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NIMCET 2008
Question 01: If \(f(x)\) is a polynomial satisfying \(f(x)f\left( {\frac{1}{x}} \right) = f(x) + f\left( {\frac{1}{x}} \right)\) and f(3) = 28, then f(4) is given by
(A) 63
(B) 65
(C) 67
(D) 68
65
Given that\(f(x)f\left( {\frac{1}{x}} \right) = f(x) + f\left( {\frac{1}{x}} \right)\)
\(f(x) = x^n +1\)
Since \(f(4) = 65\), hence \(f(x) = 4^3 +1\)
so \(f(x) = x^3 +1\)
Hence \(f(4) = 4^3 + 1 = 65\).
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Question 02: Suppose \(P_1, P_2, …….. P_{30}\) are thirty sets each having 5 elements and \(Q_1, Q_2, …….. P_n\) are \(n\) sets with 3 elements each. Let \(\bigcup\limits_{i = 1}^{30} {{P_i}} = \bigcup\limits_{j = 1}^{30} {{Q_j}} = S\) and each element of \(S\) belong to exactly 10 of \({P_i}s\) and exactly 9 of the \({Q_j}s\). Then, \(n\) is equal to
(A) 15
(B) 3
(C) 45
(D) None of these
45
Suppose the set \(S\) has \(k\) elements, then
\(10k = 30 \times 5 \Rightarrow k = 15\)
(as each element of S belong to exactly 10 of \({P_i}s\))
and \(9k = 3n \Rightarrow n = 45\)
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Question 03: The number of functions f from the set A = {0, 1, 2} into the set B = {0, 1, 2, 3, 4, 5, 6, 7} such that f(i) ≤ f(j) for i < j and i, j belong to A is
(A) \(^8{{\rm{C}}_3}\)
(B) \(^8{{\rm{C}}_3} + 2\left( {^8{{\rm{C}}_2}} \right)\)
(C) \(^{10}{{\rm{C}}_3}\)
(D) None of these
\(^{10}{{\rm{C}}_3}\)
Given that \(f(0) \le f(1) \le f(2)\)and \(f(0),\,f(1),\,f(2)\) can take values from the set {0, 1, 2, 3,….7}, then there are three possible cases:
Case 1: \(f(0),\,f(1),\,f(2)\) all three are different; this can be done in \(^{\rm{8}}{{\rm{C}}_{\rm{3}}}\) ways.
Case 2: In \(f(0),\,f(1),\,f(2)\) two of them are same, this can be done in \(^{\rm{8}}{{\rm{C}}_{\rm{2}}} \times 2\) ways.
Case 3: All \(f(0),\,f(1),\,f(2)\) are same, this can be done in \(^{\rm{8}}{{\rm{C}}_{\rm{1}}}\)ways.
Number of such mappings = \(^{\rm{8}}{{\rm{C}}_{\rm{3}}}{ + ^{\rm{8}}}{{\rm{C}}_{\rm{2}}} \times 2{ + ^{\rm{8}}}{{\rm{C}}_{\rm{1}}}\)
= \(\underbrace {^{\rm{8}}{{\rm{C}}_{\rm{3}}}{ + ^{\rm{8}}}{{\rm{C}}_{\rm{2}}}}_{} + \underbrace {^{\rm{8}}{{\rm{C}}_{\rm{2}}}{ + ^{\rm{8}}}{{\rm{C}}_{\rm{1}}}}_{}\)= \(^{\rm{9}}{{\rm{C}}_{\rm{3}}}{ + ^{\rm{9}}}{{\rm{C}}_{\rm{2}}} = {\,^{{\rm{10}}}}{{\rm{C}}_{\rm{3}}}\).
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Question 04: If \(f(x) + f(1-x) = 2\), then the value of \(f\left( {\frac{1}{{2001}}} \right) + f\left( {\frac{2}{{2001}}} \right) + ....f\left( {\frac{{2000}}{{2001}}} \right)\,\) is
(A) 2000
(B) 2001
(C) 1999
(D) 1998
2000
Given that \(f(x) + f(1 - x) = 1\)
Hence \(f\left( {\frac{1}{{2000}}} \right) + f\left( {\frac{{1999}}{{2000}}} \right) = 2\)
Similarly \(f\left( {\frac{2}{{2000}}} \right) + f\left( {\frac{{1998}}{{2000}}} \right) = 2\) and so on. Since there are 2000 terms, thus there will be 1000 pairs and their sum is 1000×2 = 2000
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NIMCET 2009
Question 05: A set contains \((2n + 1)\) elements. If the number of subsets which contain at most \(n\) elements is 4096, then the value of \(n\) is.
(1) 28
(2) 21
(3) 15
(4) 6
6
number of subsets which contain at most \(n\) elements is
\(^{2n + 1}{{\rm{C}}_0}{ + ^{2n + 1}}{{\rm{C}}_1} + ......{ + ^{2n + 1}}{{\rm{C}}_n}\)
We know that \(^{2n + 1}{{\rm{C}}_0}{ + ^{2n + 1}}{{\rm{C}}_1} + ......{ + ^{2n + 1}}{{\rm{C}}_n}\)is exactly half of the sum \(^{2n + 1}{{\rm{C}}_0}{ + ^{2n + 1}}{{\rm{C}}_1} + ......{ + ^{2n + 1}}{{\rm{C}}_{2n + 1}}\)
\( \Rightarrow \,{\,^{2n + 1}}{{\rm{C}}_0}{ + ^{2n + 1}}{{\rm{C}}_1} + ......{ + ^{2n + 1}}{{\rm{C}}_n} = \frac{{{2^{2n + 1}}}}{2} = {2^{2n}}\)
Given that \({2^{2n}} = 4096 = {2^{12}} \Rightarrow n = 6\)
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Question 06: The total number of relations that exist from the set A with m elements into the set A×A is.
