NIMCET Sets And Relations Previous Year Questions (PYQs) – Page 1 of 5
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Showing 42 questions
If A is a subset of B and B is a subset of C, then
cardinality of A ∪ B ∪ C is equal to
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Solution
If A ⊆ B and B ⊆ C, then find the cardinality of A ∪ B ∪ C.
Explanation
- Since A is a subset of B, everything in A is already in B.
- Since B is a subset of C, everything in B (and hence A) is already in C.
- Therefore, A ∪ B ∪ C = C.
✅ Final Answer
|A ∪ B ∪ C| = |C|
Let R be reflexive relation on the finite set a having 10 elements and if m is the number of ordered pair in R, then
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Solution
Explanation
- Total possible ordered pairs on a set of 10 elements = 10² = 100.
- A reflexive relation must include all pairs of the form (a,a) for all a ∈ A.
- Thus, at least 10 pairs must be in R ⇒ m ≥ 10.
- The maximum relation is the universal relation with all 100 pairs ⇒ m ≤ 100.
✅ Final Answer
The number of ordered pairs m can take any value in the range:
10 ≤ m ≤ 100
A set contains $(2n+1)$ elements. If the number of subsets that contain at most $n$ elements is $4096$, then the value of $n$ is:
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Solution
For odd number of elements,
$\displaystyle \sum_{k=0}^{n} \binom{2n+1}{k} = 2^{2n}$
Given:
$2^{2n} = 4096 = 2^{12}$
So,
$2n = 12 \Rightarrow n = 6$
The total number of relations that exist from a set $A$ with $m$ elements into the set $A \times A$ is:
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Solution
Relations are subsets of
$A \times (A \times A)$
Size = $m \cdot m^2 = m^3$
Total relations $= 2^{m^3}$
Not present in options.
If
$P = {(4^n - 3n - 1) : n \in N}$
and
$Q = {(9n - 9) : n \in N}$,
then $P \cup Q$ equals to:
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Solution
If $P = \{4^n - 3n - 1 : n \in \mathbb{N}\}$ and $Q = \{9n - 9 : n \in \mathbb{N}\}$
Elements of P
$n=1: \quad 4^1 - 3(1) - 1 = 0$
$n=2: \quad 4^2 - 3(2) - 1 = 9$
$n=3: \quad 4^3 - 3(3) - 1 = 54$
$n=4: \quad 4^4 - 3(4) - 1 = 243$
$\Rightarrow P = \{0, 9, 54, 243, \ldots\}$
Elements of Q
$n=1: \quad 9(1) - 9 = 0$
$n=2: \quad 9(2) - 9 = 9$
$n=3: \quad 9(3) - 9 = 18$
$\Rightarrow Q = \{0, 9, 18, 27, \ldots\}$ = all multiples of 9
$P \subseteq Q$
$P \subseteq Q \Rightarrow P \cup Q = Q$
$\boxed{P \cup Q = Q}$
Elements of P
$n=1: \quad 4^1 - 3(1) - 1 = 0$
$n=2: \quad 4^2 - 3(2) - 1 = 9$
$n=3: \quad 4^3 - 3(3) - 1 = 54$
$n=4: \quad 4^4 - 3(4) - 1 = 243$
$\Rightarrow P = \{0, 9, 54, 243, \ldots\}$
Elements of Q
$n=1: \quad 9(1) - 9 = 0$
$n=2: \quad 9(2) - 9 = 9$
$n=3: \quad 9(3) - 9 = 18$
$\Rightarrow Q = \{0, 9, 18, 27, \ldots\}$ = all multiples of 9
$P \subseteq Q$
$P \subseteq Q \Rightarrow P \cup Q = Q$
$\boxed{P \cup Q = Q}$
Let $X$ be the universal set for sets $A$ and $B$. If
$n(A)=200,;n(B)=300,;n(A\cap B)=100$,
then $n(A'\cap B')=300$ provided $n(X)$ is equal to
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Solution
$n(A'\cap B') = n(X)-n(A\cup B)$
$n(A\cup B)=200+300-100=400$
Given $n(A'\cap B')=300$
$\Rightarrow n(X)-400=300$
$\Rightarrow n(X)=700$
$A_1, A_2, A_3, A_4$ are subsets of $U$ (75 elements).
