Thinking Process

Solve the equation and get the value of $x$.

Solution

(i) We have,

$\begin{array}{rl}A& =\{x:x\in R,2x+11=15\}\\ \therefore 2x+11& =15\\ 2x& =15-11\Rightarrow 2x=4\\ x& =2\\ A& =\{2\}\\ \text{(ii) We have},B& =\{x\mid x^2=x,x\in R\}\\ x^2& =x\\ \Rightarrow x^2-x& =0\Rightarrow x(x-1)=0\\ \Rightarrow x& =0,1\\ \Rightarrow B& =\{0,1\}\end{array}$

(iii) We have, $C=\{x\mid x$ is a positive factor of prime number $p\}$.

Since, positive factors of a prime number are 1 and the number itself.

$\therefore C=\{1,p\}$

Thinking Process

Solve the given equation and get the value of respective variable.

Solution

(i) We have,

$\begin{array}{cc}\text{ We have, }& D=\{t\mid t^3=t,t\in R\}\\ \therefore & t^3=t\\ \Rightarrow & t^3-t=0\Rightarrow t(t^2-1)=0\\ \Rightarrow & t(t-1)(t+1)=0\Rightarrow t=0,1,-1\\ \therefore & D=\{-1,0,1\}\end{array}$

$\begin{array}{rl}& \begin{array}{cc}\therefore & t^3=t\\ \Rightarrow & t^3-t=0\Rightarrow t(t^2-1)=0\end{array}\\ & \therefore D=\{-1,0,1\}\end{array}$

(ii)

$\begin{array}{rl}& \text{ We have, }E=\{w|,\frac{w-2}{w+3}=3.,w\in R\}\\ & \therefore \frac{w-2}{w+3}=3\\ & \Rightarrow w-2=3w+9\Rightarrow w-3w=9+2\\ & \Rightarrow -2w=11\Rightarrow w=\frac{-11}{2}\\ & \therefore E=\{\frac{-11}{2}\}\end{array}$

(iii) We have,

$F=\{x\mid x^4-5x^2+6=0,x\in R\}$

$\therefore x^4-5x^2+6=0$

$\Rightarrow x^4-3x^2-2x^2+6=0$

$\Rightarrow x^2(x^2-3)-2(x^2-3)=0$

$\Rightarrow (x^2-3)(x^2-2)=0$

$\Rightarrow x=\pm \sqrt{3},\pm \sqrt{2}$

$\therefore F=\{-\sqrt{3},-\sqrt{2},\sqrt{2},\sqrt{3}\}$

Note in roaster form, the order in which elements are listed is immaterial. Thus, we can also write $F=\{-\sqrt{3},\sqrt{2},-\sqrt{2},\sqrt{3}\}$.

Thinking Process

First, write all the factors of $2^{p-1}$, where $p=1,2,3,\dots ,p$ and then get $y$.

Solution

$Y=\{x\mid x.$ is a positive factor of the number $2^{p-1}(2^p-1)$, where $2^p-1$ is a prime number $\}$. So, the factor of $2^{p-1}$ are $1,2,2^2,2^3,\dots ,2^{p-1}$.

$\therefore Y=\{1,2,2^2,2^3,\dots ,2^{p-1},2^p-1\}$

Solution

(i) Since, the factors of 35 are $1,5,7$ and 35 . So, statement (i) is true.

(ii) Since, the factors of 128 are 1, 2, 4, 8, 16, 32, 64 and 128.

$\begin{array}{rl}\therefore \text{ Sum of factors }& =1+2+4+8+16+32+64+128\\ & =255\ne 2\times 128\end{array}$

So, statement (ii) is false. (iii) We have,

$x^4-5x^3+2x^2-112x+6=0$

$\therefore$ For $x=3$,

$\Rightarrow 81-135+18-336+6=0$

$\Rightarrow$ $-346=0$

which is not true.

Hence, statement (iii) is true.

(iv) $\because$ $496=2^4\times 31$

So, the factors of 496 are $1,2,4,8,16,31,62,124,248$ and 496.

