Answer with explanation:
a. f(x)=3x-4
Let ![f(x_1)=f(x_2)](https://tex.z-dn.net/?f=f%28x_1%29%3Df%28x_2%29)
![3x_1-4=3x_2-4](https://tex.z-dn.net/?f=3x_1-4%3D3x_2-4)
![3x_1=3x_2-4+4](https://tex.z-dn.net/?f=3x_1%3D3x_2-4%2B4)
![3x_1=3x_2](https://tex.z-dn.net/?f=3x_1%3D3x_2)
![x_1=x_2](https://tex.z-dn.net/?f=x_1%3Dx_2)
Hence, the function one-one.
Let f(x)=y
![y=3x-4](https://tex.z-dn.net/?f=y%3D3x-4)
![3x=y+4](https://tex.z-dn.net/?f=3x%3Dy%2B4)
![x=\frac{y+4}{3}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7By%2B4%7D%7B3%7D)
We can find pre image in domain R for every y in range R.
Hence, the function onto.
b.g(x)=![x^2-2](https://tex.z-dn.net/?f=x%5E2-2)
Substiute x=1
Then ![g(x)=1-2=-1](https://tex.z-dn.net/?f=g%28x%29%3D1-2%3D-1)
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.![h(x)=\frac{2}{x}](https://tex.z-dn.net/?f=h%28x%29%3D%5Cfrac%7B2%7D%7Bx%7D)
The function is not defined for x=0 .Therefore , it is not a function on domain R.
Let ![h(x_1)=h(x_2)](https://tex.z-dn.net/?f=h%28x_1%29%3Dh%28x_2%29)
![\frac{2}{x_1}=\frac{2}{x_2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7Bx_1%7D%3D%5Cfrac%7B2%7D%7Bx_2%7D)
By cross mulitiply
![x_1= \frac{2\times x_2}{2}](https://tex.z-dn.net/?f=x_1%3D%20%5Cfrac%7B2%5Ctimes%20x_2%7D%7B2%7D)
![x_1=x_2](https://tex.z-dn.net/?f=x_1%3Dx_2)
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![k(x_1)=k(x_2)](https://tex.z-dn.net/?f=k%28x_1%29%3Dk%28x_2%29)
![ln(x_1)=ln(x_2)](https://tex.z-dn.net/?f=%20ln%28x_1%29%3Dln%28x_2%29)
Cancel both side log then
![x_1=x_2](https://tex.z-dn.net/?f=x_1%3Dx_2)
Hence, the given function one- one on positive R.
We can find pre image in positive R for every value of
.
Therefore, the function k(x) is one-one and onto on
but not on R.
e.l(x)=![e^x](https://tex.z-dn.net/?f=e%5Ex)
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.