Maths M.C.Q (1 Marks)

$\mathrm{B}=(-\infty,-2) \cup(2, \infty)$

$\mathrm{C}=(-\infty, 2] \cup[6, \infty)$

So, $\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}=(-\infty,-2) \cup[6, \infty)$

$\mathrm{z} \cap(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})^{\prime}=\{-2,-1,0,-1,2,3,4,5\}$

Hence no. of its subsets $=2^{8}=256$.

$m =7, n =4$

$\left(2^{7}-2^{4}=112\right)$

$m \times n =7 \times 4=28$

$\mathrm{n}(\mathrm{B})=7$

$\mathrm{n}(\mathrm{A} \cap \mathrm{B})=3$

$n(A \cup B)=25+7-3=29$

$\Rightarrow B-A=R-(-2,5)$

each element of $T$ belongs to exactly 20

elements of $X _{ i } \Rightarrow \frac{500}{20}=25$ distinct elements

so $\frac{5 n}{6}=25 \Rightarrow n=30$

$\Rightarrow 76 \leq 76+63-x \leq 100$

$\Rightarrow-63 \leq-x \leq-39$

$\Rightarrow 63 \geq x \geq 39$

$T \rightarrow$ person like Tea

$n(C)=73$

$n(T)=65$

$n(C \cup T) \leq 100$

$n(C)+n(T)-n(C \cap T) \leq 100$

$73+65-x \leq 100$

$x \geq 38$

$73-x \geq 0 \Rightarrow x \leq 73$

$65-x \geq 0 \Rightarrow x \leq 65$

$38 \leq x \leq 65$

$\Rightarrow \quad(m+1)^{2}-4(m+4) \geq 0$

$\Rightarrow \quad m^{2}+2 m+1-4 m-16 \geq 0$

$\Rightarrow \quad m^{2}-2 m-15 \geq 0$

$\Rightarrow \quad(m-5)(m+3) \geq 0$

$\Rightarrow \quad m \in(-\infty,-3] \cup[5, \infty)$

$\therefore \quad A=(-\infty,-3] \cup[5, \infty)$

$B=[-3,5)$

$A-B=(-\infty,-3) \cup[5, \infty)$

$A \cap B=\{-3\}$

$B-A=(-3,5)$

$A \cup B=R$

$ \Rightarrow \phi \, \subseteq \,B$

But $A\, \subseteq \,B$

$ \Rightarrow \,$ option $A$ is NOT true

Let $x\, \in \,\,(C\,x\, \in \,(C\, \cup \,A)\,\, \cap (C\, \cup \,B)\,)$

$ \Rightarrow \,x\,(C\, \cup \,A)$ and $x\, \in \,\,(C\, \cup \,B)$

$ \Rightarrow \,(x\, \in \,C$ or $x\, \in \,A)$ and  $(x\, \in \,C$ or $x\, \in \,B)$

$ \Rightarrow \,(x\, \in \,C$ or $x\, \in \,\,(A\, \cap \,B)$

$ \Rightarrow \,(x\, \in \,C$   or  $x\, \in \,C$  (as $A\, \cup \,B \subseteq \,C\,$ )

$ \Rightarrow \,x\, \in \,C$

$ \Rightarrow (C\, \cup \,A)\,\, \cap (C\, \cup \,B)\, \subseteq \,C\,\,\,(1)$

Now $x\, \in \,C\, \Rightarrow \,x\, \in \,(C\, \cup \,A)$ and $x\, \in \,\,(C\, \cup \,B\,)$

$ \Rightarrow x\, \in (C\, \cup \,A)\,\, \cap (C\, \cup \,B)$

$ \Rightarrow C\, \subseteq \,(C\, \cup \,A)\,\, \cap (C\, \cup \,B)\, \subseteq \,C\,\,\,(2)$

$ \Rightarrow $ from $(1)$ and $(2)$

$C\, = \,(C\, \cup \,A)\,\, \cap (C\, \cup \,B)$

$ \Rightarrow $ option $B$ is true

Let $x\, \in \,A$ and $x\, \notin \,B$

$\Rightarrow \,x\, \in \,(A - B)$

$ \Rightarrow \,x\, \in \,C$ (as $A - B \subseteq \,C$ )

Let $x\, \in \,A$ and $x\, \in \,B$

$ \Rightarrow \,x\, \in \,(A \cap B)$

$ \Rightarrow \,x\, \in \,C$ (as $A \cap B \subseteq \,C$ )

Hence $x\, \in \,A\,\, \Rightarrow \,x\, \in \,C$

$ \Rightarrow A\, \subseteq \,C$

$ \Rightarrow $ option $C$ is true

As $C \supseteq \,(A \cap B)$

$ \Rightarrow B \cap C \supseteq \,(A \cap B)$

As $A \cap B \ne \phi $

$ \Rightarrow B \cap C \ne \phi $

Hence the correct answer is option $(A)$

$n(C)\, = \,\left[ {\frac{{140}}{5}} \right]\, = \,28$

$n(M)\, = \,\left[ {\frac{{140}}{2}} \right]\, = \,70$

$n(p\, \cup \,C\, \cup \,M)\, = \,n(P)\, + \,n(C)\, + \,n(M)$ $ - \,n(P \cap C) - \,n(C \cap M) - $ $n(M\, \cap \,P)\, + \,n(P \cap M \cap C)$

