IBA ফ্যাকাল্টি ভিত্তিক প্রস্তুতি

Question: Find 

Solution:

Question: If 1/Q > 1, which of the following must be true?

Solution:
Given, 1/Q > 1
Since 1/Q > 1 and 1 > 0,  we know that 1/Q is positive, hence Q must also be positive.

Now, multiply both sides of the inequality by Q :
Q × (1/Q) > Q × 1
⇒ 1 > Q
⇒ 1 × Q > Q × Q
⇒ Q > Q²
⇒ 1 > Q² [Since 1 > Q and Q > Q²]

Therefore, 1 > Q² must be true.

Question: If b is one-fourth of a, then what is the value of  

Solution:
Given, b = 1/4 of a = a/4

Now,

Question: If 2 < x < 5 and 3 < y < 5, which of the following best describes x - y?

Solution:
দেয়া আছে,
2 < x < 5
3 < y < 5

এখন, আমরা x - y এর সীমা বের করতে চাই। এর জন্য, y এর অসমতাকে - y এর অসমতায় রূপান্তর করতে হবে।
3 < y < 5
⇒ - 3 > - y > - 5 [- 1 দ্বারা গুণ করে]
⇒ - 5 < - y < - 3 

এইবার x এবং - y এর অসমতা দুটি যোগ করি,
⇒ (2 < x < 5) + (- 5 < - y < - 3)
⇒ −3 < x - y < 2

Question: If x = 10^1.4, y = 10^0.7 and x^z = y³, then what is the value of z?

Solution:
Given,
x = 10^1.4, y = 10^0.7

Now,
x^z = y³
⇒ (10^1.4)^z = (10^0.7)³
⇒ 10^1.4z = 10^2.1
⇒ 1.4z = 2.1
⇒ z = 2.1/1.4
⇒ z  = (2.1 × 10)/(1.4 × 10)
⇒ z = 21/14
∴ z = 3/2

Question: , then what is the value of m?

Solution:

We have 4^m > 1 
Now, if m = -1, then 4^m = 4 ^-1 = 1/4 = 0.25 < 1, so incorrect.
if m = 1, then 4^m = 4¹ = 4 > 1, correct.

∴ m = 1

Question: If X ∈ N and 31 < x < 37, and x is a prime number, then which of the following represents the list form of the set of such numbers?

Solution:
The natural numbers between 31 and 37 are:
32, 33, 34, 35, 36

Now, check which of these are prime:
32: divisible by 2 → not prime
33: divisible by 3 and 11 → not prime
34: divisible by 2 → not prime
35: divisible by 5 and 7 → not prime
36: divisible by 2, 3, etc. → not prime

So, there are no prime numbers between 31 and 37.

Therefore, the correct answer is the empty set: { }

Question: If x²b⁴ = ab^{- 1}, what is a in terms of b and x ?

Solution:
x²b⁴ = ab^{- 1}
⇒ a/b = x²b⁴ 
⇒ a = x²b⁴.b
⇒ a = x²b^{4 + 1}
⇒ a = x²b⁵

Question: What is the solution of the inequality: - 6 ≤ 3x + 3 < 27?

Solution:
- 6 ≤ 3x + 3 < 27
⇒ - 6 - 3 ≤ 3x + 3 - 3 < 27 - 3
⇒ - 9 ≤ 3x < 24
⇒ - 9/3 ≤ 3x/3 < 24/3
⇒ - 3 ≤ x < 8

∴ solution of the inequality: [-3, 8)

Question: If P = 216^{- 1/3} + 243^{- 2/5} + 256^{- 1/4}, then which one of the following is an integer?

Solution:
P = 216^{- 1/3} + 243^{- 2/5} + 256^{- 1/4}
= (6³)^{- 1/3} + (3⁵)^{- 2/5} + (4⁴)^{- 1/4
=} 6^{3(- 1/3)} + 3^{5(- 2/5)} + 4^{4(- 1/4)}
= 6^{- 1}+ 3^{- 2}+ 4^{- 1}
= (1/6)+ (1/9) + (1/4)
= (6 + 4 + 9)/36
∴ P = 19/36

Now,
Option (A): P/19 = (19/36)/19 = 1/36, not an integer. Reject.
Option (B): P/36 = (19/36)/36 = 19/36², not an integer. Reject.
Option (C): 36/P = 36/(19/36) = 36²/19, not an integer. Reject.
Option (D): 19/P = 19/(19/36) = 36, an integer. Correct.
Option (E): P = 19/36, not an integer. Reject.

