Students can Download Basic Maths Chapter 2 Sets, Relations and Functions Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## 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/x^{2} – 16 = 0 and x < 0}. FindB x A,

Answer:

⇒ A = {1,2), since n^{2} – 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(x_{1}) = f(x_{2}) = x_{1} = x_{2} ⇒ 9x_{1} – 1 = 9x_{2} – 11

⇒ 9x_{1} = 9x_{2} ⇒ x_{1} = x_{2} 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) = x^{2} + 2x + 1, g(x) = 2x – 3. Find (i) fog, (ii) gof.

Answer:

By data f(x) = x^{2}(2x + 1), gx = 2x – 3

(i) fog fg(x) = f(2x – 3) where k = 2x – 3

= k^{2} + 2k + 1 = (2x – 3)^{2} + 2(2x – 3) + 1

= 4x^{2} 9 – 12x + 4x – 6 + 1 = 4x^{2} – 8x + 4

(ii) gof = gf(x) = g(n^{2} + 2x + 1) = g(k)

= 2k – 3 = 2(n^{2} + 2x + 1) – 3

= 2x^{2} + 4x + 2 – 3 = 2x^{2} + 4x – 1

∴ gof(2) = 2(2^{2}) + 4(2) – 1 = 8 + 8 – 1 = 15