Answer with explanation:
a. f(x)=3x-4
Let [tex]f(x_1)=f(x_2)[/tex]
[tex]3x_1-4=3x_2-4[/tex]
[tex]3x_1=3x_2-4+4[/tex]
[tex]3x_1=3x_2[/tex]
[tex]x_1=x_2[/tex]
Hence, the function one-one.
Let f(x)=y
[tex]y=3x-4[/tex]
[tex]3x=y+4[/tex]
[tex]x=\frac{y+4}{3}[/tex]
We can find pre image in domain R for every y in range R.
Hence, the function onto.
b.g(x)=[tex]x^2-2[/tex]
Substiute x=1
Then [tex]g(x)=1-2=-1[/tex]
Substitute x=-1
Then g(x)=1-2=-1
Hence, the image of 1 and -1 are same . Therefore, the given function g(x) is not one-one.
The given function g(x) is not onto because there is no pre image of -2, -3,-4...... R.
Hence, the function neither one-one nor onto on given R.
c.[tex]h(x)=\frac{2}{x}[/tex]
The function is not defined for x=0 .Therefore , it is not a function on domain R.
Let [tex]h(x_1)=h(x_2)[/tex]
[tex] \frac{2}{x_1}=\frac{2}{x_2}[/tex]
By cross mulitiply
[tex]x_1= \frac{2\times x_2}{2}[/tex]
[tex]x_1=x_2[/tex]
Hence, h(x) is a one-one function on R-{0}.
We can find pre image for every value of y except zero .Hence, the function
h(x) is onto on R-{0}.
Therefore, the given function h(x) is both one- one and onto on R-{0} but not on R.
d.k(x)= ln(x)
We know that logarithmic function not defined for negative values of x. Therefore, logarithmic is not a function R.Hence, the given function K(x) is not a function on R.But it is define for positive R.
Let[tex]k(x_1)=k(x_2)[/tex]
[tex] ln(x_1)=ln(x_2)[/tex]
Cancel both side log then
[tex]x_1=x_2[/tex]
Hence, the given function one- one on positive R.
We can find pre image in positive R for every value of [tex]y\in R^+[/tex].
Therefore, the function k(x) is one-one and onto on [tex]R^+[/tex] but not on R.
e.l(x)=[tex]e^x[/tex]
Using horizontal line test if we draw a line y=-1 then it does not cut the graph at any point .If the horizontal line cut the graph atmost one point the function is one-one.Hence, the horizontal line does not cut the graph at any point .Therefore, the function is one-one on R.
If a horizontal line cut the graph atleast one point then the function is onto on a given domain and codomain.
If we draw a horizontal line y=-1 then it does not cut the graph at any point .Therefore, the given function is not onto on R.
