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MATHS 2 Marks Questions

Sets · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 21 questions.

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Question 12 Marks
For all sets A, B and C, if $\text{A}\subset\text{B},$ then $\text{A}\cap\text{C}\subset\text{B}\cap\text{C}$
Answer
True.
Let $\text{x}\in\text{A}\cap\text{C}$
$\Rightarrow \text{x}\in \text{A}$ and $\text{x}\in\text{C}$
$\Rightarrow \text{x}\in\text{B}$ and $\text{x}\in\text{C}\ \big[\because \text{A}\subset\text{B}\big]$
$\Rightarrow \text{x}\in(\text{B}\cap\text{C})$
$\Rightarrow (\text{A}\cap\text{C})\subset(\text{B}\cap\text{C})$
Hence, given statement is true.
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Question 22 Marks
For all sets A, and B, $\text{A}-(\text{A}\cap\text{B})=\text{A} - \text{B}.$
Answer
$\text{L.H.S.}=\text{A}-(\text{A}\cap\text{B})$
$=\text{A}\cap(\text{A}\cap\text{B})'\ \big[\because \text{A}-\text{B}=\text{A}\cap\text{B}'\big]$
$=\text{A}\cap(\text{A}'\cup\text{B}')\ \big[\because(\text{A}\cap\text{B})'=\text{A}'\cup\text{B}'\big]$
$=(\text{A}\cap\text{A}')\cup(\text{A}\cap\text{B}')\ \big[\text{distributive law}\big]$
$=\phi\cup(\text{A}-\text{B})\big[\because \text{A}\cap\text{A}'=\phi\text{ and A}-\text{B}=\text{A}\cap\text{B}'\big]$
$=(\text{A}-\text{B})=\text{R.H.S.}$
$\text{L.H.S.}=\text{R.H.S.}$
Hence proved.
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Question 32 Marks
Given L = {1, 2, 3, 4}, M = {3, 4, 5, 6} and N = {1, 3, 5}
Verify that $\text{L}-(\text{M}\cup\text{N})=(\text{L}-\text{M})\cap(\text{L}-\text{N})$
Answer
Given L, = {1, 2, 3, 4}, M = {3, 4, 5, 6} and N = {1, 3, 5}
$\text{M}\cap\text{N}=\{1,3,4,5,6\}$
$\text{L}-(\text{M}\cap\text{N})=\{2\}$
Now, L-M = {1, 2} and L-N = {2, 4}
$\{\text{L}-\text{M})\cap\{\text{L}-\text{N})=\{2\}$
Hence, $\text{L}-\{\text{M}\cup\text{N})=\{\text{L}-\text{M})\cap(\text{L}-\text{N}).$
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Question 42 Marks
If Y = {x | x is a positive factor of the number 2p - 1 (2p - 1), where 2p - 1 is a prime number}.Write Y in the roaster form.
Answer
Y - {x | x is a position factor of the number 2p-1 (2P - 1) is a prime number}
So, the factors of 2p-1 are 1, 2, 22, 23, ..... 2p-1
Y = {1, 2, 22, 23, ....., 2p-1, 2p - 1}
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Question 52 Marks
State the following statement is true and false. Justify your answer.
496 $\notin$ {y | the sum of all the positive factors of y is 2y}.
Answer
Given that: 496 $\notin$ {y | the sum of all the positive factors of y is 2y}.
$\therefore$ The positive factors of 496 are 1, 2, 4, 8, 16, 31, 62, 124, 248 and 496
$\therefore$ The sum of all the positive factor of 496 = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 + 496 = 992 = 2 × 496
So 496 is an element of the given set.
Hence, the given statement is 'False'.
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Question 62 Marks
For all sets A, and B, $(\text{A}\cup\text{B})-\text{B}=\text{A} - \text{B}.$
Answer
$\text{L.H.S.}=(\text{A}\cup\text{B})-\text{B}$
$=(\text{A}\cap\text{B})\cap\text{B}'\ \big[\because \text{A}-\text{B}=\text{A}\cap\text{B}'\big]$
$=(\text{A}\cap\text{B}')\cup(\text{B}\cap\text{B}')$
$=(\text{A}\cap\text{B}')\cup\phi$
$=\text{A}\cap\text{B}'=\text{A}-\text{B}=\text{R.H.S.}$
$\text{L.H.S.}=\text{R.H.S.}$
Hence proved.
