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NCERT Exemplar · Class 11 Mathematics Sets

58 questions · 58 still being checked

EXERCISE 1.3 11–20 (part 2 of 6)

  1. Exercise 11

    Let U be the set of all boys and girls in a school, G be the set of all girls in the school, B be the set of all boys in the school, and S be the set of all students in the school who take swimming. Some, but not all, students in the school take swimming. Draw a Venn diagram showing one of the possible interrelationship among sets U, G, B and S.

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    Every student is a girl or a boy, not both: \[U = G\cup B, \qquad G\cap B = \varnothing \] Some but not all swim: \[\varnothing \ne S \subsetneq U \] \(\displaystyle S\) cuts each of \(\displaystyle G\), \(\displaystyle B\) in two: \[U = (G\cap S)\cup(G-S)\cup(B\cap S)\cup(B-S) \] One possibility: some girls and some boys swim, and some of each do not. \[G\cap S\ne\varnothing,\quad G-S\ne\varnothing,\quad B\cap S\ne\varnothing,\quad B-S\ne\varnothing \] Answer: \(\displaystyle U=G\cup B\) with \(\displaystyle G\cap B=\varnothing\); \(\displaystyle S\) overlaps both \(\displaystyle G\) and \(\displaystyle B\) without containing either, so \(\displaystyle G\cap S,\ G-S,\ B\cap S,\ B-S\) are all non-empty. NCERT prints: a Venn diagram labelled \(\displaystyle A\), \(\displaystyle S\), \(\displaystyle G\) -- its \(\displaystyle A\) is the set of boys, \(\displaystyle B\).

  2. Exercise 12

    For all sets A,B\displaystyle \mathrm{A}, \mathrm{B} and C, show that (A−B)∩(C−B)=A−(B∪C)\displaystyle (\mathrm{A}-\mathrm{B}) \cap(\mathrm{C}-\mathrm{B})=\mathrm{A}-(\mathrm{B} \cup \mathrm{C})

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    The identity is false as printed. Counterexample: \[A=\{1\},\quad B=C=\phi \] \[(A-B)\cap(C-B)=\{1\}\cap\phi=\phi \] \[A-(B\cup C)=\{1\}-\phi=\{1\}\neq\phi \] What each side really is: \[x\in(A-B)\cap(C-B) \iff x\in A,\ x\in C,\ x\notin B \iff x\in(A\cap C)-B \] \[x\in A-(B\cup C) \iff x\in A,\ x\notin B\cup C \] \[\iff x\in A,\ x\notin B,\ x\notin C \quad \text{(De Morgan's law)} \] \[\iff x\in A-B,\ x\in A-C \iff x\in(A-B)\cap(A-C) \] So \[(A-B)\cap(C-B)=(A\cap C)-B, \qquad A-(B\cup C)=(A-B)\cap(A-C) \] Every element of the left side is in \(\displaystyle C\) and no element of the right side is, so the two sides are disjoint: \[\big((A\cap C)-B\big)\cap\big(A-(B\cup C)\big)=\phi \] \[\big((A\cap C)-B\big)\cup\big(A-(B\cup C)\big)=A-B \quad \text{(split } A-B \text{ by } x\in C,\ x\notin C) \] Disjoint sets are equal only when both are empty, that is, when their union is empty: \[(A-B)\cap(C-B)=A-(B\cup C) \iff A-B=\phi \iff A\subseteq B \]Answer: False as printed (e.g. \(\displaystyle A=\{1\},\ B=C=\phi\)); in fact \(\displaystyle (A-B)\cap(C-B)=(A\cap C)-B\) and \(\displaystyle A-(B\cup C)=(A-B)\cap(A-C)\), and the two are equal only when \(\displaystyle A\subseteq B\).
  3. Determine whether each of the statement in Exercises $\displaystyle 13$ - $\displaystyle 17$ is true or false. Justify your answer.

    Exercise 13

    For all sets A and B,(A−B)∪(A∩B)=A\displaystyle \mathrm{B},(\mathrm{A}-\mathrm{B}) \cup(\mathrm{A} \cap \mathrm{B})=\mathrm{A}

    Matches the book, not yet reviewed

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

    NCERT’s answer
    True
    True. With \(\displaystyle A-B = A\cap B'\): \[(A-B)\cup(A\cap B) = (A\cap B')\cup(A\cap B) \] \[= A\cap(B'\cup B) \quad \text{(distributive law)} \] \[= A\cap U = A \]

  4. Exercise 14

    For all sets A,B\displaystyle \mathrm{A}, \mathrm{B} and C,A−(B−C)=(A−B)−C\displaystyle \mathrm{C}, \mathrm{A}-(\mathrm{B}-\mathrm{C})=(\mathrm{A}-\mathrm{B})-\mathrm{C}

