Past JEE Main Entrance Paper

Hint

(b) Q Product of two even number is always even and product of two odd numbers is always odd.
\(\therefore\) Number of required subsets
\(=\) Total number of subsets – Total number of subsets having only odd numbers
\(
=2^{100}-2^{50}=2^{50}\left(2^{50}-1\right)
\)

Hint

(b) Case-I: \(x \in[0,9]\)
\(
\begin{aligned}
&2(3-\sqrt{x})+x-6 \sqrt{x}+6=0 \\
&\Rightarrow x-8 \sqrt{x}+12=0 \Rightarrow \sqrt{x}=4,2 \Rightarrow x=16,4
\end{aligned}
\)
Since \(x \in[0,9]\)
\(
\begin{aligned}
&\therefore \quad x=4 \\
&\text { Case-II: } x \in[9, \infty] \\
&2(\sqrt{x}-3)+x-6 \sqrt{x}+6=0 \Rightarrow x-4 \sqrt{x}=0 \Rightarrow x=16,0
\end{aligned}
\)
Since \(x \in[9, \infty]\)
\(\therefore \mathrm{x}=16\)
Hence, \(x=4 \& 16\)

Hint

(a) \(f(x)+2 f\left(\frac{1}{x}\right)=3 x \text { ..(i) }\)
\(f\left(\frac{1}{x}\right)+2 f(x)=\frac{3}{x} \text { ..(ii) }\)
Adding (i) and (ii)
\(
\Rightarrow f(x)+f\left(\frac{1}{x}\right)=x+\frac{1}{x} \text { ..(iii) }
\)
Substracting (i) from (ii)
\(
\Rightarrow f(x)-f\left(\frac{1}{x}\right)=\frac{3}{x}-3 x \text { ..(iv) }
\)
On adding (iii) and (iv)
\(
\begin{aligned}
&\Rightarrow f(x)=\frac{2}{x}-x \\
&f(x)=f(-x) \Rightarrow \frac{2}{x}-x=\frac{-2}{x}+x \Rightarrow x=\frac{2}{x} \\
&x^{2}=2 \quad \text { or } x=\sqrt{2},-\sqrt{2}
\end{aligned}
\)

Hint

(d) \(\mathrm{P}=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\}\)
\(
\Rightarrow \sin \theta=(\sqrt{2}+1) \cos \theta \Rightarrow \tan \theta=\sqrt{2}+1
\)
Now, \(Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\}\)
\(
\begin{aligned}
\Rightarrow \cos \theta=(\sqrt{2}-1) \sin \theta \Rightarrow \tan \theta=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} \\
\Rightarrow \tan \theta=\sqrt{2}+1 \\
\therefore \quad \mathrm{P}=\mathrm{Q}
\end{aligned}
\)

Hint

(d) \(\mathrm{S}=\{1,2,3,4\}\)
Let \(\mathrm{A}\) and \(\mathrm{B}\) be disjoint subsets of \(\mathrm{S}\)
Now for any element \(a \in S\), has got three possibilities either, it is in A or B
or none
\(\Rightarrow\) For every element out of 4 elements there are three choices
\(\therefore\) Total options \(=3^{4}=81\)
Here \(\mathrm{A} \neq \mathrm{B}\) except when \(\mathrm{A}=\mathrm{B}=\phi\)
\(\therefore 81-1=80\) ordered pairs \((\mathrm{A}, \mathrm{B})\) are there for which \(\mathrm{A} \neq \mathrm{B}\)
Hence total number of unordered pairs of disjoint subsets \(=\frac{80}{2}+1=41\)

Hint

(b) \(2^{m}=112+2^{n} \Rightarrow 2^{m}-2^{n}=112\)
\(
\Rightarrow 2^{n}\left(2^{m-n}-1\right)=2^{4}\left(2^{3}-1\right)
\)
\(
\therefore m=7, n=4 \Rightarrow m n=28
\)

