Ask & Answer Question

Sets Ncert Solutions Maths class 11th Maths Class 11th Sets

27. If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find

(i) A $\cap$ B

(ii) B $\cap$ C

(iii) A $\cap$ C $\cap$ D

(iv) A $\cap$ C

(v) B $\cap$ D

(vi) A $\cap$ (B $\cup$ C)

(vii) A $\cap$ D

(viii) A $\cap$ (B $\cup$ D)

(ix) (A $\cap$ B) $\cap$ (B $\cup$ C )

(x) (A $\cup$ D) $\cap$ (B $\cup$ C)

6 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

4 months ago

27. (i) A ∩ B = {3,5,7,9,11} ∩ {7,9,11,13}

= {7, 9, 11}

(ii) B ∩ C = {7,9,11,13} ∩ {11,13,15}

= {11,13}

(iii) A ∩ C ∩ D = (A ∩ C) ∩ D

= [ {3,5,7,9,11} ∩ {11,13,15}] ∩ {15,17}.

= {11} ∩ {15,17} = $?$ .

(iv) A ∩ C = {3,5,7,9,11} ∩ {11,13,15}.

= {11}

(v) B ∩ D = {7,9,11,13} ∩ {15,17}= $?$

(vi) A ∩ (B∪ C) = {3,5,7,9,11} ∩ [ {7,9,11,13}∪ {11,13,15}]

= {3,5,7,9,11} ∩ {7,9,11,13,15}.

= {7,9,11}

(vii) A ∩ D = {3,5,7,9,11} ∩ {15,17} = $?$ .

(viii) A ∩ (B ∪ D) = {3,5,7,9,11} ∩ [ {7,9,11,13} ∪ {15,17}]

= {3,5,7,9,11} ∩ {7,9,11,13

...more

27. (i) A ∩ B = {3,5,7,9,11} ∩ {7,9,11,13}

= {7, 9, 11}

(ii) B ∩ C = {7,9,11,13} ∩ {11,13,15}

= {11,13}

(iii) A ∩ C ∩ D = (A ∩ C) ∩ D

= [ {3,5,7,9,11} ∩ {11,13,15}] ∩ {15,17}.

= {11} ∩ {15,17} = $?$ .

(iv) A ∩ C = {3,5,7,9,11} ∩ {11,13,15}.

= {11}

(v) B ∩ D = {7,9,11,13} ∩ {15,17}= $?$

(vi) A ∩ (B∪ C) = {3,5,7,9,11} ∩ [ {7,9,11,13}∪ {11,13,15}]

= {3,5,7,9,11} ∩ {7,9,11,13,15}.

= {7,9,11}

(vii) A ∩ D = {3,5,7,9,11} ∩ {15,17} = $?$ .

(viii) A ∩ (B ∪ D) = {3,5,7,9,11} ∩ [ {7,9,11,13} ∪ {15,17}]

= {3,5,7,9,11} ∩ {7,9,11,13,15,17}

= {7,9,11}

(ix) (A ∩ B) ∩ (B ∪ C) = [ {3,5,7,9,11} ∩ {7,9,11,13}] ∩ [ {7,9,11, 13} ∪ {11,13, 15}]

= {7,9,11} ∩ {7,9,11,13,15}

= {7,9,11}.

(x) (A ∪ D) ∩ (B ∪ C) = [ {3,5,7,9,11} ∪ {15,17}] ∩ [ {7, 9, 11, 13} ∪ {11, 13, 15}

= {3,5,7,9,11,15,17} ∩ {7,9,11,13,15}

= {7,9,11,15}.

less

0 Upvotes 0 Downvotes

Similar Questions for you

The number of elements in the set {x ∈ R : (|x| - 3)|x + 4| = 6} is equal to :

Raj Pandey

(|x| - 3)|x + 4| = 6

Case (i) x < -4
(-x - 3) (- (x + 4) = 6
(x + 3) (x + 4) = 6 ⇒ x² + 7x + 12 = 6 ⇒ x² + 7x + 6 = 0
(x + 1) (x + 6) = 0 ⇒ x = -6 (since x < -4)

Case (ii) -4 ≤ x < 0
(-x - 3) (x + 4) = 6
⇒ -x² - 7x - 12 = 6
⇒ x² + 7x + 18 = 0
The discriminant is D = 7² - 4 (1) (18) = 49 - 72 < 0, so no real solution.