Question: Let U = {1,2,3,4,5,6,7,8}, A = {2,3,6}, and B = {1,4,5}. Find A^c ∪ B^c.

Solution:
Complement of A:
A = {2,3,6}
U = {1,2,3,4,5,6,7,8}
A^c = U - A = {1,4,5,7,8}

Complement of B:
B = {1,4,5}
B^c = U - B = {2,3,6,7,8}

Union of Complements:
A^c ∪ B^c = {1,4,5,7,8} ∪ {2,3,6,7,8} = {1,2,3,4,5,6,7,8} = U

∴A^c ∪ B^c = {1,2,3,4,5,6,7,8}

Question: If (2 + √x) > 2√x, which of the following must be true?

Solution:
2 + √x > 2√x
⇒ 2 > 2√x - √x
⇒ 2 > √x
⇒ 4 > x
∴ x < 4

Question: If x = 3 + 2√2, find the value of .

Solution:
Given,

Question: If |x + 2| ≤ 6, then which of the following intervals represents all possible values of the expression 3x - 4 ?

Solution:
|x + 2| ≤ 6
⇒ - 6 ≤ x + 2 ≤ 6
⇒ - 6 - 2 ≤ x + 2 - 2 ≤ 6 - 2
⇒ - 8 ≤ x ≤ 4
⇒ - 24 ≤ 3x ≤ 12
⇒ - 24 - 4 ≤ 3x - 4 ≤ 12 - 4
⇒ - 28 ≤ 3x - 4 ≤ 8

∴ All possible values of 3x - 4 lie in the interval [- 28, 8].

Question: What is the value of the following expression?

Solution:

= log₆₀ 3 + log₆₀ 4 + log₆₀ 5 [∵ 1/log_ba = log_ab]
= log₆₀(3 × 4 × 5) [∵ log_b(m) + log_b(n) = log_b(m × n)]
= log₆₀ 60
= 1

Question: In a class, 25 students play cricket, 25 students play football, and 10 students play both. 10 students play neither cricket nor football. What is the total number of students in the class?

Solution:
Number of students who play cricket, n(C) = 25
Number of students who play football, n(F) = 25
Number of students who play both cricket and football, n(C ∩ F) = 10
Number of students who play neither = 10

n(C ∪ F) = n(C) + n(F) - n(C ∩ F)
= 25 + 25 − 10 = 40

Total students in the class = students who play cricket or football + students who play neither
n(U) = n(C ∪ F) + neither = 40 + 10 = 50

∴ Total 50 students in the class.

Question: If

Solution:

Question: If log₁₀5+ log₁₀(5x + 1) = log₁₀(x + 5) + 1, then what is the value of x ?

Solution:
log₁₀5+ log₁₀(5x + 1) = log₁₀(x + 5) + 1
⇒ log₁₀5+ log₁₀(5x + 1) = log₁₀(x + 5) + log₁₀10
⇒ log₁₀[5(5x + 1)] = log₁₀[10(x + 5)
⇒ 5(5x + 1) = 10(x + 5)
⇒ 5x + 1 = 2x + 10
⇒ 3x = 9
∴ x = 3

Question: In a class of 60 students, 20 students like Math, 25 students like English, and 30 students like Science. If 5 students like both Math and English, 7 students like both Math and Science, 8 students like both English and Science, and 3 students like neither of these subjects, how many students like all three subjects?

Solution:
Total students, n(U) = 60
Number who like Math, n(M) = 20
Number who like English, n(E) = 25
Number who like Science, n(S) = 30
Number who like both Math and English, n(M ∩ E) = 5
Number who like both Math and Science, n(M ∩ S) = 7
Number who like both English and Science, n(E ∩ S) = 8
Number who like neither subject = 3

n(M ∪ E ∪ S) = n(U) - neither
= 60 - 3 = 57

∴ n(M ∪ E ∪ S) = n(M) + n(E) + n(S) - n(M ∩ E) - n(M ∩ S) - n(E ∩ S) + n(M ∩ E ∩ S)
⇒ 57 = 20 + 25 + 30 - 5 - 7 - 8 + n(M ∩ E ∩ S)
⇒ 57 = 75 - 20 + n(M ∩ E ∩ S)
⇒ 57 = 55 + n(M ∩ E ∩ S)
⇒ n(M ∩ E ∩ S) = 57 - 55
⇒ n(M ∩ E ∩ S) = 2

∴ 2 Students like all three subjects.