$\therefore$ Sum of factors $=1+2+4+8+16+31+62+124+248+496$

$=992=2(496)$

So, $496\in \{y\mid$ the sum of all the positive factor of $y$ is $2y\}$.

Hence, statement (iv) is false.

Solution

Given, $L=\{1,2,3,4\},M=\{3,4,5,6\}$ and $N=\{1,3,5\}$

$\therefore M\cup N=\{1,3,4,5,6\}$

$L-(M\cup N)=\{2\}$

Now,

$L-M=\{1,2\},L-N=\{2,4\}$

$\therefore (L-M)\cap (L-N)=\{2\}$

Hence, $L-(M\cup N)=(L-M)\cap (L-N)$.

Solution

(i) Let $x\in A$

$\Rightarrow x\in A$ or $x\in B\Rightarrow x\in A\cup B$

Hence, $\subset A\cup B$

(ii) If

$\begin{array}{c}A\subset B\\ x\in A\cup B\end{array}$

Let

$\Rightarrow$ $x\in A\text{ or }x\in B\Rightarrow x\in B$

$\Rightarrow$ $A\cup B\subset B$

But $B\subset A\cup B$

From Eqs. (i) and (ii),

$\Rightarrow A\cup B=B$

If $A\cup B=B$

Let $y\in A$

$\Rightarrow y\in A\cup B\Rightarrow y\in B$

$\Rightarrow A\subset B$

Hence, $A\subset B\Rightarrow A\cup B=B$

(iii) Let $\Rightarrow x\in A\cap B$

$\Rightarrow x\in A\text{and}x\in B\Rightarrow x\in A$

Hence, $A\cap B\subset A$

Solution

We have,

$N=\{1,2,3,4,\dots ,100\}$

(i) Required subset $=\{2,4,6,8,\dots ,100\}$

(ii) Required subset $=\{1,4,9,16,25,36,49,64,81,100\}$

Solution

Given,

$X=\{1,2,3\}$

(i) $\{4n\mid n\in X\}=\{4,8,12\}$

(ii) $\{n+6\mid n\in X\}=\{7,8,9\}$

(iii) $\frac{n}{2}|,n\in X=\frac{1}{2}.,1,\frac{3}{2}$

(iv) $\{n-1\mid n\in X\}=\{0,1,2\}$

Solution

Given,

$Y=\{1,2,3,\dots ,10\}$

(i) $\{a:a\in Y.$ and $.a^2\notin Y\}=\{4,5,6,7,8,9,10\}$

(ii) $\{a:a+1=6,a\in Y\}=\{5\}$

(iii) is less than 6 and $a\in Y=\{1,2,3,4,5\}$,

Solution

Solution

Thinking Process

To prove this we have to show that $(A-B)\cap (A-C)\subseteq A-(B\cup C)$ and $A-(B\cup C)\subseteq (A-B)\cap (A-C)$

Solution

Let $x\in (A-B)\cap (A-C)$

$\Rightarrow x\in (A-B)$ and $x\in (A-C)$

$\Rightarrow (x\in A$ and $x\notin B)$ and $(x\in A$ and $x\notin C)$

$\Rightarrow x\in A$ and $(x\notin B$ and $x\notin C)$

$\Rightarrow x\in A$ and $x\notin (B\cup C)$

$\Rightarrow x\in A-(B\cup C)$

$\Rightarrow (A-B)\cap (A-C)\subset A-(B\cup C)$

Now, let $y\in A-(B\cup C)$

$\Rightarrow$ $y\in A$ and $y\notin (B\cup C)$

$\Rightarrow y\in A$ and $(y\notin B$ and $y\notin C)$

$\Rightarrow (y\in A$ and $y\notin B)$ and $(y\in A$ and $y\notin C)$

$\Rightarrow y\in (A-B)$ and $y\in (A-C)$

$\Rightarrow y\in (A-B)\cap (A-C)$

$\Rightarrow A-(B\cup C)\subset (A-B)\cap (A-C)$

From Eqs. (i) and (ii),

$A-(B\cup C)=(A-B)\cap (A-C)$

Thinking Process

To solve the above problem, use distributive law on sets

i.e., $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$.

