Algebra

Question: Which of the following is irrational?

Solution:
অমূলদ সংখ্যা (irrational number):
- যে সংখ্যাকে p/q আকারে প্রকাশ করা যায় না, যেখানে p ও q পূর্ণসংখ্যা এবং q ≠ 0, সে সংখ্যাকে অমূলদ সংখ্যা বলা হয়।
- পূর্ণবর্গ নয় এরূপ যে কোনো স্বাভাবিক সংখ্যার বর্গমূল কিংবা তার ভগ্নাংশ একটি অমূলদ সংখ্যা। যেমন, √2 = 1.414213..., √6 = 2.229489... ইত্যাদি অমূলদ সংখ্যা।
- কোনো অমূলদ সংখ্যাকে দুইটিপূর্ণ সংখ্যার অনুপাত হিসেবে প্রকাশ করা যায় না।
-অমূলদ সংখ্যাকে একটি মূলদ সংখ্যা দ্বারা গুণ করলে অমূলদ সংখ্যা পাওয়া যায়।
অর্থাৎ, non zero rational number × irrational number = irrational number.

অপশন গুলো ব্যাখ্যা করে,
ক) 3/5​  ; এটি একটি ভগ্নাংশ, তাই rational
খ) √2 ; এটি কোন পূর্ণসংখ্যার বর্গমূল নয়, তাই irrational
গ) 0.75 = 3/4  ; ভগ্নাংশের সমান, তাই rational
ঘ) 1.2 = 6/5 ; ভগ্নাংশের সমান, তাই rational 
 
সুতরাং, √2 একটি অমূলদ সংখ্যা। 

ধারাটিঃ
3 + 5 = 8
8 + 7 = 15
15 + 9 = 24
24 + 11 = 35
35 + 13 = 48
48 + 15 = 63
63 + 17 = 80
80 + 19 = 99

Question: 

Solution: 

Question: 

Solution: 

Question: If x = 2 + √3 and y = 2 - √3, find the value of (x² + y²)²

Solution:
We are given:
x = 2 + √3, y = 2 - √3

Now,
x + y
= (2 + √3) + (2 - √3)
= 4

And,
⇒ xy
= (2 + √3)(2 - √3)
= 2² - (√3)²
= 4 - 3 = 1

We know,
x² + y² = (x + y)² - 2xy
⇒ x² + y² = (4)² - 2(1) [Substitute the values]
⇒ x² + y² = 16 - 2
⇒ x² + y² = 14

∴ (x² + y²)² = 14² = 196

Given, -1 ≤ 3 - 2x ≤ 3
⇒ -4 ≤ - 2x ≤ 0 [Adding - 3 in every section]
⇒ 2 ≥ x ≥ 0
⇒ 0 ≤ x ≤ 2

Question: If a set has 5 elements, then the power set of that set has ______ elements.

Solution:
The number of elements in the power set of a set with n elements is 2ⁿ.
Here, n = 5
∴ Number of elements in the power set = 2⁵ = 32

So the power set has 32 elements.

Question: Find the determinant of the matrix:
 

Solution:
2 × 2 ম্যাট্রিক্সের জন্য: 

⇒ det(A) = ad - bc

Given matrix,

∴ det(A) = 1 × 4 - 2 × 3 = 4 - 6 = - 2

Question: If x² - 6x + 8 < 0, then solve the inequality.

Solution:
x² - 6x + 8 < 0
⇒ x² - 2x - 4x + 8 < 0
⇒ x(x - 2) - 4(x - 2) < 0
⇒ (x - 2)(x - 4) < 0

দুটি রাশির গুণফল শূন্যের চেয়ে ছোট হওয়ার শর্ত হলো, একটি রাশি ধনাত্মক এবং অন্যটি ঋণাত্মক হতে হবে।
∴ অসমতাটি সত্য হবে যদি x - 2 > 0 এবং x - 4 < 0 হয়।

x - 2 > 0 অর্থাৎ, x > 2
x - 4 < 0 অর্থাৎ, x < 4

∴ অসমতার সমাধান হলো 2 < x < 4

If g(m) > g(n)
∴ m > n
So, m ≠ n

x² - 5x - 6
x² - 6x + x - 6
x(x - 6) + 1(x - 6)
(x - 6)(x + 1).

