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NCERT Exemplar · Class 11 Mathematics Sets

58 questions · 58 still being checked

EXERCISE 1.3 51–58 (part 6 of 6)

  1. Fill in the blanks in each of the Exercises from $\displaystyle 44$ to $\displaystyle 51$ :

    Exercise 51

    For all sets A and B,A−(A∩B)\displaystyle \mathrm{B}, \mathrm{A}-(\mathrm{A} \cap \mathrm{B}) is equal to ____\displaystyle \_\_\_\_.

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    \(\displaystyle A-B\) (that is, \(\displaystyle A\cap B'\)).\[A-(A\cap B)=A\cap(A\cap B)' \]\[=A\cap(A'\cup B') \]\[=(A\cap A')\cup(A\cap B') \]\[=\varnothing\cup(A\cap B')=A\cap B'=A-B \]

  2. Exercise 52

    Match the following sets for all sets A, B and C
    (i) ((A′∪B′)−A)′\displaystyle \left(\left(\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}\right)-\mathrm{A}\right)^{\prime}(a) A−B\displaystyle \mathrm{A}-\mathrm{B}
    (ii) [B′∪(B′−A)]′\displaystyle \left[\mathrm{B}^{\prime} \cup\left(\mathrm{B}^{\prime}-\mathrm{A}\right)\right]^{\prime}(b) A
    (iii) (A−B)−(B−C)\displaystyle (\mathrm{A}-\mathrm{B})-(\mathrm{B}-\mathrm{C})(c) B
    (iv) (A−B)∩(C−B)\displaystyle (\mathrm{A}-\mathrm{B}) \cap(\mathrm{C}-\mathrm{B})(d) (A×B)∩(A×C)\displaystyle (\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})
    (v) A×(B∩C)\displaystyle \mathrm{A} \times(\mathrm{B} \cap \mathrm{C})(e) (A×B)∪(A×C)\displaystyle (\mathrm{A} \times \mathrm{B}) \cup(\mathrm{A} \times \mathrm{C})
    (vi) A×(B∪C)\displaystyle \mathrm{A} \times(\mathrm{B} \cup \mathrm{C})(f) (A∩C)−B\displaystyle (\mathrm{A} \cap \mathrm{C})-\mathrm{B}

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    NCERT’s answer
    (i)
    \(\displaystyle \leftrightarrow\) (b)
    (ii)
    \(\displaystyle \leftrightarrow\) (c)
    (iii)
    \(\displaystyle \leftrightarrow\) (a)
    (iv)
    \(\displaystyle \leftrightarrow\) (f)
    (v)
    \(\displaystyle \leftrightarrow\) (d)
    (vi)
    \(\displaystyle \leftrightarrow\) (e)
    (i)-(b), (ii)-(c), (iii)-(a), (iv)-(f), (v)-(d), (vi)-(e)
    (i)
    Since \(\displaystyle A\cap B\subseteq A\), we have \(\displaystyle A'\subseteq(A\cap B)'\):
    \[\big((A'\cup B')-A\big)'=\big((A\cap B)'\cap A'\big)'=(A')'=A \]
    (ii)
    Since \(\displaystyle B'-A\subseteq B'\):
    \[\big[B'\cup(B'-A)\big]'=(B')'=B \]
    (iii)
    Since \(\displaystyle B'\subseteq B'\cup C\):
    \[(A-B)-(B-C)=A\cap B'\cap(B\cap C')'=A\cap B'\cap(B'\cup C)=A\cap B'=A-B \]
    (iv)
    \[(A-B)\cap(C-B)=(A\cap B')\cap(C\cap B')=(A\cap C)\cap B'=(A\cap C)-B \]
    (v)
    \(\displaystyle (x,y)\) with \(\displaystyle x\in A\) and \(\displaystyle y\in B\cap C\) means \(\displaystyle x\in A,\ y\in B\) and \(\displaystyle x\in A,\ y\in C\):
    \[A\times(B\cap C)=(A\times B)\cap(A\times C) \]
    (vi)
    \(\displaystyle x\in A\) and \(\displaystyle y\in B\cup C\) means \(\displaystyle (x,y)\in A\times B\) or \(\displaystyle (x,y)\in A\times C\):
    \[A\times(B\cup C)=(A\times B)\cup(A\times C) \]
  3. State True or False for the following statements in each of the Exercises from $\displaystyle 53$ to $\displaystyle 58$ :