Solution

$\begin{array}{rl}LHS& =(A-B)\cup (A\cap B)\\ & =[(A-B)\cup A]\cap [(A-B)\cup B]\\ & =A\cap (A\cup B)=A=RHS\end{array}$

Hence, given statement is true.

Solution

See the Venn diagrams given below, where shaded portions are representing $A-(B-C)$ and $(A-B)-C$ respectively.

Clearly, $A-(B-C)\ne (A-B)-C$

Hence, given statement is false.

Solution

Let

$x\in A\cap C$

$\Rightarrow x\in A$ and $x\in C$

$\Rightarrow x\in B$ and $x\in C[\because A\subset B]$

$\Rightarrow x\in (B\cap C)\Rightarrow (A\cap C)\subset (B\cap C)$

Hence, given statement is true.

Solution

Let

$x\in A\cup C$

$\Rightarrow$ $x\in A$ and $x\in C$

$\Rightarrow x\in B$ and $x\in C[\because A\subset B]$

$\Rightarrow x\in B\cup C\Rightarrow A\cup C\subset B\cup C$

Hence, given statement is true.

Solution

Let $x\in A\cup B$

$\Rightarrow$ $x\in A$ and $x\in B$

$\Rightarrow x\in C$ and $x\in C$ $[\because A\subset C$ and $B\subset C]$

$\Rightarrow$ $x\in C\Rightarrow A\cup B\subset C$

Hence, given statement is true.

Thinking Process

To solve the above problem, use distributive law i.e., $A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$.

Solution

$LHS=A\cup (B-A)=A\cup (B\cap A^{\prime })$

$[\because A-B=A\cap B^{\prime }]$

$\begin{array}{cc}=(A\cup B)\cap (A\cup A^{\prime })=(A\cup B)\cap U& [\because A\cup A^{\prime }=\cup ]\\ =A\cup B=RHS& [\because A\cap U=A]\end{array}$

Solution

$\begin{array}{rl}LHS& =A-(A-B)=A-(A\cap B^{\prime })\\ & =A\cap (A\cap B^{\prime }{)}^{\prime }=A\cap [A^{\prime }\cup (B^{\prime }{)}^{\prime }]\\ & =A\cap (A^{\prime }\cup B)\\ & =(A\cap A^{\prime })\cup (A\cap B)=\phi \cup (A\cap B)\\ & =A\cap B=RHS\end{array}$

Solution

$\begin{array}{rl}LHS& =A-(A\cap B)=A\cap (A\cap B{)}^{\prime }[\because A-B=A\cap B^{\prime }]\\ & =A\cap (A^{\prime }\cup B^{\prime })[\because (A\cap B{)}^{\prime }=A^{\prime }\cup B^{\prime }]\\ & =(A\cap A^{\prime })\cup (A\cap B^{\prime })=\varphi \cup (A\cap B^{\prime })[\because (A^{\prime }{)}^{\prime }=A]\\ & =A\cap B^{\prime }\\ & =A-B=RHS\end{array}$

Solution

$\begin{array}{rl}LHS& =(A\cup B)-B=(A\cup B)\cap B^{\prime }[\because A-B=A\cap B^{\prime }]\\ & =(A\cap B^{\prime })\cup (B\cap B^{\prime })=(A\cap B^{\prime })\cup \varphi [\because B\cap B^{\prime }=\varphi ]\\ & =A\cap B^{\prime }[\because A\cup \varphi =A]\\ & =A-B=RHS\end{array}$

Thinking Process

First of all solve the given equation and get the value of $x$.