Hint

(d) Here set S contains 5 odd and 4 even numbers. Since each of \(\mathrm{N}_{\mathrm{K}}\) containing five elements out of which exactly are odd.
\(\begin{aligned}
&\therefore \mathrm{N}_{1}={ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{4}=5 \\
&\mathrm{~N}_{2}={ }^{5} \mathrm{C}_{2} \times{ }^{4} \mathrm{C}_{3}=40 \\
&\mathrm{~N}_{3}={ }^{5} \mathrm{C}_{3} \times{ }^{4} \mathrm{C}_{2}=60 \\
&\mathrm{~N}_{4}={ }^{5} \mathrm{C}_{4} \times{ }^{4} \mathrm{C}_{1}=20 \\
&\mathrm{~N}_{5}={ }^{5} \mathrm{C}_{5}=1 \\
&\therefore \mathrm{N}_{1}+\mathrm{N}_{2}+\mathrm{N}_{3}+\mathrm{N}_{4}+\mathrm{N}_{5}=126
\end{aligned}\)

Hint

\(
\begin{aligned}
&\text { (b) Given, } n(C)=73, n(T)=65, n(C \cap T)=x \\
&\therefore 65 \geq n(C \cap T) \geq 65+73-100 \\
&\Rightarrow 65 \geq x \geq 38 \Rightarrow x \neq 36 .
\end{aligned}
\)

Hint

(d) Let \(n(U)=100\), then \(n(A)=63, n(B)=76\)
\(
n(A \cap B)=x
\)
Now,
\(
\begin{aligned}
&n(A \cup B)=n(A)+n(B)-n(A \cap B) \leq 100 \\
&=63+76-x \leq 100 \\
&\Rightarrow x \geq 139-100 \Rightarrow x \geq 39 \\
&\because n(A \cap B) \leq n(A) \Rightarrow x \leq 63 \\
&\therefore 39 \leq x \leq 63
\end{aligned}
\)

Hint

Hint

\(
\begin{aligned}
A=\{x \in Z, \quad& 2^{(x+2)\left(x^{2}-5 x+6\right)}=2^{0} \\
&(x+2)\left(x^{2}-5 x+6\right)=0 \\
& x=-2,2,3
\end{aligned}
\)

\(
\begin{aligned}
B=\{x \in Z, \quad&-3<2 x-1<9 \\
\Rightarrow &-2<2 x<10 \\
\Rightarrow &-1<x<5
\end{aligned}
\)

\(
\begin{aligned}
&x=0,1,2,3,4 \\
&A \times B=\{(-2,0),(-2,1)(-2,2)(-2,3)(-2,4),(2,0),(2,1)(2,2),(2,3),(2,4)(3,0),(3,1) \\
&(3,2)(3,3),(3,4)] \\
&A \times B \text { elements }=15 \\
&\text { total substent }=2^{15}
\end{aligned}
\)

Hint

(d) Let \(\mathrm{n}(\mathrm{A})=\) number of students opted Mathematics \(=70\),
\(\mathrm{n}(\mathrm{B})=\) number of students opted Physics \(=46\),
\(\mathrm{n}(\mathrm{C})=\) number of students opted Chemistry \(=28\),
\(\mathrm{n}(\mathrm{A} \cap \mathrm{B})=23\)
\(\mathrm{n}(\mathrm{B} \cap \mathrm{C})=9\)
\(\mathrm{n}(\mathrm{A} \cap \mathrm{C})=14\),
\(\mathrm{n}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=4\)
Nown( \(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})\)
\(=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})+\mathrm{n}(\mathrm{C})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})-\mathrm{n}(\mathrm{B} \cap \mathrm{C})-\mathrm{n}(\mathrm{A} \cap \mathrm{C})+\mathrm{n}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})\) \(=70+46+28-23-9-14+4=102\)

So number of students not opted for any course \(=\) Total \(-\mathrm{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})\) \(=140-102=38\).

Hint

(d) Let \(A=\{1,2,3,4\}\)
\(B=\{3,4,5,6\}\)
\(C=\{1,2,3,4,7,8\}\)
Here \(A \cap B=\{3,4\} \subseteq C\)
\(A-C=\phi \subseteq B\)
but \(\mathrm{A} \nsubseteq \mathrm{B}\)
So not true (wrong) statement is d, If \(A-C \subseteq B\) then \(A \subseteq B\)

Hint

(c) Step 1: Define the Variables
Let \(x\) be the total number of families in the town.