    Matches the book, not yet reviewed

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

    NCERT’s answer
    False
    False. The two sides differ in general: \[A-(B-C) = A\cap(B\cap C')' = A\cap(B'\cup C) = (A-B)\cup(A\cap C) \] \[(A-B)-C = A\cap B'\cap C' \] Counterexample: \(\displaystyle A=\{1\},\ B=\varnothing,\ C=\{1\}\). \[A-(B-C) = \{1\}-\varnothing = \{1\} \] \[(A-B)-C = \{1\}-\{1\} = \varnothing \] \[\{1\}\ne\varnothing \]

  5. Exercise 15

    For all sets A,B\displaystyle \mathrm{A}, \mathrm{B} and C, if A⊂B\displaystyle \mathrm{A} \subset \mathrm{B}, then A∩C⊂B∩C\displaystyle \mathrm{A} \cap \mathrm{C} \subset \mathrm{B} \cap \mathrm{C}

    Matches the book, not yet reviewed

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

    NCERT’s answer
    True
    True. Take any \(\displaystyle x\in A\cap C\). \[x\in A \ \text{and}\ x\in C \] \[A\subset B \Rightarrow x\in B \] \[x\in B \ \text{and}\ x\in C \Rightarrow x\in B\cap C \] \[\therefore\ A\cap C\subset B\cap C \]

  6. Exercise 16

    For all sets A,B\displaystyle \mathrm{A}, \mathrm{B} and C, if A⊂B\displaystyle \mathrm{A} \subset \mathrm{B}, then A∪C⊂B∪C\displaystyle \mathrm{A} \cup \mathrm{C} \subset \mathrm{B} \cup \mathrm{C}

    Matches the book, not yet reviewed

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

    NCERT’s answer
    True
    True. Take any \(\displaystyle x\in A\cup C\). \[x\in A \ \text{or}\ x\in C \] \[x\in A \Rightarrow x\in B \Rightarrow x\in B\cup C \quad (A\subset B) \] \[x\in C \Rightarrow x\in B\cup C \] \[\therefore\ A\cup C\subset B\cup C \]

  7. Exercise 17

    For all sets A,B\displaystyle \mathrm{A}, \mathrm{B} and C, if A⊂C\displaystyle \mathrm{A} \subset \mathrm{C} and B⊂C\displaystyle \mathrm{B} \subset \mathrm{C}, then A∪B⊂C\displaystyle \mathrm{A} \cup \mathrm{B} \subset \mathrm{C}.

    Matches the book, not yet reviewed

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

    NCERT’s answer
    True
    True. Every element of \(\displaystyle A\cup B\) lands in \(\displaystyle C\). \[x\in A\cup B \Rightarrow x\in A \ \text{or}\ x\in B \] \[x\in A,\ A\subset C \Rightarrow x\in C \] \[x\in B,\ B\subset C \Rightarrow x\in C \] \[x\in A\cup B \Rightarrow x\in C \quad\Rightarrow\quad A\cup B\subset C \]
  8. Using properties of sets prove the statements given in Exercises $\displaystyle 18$ to $\displaystyle 22$

    Exercise 18

    For all sets A and B,A∪(B−A)=A∪B\displaystyle \mathrm{B}, \mathrm{A} \cup(\mathrm{B}-\mathrm{A})=\mathrm{A} \cup \mathrm{B}

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[A\cup(B-A)=A\cup(B\cap A') \quad \text{(} B-A=B\cap A' \text{)} \] \[=(A\cup B)\cap(A\cup A') \quad \text{(distributive law)} \] \[=(A\cup B)\cap U \quad \text{(} A\cup A'=U \text{)} \] \[=A\cup B \] Answer: \(\displaystyle A\cup(B-A)=A\cup B\).

  9. Exercise 19

    For all sets A and B,A−(A−B)=A∩B\displaystyle \mathrm{B}, \mathrm{A}-(\mathrm{A}-\mathrm{B})=\mathrm{A} \cap \mathrm{B}

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[A-(A-B)=A-(A\cap B') \quad \text{(} A-B=A\cap B' \text{)} \] \[=A\cap(A\cap B')' \quad \text{(} X-Y=X\cap Y' \text{)} \] \[=A\cap(A'\cup B) \quad \text{(De Morgan, } B''=B \text{)} \] \[=(A\cap A')\cup(A\cap B) \quad \text{(distributive law)} \] \[=\varnothing\cup(A\cap B) \quad \text{(} A\cap A'=\varnothing \text{)} \] \[=A\cap B \] Answer: \(\displaystyle A-(A-B)=A\cap B\).

  10. Exercise 20

    For all sets A and B,A−(A∩B)=A−B\displaystyle \mathrm{B}, \mathrm{A}-(\mathrm{A} \cap \mathrm{B})=\mathrm{A}-\mathrm{B}

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[A-(A\cap B)=A\cap(A\cap B)' \quad \text{(} X-Y=X\cap Y' \text{)} \] \[=A\cap(A'\cup B') \quad \text{(De Morgan)} \] \[=(A\cap A')\cup(A\cap B') \quad \text{(distributive law)} \] \[=\varnothing\cup(A\cap B') \quad \text{(} A\cap A'=\varnothing \text{)} \] \[=A\cap B'=A-B \] Answer: \(\displaystyle A-(A\cap B)=A-B\).