Solution

Since

$\begin{array}{rl}T=\{x|,\frac{x+5}{x-7}-5.& =\frac{4x-40}{13-x}\}\\ \because \frac{x+5}{x-7}-5& =\frac{4x-40}{13-x}\end{array}$

$\begin{array}{rl}& \Rightarrow \frac{x+5-5(x-7)}{x-7}=\frac{4x-40}{13-x}\\ & \Rightarrow \frac{x+5-5x+35}{x-7}=\frac{4x-40}{13-x}\\ & \Rightarrow \frac{-4x+40}{x-7}=\frac{4x-40}{13-x}\\ & \Rightarrow -(4x-40)(13-x)=(4x-40)(x-7)\\ & \Rightarrow (4x-40)(x-7)+(4x-40)(13-x)=0\\ & \Rightarrow (4x-40)(x-7+13-x)=0\\ & \Rightarrow 4(x-10)6=0\\ & \Rightarrow 24(x-10)=0\\ & \Rightarrow x=10\\ & \therefore T=\{10\}\end{array}$

Hence, $T$ is not an empty set.

Solution

Let

$\Rightarrow x\in A\cap (B\cup C)$

$\Rightarrow x\in A$ and $x\in (B\cup C)$

$\Rightarrow x\in A$ and $(x\in B$ or $x\in C)$

$\Rightarrow (x\in A$ and $x\in B)$ or $(x\in A$ and $x\in C)$

$\Rightarrow x\in A\cap B$ or $x\in A\cap C$

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

$\Rightarrow A\cap (B\cup C)\subset (A\cap B)\cup (A\cap C)$

Again, let

$\Rightarrow y\in (A\cap B)\cup (A\cap C)$

$\Rightarrow y\in (A\cap B)$ or $y\in (A\cap C)$

$\Rightarrow (y\in A$ and $y\in B)$ or $(y\in A$ and $y\in C)$

$\Rightarrow y\in A$ and $(y\in B$ or $y\in C)$

$\Rightarrow y\in A$ and $y\in B\cup C$

$\Rightarrow y\in A\cap (B\cup C)$

$\Rightarrow (A\cap B)\cup (A\cap C)\subset A\cap (B\cup C)$

From Eqs. (i) and (ii)

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

Solution

Let $M$ be the set of students who passed in Mathematics, $E$ be the set of students who passed in English and $S$ be the set of students who passed in Science.

Then, $n(u)=100$

$\begin{array}{rl}n(E)=15,n(M)& =12,n(S)=8,n(E\cap M)=6,n(M\cap S)=7\\ n(E\cap S)& =4,\text{ and }n(E\cap M\cap S)=4\end{array}$

$\because n(E)=15$

$\Rightarrow a+b+e+f=15$

and $n(M)=12$

$\Rightarrow b+c+e+d=12$

On substituting the values of $d,e$ and $f$ in Eq. (iii), we get

$3+4+0+g=8$

$\Rightarrow g=1$

On substituting the value of $b,e$ and $d$ in Eq. (ii), we get

$2+c+4+3=12$

$\Rightarrow c=13$

On substituting $b,e$, and $f$ in Eq. (i), we get

$a+2+4+0=15$

$\Rightarrow$ $a=9$

(i) Number of students who passed in English and Mathematics but not in Science

$=b=2$

(ii) Number of students who passed in Mathematics and Science but not in English

$=d=3$

(iii) Number of students who passed in Mathematics only $=c=3$

(iv) Number of students who passed in more than one subject

$\begin{array}{rl}& =b+e+d+f\\ & =2+4+3+0=9\end{array}$

Alternate Method

Let $E$ denotes the set of student who passed in English. $M$ denotes the set of students who passed in Mathematics. $S$ denotes the set of students who passed in Science.

Now,

$\begin{array}{rl}n(U)& =100,n(E)=15,n(m)=12,n(S)=8\\ n(E\cap M)& =6,n(M\cap S)=7\\ n(E\cap S)& =4,n(E\cap M\cap S)=4\end{array}$

(i) Number of students passed in English and Mathematics but not in Science

$\text{ i.e., }\begin{array}{rl}n(E\cap M\cap S^{\prime })& =n(E\cap M)-n(E\cap M\cap S)[\because A\cap B^{\prime }=A-(A\cap B)]\\ & =6-4=2\end{array}$

(ii) Number of students passed in Mathematics and Science but not in English.