Step 2: Identify the Percentages
– Families owning a phone: \(25 \%\) of \(x\)
– Families owning a car: \(15 \%\) of \(x\)
– Families owning neither a phone nor a car: \(65 \%\) of \(x\)
– Families owning both a car and a phone: 2000 families

Step 3: Calculate Families Owning Either a Phone or a Car
Since \(65 \%\) of families own neither a phone nor a car, the percentage of families that own either a phone or a car is:
\(
100 \%-65 \%=35 \%
\)
Thus, the number of families owning either a phone or a car is:
\(
0.35 x
\)
Step 4: Use the Principle of Inclusion-Exclusion
According to the principle of inclusion-exclusion for two sets:
\(
n(P \cup C)=n(P)+n(C)-n(P \cap C)
\)
Where:
\(-n(P)=0.25 x\) (families with a phone)
\(-n(C)=0.15 x\) (families with a car)
\(-n(P \cap C)=2000\) (families with both)
Substituting the values we have:
\(
0.35 x=0.25 x+0.15 x-2000
\)
Step 5: Simplify the Equation
Combine like terms:
\(
0.35 x=0.40 x-2000
\)
Rearranging gives:
\(
\begin{aligned}
& 0.35 x-0.40 x=-2000 \\
& -0.05 x=-2000
\end{aligned}
\)
Step 6: Solve for \(x\)
Dividing both sides by -0.05 :
\(
x=\frac{2000}{0.05}=40000
\)
Step 7: Verify Each Statement
1. Statement A: \(5 \%\) of families own both a car and a phone.
– Calculation:
\(
\frac{2000}{40000} \times 100=5 \%
\)
– True.
2. Statement B: \(35 \%\) of families own either a car or a phone.
– Calculation:
\(0.35 \times 40000=14000\) families
– True.
3. Statement C: 40,000 families live in the town.
– True.

Conclusion
All three statements (A, B, and C) are correct.

Hint

(c) We have \(\mathrm{X} \cap(\mathrm{Y} \cup \mathrm{X})^{\prime}=\mathrm{X} \cap\left(\mathrm{Y}^{\prime} \cap \mathrm{X}^{\prime}\right)\)
\(
\begin{aligned}
&=\mathrm{X} \cap \mathrm{Y}^{\prime} \cap \mathrm{X}^{\prime}=\mathrm{X} \cap \mathrm{X}^{\prime} \cap \mathrm{Y}^{\prime} \\
&=\phi \cap \mathrm{Y}^{\prime}=\phi
\end{aligned}
\)

Hint

(c) From the given conditions,
\(
\begin{aligned}
&n(A)=25, n(B)=7 \text { and } n(A \cap B)=3 \\
&n(A \cup B)=n(A)+n(B)-n(A \cap B)=25+7-3=29
\end{aligned}
\)

Hint

(c) Let the number of newspapers which are read be \(\mathrm{n}\). Then \(60 n=\) \((300)(5) \Rightarrow n=25\)

Hint

(b) Given that \(\bigcup_{i=1}^{30} A_{i}=\cup_{j=1}^{n} B_{j}=S \text { …(i) }\)
and each \(A_{i}^{\text {s }}\) contain 5 elements
So, total number of elements in \(A_{i}=5 \times 30=150\).
Since each element of \(S\) belongs to exactly ten of the \(A_{i}\) ‘s.
\(
\therefore n(S)=n\left[\bigcup_{i=1}^{30} A_{i}\right]=\frac{150}{10}=15 \text { …(ii) }
\)
Now, each \(\mathrm{B}_{\mathrm{j}}^{\text {s }}\) contain 3 elements
So, total number of elements in \(B_{j}=3 \times n=3 n\).
Since each element of \(\mathrm{S}\) belongs to exactly nine of the \(\mathrm{B}_{\mathrm{j}}{ }^{\prime {\text {s }}}\).
\(
\therefore n(S)=n\left[\bigcup_{j=1}^{n} B_{j}\right]=\frac{3 n}{9} \text { …(iii) }
\)
from (ii) and (iii)
\(
\frac{3 n}{9}=15 \Rightarrow n=45 \text {. }
\)

