Solution to Find out which of the following functions from R to R are (i) One-to-one, (ii) … - Sikademy
Author Image

Archangel Macsika

Find out which of the following functions from R to R are (i) One-to-one, (ii) Onto, (iii) One-to-one corre￾spondence. (a) f: R—>R defined by f(x) = x (b) f: R—>R defined by f(x) = |x| (c) f: R—>R defined by f(x) = x + 1 (d) f: R—>R defined by f(x) = x^2 (e) f: R—>R defined by f(x) = x^3 (f) f: R—>R defined by f(x) = x – x^2 (g) f: R—>R defined by f(x) = Floor(x) (h) f: R—>R defined by f(x) = Ceiling(x) (i) f: R—>R defined by f(x) = – 3x+4 (j) f: R—>R defined by f(x)= – 3x^2 +7

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

(a)

If y_1=y_2, then x_1=x_2=> One-to-one.

For every y\in \R exists x\in \R such that x=y=> Onto

f(x)=x is One-to-one correspondence (a bijection).


(b)

|-2|=2=|2|

If y=-2, then there is no x\in \R such that |x|=-2.

f(x)=|x| is neither One-to-one nor Onto.


(c)

If y_1=y_2, then x_1+1=x_2+1=>x_1=x_2=> One-to-one.

For every y\in \R exists x\in \R such that x=y-1=> Onto

f(x)=x+1 is One-to-one correspondence (a bijection).


(d)


(-2)^2=4=(2)^2


If y=-4, then there is no x\in \R such that x^2=-4.

f(x)=x^2 is neither One-to-one nor Onto.


(e)

If y_1=y_2, then x_1^3=x_2^3=>(x_1-x_2)(x_1^2+x_1x_2+x_2^2)=0

=>x_1=x_2=> One-to-one.

For every y\in \R exists x\in \R such that x=\sqrt[3]{y}=> Onto

f(x)=x^3 is One-to-one correspondence (a bijection).


(f)


0-(0)^2=0=1-(1)^2

If y=1, then there is no x\in \R such that x-x^2=1.

f(x)=x-x^2 is neither One-to-one nor Onto.


(g)


Floor(0.2)=0=Floor(0.5)


If y=0.5, then there is no x\in \R such that Floor(x)=0.5.

f(x)=Floor(x) is neither One-to-one nor Onto.


(h)


Ceiling(0.2)=1=Ceiling(0.5)


If y=0.5, then there is no x\in \R such that Ceiling(x)=0.5.

f(x)=Ceiling(x) is neither One-to-one nor Onto.


(i)

If y_1=y_2, then -3x_1+4=-3x_2+4=>-3x_1=-3x_2

=>x_1=x_2=> One-to-one.

For every y\in \R exists x\in \R such that x=\dfrac{-y+4}{3}=> Onto

f(x)=-3x+4 is One-to-one correspondence (a bijection).


(j)


-3(-1)^2+7=4=-3(1)^2+7


If y=8, then there is no x\in \R such that x^2=8.

f(x)=-3x^2+7 is neither One-to-one nor Onto.


Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-5-stid-8-sqid-2836-qpid-1393