$\text{ i.e., }n(M\cap S\cap E^{\prime })=n(M\cap S)-n(M\cap S\cap E)$

$=7-4=3$

(iii) Number of students passed in mathematics only

$\text{ i.e., }\begin{array}{rl}n(M\cap S^{\prime }\cap E^{\prime })& =n(M)-n(M\cap S)-n(M\cap E)+n(M\cap S\cap E)\\ & =12-7-6+4=3\end{array}$

(iv) Number of students passed in more than one subject only

$\text{ i.e., }\begin{array}{rl}n(E\cap M)+n(M& \cap S)+n(E\cap S)-3n(E\cap M\cap S)+n(E\cap M\cap S)\\ & =6+7+4-4\times 3+4\\ & =17-12+4=5+4=9\end{array}$

Solution

Let $C$ be the set of students who play cricket and $T$ be the set of students who play tennis. Then

$n(U)=60,n(C)=25,n(T)=20\text{, and }n(C\cap T)=10$

$\therefore$ $\begin{array}{rl}n(C\cup T)& =n(C)+n(T)-n(C\cap T)\\ & =25+20-10=35\end{array}$

$\therefore$ Number of students who play neither $=n(U)-n(C\cup T)$

$=60-35=25$

Thinking Process

To solve this problem, use the formula for all the three subjects $n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C)$.

Solution

Let $M$ be the set of students who study Mathematics, $P$ be the set of students who study Physics and $C$ be the set of students who study Chemistry.

Then, $n(U)=200,n(M)=120,n(P)=90$

$\begin{array}{c}n(C)=70,n(M\cap P)=40,n(P\cap C)=30,\\ \\ n(C\cap M)=50,n(M^{\prime }\cap P^{\prime }\cap C^{\prime })=20,\\ \\ n(U)-n(M\cup P\cup C)=20,\\ \\ \because n(M\cup P\cup C)=200-20=180\\ \\ \Rightarrow n(M\cup P\cup C)=n(M)+n(P)+n(C)-n(C\cap M)+n(M\cap P\cap C)\\ \\ \Rightarrow 180=120+90+70-40-30-50+n(M\cap P\cap C)\\ \\ \Rightarrow n(M\cap P\cap C)=180-160=20\end{array}$

So, the number of students who study all the three subjects is 20 .

Solution

Let $A$ be the set of families which buy newspaper $A,B$ be the set of families which buy newspaper $B$ and $C$ be the set of families which buy newspaper $C$.

Then,

$n(U)=10000,n(A)=40$% n(B)=20% and n(C)=10%

$n(A\cap B)=5$ %

$n(B\cap C)=3$%

$n(A\cap C)=4$ %

$n(A\cap B\cap C)=2$%

(i) Number of families which buy newspaper $A$ only

$=n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)$

=(40-5-4+2)%=33 %

$10000\times 33/100=3300$

(ii) Number of families which buy none of $A,B$ and $C$

$=n(U)-n(A\cup B\cup C)$

$=n(U)-[n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)$

=100-[40+20+10-5-3-4+2] =100-60 %=40 %

$=10000\times \frac{40}{100}=4000$

Solution

Let $F$ be the set of students who study French, $E$ be the set of students who study English and $S$ be the set of students who study Sanskrit.

Then,

$\begin{array}{rl}n(U)=50,n(F)& =17,n(E)=13,\text{ and }n(S)=15,\\ n(F\cap E)& =9,n(E\cap S)=4,n(F\cap S)=5,\\ n(F\cap E\cap S)& =3\end{array}$

$\because n(F)=17$

$\begin{array}{cc}\Rightarrow & a+b+e+f=17\\ \Rightarrow & n(E)=13\\ \Rightarrow & b+c+d+e=13\\ \Rightarrow & n(S)=15\\ \Rightarrow & d+e+f+g=15\\ \Rightarrow & n(F\cap E)=9\\ & b+e=9\\ & n(E\cap S)=4\end{array}$

