All under in one

PLUS TWO
MATHEMATICS
IMPORTANT
QUESTIONS

, CH 1: RELATIONS & FUNCTIONS
1. What is the minimum number of ordered pairs to form a non–zero reflexive
relation on a set of n elements?


Ans: ‘n’ ordered pairs.

2. (a) The function given 𝑓 : N → N, by
𝑓(𝑥) = 2𝑥 is
(i) one – one and onto
(ii) one – one but not onto
(iii) not one – one and not onto
(iv) onto, but not one – one
(b) Find go𝑓(𝑥), if 𝑓(𝑥) = 8𝑥 3 and 𝑔(𝑥) = 𝑥 1⁄3
Ans:(a) (ii) one – one but not onto
(b) 𝑓(𝑥) = 8𝑥 3
1⁄
𝑔(𝑥) =𝑥 3

𝑔of (𝑥) = 𝑔[𝑓(𝑥)]
= 𝑔 (8𝑥 3 )
1⁄
= (8𝑥 3 ) 3= 2𝑥
3. Consider the function 𝑓 ∶ 𝑁 → N, given by 𝑓(𝑥) = 𝑥 3 . Show that the function f
is injective but not surjective.
Ans: 𝑓 ∶ 𝑁 → N defined by 𝑓(𝑥) = 𝑥 3
One – one (Injective)
Let 𝑥1 , 𝑥2 ∈ N
Let 𝑓(𝑥1 ) = 𝑓(𝑥2 )
𝑥13 = 𝑥23
⟹ 𝑥1 = 𝑥2
∴ 𝑓 is injective.
2,3,4 in the co–domain N are not images of any element of its domain N.
∴ f is not Surjective.

,4. Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by
𝑥−2
𝑓(𝑥) = .
𝑥 −3

(i) Is f one – one and onto? Justify your answer.
(ii) Is it invertible? Why?
𝑥−2
Ans: (i) 𝑓 : A → B, defined as 𝑓(𝑥) = 𝑥 −3

One – one
Let 𝑥, 𝑦 ∈ A such that 𝑓(𝑥1 ) = 𝑓(𝑥2 )
𝑥1 − 2 𝑥2 − 2
=
𝑥1 − 3 𝑥2 − 3

( 𝑥1 − 2) ( 𝑥2 − 3) = ( 𝑥2 − 2) ( 𝑥1 − 3)
⇒ 𝑥1 𝑥2 − 3𝑥1 − 2𝑥2 + 6 = 𝑥1 𝑥2 − 3𝑥2 − 2𝑥1 + 6
⇒ − 3𝑥1 − 2𝑥2 = − 3𝑥2 − 2𝑥1
⇒ 𝑥1 = 𝑥2 ∴ 𝒇 is one – one


Onto
Let y ∈ A = R – {1}
Now 𝑓(𝑥) = y
𝑥−2
=y
𝑥−3

𝑥−2 = y (𝑥 − 3)
𝑥–2 = y𝑥 – 3y
𝑥 − 𝑥𝑦 = –3y + 2
𝑥(1 − 𝑦) = –3y + 2
2−3𝑦
𝑥 = ∈A
1−𝑦

Thus for any y ∈ B, there exist 𝑥 ∈ A
∴ f is onto
∴ It is bijective function.
(ii) Every bijective function is invertible.
∴ It is invertible.


5. Determine whether the relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y) ; y is divisible by x} is reflexive, symmetric and transitive.

, Ans: (i) Given that
A = {1, 2, 3, 4, 5, 6}
R = {(𝑥, 𝑦) : y is divisible by 𝑥}
Reflexive
Let 𝑎 ∈ A
𝑎 is divisible by a itself
∴ (𝑎, 𝑎) ∈ R for all 𝑎 ∈ A
∴ R is Reflexive
Symmetric
Let (2, 4) ∈ R
ie, 4 is divisible by 2
but 2 is not divisible by 4
∴ (4, 2) ∉ R
∴ R is not Symmetric
Transitive
Let (𝑎, 𝑏), (b, c) ∈ R
ie, b is divisible by 𝑎 and c is divisible by b
∴ c is divisible by 𝑎
ie, (𝑎, 𝑐) ∈ R
∴ R is transitive
∴ R is reflexive and transitive, but not Symmetric.


6. R = {(x, y) : 𝑥, 𝑦 ∈ Z, (x – y) is an integer}. Show that R is an equivalence
relation.
Ans: Let 𝑥 ∈ Z
Reflexive
R = {(x, y) : 𝑥 − 𝑦 is an integer}
𝑥 − 𝑥 = 0 ; is an integer
∴ R is Reflexive
Symmetric
Let (𝑥, 𝑦) ∈ R where 𝑥, 𝑦 ∈ Z