NCERT Solutions for Class 12 Maths Chapter 1 - Relations and Functions
Q. Let us define a relation R in R as a aRb if a ≥ b. Then R is
Explanation :
Given that, aRb if a ≥ b
⇒ aRa ⇒ a ≥ a which is true.
Let aRb, a ≥ b, then b ≥ a which is not true, so R is
not symmetric.
But aRb and bRc
⇒ a ≥ b and b ≥ c
⇒ a ≥ c
Hence, R is transitive.
Q. Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}Then R is :
Explanation :
Given that, A = {1, 2, 3}
and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}
·.· (1, 1), (2, 2), (3, 3) ∈ R
Hence, R is reflexive.
(1, 2) ∈ R but (2, 1) ∉ R
Hence, R is not symmetric.
(1, 2) ∈ R and (2, 3) ∈ R
⇒ (1, 3) ∈ R
Hence, R is transitive.
Q. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
It is seen that (a, a) ∈ R, for every a ∈ {1, 2, 3, 4}.
∴ R is reflexive.
It is seen that (1, 2) ∈ R, but (2, 1) ∉ R.
∴ R is not symmetric.
Also, it is observed that (a, b), (b, c) ∈ R ⇒ (a, c) ∈
R for all a, b, c ∈ {1, 2, 3, 4}.
∴ R is transitive.
Hence, R is reflexive and transitive but not
symmetric.
Q. Let f : R → R be defined as f(x) = x4. Choose the correct answer :
f : R → R is defined as f(x) = x4
Let x, y ∈ R such that
f(x) = f(y)
⇒ x4 = y4
⇒ x4 – y4 = 0
⇒ (x2 + y2) (x2 – y2) = 0
⇒ x2 = y2
⇒ x = ± y
∴ f(x) = f(y) does not imply that x = y
For instance,
f(1) = f(– 1) = 1.
Here f(1) = f(–1) but – 1 ≠ 1, Hence not one-one
check onto :
f(x) = x4
Let f (a) = y such that y∈R
x4 = y
⇒ x = ± y¼
Note that y is a real number, but it can be –ve
also,
Ex. put y = – 3
x = (± 3)¼
$$x =±(-\sqrt{3})^\frac{1}{2} $$