Hint

Concept:
If \(A\) and \(B\) are two sets then, \(n(A \cup B)=n(A)+n(B)-n(A \cap B)\)
Calculation:
Given: \(n(A)=3\) and \(n(B)=6\).
As we know that, if \(A\) and \(B\) are two sets then, \(n(A \cup B)=n(A)+n(B)-n(A \cap B)\)
\(
\Rightarrow n(A \cup B)=3+6-n(A \cap B)
\)
In order to minimize \(n(A \cup B)\) we have to maximize \(n(A \cap B)\).
If \(A\) is a subset of \(B\), then \(A \cap B=A \Rightarrow n(A \cap B)=n(A)=3\)
\(
\Rightarrow n(A \cup B)=3+6-3=6 \text {. }
\)
Hence, the minimum number of elements that \((A \cup B)\) can have is 6 .

Hint

(d) We have
\(
\begin{aligned}
&n(U)=100, \text { where } U \text { stands for universal set } \\
&n(M \cap C \cap T)=d=10 ; n(M \cap C)=b+d=20 ; \\
&n(C \cap \mathrm{T})=d+f=30 ; n(M \cap T)=d+e=25 \\
&\quad \Rightarrow b=10, f=20 \text { and } e=15 \\
&\quad n(\text { only } M)=a=12 ; n(\text { only } C)=c=5 ; n(\text { only } T)=\mathrm{g}=8
\end{aligned}
\)
Filling all the entries we obtain the Venn diagram as shown:

\(\begin{gathered}
\quad \therefore \quad n(M \cap C \cup T)=\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}+\mathrm{g} \\
=12+10+5+15+10+20+8=80 \\
\therefore \quad n(M \cup C \cup T)^{\prime}=100-80=20
\end{gathered}\)

Hint

\(
\text { (b) } A =\{x:|x|<3, x \in Z\}
\)
\(
\begin{aligned}
& A =\{-2,-1,0,1,2\} \\
& R =\{(x, y): y=|x|, x \neq-1\} \\
& R =\{(-2,2),(0,0),(1,1),(2,2)\}
\end{aligned}
\)
\(R\) has four elements
Number of elements in the power set of \(R\)
\(
=2^4=16
\)

Hint

(b) Let \(X=\{1,2,3,4,5\}\)
Total no. of elements \(=5\)
Each element has 3 options. Either set \(Y\) or set \(Z\) or none. \((\because Y \cap Z=\phi)\)
So, number of ordered pairs \(=3^5\)

Hint

(b) Let \(x \in A\) and \(x \in B \Leftrightarrow x \in A \cup B\)
\(
\begin{aligned}
& \Leftrightarrow x \in A \cup C \quad(\because A \cup B=A \cup C) \\
& \Leftrightarrow x \in C
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \quad B=C . \\
& \text { Let } x \in A \text { and } x \in B \Leftrightarrow x \in A \cap B \\
& \Leftrightarrow x \in A \cap C \quad(\because A \cap B=A \cap C) \\
& \Leftrightarrow x \in C \\
& \therefore B=C
\end{aligned}
\)

Hint

For disoint sets, \(A \cap B=\phi\)
Each element in either \(A\) or \(B\) or neither.
\(\therefore\) Total ways \(=3^4=81 ; A=B\) iff \(A=B=\phi\)
Otherwise, \(A\) and \(B\) are interchangable
\(\therefore\) Number of unordered pair for disoint subsets of
\(
S=\frac{3^4+1}{2}=41
\)

Hint

\(\because A \times B\) has 8 elements.
\(\therefore\) Number of subsets \(=2^8=256\)
Number of subsets with zero element \(={ }^8 C_0=1\)
Number of subsets with one element \(={ }^8 C_1=8\)
Number of subsets with one elements \(={ }^8 C_2=28\)
Hence, Number of subsets of \(A \times B\) having 3 or more elements
\(
=256-(1+8+28)=256-37=219
\)

