NCERT Exemplar MCQs

Hint

(c) If elements are not repeated, then number of elements in \(A_{1} \cup A_{2} \cup A_{3} \cdots\) \(\cup A_{30}\) is \(30 \times 5\).
But each element is used 10 times.
So, \(\quad S=\frac{30 \times 5}{10}=15\)
If elements in \(B_{1}, B_{2}, \ldots, B_{n}\), are not repeated, then total number of elements in \(B_{1} \cup B_{2} \cup B_{3} \ldots \cup B_{n}\) is \(3 n\)
But each element is repeated 9 times.
So, \(\quad S=\frac{3 n}{9} \Rightarrow 15=\frac{3 n}{9} \Rightarrow n=45\)

Hint

(b) According to the question,
\(
\begin{array}{ll}
& 2^{m}-2^{n}=112 \\
\Rightarrow & 2^{n}\left(2^{m-n}-1\right)=2^{4} \cdot 7 \\
\Rightarrow & 2^{n}=2^{4} \text { and } 2^{m-n}-1=7 \\
\Rightarrow & n=4 \text { and } 2^{m-n}=8 \\
\Rightarrow & 2^{m-n}=2^{3} \Rightarrow m-n=3 \Rightarrow m-4=3 \Rightarrow m=7
\end{array}
\)

Hint

(b)

\(
\begin{aligned}
\left(A \cap B^{\prime}\right)^{\prime} \cup(B \cap C) =\left(\left(A^{\prime} \cup\left(B^{\prime}\right)^{\prime}\right) \cup(B \cap C)\right.\\
=\left(A^{\prime} \cup B\right) \cup(C \cap B)=A^{\prime} \cup(B \cup(C \cap B))=A^{\prime} \cup B
\end{aligned}
\)

Hint

(d) We know that every rectangle, rhombus, and square in a plane is a parallelogram but every trapezium is not a parallelogram.
So, \(F_{1}\) is either of \(F_{1}\) or \(F_{2}\) or \(F_{3}\) or \(F_{4}\).
\(
\therefore F_{1}=F_{1} \cup F_{2} \cup F_{3} \cup F_{4}
\)

Hint

(c) The given sets are represented in Venn diagram as shown below:

\(
\text { It is clear from the diagram that, } S \cup T \cup C=S \text {. }
\)

Hint

(d) Since, \(R\) be the set of points inside the rectangle.
\(
\therefore \quad R=\{(x, y): 0<x<a \text { and } 0<y<b\}
\)

Hint

(b)
Let ‘ \(\mathrm{A}\) ‘ be the set of students who play cricket, ‘ \(\mathrm{B}\) ‘ be the set of students who play tennis.

It is given that, \(n(A)=25, n(B)=20\) and \(n(A \cap B)=10=\) students who play both the games
Thus \(n(\mathrm{~A} \cup \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})=20+25-10=35\)
Therefore the students who play neither of the cricket and tennis \(=60-35=25\)

Hint

(b) Let \(H\) and \(E\) denote the number of persons who read Hindi and English newspaper respectively.
Given, total number of persons, \(n(U)=840\)
Total number of persons who read Hindi newspaper,
\(
n(H)=450
\)
Total number of persons who read English newspaper,
\(
n(E)=300
\)
and \(n(H \cap E)=200\)
\(
\because n(H \cup E)=n(H)+n(E)-n(H \cap E)=550
\)
\(\therefore\) The number of persons who read neither of the newspaper
\(
=n\left(H^{\prime} \cap E^{\prime}\right)=n(U)-n(H \cup E)=290
\)

Hint

(a) \(\begin{aligned}
X &=\left\{8^{n}-7 n-1 \mid n \in N\right\}=\{0,49,490, \ldots\} \\
Y &=\{49 n-49 \mid n \in N\}=\{0,49,98,147, \ldots, 490, \ldots\}
\end{aligned}\)

Clearly, every element of \(X\) is in \(Y\) but every element of \(Y\) is not in \(X\).
\(
\therefore \quad X \subset Y
\)

