In this page, you can find the complete solutions of the second exercise of Relations, Functions, and Graphs chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, Relations, Functions, and Graphs is the 2nd chapter and has five exercises only. Out of which, this is the solution of the second exercise.

Check: Basic Mathematics Grade 11 (Sukunda Publication) Guide:
Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal

 

Disclaimer:

Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here. Few questions have been typed while most of them have been updated as pictures.

1) Let X = {a,b,c} and Y = {p,r,s}. Determine which of the following relations from X to Y are functions. Give reasons for your answers.

a) R1 = {(a,p), (a,r), (b,r),(c,s)}

This is not a function.
Reason: Domain in X cannot have more than one range. Here, element 'a' in X has two images in Y.

b) R2 = {(a,p),(b,r)}

This is not a function.
Reason: All pre-images in X must have at least one image in Y. Here, c has no image.

c) R3 = {(a,s),(b,s),(c,s)}

This is a function.
Reason: All pre-images in X have at least one image in Y (s).

d) R4 = {(a,p), (b,p), (c,s)}

This is a function.
Reason: All pre-images in X have at least one image in Y.

e) R5 = {(a,p),(b,r),(c,s)}

This is a function.
Reason: All pre-images n X have at least one image in Y.

2) If $\rm f: A \rightarrow B$ where A and B $\in$ R, is defined by $\rm f(x) = 1- x$, find the images of 1,$\frac{3}{2}$, -1, 2 $\in$ A.

Solution:
Given,

$$\rm f(x) = 1 - x$$

For x = 1, f(1) = 1 - 1 = 0

For x = $\rm \frac{3}{2}$, f $\rm \left ( \frac{3}{2} \right ) = 1 - \frac{3}{2}$
$\rm f \left ( \frac{3}{2} \right ) = - \frac{1}{2}$

For x = -1, f(-1) = 1 - (-1) = 1 + 1 = 2

For x = 2, f(2) = 1 - 2 = -1

Hence, the required images are found.

3. Find a formula that defines the inverse function f$\rm ^{-1}$. Let a function f: R $\rightarrow$ R be defined by

a) f(x) = x + 1, x $\in$ R

Solution:

Let y = f(x) = x + 1

Interchanging x and y

$$\rm x = y + 1$$

$$\rm y = x - 1$$

$$\rm \therefore f^{-1} (x) = y = (x - 1)$$


b) f(x) = 2x + 3, x $\in$ R

Solution:

Let y = f(x) = 2x + 3

Interchanging x and y

$$\rm x = 2y + 3$$

$$\rm y = \frac{x-3}{2}$$

$$\rm \therefore f^{-1}(x) = y = \frac{x-3}{2}$$


c) f(x) = 2x + 5, x $\in$ R

Solution:

Let y = f(x) = 2x + 5

Interchanging x and y

$$\rm x = 2y + 5$$

$$\rm y = \frac{x-5}{2}$$

$$\rm \therefore f^{-1}(x) = \frac{x-5}{2}$$


d) f(x) = x$^3$ + 5, x $\in$ R

Solution:

Let y = f(x) = x$^3$ + 5

Interchanging x and y

$$\rm x = y^3 + 5$$

$$\rm y = \sqrt[3]{x - 5}$$

$$\rm \therefore f^{-1}(x) = y = \sqrt[3] {x-5}$$


e) f(x) = 3x - 2, x $\in$ R

Solution:

Let y = f(x) = 3x - 2

Interchanging x and y

$$\rm x = 3y - 2$$

$$\rm y = \frac{x+2}{3}$$

$$\rm \therefore f^{-1}(x) = y = \frac{x+2}{3}$$

4) Let f:R $\rightarrow$R be defined by. Find gof(x) and fog(x).

a) f(x = 2x + 1 and g(x) = 3x - 1

Solution:

$$\rm gof(x) = g[f(x)] = g[2x + 1]$$

$$\rm gof(x) = 3(2x + 1) - 1$$

$$\rm \therefore gof(x) = 6x + 2$$


$$\rm fog(x) = f[g(x)] = f[3x - 1]$$

$$\rm fog(x) = 2(3x - 1) + 1$$

$$\rm \therefore fog(x) = 6x - 1$$


b) f(x) = 3x$^2$ - 4 and g(x) = 2x - 5

Solution:

$$\rm gof(x) = g[f(x)] = g[3x^2 - 4]$$

$$\rm gof(x) = 2(3x^2 - 4) - 5$$

$$\rm \therefore gof(x) = 6x^2 - 13$$


$$\rm fog(x) = f[g(x)] = f[2x - 5]$$

$$\rm fog(x) = 3(2x - 5)^2 - 4$$

$$\rm fog(x) = 3(4x^2 - 20x + 25) - 4$$

$$\rm \therefore fog(x) = 12x^2 - 60x + 71$$


c) f(x) = x$^3$ - 1 and g(x) = x$^2$

Solution:

$$\rm gof(x) = g[f(x)] = g[x^3 - 1]$$

$$\rm \therefore gof(x) = (x^3 - 1)^2$$


$$\rm fog(x) = f[g(x)] = f[x^2]$$

$$\rm fog(x) = (x^2)^3 - 1$$

$$\rm \therefore fog(x) = x^6 - 1$$


d) f(x) = x$^2$ + 1 and g(x) = x$^5$

Solution:

$$\rm gof(x) = g[f(x)] = g[x^2 + 1]$$

$$\rm \therefore gof(x) = (x^2 +1)^5$$


$$\rm fog(x)  = f[g(x)] = f[x^5]$$

$$\rm fog(x) = (x^5)^2 + 1$$

$$\rm \therefore fog(x) = x^10 + 1$$




About the Textbook:

Name: Basic Mathematics Grade XI
Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya
Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)
Telephone: 5320379, 5353537
Price: 695 /- (2078 BS)