Free AIOU Solved Assignment Code 1307 Spring 2023

Free AIOU Solved Assignment Code 1307 Spring 2023

Download Aiou solved assignment code1307 free autumn/spring 2023, aiou updates solved assignments. Get free AIOU All Level Assignment from aiousolvedassignment.

Title Name Mathematics-I (1307)
University AIOU
Service Type Solved Assignment (Soft copy/PDF)
Course FA
Language ENGLISH
Semester 2023-2023
Assignment Code 1307/2020-2023
Product Assignment of MA 2023-2023 (AIOU)

Course: Mathematics-I (1307)
Semester: Spring 2023
ASSIGNMENT No. 1

  1. 1

(a)     If x, y are real numbers and 3x + y + (2x+2y) i=13 +10i

3x + y = 13……………….i

2x + 2y = 10………………ii

2(i)-ii

4x = 16

x = 4

from i

12 + y = 13

y = 1

What is ‘Z’ if

Z = 16 – ¼

Z = 64-1 / 4

Z = 63 / 4

AIOU Solved Assignment Code 1307 Spring 2023

(b)    Solve for ‘a’, ‘b’: ai93 + bi35 + ai24 – bi86 = 40 + 20i

ai93 + bi35 + ai24 – bi86 = 40 + 20i

a + bi2 + a – bi2 = 40 + 20i

2a + bi2 – bi2 = 40 + 20i

2a +0i = 40 + 20i

2a = 40

a = 20

b is missing.

AIOU Solved Assignment Code 1307

Q No.2 (a) Prove that: p˄ (q ˅ r) ⎔ (p ˄ q) v (q ˄ r)

p q r ((p ∧ (q ∨ r)) ⎔ ((p ∧ q) ∨ (q ∨ r)))
F F F T
F F T F
F T F F
F T T F
T F F T
T F T T
T T F T
T T T T

AIOU Solved Assignment 1 Code 1307 Spring 2023

(b)    Show that: G = {l, –l,  i, – i} form a group w.r.t. multiplication.

Since for every pair a,b,E, G

code 1307
code 1307

G1⇒ since multiplication of complex number is associative, the multiplication associative in G

G2⇒ From the first column (or row), we see that l is an identity element. Hence, 1 E G is an identity element.

G3⇒ G3⇒ Since

11=1, _1-1=1, i * -i=1

-i * i=1, inverse exists

for every elements in G

and, we have

1^{-1} =1, -1^{-1} =-1, i^{-1} =-i, -i^{-i} = i

Hence G is a group under multiplication.

AIOU Solved Assignment 2 Code 1307 Spring 2023

Q No.3 (a) Solve the equation: x3 – 6x2 – x + 30 = 0

Polynomial Long Division

Dividing :  x3-6x2-x+30

(“Dividend”)

By         :    x-5    (“Divisor”)

dividend  x3  6x2  x +  30
– divisor  * x2  x3  5x2
remainder  x2  x +  30
– divisor  * -x1  x2 +  5x
remainder  6x +  30
– divisor  * -6x0  6x +  30
remainder 0

Quotient :  x2-x-6  Remainder:  0

Factoring  x2-x-6

The first term is,  x2  its coefficient is  1 .

The middle term is,  -x  its coefficient is  -1 .

The last term, “the constant”, is  -6

Step-1 : Multiply the coefficient of the first term by the constant   1 • -6 = -6

Step-2 : Find two factors of  -6  whose sum equals the coefficient of the middle term, which is   -1 .

-6    + 1    = -5
-3    + 2    = -1    That’s it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -3  and  2

x2 – 3x + 2x – 6

Step-4 : Add up the first 2 terms, pulling out like factors : x • (x-3)

Add up the last 2 terms, pulling out common factors : 2 • (x-3)

Step-5 : Add up the four terms of step 4 :(x+2)  •  (x-3)

Which is the desired factorization

(x + 2) • (x – 3) • (x – 5)  = 0

x+2 = 0

Subtract  2  from both sides of the equation :

x = -2        

x-3 = 0

Add  3  to both sides of the equation :

x = 3

x-5 = 0

Add  5  to both sides of the equation :

x = 5

AIOU Solved Assignment Code 1307 Autumn 2023

         (b)    Solve the equation:  = 350

aiou solved assignment code 1307
aiou solved assignment code 1307

Leave a Reply

Your email address will not be published. Required fields are marked *

aiou solved assignment code 207 Previous post Free AIOU Solved Assignment Code 967 Spring 2023
aiou solved assignment code 207 Next post Free AIOU Solved Assignment Code 1345 Spring 2023
Close