Hint

(c) Let \(\mathrm{p} \%\) of the people watch a channel and \(\mathrm{q} \%\) of the people watch another channel
\(\therefore \mathrm{n}(\mathrm{p} \cap \mathrm{q})=\mathrm{x} \%\) and \(\mathrm{n}(\mathrm{p} \cup \mathrm{q}) \leq 100\)
So \(n(p \cup q) \geq n(p)+n(q)-n(p \cap q)\)
\(100 \geq 63+76-x\)
\(100 \geq 139-x\)
\(\Rightarrow x \geq 139-100\)
\(\Rightarrow x \geq 39\)
Now \(n(p)=63\)
\(
\begin{aligned}
&\therefore \mathrm{n}(\mathrm{p} \cap \mathrm{q}) \leq \mathrm{n}(\mathrm{p}) \\
&\Rightarrow \mathrm{x} \leq 63
\end{aligned}
\)
So \(39 \leq x \leq 63\).

Hint

(c) Graph of set A represents the rectangular hyperbola and graph of set B represents a straight line passing through the origin.
Clearly, from the graph nothing is common between them.
\(
\therefore A \cap B=\phi
\)

Hint

(a) \(
\begin{aligned}
&A \cap(A \cup B) \\
&=(A \cap A) \cup(A \cap B) \\
&=A
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\mathrm{A} \cup \mathrm{B} \\
&=(1,2,3,4,5 \ldots 17,18) \\
&\mathrm{B}^{\prime}=(1,3,5,7,9 \ldots 17,19,20,21 \ldots) \\
&(\mathrm{A} \cup \mathrm{B}) \cap \mathrm{B}^{\prime}=(1,3,5,7,9 . .17)=\mathrm{A} \\
&\mathrm{A}^{\prime} \cup \mathrm{A} \\
&=\text { Universal Set } \\
&=\mathrm{N}
\end{aligned}
\)

Hint

(a) Given: \(S=\{x \mid x\) is a positive multiple of 3 less than 100\(\}\) and \(P=\{x \mid x\) is a prime number less than 20}
To find: \(n(S)+n(P)\)
\(S=\{x \mid x\) is a positive multiple of 3 less than 100\(\}\)
\(\Rightarrow \mathrm{S}=\{3,6,9,12,15, \ldots \ldots, 99\}\)
\(n(S)=99 / 3=33\)
\(P=\{x \mid x\) is a prime number less than 20\(\}\)
\(\Rightarrow P=\{2,3,5,7,11,13,17,19\}\)
\(\Rightarrow n(P)=8\)
\(n(S)+n(P)=33+8=41\)
Hence, the answer is 41

Hint

(c)

\(
\begin{aligned}
&X \cap(X \cup Y)^{\prime}\\
&=X \cap\left(X^{\prime} \cap Y^{\prime}\right) \quad\left[\because(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}\right]\\
&=\left(X \cap X^{\prime}\right) \cap Y^{\prime}\\
&=\phi \cap Y^{\prime}\\
&=\phi \quad[\because \phi \cap A=\phi]
\end{aligned}
\)

Hint

(b) Since, \(n\left(\mathrm{X}_{r}\right)=5, \bigcup_{r=1}^{20} \mathrm{X}_{r}=\mathrm{S}\), we get \(n(\mathrm{~S})=100\)
But each element of \(\mathrm{S}\) belong to exactly 10 of the \(\mathrm{X}_{r}\) ‘ \(s\)
So, \(\frac{100}{10}=10\) are the number of distinct elements in \(\mathrm{S}\).
Also each element of \(\mathrm{S}\) belong to exactly 4 of the \(\mathrm{Y}_{r}{ }^{\prime} s\) and each \(\mathrm{Y}_{r}\) contain 2 elements. If \(\mathrm{S}\) has \(n\) number of \(\mathrm{Y}_{r}\) in it. Then
\(
\frac{2 n}{4}=10
\)
which gives
\(
n=20
\)