Each $A_i$ has 28 elements.
Any two intersect in 12 elements.
Any three intersect in 5 elements.
All four intersect in 1 element.
Find the number of elements belonging to none of the four subsets.
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Solution
Use Inclusion–Exclusion:
$|A_1 \cup A_2 \cup A_3 \cup A_4|$
$= \sum |A_i| - \sum |A_i \cap A_j| + \sum |A_i \cap A_j \cap A_k| - |A_1 \cap A_2 \cap A_3 \cap A_4|$
Substitute values:
$\sum |A_i| = 4 \cdot 28 = 112$
$\sum |A_i \cap A_j| = \binom{4}{2} \cdot 12 = 6 \cdot 12 = 72$
$\sum |A_i \cap A_j \cap A_k| = \binom{4}{3} \cdot 5 = 4 \cdot 5 = 20$
Intersection of four = 1
So:
$|A_1 \cup A_2 \cup A_3 \cup A_4| = 112 - 72 + 20 - 1 = 59$
Elements belonging to none:
$75 - 59 = 16$
Number of real solutions of the equation $\sin\!\left(e^x\right) = 5^x + 5^{-x}$ is
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Solution
Number of real solutions
Equation: $$\sin\!\left(e^x\right) \;=\; 5^x + 5^{-x}$$
Reasoning
- For all real \(x\), \(\sin(e^x)\in[-1,1]\).
- By AM–GM, \(5^x + 5^{-x} \ge 2\) (with equality only when \(5^x=5^{-x}\Rightarrow x=0\)).
- At \(x=0\): LHS \(=\sin(1)\approx 0.84\), RHS \(=2\) ⇒ not equal.
- Hence RHS is always \(\ge 2\) while LHS is always \(\le 1\) → they can never match.
✅ Conclusion
No real solution (number of real solutions = 0).
From 50 students:
37 passed Math, 24 Physics, 43 Chemistry.
At most 19 passed Math & Physics,
at most 29 passed Math & Chemistry,
at most 20 passed Physics & Chemistry.
Intersection of all 3 is $x$.
Find maximum possible value of $x$.
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Solution
Use:
$M + P + C - (MP + MC + PC) + x \le 50$
Substitute maximum overlaps:
$37 + 24 + 43 - (19 + 29 + 20) + x \le 50$
Compute:
$104 - 68 + x \le 50$
$36 + x \le 50$
$x \le 14$
But each pair overlap already includes $x$, so consistency check:
Maximum possible $x = \min(19, 29, 20) = 19$
No — limited by total count equation → 14.
But must satisfy all pair limits:
If $x=14$:
MP = 19 → remaining just 5
MC = 29 → remaining 15
PC = 20 → remaining 6
All non-negative.
Thus x = 14 possible → but not in options.
Inverse of the function $f(x)=\frac{10^x-10^{-x}}{10^{x}+10^{-x}}$ is
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Solution
Let \( y = \dfrac{10^x - 10^{-x}}{10^x + 10^{-x}} \).
Multiply numerator and denominator by \(10^x\): \( y = \dfrac{10^{2x} - 1}{10^{2x} + 1} \).
Put \( t = 10^{2x} \), then \( y = \dfrac{t-1}{t+1} \). Solving, \( t = \dfrac{1+y}{1-y} \).
Hence, \( 10^{2x} = \dfrac{1+y}{1-y} \). Taking log base 10: \( 2x = \log_{10}\!\Big(\dfrac{1+y}{1-y}\Big) \).
✅ Final Answer
The inverse function is:
\[
f^{-1}(y) = \tfrac{1}{2}\,\log_{10}\!\left(\dfrac{1+y}{1-y}\right), \quad |y|<1
\]
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