Question: If (x/y) + (y/x) = √7 then what is the value of [(x⁴/y⁴) + (y⁴/x⁴)]³ ?
(Janata RC 22 এর অনুরূপ)

Solution:
দেওয়া আছে,
(x/y) + (y/x) = √7

∴ x⁴/y⁴ + y⁴/x⁴
= (x/y)⁴ + (y/x)⁴
= {(x/y)²}² + {(y/x)²}²
= {(x/y)² + (y/x)²}² - 2.(x²/y²).(y²/x²)
= {(x/y)² + (y/x)²}² - 2
= [{(x/y) + (y/x)}² - 2.(x/y).(y/x)]² - 2
= {(√7)² - 2}² - 2
= (7 - 2)² - 2
= 5² - 2
= 25 - 2
= 23

[(x⁴/y⁴) + (y⁴/x⁴)]³ = (23)³ = 12,167

Question: The value ofis:
(Officer General 22 এর অনুরূপ) 

Solution:
1 - cosx
≈ x²/2 for small x

Again,
1 - cosx/x²

= (x²/2)/x²

= 1/2

a² + b² = 5ab
a²/ab + b²/ab = 5
a/b + b/a = 5
Squaring the both sides
(a/b)² + (b/a)² = (5)²
a²/b² + b²/a² + 2 × (a/b) × (b/a)= 25
a²/b² + b²/a² + 2 = 25
a²/b² + b²/a² = 25 -2
a²/b² + b²/a² = 23.

এখানে দুটি ধারা বিদ্যমান।
বিজোড় অবস্থানের ধারা = 4, 6, 8, 10
জোড় অবস্থানের ধারা = 9, 11, 13, 15

Solution:
Given that,
x² + 2xy + y² = 25 and xy = 6
⇒ (x + y)² = 25
⇒ x + y = ± 5

Question: What is the mean of the range, mode and median of the data given below? 
5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4

Solution:
Given data = 5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4
Arranging in ascending order = 2, 3, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10, 11, 15, 19
Here,
Most frequent data is 4
So, Mode = 4

Total terms in the given data, (n) = 15 (It is odd)
∴ Median = {(n + 1)/2}^th term
= {(15 + 1)/2}^th term
= (8)^th term
= 6

Now,
Range = (Maximum value - Minimum value) + 1
= (19 - 2) + 1
= 18

∴ Mean of Range, Mode and Median = (Range + Mode + Median)/3
= (18 + 4 + 6)/3 = 28/3 = 9.33

Question: If x : y = 5 : 2 then the value of (xy + y²)/(x² - y²) ?

Solution: Let x = 5k and y = 2k

Now,
(xy + y²)/(x² - y²) 
= {(5k)(2k)+(2k)²}/{(5k)² - (2k)²}
= (10k² + 4k²)/(25k² - 4k²)
= 14k²/21k²
= 14/21
= 2/3

Question:

Solution:

Let, L = Lalon’s current weight, in pounds
S = Sister’s current weight, in pounds

We are told that “If Lalon loses 8 pounds, he will weigh twice as much as his sister.'' We put this into an equation:
L – 8 = 2S
∴ L = 2S + 8...... (i)

Next, we are told that “Together they now weigh 278 pounds.” We can also put this into an equation.
L + S = 278........ (ii)
To solve this equation, we can substitute 2S + 8 from Equation (i) for the variable L in Equation 2:
2S + 8 + S = 278
3S = 270
S = 90

From equation (ii) we can find,
L = 278 - 90 = 188

Question: If P^x = Q^y = R^z and Q/P = R/Q then [2z/(x + z)]³ = ?
(Officer Cash 22 এর অনুরূপ) )

Solution:
ধরি,
P^x = Q^y = R^z = k
এখন,
P^x = k
∴ P = k^(1/x)