Hint

(c) Since, let \(\mathrm{A}\) and \(\mathrm{B}\) be such sets, i.e., \(n(\mathrm{~A})=m, \quad n(\mathrm{~B})=n\)
So \(n(\mathrm{P}(\mathrm{A}))=2^{m}, n(\mathrm{P}(\mathrm{B}))=2^{n}\)
\(\Rightarrow\)
Thus, \(n(\mathrm{P}(\mathrm{A}))-n(\mathrm{P}(\mathrm{B}))=56\), i.e., \(2^{m}-2^{n}=56\)
\(2^{n}\left(2^{m-n}-1\right)=2^{3} 7\)
\(\Rightarrow\)
\(n=3,2^{m-n}-1=7\)
\(\Rightarrow\)
\(m=6\)

Hint

(a) \(\text { Since }(A \cup B \cup C) \cap\left(A \cap B^{\prime} \cap C^{\prime}\right)^{\prime} \cap C^{\prime}\)
\(\begin{aligned}
&=(\mathrm{A} \cup(\mathrm{B} \cup \mathrm{C})) \cap\left(\mathrm{A}^{\prime} \cup(\mathrm{B} \cup \mathrm{C})\right) \cap \mathrm{C}^{\prime} \\
&=\left(\mathrm{A} \cap \mathrm{A}^{\prime}\right) \cup(\mathrm{B} \cup \mathrm{C}) \cap \mathrm{C}^{\prime} \\
&=\phi \cup(\mathrm{B} \cup \mathrm{C}) \cap \mathrm{C}^{\prime} \\
&=\mathrm{B} \cap \mathrm{C}^{\prime} \cup \phi=\mathrm{B} \cap \mathrm{C}^{\prime}
\end{aligned}\)

Hint

(a) Since \(n(\mathrm{~A} \cup \mathrm{B})=n(\mathrm{~A})+n(\mathrm{~B})-n(\mathrm{~A} \cap \mathrm{B})\)
So,  \(n(\mathrm{~A})+n(\mathrm{~B})=n(\mathrm{~A} \cup \mathrm{B})+n(\mathrm{~A} \cap \mathrm{B})\)

Hint

(c) \(2^{n}\)

Hint

(b) False
Since 6 is divisible by both 3 and 2.
Thus \(\quad R \cap S \neq \phi\)

Hint

(a) True, Since\(\mathbf{Q} \subset \mathbf{R}\)
So, \(\mathbf{Q} \cap \mathbf{R}=\mathbf{Q}\)

Hint

(a) Glven: \(\{x \in R: 1 \leq x<2\}\)
To find: roster form of given set
Let \(A=\{x \in R: 1 \leq x<2\}\)
\(
\Rightarrow A=[1,2)
\)

Hint

(d) \(P(A)\) is the power set of set \(A\)
\(
\mathrm{P}(\mathrm{A})=2^{\mathrm{n}}(\mathrm{n}=0)
\)
\(P(A)=2^{0} =1 \)

Hint

(a) \(A\) and \(B\) are finite set and \(A \subset B\)
\(A \subset B\) means \(A\) is a subset of \(B\)
If \(A\) is subset of \(B\), then all elements of \(A\) are present in \(B\)
\(
\therefore \mathrm{n}(\mathrm{A} \cup \mathrm{B})=\mathrm{n}(\mathrm{B})
\)

Hint

(c) From Venn diagram you can clearly see,

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

Hint

(d) Given: set \(A=\{1,2\}\)
To find: power set of \(A\)
A power set is a set of all the subsets of set \(A\)
Subsets of \(A\) are \(\{1\},\{2\},\{1,2\}, \Phi\)

Hint

(a) Given: \(A=\{1,3,5\} . B=\{2,4,6\}\) and \(C=\{0,2,4,6,8\}\)
To find: Universal set of all the three sets
Universal set is a set which contains elements of all sets
\(
\Rightarrow U=\{0,1,2,3,4,5,6,8\}
\)

