## Free AIOU Solved Assignment Code 1307 Spring 2024

Download *Aiou solved assignment* code1307 free autumn/spring 2024, 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 |
2024-2024 |

Assignment Code |
1307/2020-2024 |

Product |
Assignment of MA 2024-2024 (AIOU) |

**Course: Mathematics-I (1307)
**

**Semester: Spring 2024**

**ASSIGNMENT No. 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 2024

**(b) Solve for ‘a’, ‘b’: ai ^{93} + bi^{35} + ai^{25} – bi^{86} = 40 + 20i**

ai^{93} + bi^{35} + ai^{25} – bi^{86} = 40 + 20i

a + bi^{2} + a – bi^{2} = 40 + 20i

2a + bi^{2} – bi^{2} = 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 2024

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

Since for every pair a,b,E, G

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

1*1=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 2024

**Q No.3 (a) Solve the equation: x ^{3} – 6x^{2} – x + 30 = 0**

Polynomial Long Division

Dividing : x^{3}-6x^{2}-x+30

(“Dividend”)

By : x-5 (“Divisor”)

dividend | x^{3} |
– | 6x^{2} |
– | x | + | 30 | ||

– divisor | * x^{2} |
x^{3} |
– | 5x^{2} |
|||||

remainder | – | x^{2} |
– | x | + | 30 | |||

– divisor | * -x^{1} |
– | x^{2} |
+ | 5x | ||||

remainder | – | 6x | + | 30 | |||||

– divisor | * -6x^{0} |
– | 6x | + | 30 | ||||

remainder | 0 |

Quotient : x^{2}-x-6 Remainder: 0

Factoring x^{2}-x-6

The first term is, x^{2} 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

x^{2} – 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 2024

** (b) Solve the equation: = 350**