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Question 72 Marks
Write the following sets in the roaster form:
F = {x | x4 - 5x2 + 6 = 0, x ∈ R}
Answer
Given that: F = {x | x4 - 5x2 + 6 = 0, x ∈ R}
$\therefore$ x4 - 5x2 + 6 = 0
⇒ x4 - 3x2 - 2x2 + 60 = 0
⇒ x2(x2 - 3) - 2(x2 - 3) = 0
⇒ (x2 - 2)(x2 - 3) = 0
⇒ x2 - 2 = 0 and x2 - 3 = 0
$\Rightarrow \text{x}=\pm\sqrt{2}$ and $\text{x}=\pm\sqrt{3}$
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Question 82 Marks
A, B and C are subsets of Universal Set U. If A = {2, 4, 6, 8, 12, 20} B = {3, 6, 9, 12, 15}, C = {5, 10, 15, 20} and U is the set of all whole numbers, draw a Venn diagram showing the relation of U, A, B and C.
Answer
Given that: A, B, and C are the subsets of a universal set U.

Where A = {2, 4, 6, 8, 12, 20} B = {3, 6, 9, 12, 15} and C = {5, 10, 15, 20}.
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Question 92 Marks
Write the following sets in the roaster form:
$\text{E}=\Big\{\text{w}|\frac{\text{w}-2}{\text{w}+2}=3,\text{w}\in\text{R}\Big\}$
Answer
$\text{E}=\Big\{\text{w}|\frac{\text{w}-2}{\text{w}+2}=3,\text{w}\in\text{R}\Big\}$
$\therefore\frac{\text{w}-2}{\text{w}+2}=3$
$\Rightarrow 3\text{w}+9=\text{w}-2$
$\Rightarrow 3\text{w}-\text{w}=-2-9$
$\Rightarrow 2\text{w}=-11$
$\Rightarrow \text{w}=\frac{-11}{2}\in\text{R}$
Hence, $\text{E}=\Big\{\frac{-11}{2}\Big\}$
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Question 102 Marks
For all sets A, and B, $\text{A}\cup(\text{B}-\text{A})=\text{A}\cup\text{B}.$
Answer
$\text{L.H.S.}=\text{A}\cup(\text{B}-\text{A})$
$=\text{A}\cup(\text{B}\cap\text{A}')\ \big[\because \text{A}-\text{B}=\text{A}\cap\text{B}'\big]$
$=(\text{A}\cup\text{B})\cap(\text{A}\cup\text{A}')$ [distributive law]
$=(\text{A}\cup\text{B})\cap\text{U}\ [\because \text{A}\cup\text{A}'=\text{U}]$
$=(\text{A}\cup\text{B})=\text{R.H.S.} \ \big[\because \text{A}\cap\text{U}=\text{A}\big]$
$\text{L.H.S.}=\text{R.H.S.}$
Hence, the given statement is proved.
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Question 112 Marks
State which the following statement is true and false. Justify your answer.
128 ∈ {y | the sum of all the positive factors of y is 2y}.
Answer
128 ∈ {y | the sum of all the positive factors of y is 2y}
$\therefore$ Factors of 128 are 1, 2, 4, 8, 16, 32, 64, 128.
Sun of all the factors = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128
$=255\neq2\times128$
Hence the given statement is 'False'.
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Question 122 Marks
State the following statement is true and false. Justify your answer.
$3\notin\{\text{x}|\text{x}^4-5\text{x}^3+2\text{x}^2-112\text{x}+6=0\}$
Answer
Given that: $3\notin\{\text{x}|\text{x}^4-5\text{x}^3+2\text{x}^2-112\text{x}+6=0\}$
$\therefore\text{x}^4-5\text{x}^3+2\text{x}^2-112\text{x}+6=0$
Now for x = 3, we have
(3)4 - 5(3)3 + 2(3)2 - 112(3) + 6 = 0
⇒ 81 - 135 + 18 - 336 + 6 = 0 which is not true
So 3 cannot be an element of the given set.
Hence, statement (iii) is 'True'.
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Question 132 Marks
State the following statement is true and false. Justify your answer.
35 ∈ {x | x has exactly four positive factors}.
Answer
Given that: 35 ∈ {x | x has exactly four positive factors}.
$\therefore$ Factors of 35 are 1, 5, 7, 35
Hence, the statement (i) is 'True'.
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Question 142 Marks
For all sets A, B and C, if $\text{A}\subset\text{C},$ and $\text{B}\subset\text{C},$ then $\text{A}\cup\text{B}\subset\text{C}.$
Answer
True
Let $\text{x}\in\text{A}\cup\text{B}$
$\Rightarrow \text{x}\in \text{A}$ or $\text{x}\in\text{B}$
$\Rightarrow \text{x}\in\text{C}$ and $\text{x}\in\text{C}\ \big[\because \text{A}\subset\text{C and B}\subset\text{C}\big]$
$\Rightarrow \text{x}\in\text{C}$
$\Rightarrow \text{A}\cup\text{B}\subset\text{C}$
Hence, given statement is 'True'.
