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Ranak Sharma
Subject: Maths
, asked on 24/5/18
Solve this :
P
rove using properties of sets and their complements .
$\left(1\right)\left(\mathrm{A}\cup \mathrm{B}\right)\cap \left(\mathrm{A}\cup \mathrm{B}\text{'}\right)=\mathrm{A}\phantom{\rule{0ex}{0ex}}\left(2\right)\mathrm{A}-\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{A}-\mathrm{B}\phantom{\rule{0ex}{0ex}}\left(3\right)\left(\mathrm{A}\cup \mathrm{B}\right)-\mathrm{C}=\left(\mathrm{A}-\mathrm{C}\right)\cup \left(\mathrm{B}-\mathrm{C}\right)\phantom{\rule{0ex}{0ex}}\left(4\right)\mathrm{A}-\left(\mathrm{B}\cup \mathrm{C}\right)=\left(\mathrm{A}-\mathrm{B}\right)\cap \left(\mathrm{A}-\mathrm{C}\right)\phantom{\rule{0ex}{0ex}}\left(5\right)\mathrm{A}\cap \left(\mathrm{B}-\mathrm{C}\right)=\left(\mathrm{A}\cap \mathrm{B}\right)-\left(\mathrm{A}\cap \mathrm{C}\right).$
Answer
1
Shikhar
Subject: Maths
, asked on 24/5/18
Good Evening Sir.
I had a doubt regarding sets chapter. I did'nt understand question no. 15 & 16 in miscellaneous exercise of sets in NCERT on page number 27. Can you please explain that.
Answer
1
Ranak Sharma
Subject: Maths
, asked on 22/5/18
Please answer questing 24
Q24. If
$\mathrm{\theta}$
= { x : x
_{$\in $}
N and x
$\le $
10}
A = { x : x is prime and x
$\le $
10}
B = { x : x is a factor of 24}
Verify the following result
(i) A – B = A ⌒ B'
(ii) (A
$\cup $
B)' = A' ⌒ B'
(iii) (A ⌒B)' = A'
$\cup $
B'
Answer
1
Vaishnavi Gupta
Subject: Maths
, asked on 19/5/18
E={x:x^2+x-4}
Answer
1
Sanitya Srivastava
Subject: Maths
, asked on 17/5/18
If R is a set of all real numbers and Q is the set of all rational numbers then what is R-Q?
Answer
2
Jeosph
Subject: Maths
, asked on 17/5/18
Prove that if a is a subset of b; the a intersection b is a
Answer
1
Arvind
Subject: Maths
, asked on 14/5/18
Guys... any1 can solve the 7th question plz... its urgent..
Answer
2
Dipesh Kumar Karesia
Subject: Maths
, asked on 13/5/18
PLZ TELL ME WHAT ( ) AND [ ] SIGNIFIES, INCLUDED OR NOT INCLUDED?? WITH EXAMPLES VERY CONFUSING
Answer
1
Harini
Subject: Maths
, asked on 12/5/18
in an ifinite set , we have infinite elements , but why the cardinality is not infinity rather not defined ????
Answer
1
Vaishnavi Gupta
Subject: Maths
, asked on 12/5/18
A={x:x?=Z and |x|
Answer
1
Vaishnavi Gupta
Subject: Maths
, asked on 12/5/18
E={x:x^2+x-4=0}
Answer
1
Vaishnavi Gupta
Subject: Maths
, asked on 12/5/18
C={1,-1,i,-i}
Answer
1
Samridhi Sinha
Subject: Maths
, asked on 11/5/18
prove A-(A intersection B)= A-B
Answer
1
अर्कज
Subject: Maths
, asked on 11/5/18
Explain 4 ,5, 11, 12 ,13 ,14, 15 and the new signs ,:
$4.A~\left(\underset{i=1}{\overset{n}{\cap}}{A}_{i}\right)=\underset{i=1}{\overset{n}{\cup}}\left(A~{A}_{1}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}5.A~\left(\underset{i=1}{\overset{n}{\cup}}{A}_{i}\right)=\underset{i=1}{\overset{n}{\cap}}\left(A~{A}_{1}\right).\phantom{\rule{0ex}{0ex}}11.A~B=A\cap {B}^{C}$
Answer
1
अर्कज
Subject: Maths
, asked on 11/5/18
Example 12
Answer
1
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Prove using properties of sets and their complements .Solve this :$\left(1\right)\left(\mathrm{A}\cup \mathrm{B}\right)\cap \left(\mathrm{A}\cup \mathrm{B}\text{'}\right)=\mathrm{A}\phantom{\rule{0ex}{0ex}}\left(2\right)\mathrm{A}-\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{A}-\mathrm{B}\phantom{\rule{0ex}{0ex}}\left(3\right)\left(\mathrm{A}\cup \mathrm{B}\right)-\mathrm{C}=\left(\mathrm{A}-\mathrm{C}\right)\cup \left(\mathrm{B}-\mathrm{C}\right)\phantom{\rule{0ex}{0ex}}\left(4\right)\mathrm{A}-\left(\mathrm{B}\cup \mathrm{C}\right)=\left(\mathrm{A}-\mathrm{B}\right)\cap \left(\mathrm{A}-\mathrm{C}\right)\phantom{\rule{0ex}{0ex}}\left(5\right)\mathrm{A}\cap \left(\mathrm{B}-\mathrm{C}\right)=\left(\mathrm{A}\cap \mathrm{B}\right)-\left(\mathrm{A}\cap \mathrm{C}\right).$

I had a doubt regarding sets chapter. I did'nt understand question no. 15 & 16 in miscellaneous exercise of sets in NCERT on page number 27. Can you please explain that.

Q24. If $\mathrm{\theta}$ = { x : x

_{$\in $}N and x $\le $10}A = { x : x is prime and x $\le $10}

B = { x : x is a factor of 24}

Verify the following result

(i) A – B = A ⌒ B'

(ii) (A$\cup $ B)' = A' ⌒ B'

(iii) (A ⌒B)' = A' $\cup $ B'

$4.A~\left(\underset{i=1}{\overset{n}{\cap}}{A}_{i}\right)=\underset{i=1}{\overset{n}{\cup}}\left(A~{A}_{1}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}5.A~\left(\underset{i=1}{\overset{n}{\cup}}{A}_{i}\right)=\underset{i=1}{\overset{n}{\cap}}\left(A~{A}_{1}\right).\phantom{\rule{0ex}{0ex}}11.A~B=A\cap {B}^{C}$