Hint

(a) Given: \(U=\{1,2,3,4,5,6,7,8,9,10\}\)
\(
A=\{1,2,3,5\}, B=\{2,4,6,7\}
\)
And \(C=\{2,3,4,8\}\)
Given that: \(U=\{1,2,3,4,5,6,7,8,9,10\}, A=\{1,2,3,5\}, B=\{2,4,6,7\}\) and \(C=\{2,3,4,8\}\)
(i) \((B \cup C)^{\prime}=\{2,3,4,6,7,8\}^{\prime}=\{1,5,9,10\}\)
(ii) \((C-A)^{\prime}=\{1,2,3,5,6,7,9,10\}\)

Hint

(d) Since \(A-B=A \cap B^{\prime}\)
\(A-(A \cap B)=A \cap B^{\prime}\)

Hint

(c) (I)
\(\begin{aligned}
\left(\left(A^{\prime} \cup B^{\prime}\right)-A\right)^{\prime} &=\left[A^{\prime} \cup B^{\prime} \cap A\right]^{\prime} & &\left.\quad \text { As } A-B=A \cap B^{\prime}\right] \\
&\left.=[A \cap B)^{\prime} \cap A^{\prime}\right]^{\prime} & & \\
&=\left[(A \cap B)^{\prime}\right]^{\prime} \cup\left(A^{\prime}\right)^{\prime} & & {\left[\text { As }\left(A^{\prime}\right)^{\prime}=A\right] } \\
&=(A \cup B) \cup A & \\
&=A
\end{aligned}\)
(II)
\(\begin{aligned}
{\left[B^{\prime} \cup\left(B^{\prime}-A^{\prime}\right)\right]^{\prime} } &=\left[B^{\prime} \cup\left(B^{\prime} \cap A^{\prime}\right)\right]^{\prime} \\
&=\left(B^{\prime}\right)^{\prime} \cap\left(B^{\prime} \cap A^{\prime}\right)^{\prime} \\
&=B \cap(B \cup A) \\
&=B
\end{aligned}\)
(iii)
\(
\begin{aligned}
(A-B)-(B-C) &=\left(A \cap B^{\prime}\right)-\left(B \cap C^{\prime}\right) \\
&=\left(A \cap B^{\prime}\right) \cap\left(B \cap C^{\prime}\right)^{\prime} \\
&=\left(A \cap B^{\prime}\right) \cap\left(B^{\prime} \cap C\right) \\
&=\left[A \cap\left(B^{\prime} \cap C\right)\right] \cap\left[B^{\prime} \cap\left(B^{\prime} \cap C\right)\right] \\
&=\left[A \cap\left(B^{\prime} \cap C\right)\right] \cap B^{\prime} \\
&=\left(A \cap B^{\prime}\right) \cap B^{\prime} \\
&=\left(A \cap B^{\prime}\right) \\
&=A-B
\end{aligned}\)
(iv)
\(
\begin{aligned}
(A-B) \cap(C-B) &=\left(A \cap B^{\prime}\right) \cap\left(C \cap B^{\prime}\right) \\
&=(A \cap C) \cap B^{\prime} \\
&=(A \cap C)-B
\end{aligned}
\)
(v) \(A \times(B \cap C)=(A \times B) \cap(A \times C)\)
(vi) \(A \times(B \cup C)=(A \times B) \cup(A \times C)\)

Hint

(b) Every set is a subset of itself. Hence, A \(\subset\) A Is true

Hint

(a)

\(
\begin{aligned}
&M=\{1,2,3,4,5,6,7,8,9\} \\
&B=\{1,2,3,4,5,6,7,8,9\}
\end{aligned}
\)
Every element of \(M\) and \(B\) is same
Hence, \(M=B\) and \(B \subset M\)
So, the given statement Is false.

Hint

(d) Sets A and B have the same number of elements but all the elements are not equal. Therefore, A and B are equivalent sets, not equal sets.