অনুরুপভাবে,
Q^y = k
∴ Q = k^(1/y)
এবং
R^z = k
∴ R = k^(1/z)

আবার,
⇒ Q/p = R/Q
⇒ Q² = PR
⇒ {k^(1/y)}² = k^(1/x) × k^(1/z)
⇒ k^(2/y) = k^{(z + x)/xz}
⇒ 2/y = (z + x)/xz
⇒ 2xz = y(z + x)
∴ 2z/(x + z) = y/x

∴ [2z/(x + z)]³ = y³/x³

Question: In a tourist group of 100 people, 55 speak French, 40 speak Spanish, and 20 speak none of the languages. How many of them speak just one language?

Solution:
 

Let,
Number of people who can speak both languages = x persons
∴ Number of people who speak only French = (55 - x) persons
∴ Number of people who speak only Spanish = (40 - x) persons

Given that,
Number of people who speak none of the languages = 20 persons

According to the question,
Only French + Both + Only Spanish = Total students - Those who speak none
⇒ (55 - x) + x + (40 - x) = 100 - 20 
⇒ 95 - x = 80
⇒ x = 95 - 80
∴ x = 15

∴ Only French = (55 - 15) = 40 persons
∴ Only Spanish = (40 - 15) = 25 persons

∴ Number of people who speak only one language (French or Spanish) = (40 + 25) = 65 persons

দেয়া আছে, y/x = 1/5
∴ x = 5y

এবং x + 2y = 21
⇒ 5y + 2y = 21
⇒ 7y = 21
∴ y = 3

∴ x = 5y
= 5 × 3
= 15
Answer: 15

5p - 3q  = 42
5p + 3q = 18
____________
10p        = 60
⇒ p        = 6

So, Value of q = -4

∴ |p| + |q| = |6| + |-4| = 6 + 4 = 10

প্রশ্ন: If x + 4/x = 4, then x/(x² + x - 1) =?

সমাধান:
দেওয়া আছে
x + 4/x = 4
⇒ (x² + 4)/x = 4
⇒ x² + 4 = 4x
⇒ x² - 4x + 4 = 0
⇒ (x - 2)² = 0
⇒ x - 2 = 0
⇒ x = 2

প্রদত্ত রাশি
x/(x² + x - 1)
= 2(2² + 2 - 1)
= 2/(4 + 2 - 1)
= 2/5

Question: If a² + b² = 45 and ab = 18, find (1/a) + (1/b)

Solution:
দেওয়া আছে,
ab = 18
a² + b² = 45 
⇒ (a + b)² - 2ab = 45
⇒ (a + b)² - 36 = 45
⇒ (a + b)² = 81
∴ a + b = 9

এখন,
(1/a) + (1/b)
=(b + a)/ab
= 9/18
= 1/2

4x² + 3x² − 4x + k = 0
⇒ 4(1)² + 3(1)² - 4(1) + k = 0 [As, x - 1 is a factor]
⇒ 4 + 3 - 4 + k = 0
⇒ k = - 3

Question: A survey in a class shows that 20 of the pupils play hockey, 18 play basketball and 8 play both hockey and basketball. How many pupils are there in the class, if everyone plays at least one of these games?

Solution:
ধরি, যারা হকি খেলে তাদের সেট হলো H এবং যারা বাস্কেটবল খেলে তাদের সেট হলো B।
দেওয়া আছে,
হকি খেলে, n(H) = 20 জন
বাস্কেটবল খেলে, n(B) = 18 জন
উভয় খেলা খেলে, n(H ∩ B) = 8 জন

যেহেতু ক্লাসের সবাই অন্তত একটি খেলা খেলে, তাই ক্লাসের মোট শিক্ষার্থীর সংখ্যা হবে n(H U B)।

আমরা জানি,
n(H U B) = n(H) + n(B) - n(H ∩ B)
= 20 + 18 - 8
= 38 - 8
= 30

সুতরাং, ক্লাসে মোট শিক্ষার্থীর সংখ্যা 30 জন।

ধারা :   1    1.5    2.5      4
অন্তর :   .5       1      1.5
∴ পরিবর্তী সংখ্যটি (4 + 2) = 6

