Set Theory(NCERT Exemplar )

Here we have given the solutions of  some important questions of chapter 1 (Set Theory) of NCERT Exemplar book .

Short answer type questions

1. Write the following sets in the roaster  form

(i) \large A=\left \{ x: x\in \mathbb{R}, 2x+11=15 \right \}

(ii)\large B=\left \{ x|x^{2} =x, x\in \mathbb{R}\right \}

(iii) \large C=\left \{ x|x \, is \, a\, positive \, factor \, of\, a\, prime\, number\, p \right \}

Sol. (i) On solving the equation     \large 2x+11=15 , we get 

\large 2x=4 \Rightarrow x=2

Thus     A=\left \{ 2 \right \}.

(ii)      \large x^{2}=x  gives  \large x(x-1)=0  which implies \large x=0, 1

Thus B=\left \{ 0,1 \right \}

(iii) Since every prime no. has  only two factors 1 and number itself. This implies

       \large C=\left \{ 1,p \right \}

2.  Write the following sets in the roaster  form:

(i)  \large D=\left \{ t|t^{3}=t, t\in \mathbb{R} \right \}

(ii) E=\left \{ w|\frac{w-2}{w+3} =3, w\in \mathbb{R}\right \}

(iii) F=\left \{ x|x^{4}-5x^{2}+6=0, x\in \mathbb{R} \right \}

Sol. (i) On solving   t^{3}=t, we get     t(t^{2}-1)=0,  which implies  either t=0 or  t^{2}-1=0.

\Rightarrow t=-1,0,+1

Thus D=\left \{ -1,0,+1 \right \}

(ii)           \frac{w-2}{w+3}=3     implies

w-2=3w+9

It gives  w=-\frac{11}{2}.

Thus E= \left \{-\frac{11}{2} \right \}.

(iii)   We have  x^{4}-5x^{2}+6=0

This implies      x^{4}-3x^{2}-2x^{2}+6=0\Rightarrow (x^{2}-3)(x^{2}-2)=0 \Rightarrow x=\pm \sqrt{3}, \pm \sqrt{2}.

Therefore  F=\left \{ -\sqrt{3},-\sqrt{2},\sqrt{2} ,\sqrt{3}\right \}.

Q3. If Y = \left \{ x: x \, is \, a \, positive \, factor \, of \, the \, number\, 2^{p-1}\left(2^{p}-1\right),\, where\, 2^{p} - 1 \, is \, a \, prime \, number\right \}. Write Y in the roaster form.

sol. Y = \left \{ x: x \, is \, a \, positive \, factor \, of \, the \, number\, 2^{p-1}\left(2^{p}-1\right),\, where\, 2^{p} - 1 \, is \, a \, prime \, number\right \}.

Since 2^{p}-1  is prime number , it has only two factors 1 and ( 2^{p}-1) and  the factors of 2^{p-1} are 1,2,22,23,…,2^{p-1} .   Therefore Y= \left\{1,2,2^{2},2^{3},.....,2^{p-1},2^{p}-1\right \}.

Q4. A, B   and C are subsets of Universal Set U.  IfA = \left \{2, 4, 6, 8, 12, 20 \right \}, B= \left \{3,6,9,12,15\right \}, C= \left \{5,10,15,20 \right \}  and U is the set of all whole numbers, draw a Venn diagram showing the relation of U, A, B and C.

Sol. Here  A = \left \{2, 4, 6, 8, 12, 20 \right \}, B= \left \{3,6,9,12,15\right \}, C= \left \{5,10,15,20 \right \}

Therefore A\cap B=\left \{ 6,12 \right \}, B\cap C=\left \{ 15 \right \}A\cap C=\left \{ 20 \right \} and A\cap B\cap C=\phi.

Hence, the Venn diagram showing relation of given sets is:

                                                   

5. Determine whether  the statement given below   is true or false. Justify your answer.

For all sets A, B and C, A - \left ( B - C \right ) = (A - B) - C

Solution:  Let A = \left \{2, 4, 6, 8, 12, 20 \right \}, B= \left \{3,6,9,12,15\right \}, C= \left \{5,10,15,20 \right \},

then  B-C=\left \{ 3,6,9,12 \right \}. This implies  A-\left ( B-C \right )=\left \{ 2, 4,8,20 \right \}             …………..(i)

Now A-B=\left \{ 2,4,8,20 \right \}, this implies  \left ( A-B \right )-C=\left \{ 2,4,8 \right \}                    …………..(ii)

From (i) and (ii) , it is clear that A - \left ( B - C \right ) \neq (A - B) - C.

We can also show that the given expression is false by using Venn Diagrams

Now, from the Venn diagrams,  it is clear that A - \left ( B - C \right ) \neq (A - B) - C.

 

 

 

 

 

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