Hint

(b) Since every integer is a rational number
\(\therefore Z \subset Q \Rightarrow Q \cup Z=Q\)
Hence the statement is true.

Hint

(a) This statement is True.
Explanation:
\(R\) and \(T\) can be represented in roster form as follows:
\(
\begin{aligned}
&R=\{\ldots,-8,-6,-4,-2,0,2,4,6,8, \ldots\} \\
&T=\{\ldots,-18,-12,-6,0,6,12,18, \ldots\}
\end{aligned}
\)
Since, every element of \(\mathrm{T}\) is present in \(\mathrm{R}\).
So, \(T \subset R\)

Hint

(c) \(
A=\{0,1,2\}, B=\{x \in \mathbf{R} \mid 0 \leq x \leq 2\}
\)
Roster form of \(B\) :
\(
B=\{0,1,2\}
\)
Every element of both the sets \(A\) and \(B\) are same
\(
\Rightarrow A=B
\)
So, the given statement Is true.

Hint

(a)
(i) \(2^{x}-1\) is always an odd number for all positive integral values of \(x\). In particular, \(2^{x}-1\) is an odd number for \(x=1,2, \ldots, 9\). Thus, \(\mathrm{A}=\{1,2,3,4,5,6,7,8,9\}\).
(ii) \(x^{2}+7 x-8=0 \quad\) or \((x+8)(x-1)=0\) giving \(x=-8\) or \(x=1\)
Thus, \(\mathrm{C}=\{-8,1\}\)

Hint

(b) 

(i) False
Since, 37 has exactly two positive factors, 1 and 37,37 belongs to the set.
(ii) True
Since, the sum of positive factors of 28
\(
\begin{aligned}
&=1+2+4+7+14+28 \\
&=56=2(28)
\end{aligned}
\)
(iii) False
7,747 is not a multiple of \(37 .\)

Hint

(c)

(i) \(\mathrm{X} \cup \mathrm{Y}=\{x \mid x \in \mathrm{X}\) or \(x \in \mathrm{Y}\}\)
Thus \(x \in \mathrm{Y} \Rightarrow x \in \mathrm{X} \cup \mathrm{Y}\)
Hence, \(\mathrm{Y} \subset \mathrm{X} \cup \mathrm{Y}\)
(ii) \(\mathrm{X} \cap \mathrm{Y}=\{x \mid x \in \mathrm{X}\) and \(x \in \mathrm{Y}\}\)
Thus \(\quad x \in \mathrm{X} \cap \mathrm{Y} \Rightarrow x \in \mathrm{X}\)
Hence \(\mathrm{X} \cap \mathrm{Y} \subset \mathrm{X}\)
(iii) Note that
\(x \in \mathrm{X} \cap \mathrm{Y} \Rightarrow x \in \mathrm{X}\)
Thus \(\mathrm{X} \cap \mathrm{Y} \subset \mathrm{X}\)
Also, since \(\quad \mathrm{X} \subset \mathrm{Y}\),
\(x \in \mathrm{X} \Rightarrow x \in \mathrm{Y} \Rightarrow x \in \mathrm{X} \cap \mathrm{Y}\)
so that \(\mathrm{X} \subset \mathrm{X} \cap \mathrm{Y}\)
Hence the result \(\mathrm{X}=\mathrm{X} \cap \mathrm{Y}\) follows.

Hint

(d)
(i) \(\mathrm{A}=\{x \mid x \in \mathrm{N}\) and \(x\) is odd \(\}=\{1,3,5,7, \ldots, 99\}\)
(ii) \(\mathrm{B}=\{y \mid y=x+2, x \in \mathrm{N}\}\)
So, for
\(
\begin{gathered}
1 \in \mathrm{N}, y=1+2=3 \\
2 \in \mathrm{N}, y=2+2=4,
\end{gathered}
\)
and so on. Therefore, \(\mathrm{B}=\{3,4,5,6, \ldots, 100\}\)

Hint

(a)

