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Free AIOU Solved Assignment Code 1307 Spring 2021
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| 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
(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
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
