Real Number System MCQ Quiz in தமிழ் - Objective Question with Answer for Real Number System - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

A function f: \(\mathbb{N}\) × \(\mathbb{N}\) → \(\mathbb{N}\) is said to be injective if (m₁, n₁) ≠ (m₂, n₂) such that f((m1, n1)) ≠ f(m2, n2) for (m1, n1), (m2, n2) ∈ \(\mathbb{N}\) × \(\mathbb{N}\)

Explanation:

(1): f1(m, n) = 2m 3n

Let (m1, n1), (m2, n2) ∈ \(\mathbb{N}\) × \(\mathbb{N}\) such that 

f1(m1, n1) = f1(m2, n2)

⇒ 2m₁ 3n₁ = 2m₂ 3n₂

⇒ 2m1^-m2 = 3n2 ^-n1 

⇒ m1 - m2 = n2 - n1 = 0  

⇒ m1 = m2 and n2 = n1 

⇒ (m1, n1) = (m2, n2)

Option (1) is correct

(2): f2(m, n) = m2 + m + n

Option (2) is true

(3): f3(m, n) = m3 + n3

(2, 1) ≠ (1, 2)

but f3(2, 1) = 2³ + 13 = 9 and f3(1, 2) = 2³ + 13 = 9

So f3(2, 1) = f3(1, 2), not injective

Option (3) is false 

(4): f4(m, n) = m n3

(1, 2) ≠ (8, 1)

but f4(1, 2) = (1)(2³) = 8 and f4(8, 1) = (8)(1)3 = 8

So f4(1, 2) = f4(8, 1), not injective

Option (4) is false 

Explanation:

an − bn = (a − b) (an−1 + an−2 b + an−3 b2 + … + bn−1)

Now, xn − yn = (x − y) (xn−1 + xn−2 y + xn−3 y2 + … + yn−1) ≤ (x − y) (xn−1 + xn−2⋅x + xn−3 x2 + … + xn−1)

⇒ xn − yn ≤ (x − y) (xn−1 + xn−1 + … + xn−1) (∵ x ⩾ y)

⇒ xn − yn ≤ n xn−1 (x − y) -

option (4) correct, option (2) Incorrect.

Also, xn − yn ⩾ (x − y) (yn−1 + yn−2 ⋅ y + yn−3 y2 + … + yn−1) (: y ≤ x)

⇒ xn − yn ⩾ (x − y) (yn−1 + yn−1 + … + yn−1)

⇒ xn − yn ⩾ n yn−1(x − y) - 

option (1) correct, option (3) Incorrect.

Also, for opt (2), Take x = 3, y = 2, n = 2

then 2 ⋅ 32−1 (3 − 2) = 2 ⋅ 3 ⋅ 1 = 6 = n xn−1 (x − y)

and xn − yn = 32 − 22 = 9 − 4 = 5

then 6 \(\nleq\) 5 ⇒ n xn−1 (x − y) \(\nleq\) xn − yn. option -

(2) False.

For option (3)

n yn−1 (x − y) = 2 ⋅ 22−1 (3 − 2) = 2 ⋅ 2 ⋅ 1 = 4

and xn − yn = 32 − 22 = 9 − 4 = 5

then 4 \(\ngeq\) 5 ⇒ nyn−1 (x − y)\(\ngeq\) xn − yn

option (3) False.