$ = \,46\, + \,28\, + 70\, - \,\left[ {\frac{{140}}{{15}}} \right]\, - \,\left[ {\frac{{140}}{{10}}} \right]\, - \,\left[ {\frac{{140}}{6}} \right]\, + \,\left[ {\frac{{140}}{{30}}} \right]\,$

$=\,144\,-\,9\,-14\,-\,23+4\,=\,102$

So required number of student $=\,140\,-\,102\,=\,38$ 

$n(A)\, = \,25$

$n(B)\, = \,20$

$n(A \cap B)\, = \,8$

$n(A \cap \bar B)\, = \,17$

$n(\bar A \cap B)\, = \,12$

Now $\%$ of th population who look advertisement

$=\,\frac {30}{100}\times 17\,+$ $\frac {40}{100}\times 12\,+$ $\frac {50}{100}\times 8$

$=\,13.9$

$2(3 - \sqrt x )\, + \,x\, - \,6\sqrt x \, + \,6\, = \,0$

$ \Rightarrow \,x\, - \,8\sqrt x \, + \,12\, = \,0\, \Rightarrow \,\sqrt x \, = \,4,2\, \Rightarrow \,x\, = \,16,4$

Since $x\, \in \,[0,\,9]$

$\therefore \,\,x\,=\,4$

Case - $II$ : $x\, \in \,[9,\,\infty ]$

$2(\sqrt x  - 3)\, + \,x\, - \,6\sqrt x \, + \,6\, = \,0$

$ \Rightarrow \,x\, - \,4\sqrt x \,\, = \,0\, \Rightarrow \,x\, = \,16,0$

Since $x\, \in \,[9,\,\infty ]$

$\therefore \,\,x\,=\,16$

Hence , $x\,=\,4$ and $16$

$n(C)\, = \,15\,\% $

$n(P'\, \cup \,C')\, = \,65\,\% $

$ \Rightarrow n(P \cup \,C')\, = \,65\,\% $

$n(P \cup \,C)\, = \,35\,\% $

$n(P \cap \,C)\, = \,n(P)\, + n(C)\, - \,n(P \cup \,C)$

$25\, + \,15\, - 35\, = \,5\% $

$x\, \times \,5\,\% \, = \,2000$

$x\, = \,40,000$

$=\{0,9,54, \ldots\} $

$ Y =\{9(n-1)\} $

$=\{0,9,18,27, \ldots\} $

$ \therefore X U Y=Y $

$Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\}$

From P

$\sin \theta-\cos \theta=\sqrt{2} \cos \theta$

$\sin \theta=(\sqrt{2}+1) \cos \theta$

$\frac{\sin \theta}{\cos \theta}=(\sqrt{2}+1)$

$\tan \theta=(\sqrt{2}+1)$

$\text { from } Q$

$\sin \theta+\cos \theta=\sqrt{2} \sin \theta$

$\sin \theta(\sqrt{2}-1)=\cos \theta$

$\frac{\sin \theta}{\cos \theta}=(\sqrt{2}-1)$

$\tan \theta=(\sqrt{2}-1)$

$\tan \theta=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}=\sqrt{2}+1$

$\therefore P=Q$

Hence, option $(D)$ is correct answer.

$=\frac{3^4+1}{2}=41 \text {. }$

Since, every students reads $5$ newspapers

 $\therefore$ Numbers of students $ = {{x \times 60} \over 5} = 300$, $x = 25$.

$ \Rightarrow $ $x = \pm$ $ i x$ is not real but $x$ is real (given) No value of $x$ is possible.

$2x = 6$ ==> $x = 3$

There is no value of $x$ which satisfies both the above equations. Thus, $A = \phi $.

${1 \over y}$ can be   $1, [ y\ can\ be\ 1].$

then, total number of subsets of $S$ is ${2^n}$.

Hence, ${2^5} = 32$.

${2^4} - 1 = 16 - 1 = 15$.

Hence $A$ and complement of $B$ are always non-disjoint.

$\therefore A \cap B = \phi $.

==> $x \in A \cap B$,$[\because A \cup B = A \cap B]$

==> $x \in A$ and $x \in B$ ==> $x \in B$, $\therefore A \subseteq B$

Similarly, $x \in B$ ==> $x \in A$, $\therefore B \subseteq A$

Now $A \subseteq B,\,\,B \subseteq A$ ==> $A = B$.

$ = \{ 3,\,4,\,10\} $, $A \cap C = \{ 4\} $.

$\therefore (A \cap B) \cup (A \cap C) = \{ 3,\,4,\,10\} $.

$\therefore A \cap (B \cup C) = \{ a,\,b,\,c\} \cap \{ a,\,b,\,c,\,d,e\,\} $$ = \{ a,\,b,c\} $.

$\therefore (A - B)\, \cup (B - A) \cup (A \cap B) = A \cup B$.

So, $n(A \cup B) = n(B) = 6$.

$0.25 = 0.16 + 0.14 - n(A \cap B)$

==> $n(A \cap B) = 0.30 - 0.25 = 0.05$.

$n(A \cap B) = 0$

Now $n\,(A \cup B) = n(A) + n(B) - n(A \cap B)$

$ = n(A) + n(B) - 0$$ = n(A) + n(B)$.