(1) \(m^2\)
(2) \(m^3\)
(3) \(m\)
(4) \({2^{{m^2}}}\)
\({2^{{m^2}}}\)
Number of elements in the Cartesian product = m×m = m².
Number of relations = Number of possible subsets of A×A.
Hence total number of relations = \({2^{{m^2}}}\)
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Question 07: If \(P = \{ ({4^n} - 3n - 1):n \in N\} \) and \(Q = \{ (9n - 9):n \in N\} \) then \( P \cup Q\) is equal to.
(1) N
(2) P
(3) Q
(4) None of these
Q
\({4^n} - 3n - 1 = {(1 + 3)^n} - 3n - 1\)
Using Binomial Theorem, it is a multiple of 9. Q is also a multiple of 9. But Q contains all multiple of 9, hence P is a subset of Q.
\( \Rightarrow P \cup Q = Q\)
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Question 08:
\(A_1 A_2, A_3\) and \(A_4\) are subsets of a set U containing 75 elements with the following properties. Each subset contains 28 elements. The intersection of any two of the subsets contains 12 elements, the intersection of any three of
the subsets contains 5 elements, the intersection of all four subsets contains 1 element. The number of elements belonging to none of the four subsets is:
(1) 15
(2) 17
(3) 16
(4) 18
16
We know that \({A_1} \cup {A_2} \cup {A_3} \cup {A_4}\) is equal to
\(\sum {{A_i}} - \sum {\left( {{A_i} \cap {A_j}} \right)} + \sum {\left( {{A_i} \cap {A_j} \cap {A_k}} \right)} - {A_1} \cap {A_2} \cap {A_3} \cap {A_4}\)
\( \Rightarrow {A_1} \cup {A_2} \cup {A_3} \cup {A_4} = 4 \times 28 - 6 \times 12 + 4 \times 5 - 1\) = 59
The number of elements belonging to none of the four subsets = 75 – 59 = 16.
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Question 09:
From 50 students taking examination in Mathematics, Physics and Chemistry, 37 passed Mathematics, 24 Physics and 43 Chemistry. At most 19 passed mathematics and Physics at most 29 Mathematics and Chemistry and at most 20 Physics and
Chemistry. The largest possible number that could have passed all three examinations is:
(1) 10
(2) 12
(3) 9
(4) None of these
14
Suppose number of students passed in Mathematics, Physics and Chemistry are M, P and C, then
M∪P∪C = M + P + C – M∩P – P∩C – M∩C + M∩P∩C
Or M∪P∪C = 37 + 24 + 43 – (M∩P + P∩C + M∩C) + M∩P∩C
\( \Rightarrow \) 50 = 104 – (M∩P + P∩C + M∩C) + M∩P∩C
\( \Rightarrow \) M∩P∩C = M∩P + P∩C + M∩C – 54
Putting the maximum values of M∩P, P∩C and M∩C, we get the maximum value of M∩P∩C as
19 + 20 + 29 – 54 = 14.
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NIMCET 2010
Question 10: How many proper subsets of {1, 2, 3, 4, 5, 6, 7} contain the numbers 1 and 7?
(1) 7
(2) 31
(3) 32
(4) 62
31
1 and 7 are already there in the subset, we have to select remaining 5 elements, this can be done in \({2^5}\) ways. But there is one subset that will contain all the elements, this will not be the proper subset.
Number of subsets = \({2^5} - 1 = 31\)
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(1) If A and B are two sets, then \(A - B = A \cap \bar B\)
(2) If A, B and C are sets, then (A – B) – C = (A – C) – (B – C)
(3) If A and B are two sets, then \(\bar A \cup \bar B = \bar A \cap \bar B\)
(4) If A, B and C are sets, then \(A \cap B \cap \bar C = \bar C \cap A \cap B\)
\(\bar A \cup \bar B = \bar A \cap \bar B\) is incorrect.
Choice (3) is incorrect. Actually \(\bar A \cup \bar B = \overline {A \cap B} \)
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Question 12: A survey shows that 63% of the Americans like cheese where as 76% like apples. If x% of the Americans lie both cheese and apples, then we have
(1) x ≥ 39
(2) x ≤ 63
(3) 39 ≤ x ≤ 63
(4) None of these
39 ≤ x ≤ 63
Suppose percentage of the American who likes at least one of them is n%, then
n = 63 + 76 – x
\( \Rightarrow \)x = 139 – n
Since maximum value of n is 100, so minimum value of x = 39.
Maximum value of x = 63.
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Question 13: Set A has 3 elements and set B has 4 elements. The number of injection that can be defined from A to B is
(1) 144
(2) 12
(3) 24
(4) 64
24
In injective functions, each element has a unique image. Hence number of injective functions from A to B = 4×3×2 = 24.
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