1st PUC Basic Maths Question Bank Chapter 2 Sets, Relations and Functions

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Karnataka 1st PUC Basic Maths Question Bank Chapter 2 Sets, Relations and Functions

sets

Question 1.
If A = {1, 2, 3, B = {2, 3, 4}, C = {3, 4, 5, 6} and U = {1, 2, 3, 4, 5, 6, 7}. Find
(i) (A ∪ B)’
(ii) Verity (A ∩ B)’ = A’ ∪ B’
Answer:
(i) A ∪ B = {1,2, 3, 4} ∴ (A ∪ B)’ = ∪ – (A ∪ B) = {5, 6, 7}
(ii) (A ∩ B) = {2,3}
(A ∩ B)’ = ∪ – (A ∩ B) = {1,4, 5,6} …… (1)
⇒ A’ = ∪ – A = {4, 5, 6, 7}
B’ = ∪ – B = {1, 5, 6, 7}
∴ A’ ∪ B’ = (1,4, 5, 6, 7} ……. (2)
From (1) and (2) (A ∩ B)’ = A’ ∪ B’

Question 2.
n(u) = 700, n(A) = 200, n(B).= 300. n(A ∩ B) = 100. Findn(A’ ∩ B’)
Answer:
n(A’ ∩ B’) = n(A ∪ B)’
= n(U) – n(A ∪ B) = 700 – n(A ∪ B)
Now n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 200 + 300 – 100 = 400
∴ n(A’ ∩ B’) = 700 – 400 = 300

Question 3.
In a town three daily news papers X, Y, Z are published. 52% of the people read paper X, 61 % read Y and 78% read Z. 40% read X and Y; 38% read Y and Z, 46% read X and Z; 18% do not read any of the three news papers. Find the percentage of persons who read all the three papers.
Answer:
Let the number of person is the town be 100.
Now it is given n(X) = 52, n(Y) = 61, n(Z) = 78, n(X ∩ Y) = 40, n(Y ∩ Z) = 38
n(X ∩ Z) = 46, n(X ∪ Y ∪ Z) = 100 – 18 = 82
We have
n(X ∪ Y ∪ Z) = n(X) + n(Y) + n(Z) – n(X ∩ Y) – n(Y ∩ Z) – n(X ∩ Z) + n(X ∩ Y ∩ Z).
⇒ 82 = 52 + 61 + 78 – 40 – 38 – 46 + n(X ∩ Y ∩ Z)
= 191 – 124 + n(X ∩ Y ∩ Z) = 67 + (X ∩ Y ∩ Z)
∴ n(X ∩ Y ∩ Z) = 82 – 67 = 15
Here 15% of the people read all the three papers.

Question 4.
How many integers between 1 and 500 are divisible by 2, 3, or 5.
Answer:
\(\frac { 500 }{ 2 }\) = 250 No. of integer divisible by 2 is 250.
∴ n(A) = 250
\(\frac { 500 }{ 3 }\) = 16666 ∴ No. of integer divisible by 3 is 166. i.e., n(B) = 166
\(\frac { 500 }{ 5 }\) = 100 i.e., No. of integer divisibility 5 is 100 is n(C) = 100
Now \(\frac{500}{2 \times 3}=\frac{500}{6}\) = 83.33 divisible 2 and 3 is 83.
n(A ∩ B) = 83.

We have
n(A ∩ B ∩ C) = n(A) + n(B) + n(c) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
= 250 + 166 + 100 – 83 – 33 – 50 + 17 = 533 – 166 = 367

Question 5.
If A, B and C are three subsets of the universal set U, draw the venn diagram of (A’ ∩ B’)∩C’
Answer:

(A’ ∩ B’)∩C’ – A’ ∩ B’ ∩ C’ = {x/x∈A’ and x∈B’ and x∈C’}
= {x/x∉A, x∉B, x∉C}
The shaded region is the compliment of the set.

Relations and Functions

Question 1.
If A = {a, b}, B = {b, c}, C = {c, d). Find (i) (A × B) ∪ (A × C) (ii) A × (B ∩ C) .
Answer:
(i) A × B = {(a, b), (a, c),(b, b), (b, c)}
A × C = {(a,c), (a,d), (b,c) (b,d)}
(A × B) ∪ (A × C) = {(a, b) (a, c) (b, b) (b, c) (a, d) (b, d)}
(ii) B ∩ C = {c}, A × (B ∩ C) = {a, b} × {c} = {(a, c), (b, c}

Question 2.
If (x, x + y) = (6,2). Find x and y.
Answer:
2x = 6 x + y = 2
∴ x = 3 y = 2 – 3 = – 1.

Question 3.
If A = {x/x ∈ N and x < 3} and B = {x/x² – 16 = 0 and x < 0}. FindB x A,
Answer:
⇒ A = {1,2), since n² – 16 = 0, x = + 4, B = {-4}
∴ B × A = {(-4,1) (-4,2)}

Question 4.
If f(n) = 7x + 10. Find f^-1(x) and f(4), f^-1(5).
Answer:
By data f(x) = 7x + 10
Let f^-1(x) = y ⇒ f(y) = x ⇒ 7y + 10 = x
⇒ y = \(\frac{x-10}{7}\)
Then f^-1(x) = \(\frac{x-10}{7}\) Now f^-1(5) = \(\frac{5-10}{10}=\frac{5}{7}\)
Also, (x) = 7(4) + 10 = 28 + 10 = 38
f(4) . f^-1(5) = 38. \(\left(\frac{-5}{7}\right)=\frac{-190}{7}\)

Question 5.
If f: R → R, 5 defined by f(n) = 9x – 11. Show that f is bijective and find the inverse function.
Answer:
We have f(x) ⇒ 9x – 11
Let f(x₁) = f(x₂) = x₁ = x₂ ⇒ 9x₁ – 1 = 9x₂ – 11
⇒ 9x₁ = 9x₂ ⇒ x₁ = x₂ f is one-one
To show that f is on to, we have to show that for every y ∈ Q (codomain) there exists x ∈ Q (domain) such that f(x) = y.
Let f(x) = y ⇒ 9x – 11 = y
⇒ 9x = y + 11 ∴ x = \(\frac{b+11}{9}\)
Then for any y ∈ Q, we have x ∈ Q, such that f(x) = y. ie., x = \(\frac{b+11}{9}\). Hence f is on to.

Question 6.
If f(x) = x² + 2x + 1, g(x) = 2x – 3. Find (i) fog, (ii) gof.
Answer:
By data f(x) = x²(2x + 1), gx = 2x – 3
(i) fog fg(x) = f(2x – 3) where k = 2x – 3
= k² + 2k + 1 = (2x – 3)² + 2(2x – 3) + 1
= 4x² 9 – 12x + 4x – 6 + 1 = 4x² – 8x + 4

(ii) gof = gf(x) = g(n² + 2x + 1) = g(k)
= 2k – 3 = 2(n² + 2x + 1) – 3
= 2x² + 4x + 2 – 3 = 2x² + 4x – 1
∴ gof(2) = 2(2²) + 4(2) – 1 = 8 + 8 – 1 = 15