    Exercise 53

    If A is any set, then A⊂A\displaystyle \mathrm{A} \subset \mathrm{A}

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    NCERT’s answer
    True
    True. Every element of \(\displaystyle A\) lies in \(\displaystyle A\):\[x\in A\Rightarrow x\in A \]\[A\subseteq A \]

  4. Exercise 54

    Given that M={1,2,3,4,5,6,7,8,9}\displaystyle \mathrm{M}=\{1,2,3,4,5,6,7,8,9\} and if B={1,2,3,4,5,6,7,8,9}\displaystyle \mathrm{B}=\{1,2,3,4,5,6,7,8,9\}, then B⊄M\displaystyle \mathrm{B} \not \subset \mathrm{M}

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    NCERT’s answer
    False
    False. \(\displaystyle B\) and \(\displaystyle M\) have the same elements, so every element of \(\displaystyle B\) is in \(\displaystyle M\):\[B=M \Rightarrow B\subseteq M \]

  5. Exercise 55

    The sets {1,2,3,4}\displaystyle \{1,2,3,4\} and {3,4,5,6}\displaystyle \{3,4,5,6\} are equal.

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    NCERT’s answer
    False
    False. Equal sets have the same elements; here\[1\in\{1,2,3,4\},\qquad 1\notin\{3,4,5,6\} \]\[\{1,2,3,4\}\neq\{3,4,5,6\} \]

  6. Exercise 56

    Q∪Z=Q\displaystyle \mathbf{Q} \cup \mathbf{Z}=\mathbf{Q}, where Q\displaystyle \mathbf{Q} is the set of rational numbers and Z\displaystyle \mathbf{Z} is the set of integers.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    True
    True. Every integer is rational, \(\displaystyle n=\dfrac{n}{1}\), so\[\mathbf Z\subseteq\mathbf Q \]\[\mathbf Q\cup\mathbf Z=\mathbf Q \]

  7. Exercise 57

    Let sets R and T be defined as R={x∈Z∣x is divisible by 2}T={x∈Z∣x is divisible by 6}. Then T⊂R\begin{aligned} & \mathrm{R}=\{x \in \mathbf{Z} \mid x \text { is divisible by } 2\} \\ & \mathrm{T}=\{x \in \mathbf{Z} \mid x \text { is divisible by } 6\} . \text { Then } \mathrm{T} \subset \mathrm{R} \end{aligned}

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    True
    True. Every multiple of $\displaystyle 6$ is also a multiple of 2.\[x \in T \Rightarrow x = 6k,\ k \in \mathbf{Z} \] \[x = 2(3k),\quad 3k \in \mathbf{Z} \] \[\Rightarrow x \in R \] \[\therefore\ T \subset R \]

  8. Exercise 58

    Given A={0,1,2},B={x∈R∣0≤x≤2}\displaystyle \mathrm{A}=\{0,1,2\}, \mathrm{B}=\{x \in \mathbf{R} \mid 0 \leq x \leq 2\}. Then A=B\displaystyle \mathrm{A}=\mathrm{B}.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    False
    False. \(\displaystyle B\) contains non-integers, which \(\displaystyle A\) does not.\[0 \le \tfrac{1}{2} \le 2 \Rightarrow \tfrac{1}{2} \in B \] \[\tfrac{1}{2} \notin \{0,1,2\} = A \] \[\therefore\ A \ne B \]