# Free AIOU Solved Assignment Code 1307 Spring 2023

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## 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

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1. 1

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

3x + y = 13……………….i

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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

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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 * -6x – 6x + 30 remainder

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

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