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# MTH101 Assignment No 01 Solution & Discussion Due Date:06-05-2013

Assignment No. 1

MTH 101 (Spring 2013)

Maximum Marks: 15

Due Date: 6th May, 2013

DON’T MISS THESE Important instructions:

• Ø To solve this assignment, you should have good command over first

10 lectures.

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• Ø This is not a group assignment, it is an individual assignment, so be careful and avoid copying others’ work. If any assignment is found to be copy of some other, both will be awarded ZERO MARKS. It also suggests you to keep your assignment safe from others. No excuse will be accepted by anyone if assignment is found to be copied or letting others to copy.

Question#1                                                                                         Marks4+4=8

Solve the following Absolute Value Problems and write their solutions in both inequality and interval notation.

Question#2                                                                                                  Marks 7

Evaluate the following limit.

(Note: In order to get full marks, do all necessary steps)

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### Replies to This Discussion

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

Assalamu Alaikum kindly please meri help kerdein maine in teenon k solutions kerliye hain per microsoft office main mjhse equations nai horhien plzz batadein k main word main kese maths ki equations likhun

I solved Q no 1 and got the ans
-12/5<=x>=-6/5
(-infi,-12/5]u[-6/5,+infin)
2
8/3
(8/3,4)

plz come and discuss here.

yeah mare ans b yahe han

@shahzab khan, it is not the correct one, and @Saireen Yousaf if you have the same then both of your answers are not correct.

12/5 ≤ x ≤ 6/5 1st Q ka ans

2nd Q ka

4  <  x < 8/3

koi baty ga k interval mein qasy likhty hen jvb baqi ya jvb bilkul correct hen mein ny khud keya hen

Limit waly ka ZERO ans ha

zeeshannayab512@gmail.com

@Zeeshan, don't you think that 2nd one should be in form of x>a or x<-a ? How it can be in the form of 4<x<8/3? (Just to guess)

And in the scenario if you punch in the answers of 1st one in calculator you will get(according to your solution) 2.4≤x≤1.2 (How it can be possible in real world?)

ok thank u zeeshan i solved it again and got the answers

6/5 >= x >=12/5

and interval form is

[6/5,-infi)U[12/5,+infini)

and the second one is

8/3<X<4

interval is

(8/3, 4)

frnd aur jo limit wala question hy uska answer tu btain evalute limit kaisy solved krain gain

M.welcome

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