Case (iii) x ≥ 0
(x - 3) (x + 4) = 6
⇒ x² + x - 12 = 6
⇒ x² + x - 18 = 0
x = [-1 ± √ (1² - 4 (1) (-18)] / 2 = [-1 ± √73] / 2
Since x ≥ 0 ⇒ x = (√73 - 1) / 2
Only two solutions.

A natural number has prime factorization given by n = $2^{x} 3^{y} 5^{z},$ where y and z are such that y + z = 5 and $y^{- 1} + z^{- 1} = \frac{5}{6}, y > z .$ Then the number of odd divisors of n, including 1, is:

Vishal Baghel

Given n = 2^x. 3^y. 5^z . (i)

$y + z = 5 & \frac{1}{y} + \frac{1}{z} = \frac{5}{6}$

On solving we get y = 3, z = 2

So, n = 2^x. 3³. 5²

So that no. of odd divisor = (3 + 1) (2 + 1) = 12

Hence no. of divisors including 1 = 12

Let x denote the total number of one-one function from a set A with 3 elements to a set B with 5 elements and y denote the total number of one-one functions form the set A to the set A × B. Then

Payal Gupta

Let A = {a, b, c}, B = {1, 2, 3, 4, 5} n (A × B) = 15

x = number of one-one functions from A to B.

$=^{5} C_{3} . 3! = 60$

y = number of one-one functions for A to (A × B)

$=^{15} C_{3} . 3! = 15 \times 14 \times 13 = 2730$

$\frac{Y}{x} = \frac{2730}{60} \Rightarrow 2 y = 91 x$

The number of elements in the set S = ${x \in R : 2 c o s (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}} i s :$

Payal Gupta

2cos $(\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}$

$- 2 \leq L H S \leq 2 L H S = 2 & R H S = 2 \Rightarrow x = 0 o n l y \to t h e n L H S = 2 a l s o$

RHS $\geq$ 2

66. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + ...

Payal Gupta

66. Given series is 1× 2× 3 + 2× 3 ×4 + 3× 4 ×5 + … to n term

a_n = (n^th term of A. P. 1, 2, 3, …) ´× (n^th terms of A. P. 2, 3, 4) ×

i e, a = 1, d = 2- 1 = 1i e, a = 2, d = 3- 2 = 1

(nth term of A. P. 3, 4, 5)

i e, a = 3, d = 3 -4 = 1.

= [1 + (n -1) 1] ×[2 + (n -1):1]× [3 + (n- 1) 1]

= (1 + n -1)×(2 + n -1)×(3 + n -1)

= n (n + 1)(n + 2)

= n(n² + 2n + n + 2)

=n³ + 2n² + 2n.

S_n = ∑n³ + 3 ∑n² + 2 ∑n

$= {[\frac{m (n + 1)}{2}]}^{2} + \frac{3 n (n + 1) (2 n + 1)}{6} + \frac{2 n (n + 1)}{2}$

= $\frac{n (n + 1)}{2} [\frac{n (n + 1)}{2} + \frac{3}{3} (2 n + 1) + 2]$

= $\frac{n (n + 1)}{2} [\frac{n^{2} + n}{2} + 2 n + 1 + 2]$

$= \frac{n (n + 1)}{2} [\frac{n^{2} + n + 2 \times 2 n + 2 \times 3}{2}]$

$= \frac{n (n + 1)}{2} [\frac{x^{2} + n + 4 x + 6}{2}]$

$= \frac{n (n + 1)}{2} \times \frac{[n^{2} + 5 n + 6]}{2}$

$= \frac{n (n + 1)}{4} [n^{2} + 2 n + 3 n + 6]$

$= \frac{n (n + 1)}{4} [n (n + 2) + 3 (n + 2)]$

$= \frac{n (n + 1) (n + 2) (n + 3)}{4}$

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

27. If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find (i) A ∩ B (ii) B ∩ C (iii) A ∩ C ∩ D (iv) A ∩ C (v) B ∩ D (vi) A ∩ (B ∪ C) (vii) A ∩ D (viii) A ∩ (B ∪ D) (ix) (A ∩ B) ∩ (B ∪ C ) (x) (A ∪ D) ∩ (B ∪ C)

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question