$\begin{array}{rl}\Rightarrow & e+d=4\\ \Rightarrow & n(F\cap S)=5\\ \Rightarrow & n(F\cap E\cap S)=3\\ \Rightarrow & e=3\end{array}$

From Eqs. (vi) and (vii), $f=2$

From Eqs. (v)and (vii), $d=1$

From Eqs. (iv) and (vii), $b=6$

On substituting the values of $e,f$ and $d$ in Eq. (iii), we get

$\begin{array}{rl}1+3+2+g& =15\\ \Rightarrow g& =9\end{array}$

On substituting the values of $b,d$ ande in E

(ii), we get

$\begin{array}{rl}6+c+1+3& =13\\ c& =3\end{array}$

On substituting the values of $b,e$ and $f$ in E

(i), we get

$a+6+3+2=17$

$\Rightarrow a=6$

(i) Number of students who study French only, $a=6$

(ii) Number of students who study English only, $c=3$

(iii) Number of students who study Sanskrit only, $g=9$

(iv) Number of students who study English and Sanskrit but not French, $d=1$

(v) Number of students who study French and Sanskrit but not English, $f=2$

(vi) Number of students who study French and English but not Sanskrit, $b=6$

(vii) Number of students who study atleast one of the three languages

$\begin{array}{rl}& =a+b+c+d+e+f+g\\ & =6+6+3+1+3+2+9=30\end{array}$

(viii) Number of students who study none of three languages $=$ Total students - Students who study atleast one of the three languages

$=50-30=20$

Thinking Process

First find the total number of elements for the both sets, then compare them.

Solution

(c) If elements are not repeated, then number of elements in $A_1\cup A_2\cup A_3,\dots \cup A_{30}$ is $30\times 5$.

But each element is used 10 times, so

$S=\frac{30\times 5}{10}=15$

If elements in $B_1,B_2,\dots ,B_n$ are not repeated, then total number of elements is $3n$ but each element is repeated 9 times, so

$\begin{array}{rl}& S=\frac{3n}{9}\Rightarrow 15=\frac{3n}{9}\\ \therefore & n=45\end{array}$

• Option (a) 15: This option is incorrect because if ( n = 15 ), then the total number of elements in ( B_1, B_2, \ldots, B_{15} ) would be ( 3 \times 15 = 45 ). Since each element is repeated 9 times, the total number of unique elements would be ( \frac{45}{9} = 5 ), which contradicts the given condition that ( S ) has 15 unique elements.

• Option (b) 3: This option is incorrect because if ( n = 3 ), then the total number of elements in ( B_1, B_2, \ldots, B_3 ) would be ( 3 \times 3 = 9 ). Since each element is repeated 9 times, the total number of unique elements would be ( \frac{9}{9} = 1 ), which contradicts the given condition that ( S ) has 15 unique elements.

• Option (d) 35: This option is incorrect because if ( n = 35 ), then the total number of elements in ( B_1, B_2, \ldots, B_{35} ) would be ( 3 \times 35 = 105 ). Since each element is repeated 9 times, the total number of unique elements would be ( \frac{105}{9} \approx 11.67 ), which is not an integer and contradicts the given condition that ( S ) has 15 unique elements.

Thinking Process

We know that, if a set A contains n elements, then the number of subsets of $A$ is equal to $2^n$.

Solution

(a) Since, number of subsets of a set containing melements is 112 more than the subsets of the set containing $n$ elements.

$\because 2^m-2^n=112$

$\Rightarrow 2^n\cdot (2^{m-n}-1)=2^4\cdot 7$

$\Rightarrow 2^n=2^4\text{ and }{2}^{m-n}-1=7$

$\Rightarrow n=4\text{ and }{2}^{m-n}=8$

$\Rightarrow 2^{m-n}=2^3\Rightarrow m-n=3$

$\Rightarrow m-4=3\Rightarrow m=4+3$

$\therefore m=7$

• Option (b) 7,4: This option is incorrect because if ( m = 7 ) and ( n = 4 ), then the number of subsets of the first set would be ( 2^7 = 128 ) and the number of subsets of the second set would be ( 2^4 = 16 ). The difference between the number of subsets would be ( 128 - 16 = 112 ), which matches the given condition. Therefore, this option is actually correct, not incorrect.

