Question Number 7323 by rohit meena last updated on 23/Aug/16
$${find}\:{the}\:{value}\:{of} \\ $$$$\Delta'\Lambda\:{b}\:{and}\:\left({B}'{U}\:{A}\right)\:\boldsymbol{\Lambda}\:{B} \\ $$$${if}\:{U}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{10}\right\} \\ $$$${A}=\left\{\mathrm{2},\mathrm{4},\mathrm{5},\mathrm{6}\right\} \\ $$$${B}=\left\{\mathrm{1},\mathrm{5},\mathrm{8},\mathrm{9}\right\} \\ $$$$\Lambda\:{is}\:{a}\:{unoin} \\ $$
Commented by rohit meena last updated on 23/Aug/16
$${answer}\:{please} \\ $$
Commented by Rasheed Soomro last updated on 24/Aug/16
$${What}\:{is}\:\Delta? \\ $$
Commented by rohit meena last updated on 24/Aug/16
$${A}\:{hai} \\ $$
Commented by rohit meena last updated on 24/Aug/16
$${A}\:{hai} \\ $$
Answered by Rasheed Soomro last updated on 24/Aug/16
![U={1,2,3,4,5,6,7,8,9,10},A={2,4,5,6}, B={1,5,8,9} (i) A′∩B (ii) (B ′∪A)∩B −−−−−−−−−−−−−−−−−−−−−−−−−− A′∩B=(U−A)∩B =({1,2,3,4,5,6,7,8,9,10}−{2,4,5,6})∩{1,5,8,9} ={1,3,7,8,9,10}∩{1,5,8,9} ={1,8,9} (B ′∪A)∩B Way#1: First simplify using properties. (B ′∪A)∩B=(B ′∩B)∪(A∩B) [∩ is right-distributive over ∪] =φ∪(A∩B) [B ′∩B=φ] =A∩B ={2,4,5,6}∩{1,5,8,9} ={5} Way#2:Using sets directly. B ′=U−B={1,2,3,4,5,6,7,8,9,10}−{1,5,8,9} ={2,3,4,6,7,10} (B ′∪A)∩B=({2,3,4,6,7,10}∪{2,4,5,6})∩{1,5,8,9} ={2,3,4,5,6,7,10}∩{1,5,8,9} ={5}](Q7346.png)
$$\:{U}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{10}\right\},{A}=\left\{\mathrm{2},\mathrm{4},\mathrm{5},\mathrm{6}\right\},\:{B}=\left\{\mathrm{1},\mathrm{5},\mathrm{8},\mathrm{9}\right\} \\ $$$$\left({i}\right)\:{A}'\cap{B}\:\:\:\:\:\:\left({ii}\right)\:\left({B}\:'\cup{A}\right)\cap{B} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${A}'\cap{B}=\left({U}−{A}\right)\cap{B} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left(\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{10}\right\}−\left\{\mathrm{2},\mathrm{4},\mathrm{5},\mathrm{6}\right\}\right)\cap\left\{\mathrm{1},\mathrm{5},\mathrm{8},\mathrm{9}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left\{\mathrm{1},\mathrm{3},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{10}\right\}\cap\left\{\mathrm{1},\mathrm{5},\mathrm{8},\mathrm{9}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left\{\mathrm{1},\mathrm{8},\mathrm{9}\right\} \\ $$$$\left({B}\:'\cup{A}\right)\cap{B} \\ $$$$ \\ $$$${Way}#\mathrm{1}:\:{First}\:{simplify}\:{using}\:{properties}. \\ $$$$\left({B}\:'\cup{A}\right)\cap{B}=\left({B}\:'\cap{B}\right)\cup\left({A}\cap{B}\right)\:\left[\cap\:{is}\:{right}-{distributive}\:{over}\:\cup\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\phi\cup\left({A}\cap{B}\right)\:\:\left[{B}\:'\cap{B}=\phi\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={A}\cap{B} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left\{\mathrm{2},\mathrm{4},\mathrm{5},\mathrm{6}\right\}\cap\left\{\mathrm{1},\mathrm{5},\mathrm{8},\mathrm{9}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left\{\mathrm{5}\right\} \\ $$$$ \\ $$$${Way}#\mathrm{2}:{Using}\:\:{sets}\:{directly}.\: \\ $$$$\:\:\:\:{B}\:'={U}−{B}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{10}\right\}−\left\{\mathrm{1},\mathrm{5},\mathrm{8},\mathrm{9}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left\{\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{6},\mathrm{7},\mathrm{10}\right\} \\ $$$$\left({B}\:'\cup{A}\right)\cap{B}=\left(\left\{\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{6},\mathrm{7},\mathrm{10}\right\}\cup\left\{\mathrm{2},\mathrm{4},\mathrm{5},\mathrm{6}\right\}\right)\cap\left\{\mathrm{1},\mathrm{5},\mathrm{8},\mathrm{9}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left\{\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{10}\right\}\cap\left\{\mathrm{1},\mathrm{5},\mathrm{8},\mathrm{9}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left\{\mathrm{5}\right\} \\ $$$$ \\ $$