Hint

Since, \(4^n-3 n-1=(1+3)^n-3 n-1\)
\(
\begin{aligned}
& =\left(1+{ }^n C_1 \cdot 3+{ }^n C_2 \cdot 3^2+{ }^n C_3 \cdot 3^3+\ldots+{ }^n C_n \cdot 3^n\right)-3 n-1 \\
& =3^2\left({ }^n C_2+{ }^n C_3 \cdot 3+\ldots+{ }^n C_n \cdot 3^{n-2}\right)
\end{aligned}
\)
\(\Rightarrow 4^n-3 n-1\) is a multiple of 9 for \(n \geq 2\)
For \(n=1,4^n-3 n-1=4-3-1=0\)
For \(n=2,4^n-3 n-1=16-6-1=9\)
\(\therefore 4^n-3 n-1\) is multiple of 9 for all \(n \in N\).
It is clear that \(X\) contains elements, which are multiples of 9 and \(Y\) contains all multiples of 9 .
\(
\therefore X \subseteq Y \text { i.e., } X \cup Y=Y
\)

Hint

\(
n(A)=4, n(B)=2 \Rightarrow n(A \times B)=8
\)
The number of subsets of \(A \times B\) having at least three elements
\(
\begin{aligned}
& ={ }^8 C_3+{ }^8 C_4+{ }^8 C_5+\ldots+{ }^8 C_8 \\
& =2^8-\left({ }^8 C_0+{ }^8 C_1+{ }^8 C_2\right) \\
& =256-(1+8+28)=219
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \because A \cap B=A \cap C \Rightarrow B=C \text { and } A \cup B=A \cup C \Rightarrow B=C \\
& \text { Hence, } \quad B=C
\end{aligned}
\)

Hint

\(
\begin{array}{ll}
x+2 y+3 z=42, & x, y, z \geq 0 \\
z=0 & x+2 y=42 \Rightarrow 22 \\
z=1 & x+2 y=39 \Rightarrow 20 \\
z=2 & x+2 y=36 \Rightarrow 19 \\
z=3 & x+2 y=33 \Rightarrow 17 \\
z=4 & x+2 y=30 \Rightarrow 16 \\
z=5 & x+2 y=27 \Rightarrow 14 \\
z=6 & x+2 y=24 \Rightarrow 13 \\
z=7 & x+2 y=21 \Rightarrow 11 \\
z=8 & x+2 y=18 \Rightarrow 10 \\
z=9 & x+2 y=15 \Rightarrow 8
\end{array}
\)
\(
\begin{array}{ll}
z=10 & x+2 y=12 \Rightarrow 7 \\
z=11 & x+2 y=9 \Rightarrow 5 \\
z=12 & x+2 y=6 \Rightarrow 4 \\
z=13 & x+2 y=3 \Rightarrow 2 \\
z=14 & x+2 y=0 \Rightarrow 1
\end{array}
\)
Total : 169

Hint

Sum of elements in \(A \cap B\)
\(
\begin{aligned}
= & \underbrace{(2+4+6+\ldots+200)}_{\text {Multiple of } 2}-\underbrace{(6+12+\ldots+198)}_{\text {Multiple of } 2 \text { & } 3 \text { i.e. } 6} \\
& -\underbrace{(10+20+\ldots+200)}_{\text {Multiple of } 5 \text { & } 2 \text { i.e. } 10}+\underbrace{(30+60+\ldots+180)}_{\text {Multiple of } 2,5 \text { & 3 i.e. } 30} \\
= & 5264
\end{aligned}
\)

Hint

\(
\begin{aligned}
& A =\{2,3,4,5, \ldots,, 30\} \\
& ( a , b ) \simeq( c , d ) \quad \Rightarrow ad = bc \\
& (4,3) \simeq(c, d) \quad \Rightarrow \quad 4 d=3 c \\
& \Rightarrow \frac{4}{3}=\frac{c}{d} \\
& \frac{c}{d}=\frac{4}{3} \quad \& c, d \in\{2,3, \ldots \ldots, 30\} \\
& ( c , d )=\{(4,3),(8,6),(12,9),(16,12),(20, \\
& 15),(24,18),(28,21)\}
\end{aligned}
\)
No. of ordered pair \(=7\)

Hint

Let the types of games be denoted as \(A, B, C\).
In all the given Venn diagrams we can observe that there is at least one region where all \(A, B, C\) intersect i.e. \(A \cap B \cap C\)
Thus none of the given Venn diagrams represents the condition ‘Some of the students play two types of games, but none play all three games’