Free AIOU Solved Assignment Code 1307 Spring 2021

Free AIOU Solved Assignment Code 1307 Spring 2021

Download Aiou solved assignment 2021 free autumn/spring, 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 2021-2021
Assignment Code 1307/2020-2021
Product Assignment of MA 2021-2022 (AIOU)

Course: Mathematics-I (1307)
Semester: Spring 2021
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 2021

(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 2021

(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


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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 2021

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 2021

         (b)    Solve the equation:  = 350

aiou solved assignment code 1307
aiou solved assignment code 1307

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