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Question 152 Marks
If A and B are subsets of the universal set U, then show that.
$\text{A} \subset \text{A} \cup \text{B}$
Answer
Given that: $\text{A}\subset\text{U}$ and $\text{B}\subset\text{U}$
To prove $\text{A}\subset\text{A}\cup\text{B}$ it is enough to show that if $\text{x}\in\text{A}$
$\Rightarrow \text{x}\in\text{A} \cup\text{B}$
Let $\text{x}\in\text{A} \Rightarrow \text{x}\in\text{A}$ or $\text{x}\in\text{B}\Rightarrow\text{x}\in\text{A}\cup\text{B}$
Hence, $\text{A}\subset(\text{A}\cup\text{B})$
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Question 162 Marks
For all sets A, B and C, if $\text{A}\subset\text{B},$ then $\text{A}\cup\text{C}\subset\text{B}\cup\text{C}$
Answer
Suppose $\text{A}\subset\text{B}$
Let $\text{x}\in\text{A}\cup\text{C}$
$\Rightarrow \text{x}\in \text{A}$ or $\text{x}\in\text{C}$
$\Rightarrow \text{x}\in\text{B}$ and $\text{x}\in\text{C}\ \big[\because \text{A}\subset\text{B}\big]$
$\Rightarrow \text{x}\in(\text{B}\cup\text{C})$
$\Rightarrow (\text{A}\cup\text{C})\subset(\text{B}\cup\text{C})$
Hence, the given statement is 'True'.
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Question 172 Marks
For all sets A, B and C, $\text{A}-(\text{B} - \text{C}) = (\text{A} - \text{B}) - \text{C}$
Answer
Let us solve the given statement by the following Venn diagram.

Clearly from the above diagram, we calculate that
$\text{A}-(\text{B}-\text{C})\neq(\text{A}-\text{B})-\text{C}$
Hence, the given statements is not 'True'.
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Question 182 Marks
For all sets A, and B, $\text{A}-(\text{A}-\text{B})=\text{A}\cap\text{B}.$
Answer
$\text{L.H.S.}=\text{A}-(\text{A}-\text{B})$
$=\text{A}-(\text{A}\cap\text{B}')\ \big[\because \text{A}-\text{B}=\text{A}\cap\text{B}'\big]$
$=\text{A}\cap(\text{A}\cap\text{B}')'$
$=\text{A}\cap[\text{A}'\cup(\text{B}')'] \ \big[\because (\text{A}\cap\text{B})'=\text{A}'\cup\text{B}'\big]$
$=\text{A}\cap(\text{A}'\cup\text{B})$
$=(\text{A}\cap\text{A}')\cup(\text{A}\cap\text{B})$
$=\phi\cup(\text{A}\cap\text{B})$
$=\text{A}\cap\text{B}=\text{R.H.S.}$
$\text{L.H.S.}=\text{R.H.S.}$
Hence proved.
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Question 192 Marks
Write the following sets in the roaster form:
D = {t | t3 = t, t ∈ R}
Answer
Given that: D = {t | t3 = t, t ∈ R}
$\therefore$ t3 = t
⇒ t3 - t = 0
⇒ t(t2 - 1) = 0
⇒ t(t - 1)(t + 1) = 0
⇒ t = 0, t = 1, t = -1
Hence, D = {-1, 0, 1}
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Question 202 Marks
For all sets A and B, $(\text{A} - \text{B}) \cup (\text{A} \cap \text{B}) = \text{A}$
Answer
True.
$\text{L.H.S.}=(\text{A}-\text{B})\cup(\text{A}\cap\text{B})$
$=\big[(\text{A}-\text{B})\cup\text{A}\big]\cap[(\text{A}-\text{B})\cup\text{B}\big]$
$=\text{A}\cap(\text{A}-\text{B})=\text{A}=\text{R.H.S.}$
$\text{L.H.S.}=\text{R.H.S.}$
Hence, given statement is true.
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Question 212 Marks
In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Find the number of students who play neither?
Answer
Let C be the set of students who play cricket and T be the set of students who play tennis.
n(U) = 60, n(C) = 25, n(T) = 20, and $\text{n}(\text{C} \cap \text{T}) = 10 $
$\text{n}(\text{C} \cup \text{T}) = \text{n(C)} + \text{n(T)} - \text{n}(\text{C} \cap \text{T})$
= 25 + 20 - 10 = 35
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