Given \(\mathrm{E}=\{2,4,6,8,10\}\)
(i) Let \(\mathrm{A}=\{x \mid x=n+1, n \in \mathrm{E}\}\)
Thus, for
\(
\begin{aligned}
&2 \in \mathrm{E}, x=3 \\
&4 \in \mathrm{E}, x=5,
\end{aligned}
\)
and so on. Therefore, \(\mathrm{A}=\{3,5,7,9,11\}\).
(ii) Let \(\mathrm{B}=\left\{x \mid x=n^{2}, n \in \mathrm{E}\right\}\)
So, for \(\quad 2 \in \mathrm{E}, x=(2)^{2}=4,4 \in \mathrm{E}, x=(4)^{2}=16,6 \in \mathrm{E}, x=(6)^{2}=36\), and so on. Hence, \(B=\{4,16,36,64,100\}\)

Hint

(b) False
Consider the following sets \(A, B\) and \(C\):
\(
\begin{aligned}
&A=\{1,2,3\} \\
&B=\{2,3,5\} \\
&C=\{4,5,6\}
\end{aligned}
\)
Now
\((\mathrm{A} \cap \mathrm{B}) \cup \mathrm{C}=(\{1,2,3\} \cap\{2,3,5\}) \cup\{4,5,6\}\)
\(=\{2,3\} \cup\{4,5,6\}\)
\(=\{2,3,4,5,6\}\)
And
\(
\begin{aligned}
\mathrm{A} \cap(\mathrm{B} \cup \mathrm{C}) &=\{1,2,3\} \cap[\{2,3,5\} \cup\{4,5,6\}\\
&=\{1,2,3\} \cap\{2,3,4,5,6\} \\
&=\{2,3\} \\
(\mathrm{A} \cap \mathrm{B}) \cup \mathrm{C} & \neq \mathrm{A} \cap(\mathrm{B} \cup \mathrm{C})
\end{aligned}
\)

Hint

(a)

We have
\(
\begin{aligned}
\mathrm{A}-(\mathrm{A} \cap \mathrm{B}) &=\mathrm{A} \cap(\mathrm{A} \cap \mathrm{B})^{\prime} \quad\left(\text { since } \mathrm{A}-\mathrm{B}=\mathrm{A} \cap \mathrm{B}^{\prime}\right) \\
&\left.=\mathrm{A} \cap\left(\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}\right) \quad \text { [by De Morgan’s law }\right) \\
&=\left(\mathrm{A} \cap \mathrm{A}^{\prime}\right) \cup\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right) \quad \text { [by distributive law] } \\
&=\phi \cup\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right) \\
&=\mathrm{A} \cap \mathrm{B}^{\prime}=\mathrm{A}-\mathrm{B}
\end{aligned}
\)

Hint

(b) Yes
Let \(x \in(\mathrm{A}-\mathrm{B}) \cap(\mathrm{C}-\mathrm{B})\)
\(\Rightarrow \quad x \in \mathrm{A}-\mathrm{B}\) and \(x \in \mathrm{C}-\mathrm{B}\)
\(\Rightarrow \quad(x \in \mathrm{A}\) and \(x \notin \mathrm{B})\) and \((x \in \mathrm{C}\) and \(x \notin \mathrm{B})\)
\(\Rightarrow \quad(x \in \mathrm{A}\) and \(x \in \mathrm{C})\) and \(x \notin \mathrm{B}\)
\(\Rightarrow \quad(x \in \mathrm{A} \cap \mathrm{C})\) and \(x \notin \mathrm{B}\)
\(\Rightarrow \quad x \in(\mathrm{A} \cap \mathrm{C})-\mathrm{B}\)
So \(\quad(\mathrm{A}-\mathrm{B}) \cap(\mathrm{C}-\mathrm{B}) \subset(\mathrm{A} \cap \mathrm{C})-\mathrm{B} \dots ….(1)\)
Now, conversely

