June 18, 2024

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) $\large&space;A=\left&space;\{&space;x:&space;x\in&space;\mathbb{R},&space;2x+11=15&space;\right&space;\}$

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

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

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

$\large&space;2x=4&space;\Rightarrow&space;x=2$

Thus     $A=\left&space;\{&space;2&space;\right&space;\}$.

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

Thus $B=\left&space;\{&space;0,1&space;\right&space;\}$

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

$\large&space;C=\left&space;\{&space;1,p&space;\right&space;\}$

2.  Write the following sets in the roaster  form:

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

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

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

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

$\Rightarrow&space;t=-1,0,+1$

Thus $D=\left&space;\{&space;-1,0,+1&space;\right&space;\}$

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

$w-2=3w+9$

It gives  $w=-\frac{11}{2}$.

Thus $E=&space;\left&space;\{-\frac{11}{2}&space;\right&space;\}$.

(iii)   We have  $x^{4}-5x^{2}+6=0$

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

Therefore  $F=\left&space;\{&space;-\sqrt{3},-\sqrt{2},\sqrt{2}&space;,\sqrt{3}\right&space;\}$.

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

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

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 $Y=&space;\left\{1,2,2^{2},2^{3},.....,2^{p-1},2^{p}-1\right&space;\}$.

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

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

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

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 $C$, $A&space;-&space;\left&space;(&space;B&space;-&space;C&space;\right&space;)&space;=&space;(A&space;-&space;B)&space;-&space;C$

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

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

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

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

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

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

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