আবার,
1 + 1.5 = 2.5
1.5 + 2.5 = 4
∴ পরিবর্তী সংখ্যাটি = 2.5 + 4 = 6.5

প্রশ্নের অপশন কিছুটা মডিফাই করা হয়েছে।

Question: A sum of TK. 304 was divided among 80 boys and girls in such a way that each boys gets TK. 4.5 and each girls TK. 3.5. The number of girls is-

Solution:
মনে করি,
বালিকার সংখ্যা x জন
বালকের সংখ্যা (80 - x) জন

1 জন বালিকা পায় 3.5 টাকা
x জন বালিকা পায় (3.5 × x) টাকা

1 জন বালক পায় 4.5 টাকা
(80 - x) জন বালক পায় 4.5 ×(80 - x) টাকা

প্রশ্নমতে,
(3.5 × x) + 4.5 ×(80 - x) = 304
or, 3.5x + 360 - 4.5x = 304
or, - x = 304 - 360
or, - x = - 56
or, x = 56
∴ x = 56

∴ বালিকার সংখ্যা 56 জন

Question: The 2^nd and 8^th term of an arithmetic progression are 17 and - 1 respectively. What is the 14^th term?

Solution:
Let the first term be a and the common difference be d.

We know, 
n term of arithmetic progression = a + (n - 1)d
Then,
2^nd term, a + d = 17 ……(i)
8^th term, a + 7d = - 1 ……(ii)

Now, Subtract (i) from (ii),
(a + 7d) - (a + d) = - 1 - 17
⇒ 6d = - 18
∴ d = - 3
From (i) we  get,
⇒ a + ( - 3) = 17   ; [d =  - 3]
⇒ a - 3 = 17
∴ a = 20

Now, 14^th term = a + 13d
= 20 + 13( - 3)
= 20 - 39
= - 19

So the 14th term of the arithmetic progression is - 19.

Question: Find the wrong term in the following series.
1200, 1188, 1164, 1116, 1020, 828, 484

Solution:
Given series:
1200, 1188, 1164, 1116, 1020, 828, 484
The series decreases with multiples of 12, doubling each time.
1^st term = 1200
2^nd term = 1200 - 12 = 1188
3^rd term = 1188 - 24 = 1164
4^th term = 1164 - 48 = 1116
5^th term = 1116 - 96 = 1020
6^th term = 1020 - 192 = 828
7^th term = 828 - 384 = 444

The series has 484 as the last term, but it should be 444.

 Hence the wrong term in the series is 484.

Question: The next term of the series: 36, 81, 144, 225, ____ is

Solution: 
Given, 36, 81, 144, 225,
The series is = 6² , 9², 12², 15², 18²
So, next term is 18² = 324

Question: Find the solution of the inequality |2x - 3| ≤ 1.

Solution:
Given that,  
|2x - 3| ≤ 1
⇒ - 1 ≤ 2x - 3 ≤ 1  
⇒ - 1 + 3 ≤ 2x ≤ 1 + 3  ; [adding 3 to all parts]
⇒ 2 ≤ 2x ≤ 4  
⇒ 1 ≤ x ≤ 2 ; [dividing all parts by 2]

Therefore, the solution is 1 ≤ x ≤ 2 or in interval notation [1, 2]

4x⁴ + 1
= (2x²)² + 1
= (2x²)² + 2.2x².1 + 1² - 4x²
= (2x² + 1)² - (2x)²
= (2x² + 2x + 1) (2x² - 2x + 1)

Question: Find the domain of f(x) = 1/x + 3

Solution:
দেওয়া আছে,
f(x) = 1/x + 3

আমরা জানি, 
একটি ভগ্নাংশের হর (denominator) শূন্য হতে পারবে না।
অর্থাৎ,
⇒ x + 3 ≠ 0
∴ x ≠ - 3

সুতরাং, f(x) এর ডোমেইন = R - {- 3}