• Option (c) 4,4: This option is incorrect because if ( m = 4 ) and ( n = 4 ), then the number of subsets of both sets would be ( 2^4 = 16 ). The difference between the number of subsets would be ( 16 - 16 = 0 ), which does not match the given condition of 112 more subsets.

• Option (d) 7,7: This option is incorrect because if ( m = 7 ) and ( n = 7 ), then the number of subsets of both sets would be ( 2^7 = 128 ). The difference between the number of subsets would be ( 128 - 128 = 0 ), which does not match the given condition of 112 more subsets.

Solution

(b) We know that, $(A\cap B{)}^{\prime }=(A^{\prime }\cup B^{\prime })$ and $(A^{\prime }{)}^{\prime }=A$

$\begin{array}{rl}\therefore & =(A\cap B^{\prime }{)}^{\prime }\cup (B\cap C)\\ & =[A^{\prime }\cup (B^{\prime }{)}^{\prime }]\cup (B\cap C)\\ & =(A^{\prime }\cup B)\cup (B\cap C)=A^{\prime }\cup B\end{array}$

• Option (a): $A^{\prime }\cup B\cup C$ is incorrect because the original expression $(A\cap B^{\prime }{)}^{\prime }\cup (B\cap C)$ does not include the union with $C$ outside of the intersection with $B$. The correct simplification only involves $A^{\prime }\cup B$.

• Option (c): $A^{\prime }\cup C^{\prime }$ is incorrect because the original expression does not involve the complement of $C$. The correct simplification involves the union of $A^{\prime }$ and $B$, not $C^{\prime }$.

• Option (d): $A^{\prime }\cap B$ is incorrect because the original expression involves the union of $A^{\prime }$ and $B$, not their intersection. The correct simplification is $A^{\prime }\cup B$.

Solution

(d) Every rectangle, rhombus, square in a plane is a parallelogram but every trapezium is not a parallelogram.

So, $F_1$ is either of $F_1,F_2,F_3$ and $F_4$.

$\therefore F_1=F_2\cup F_3\cup F_4\cup F_1$

• (a) $F_2\cap F_3$: This option is incorrect because the intersection of rectangles ($F_2$) and rhombuses ($F_3$) is the set of squares ($F_4$), not the set of all parallelograms ($F_1$). Not all parallelograms are squares.

• (b) $F_3\cap F_4$: This option is incorrect because the intersection of rhombuses ($F_3$) and squares ($F_4$) is the set of squares ($F_4$), not the set of all parallelograms ($F_1$). Not all parallelograms are squares.

• (c) $F_2\cup F_5$: This option is incorrect because the union of rectangles ($F_2$) and trapeziums ($F_5$) does not cover all parallelograms ($F_1$). Specifically, it misses out on rhombuses that are not rectangles and parallelograms that are neither rectangles nor trapeziums.

Solution

(c) The given sets can be represented in Venn diagram as shown below

It is clear from the diagram that, $S\cup T\cup C=S$.

• (a) $S\cap T\cap C=\varphi$: This option is incorrect because the intersection of the triangle and the circle is not necessarily empty. Since both the triangle and the circle are contained within the square, there can be points that lie within all three sets $S$, $T$, and $C$.

• (b) $S\cup T\cup C=C$: This option is incorrect because the union of the square, triangle, and circle is not necessarily equal to the circle. The square $S$ contains points that are outside the circle, and the triangle $T$ also contains points that are outside the circle. Therefore, $S\cup T\cup C$ is larger than just $C$.

• (d) $S\cup T=S\cap C$: This option is incorrect because the union of the square and the triangle is not necessarily equal to the intersection of the square and the circle. The union $S\cup T$ includes all points inside the square and the triangle, while the intersection $S\cap C$ includes only the points that are inside both the square and the circle. These two sets are not equivalent.