\(
\begin{array}{ll}
\text { Let }  y \in(\mathrm{A} \cap \mathrm{C})-\mathrm{B} \\
\Rightarrow  y \in(\mathrm{A} \cap \mathrm{C}) \text { and } y \notin \mathrm{B} \\
\Rightarrow  (y \in \mathrm{A} \text { and } y \in \mathrm{C}) \text { and }(y \notin \mathrm{B}) \\
\Rightarrow  (y \in \mathrm{A} \text { and } y \notin \mathrm{B}) \text { and }(y \in \mathrm{C} \text { and } y \notin \mathrm{B}) \\
\Rightarrow  y \in(\mathrm{A}-\mathrm{B}) \text { and } y \in(\mathrm{C}-\mathrm{B}) \\
\Rightarrow  y \in(\mathrm{A}-\mathrm{B}) \cap(\mathrm{C}-\mathrm{B}) \\
\text { So }  (\mathrm{A} \cap \mathrm{C})-\mathrm{B} \subset(\mathrm{A}-\mathrm{B}) \cap(\mathrm{C}-\mathrm{B}) \dots ….(2) \\
\text { From }(1) \text { and }(2),(\mathrm{A}-\mathrm{B}) \cap(\mathrm{C}-\mathrm{B})=(\mathrm{A} \cap \mathrm{C})-\mathrm{B}
\end{array}
\)

Hint

(b)

Now the equivalent contrapositive statement of \(x \in \mathrm{S} \Rightarrow x \in \mathrm{P}\) is \(x \notin \mathrm{P} \Rightarrow\) \(x \notin \mathrm{S} .\)
Now, we will prove the above contrapositive statement by contradiction method
\(\begin{array}{ll}\text { Let } & x \notin \mathrm{P} \\ \Rightarrow & x \text { is a composite number }\end{array}\)
Let us now assume that \(x \in \mathrm{S}\)
\(\Rightarrow \quad 2^{x}-1=m \quad\) (where \(m\) is a prime number)
\(\Rightarrow \quad 2^{x}=m+1\)
Which is not true for all composite number, say for \(x=4\) because
\(2^{4}=16\) which can not be equal to the sum of any prime number \(m\) and \(1 .\)
Thus, we arrive at a contradiction
\(\Rightarrow \quad x \notin \mathrm{S}\).
Thus, when \(x \notin \mathrm{P}\), we arrive at \(x \notin \mathrm{S}\)
So \(\quad \mathrm{S} \subset \mathrm{P}\).

Hint

(d) Let \(\mathrm{M}\) be the set of students passing in Mathematics
\(\mathrm{P}\) be the set of students passing in Physics
\(C\) be the set of students passing in Chemistry
\(
\begin{array}{ll}
\text { Now, } \quad & n(\mathrm{M} \cup \mathrm{P} \cup \mathrm{C})=50, n(\mathrm{M})=37, n(\mathrm{P})=24, n(\mathrm{C})=43 \\
& n(\mathrm{M} \cap \mathrm{P}) \leq 19, n(\mathrm{M} \cap \mathrm{C}) \leq 29, n(\mathrm{P} \cap \mathrm{C}) \leq 20 \text { (Given }) \\
& n(\mathrm{M} \cup \mathrm{P} \cup \mathrm{C})=n(\mathrm{M})+n(\mathrm{P})+n(\mathrm{C})-n(\mathrm{M} \cap \mathrm{P})-n(\mathrm{M} \cap \mathrm{C}) \\
& -n(\mathrm{P} \cap \mathrm{C})+n(\mathrm{M} \cap \mathrm{P} \cap \mathrm{C}) \leq 50 \\
\Rightarrow & 37+24+43-19-29-20+n(\mathrm{M} \cap \mathrm{P} \cap \mathrm{C}) \leq 50 \\
\Rightarrow & n(\mathrm{M} \cap \mathrm{P} \cap \mathrm{C}) \leq 50-36 \\
\Rightarrow & n(\mathrm{M} \cap \mathrm{P} \cap \mathrm{C}) \leq 14
\end{array}
\)
Thus, the largest possible number that could have passed all the three examinations is \(\)14.\(\)