! & 13! 12! & 13! & 14! \end{matrix} ∣ ∣ ∣ ∣ =

__________.

Join BYJU'S Learning Program

(i) 12 and 14 have the same image hence f1 is many-one, also domain≠codomain ∴f is not onto.(ii) Every element has the same image 1 so f2 is a constant function. It is not onto.(iii) Every element has a unique image and domain=codomain hence f3 is one-one and onto.