MTH100 ASSIGNMENT NO. 1 FALL 2022 || 100% RIGHT SOLUTION || GENERAL MATHEMATICS || BY VuTech

MTH100 ASSIGNMENT NO. 1 FALL 2022

 KINDLY, DON’T COPY PASTE

MTH100 ASSIGNMENT NO. 1 FALL 2022 || 100% RIGHT SOLUTION || GENERAL MATHEMATICS || BY VuTech

SEND WHATSAPP OR E-MAIL FOR ANY QUERY

0325-6644800

kamranhameedvu@gmail.com 

Fall 2022     MTH100:

Assignment No. 1          Total Marks: 10

Due Date: 01-12- 2022

 

Please read the following instructions before attempting the solution of this assignment:

 

Ø  To solve this assignment, you should have good command on lecture no.1 - 8.

Ø  Try to consolidate your concepts that you learn in the lectures with these questions.

Ø  Upload assignments properly through VULMS. No Assignment will be accepted through Email.

Ø  No assignment will be accepted after the due date.

Ø  All students are directed to use the font and style of text as is used in this document.

Ø  Use MathType or Equation Editor etc. for mathematical symbols and equations.

Ø  Remember that you are supposed to submit your assignment in MS-Word format any other format like scanned, images, MS-Excel, HTML etc. will not be accepted.

Ø  Do not use colorful backgrounds in your solution files.

Ø  This is an individual assignment (not a group assignment). So keep in mind that you are supposed to submit your own, self-made and different assignment even if you discuss the questions with your class fellows. All similar assignments (even with some meaningless modifications) will be awarded zero marks and no excuse will be accepted. This is your responsibility to keep your assignment safe from others.

 Visit Website For More Solutions

www.vutechofficial.blogspot.com

Question No.1                                                                                                              Marks: 10

If `\f\left(x\right) = \sqrt{x - 3}` then verify `\left(fof^{-1}\right)\left(x\right) = \left(f^- 1of\right)\left(x\right) = x` 

 

 Answer: 

 

Here `f\left(x \right) = \sqrt {x - 3}` 

 

Let assume `f\left(x\right) = y` 

 

then `y = \sqrt {x - 3}` 

 

`y^2 = x - 3` 

 

`x = y^2 + 3` 

 

We know that `\(f^ - 1\left(y\right) = x` 

 

`\(f^- 1\left(y\right) = y^2 + 3` 

 

By replacing y by x then `\(f^- 1\left(x\right) = x^2 + 3` 

 

First we solve `\left(fof^- 1\right)\left(x\right)` 

 

`\left(fof^- 1\right)\left(x \right) = f\left(f^ - 1\left(x\right)\right)` 

 

`\left(fof^- 1\right)\left( x \right) = f\left(x^2 + 3\right)` 

 

`\left(fof^- 1\right)\left(x\right) = \sqrt{\left(x^2 + 3\right)- 3}`

 

`\left(fof^- 1\right)\left(x\right) = \sqrt x^2` 

 

`\left(fof^ - 1\right)\left(x\right) = x                                  - - - - - - - \left(1\right)`

 

Now we solve  `\left(f^- 1of\right)\left(x\right)` 

 

`\left(f^- 1of\right)\left(x\right) = f^- 1\left(f\left(x\right)\right)` 

 

`\left(f^- 1of\right)\left(x\right) = f^- 1\left(\sqrt {x - 3}\right)` 

 

`\left(f^- 1of\right)\left(x\right) = \left(\sqrt{x - 3}\right)^2 + 3` 

 

`\left(f^- 1of\right)\left( x \right) = x - 3 + 3` 

 

`\left(f^- 1of\right)\left(x\right) = x              - - - - - - - - - -\left( 2 \right)` 

 

By (1) and (2) we can prove that 

 

`\left(fof^- 1\right)\left(x\right) = \left(f^- 1of \right)\left(x\right) = x`

 

 

 KINDLY, DON’T COPY PASTE

SUBSCRIBE, SHARE, LIKE AND COMMENTS FOR MORE UPDATES

SEND WHATSAPP OR E-MAIL FOR ANY QUERY

0325-6644800

kamranhameedvu@gmail.com