Ask & Answer Question

Relations and Functions Ncert Solutions Maths class 11th Maths Class 11th Relations and Functions

7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that (i) A × (B∩ C) = (A × B)∩ (A × C). (ii) A × C is a subset of B × D.

5 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

4 months ago

7. Given,

A= {1, 2}, B = {1,2,3,4}, C= {5,6} and D= {5,6,7,8}

(i) L.H. S = A × (B∩ C) = {1,2} [ {1,2,3,4} ∩ {5,6}]

= {1,2}×

= .

R.H.S = (A× B)∩ (A ×C)= [ {1,2}× {1,2,3,4}]∩ [ {1,2} {5,6}]

= [ { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)]∩ [ {1,5), (1,6), (2,5), (2,6)}]

= .

Hence, L.H.S= R.H.S.

(ii) A× C = {1, 2}× {5,6}

= { (1,5), (1,6), (2,5), (2,6)}

B× D = {1,2,3,4} × {5,6,7,8}

= { (1,5), (1,6), (1,7), (1,8), (2,5), (2,6), (2,7), (2,8), (3,5), (3,6), (3,7), (3,8), (4,5), (4,6), (4,7), (4,8)}

As every element of A C is al

...more

7. Given,

A= {1, 2}, B = {1,2,3,4}, C= {5,6} and D= {5,6,7,8}

(i) L.H. S = A × (B∩ C) = {1,2} [ {1,2,3,4} ∩ {5,6}]

= {1,2}×

= .

R.H.S = (A× B)∩ (A ×C)= [ {1,2}× {1,2,3,4}]∩ [ {1,2} {5,6}]

= [ { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)]∩ [ {1,5), (1,6), (2,5), (2,6)}]

= .

Hence, L.H.S= R.H.S.

(ii) A× C = {1, 2}× {5,6}

= { (1,5), (1,6), (2,5), (2,6)}

B× D = {1,2,3,4} × {5,6,7,8}

= { (1,5), (1,6), (1,7), (1,8), (2,5), (2,6), (2,7), (2,8), (3,5), (3,6), (3,7), (3,8), (4,5), (4,6), (4,7), (4,8)}

As every element of A C is also an element of B× D.

A ×C $\subset$ B ×D

less

0 Upvotes 0 Downvotes

Similar Questions for you

The probability that a relation R from ${x, y} t o {x, y}$ is both symmetric and transitive, is equal to

alok kumar singh

Total number of possible relation = $2^{n^{2}} = 2^{4} = 16$

Favourable relations = $?, {(x, x)}, {(y, y)}$

${(x, x), (y, y)}$

${(x, x), (y, y), (x, y), (y, x)}$

Probability = $\frac{5}{16}$

Choose the incorrect statement about the two circles whose equations are given below:
x² + y² - 10x - 10y + 41 = 0 and x² + y² - 16x - 10y + 80 = 0

Vishal Baghel

Circle S? : x² + y² - 10x - 10y + 41 = 0.
Center C? = (5, 5). Radius r? = √ (5² + 5² - 41) = √ (25 + 25 - 41) = √9 = 3.
Circle S? : x² + y² - 16x - 10y + 80 = 0.
Center C? = (8, 5). Radius r? = √ (8² + 5² - 80) = √ (64 + 25 - 80) = √9 = 3.
The solution checks if the center of one circle lies on the other.
Put C? (8, 5) into S? : 8² + 5² - 10 (8) - 10 (5) + 41 = 64 + 25 - 80 - 50 + 41 = 130 - 130 = 0. So C? lies on S?
Put C? (5, 5) into S? : 5² + 5² - 16 (5) - 10 (5) + 80 = 25 + 25 - 80 - 50 + 80 = 130 - 130 = 0. So C? lies on S?
This means bo

...more

The inverse of y = 5^logx is:

Vishal Baghel

Kindly consider then following figure

Let [t] denote the greatest integer $\leq t .$ Then the value of 8 $\cdot \int_{- \frac{1}{2}}^{1} ([2 x] + | x |) d x i s________.$

Raj Pandey

$l = \int_{- 1 / 2}^{1} [2 x] d x + \int_{- 1 / 2}^{1} | 2 x | d x$

let $l_{1} = \int_{- 1 / 2}^{1} [2 x] d x p u t 2 x = t \Rightarrow d x = \frac{d t}{2}$

$∴ l = l_{1} + l_{2} = 0 + \frac{5}{8} = \frac{5}{8} \Rightarrow 8 l = 5$

36. Let A = {9,10,11,12,13} and let f : A→N be defined by f (n) = the highest prime factor of n. Find the range of f.

Payal Gupta

36. Given, A={9,10,11,12,13}.

f(x)=the highest prime factor of n.

and f: A → N.

Then, f(9)=3 [? prime factor of 9=3]

f (10)=5 [? prime factor of 10=2,5]

f(11)=11 [? prime factor of 11 = 11]

f(12)=3 [? prime factor of 12 = 2, 3]

f(13)=13 [? prime factor of 13 = 13]

?Range of f=set of all image of f(x) = {3,5,11,13}.

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that (i) A × (B∩ C) = (A × B)∩ (A × C). (ii) A × C is a subset of B × D.

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