Sets Test

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Sets Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

If P(S) denotes the set of all subsets of a given set S, then the number of one to one function from the set s={1,2,3} to the set of P(S) is

A
336

B
8

C
36

D
320

Solution

$\begin{array}{l} Given \\ P\left( s \right) =\left\{ { 1,\, 2,\, 3 } \right\} \\ P\left( s \right) =all\, subset\, of\, 5 \\ =\left\{ { \phi ,\left\{ 1 \right\} ,\left\{ 3 \right\} ,\left\{ { 1,2 } \right\} ,\left\{ { 2,3 } \right\} ,\left\{ { 3,1 } \right\} ,\left\{ { 1,2,3 } \right\} } \right\} \\ No.\, of\, element\, of\, 5=0\left( 5 \right) =3 \\ similarly \\ O\left( { p\left( s \right) } \right) =8 \\ Total\, no.\, of\, one-one\, function \\ =\frac { { n! } }{ { \left( { n-m } \right) ! } } \, ,\, where\, n=o\left( { p\left( s \right) } \right) \, and\, m=o\left( 5 \right) \\ =\frac { { 8! } }{ { \left( { 8-3 } \right) ! } } =\frac { { 8! } }{ { 5! } } \\ =336\, \, \\ No.\, of\, one-one\, function\, is\, 336 \\ Hence,\, \, option\, \, A\, \, is\, correct, \end{array}$
Question 2
1 / -0

In a group of 800 people, 550 can speak Hindi and 450 can speak English. How many can speak both Hindi and English ?

A
$100$

B
$150$

C
$200$

D
None of these

Solution

Let H denote the set of people speaking Hindi and E denote the set of people speaking English. We are given :
$n(H)=550, n(E)=450, n(H\cup E)=800$
Now,
$n(H\cup E)=n(H)+n(E)-n(H\cap E)$
$n(H \cap E)=550+450-800=200$
Hence, 200 persons can speak both Hindi and English.
Question 3
1 / -0

If n(A)=3, n(B)=4, then $n(A\times B\times C)=36 find \,n(C)$ is equal to :

A
3

B
12

C
108

D
none of these

Solution

$n(A)=3$
$n(B)=4$
$n(A\times B\times C)=36$
$n(3\times 4\times C)=36$
$n(12\times C)=36$
$\therefore n(C)=3$
So, $n(A\times B\times C)=36$
$n(3\times 4\times 3)=36$ . H.P.
Question 4
1 / -0

If $A=\left\{1,2,3,4\right\}; B=\left\{2,4,6,8\right\}; C=\left\{3,4,5,8\right\}$ then $A\cap B\cap C=$

A
$\phi$

B
${4}$

C
$\mu$

D
${2,4}$

Solution

$\begin{matrix} A=\left\{ { 1,2,3,4 } \right\} \\ B=\left\{ { 2,4,6,8 } \right\} \\ C=\left\{ { 3,4,5,8 } \right\} \\ A\cap B\cap C=\left\{ { 4 } \right\} . \\ \end{matrix}$
Question 5
1 / -0

Let A,B are two sets such that n(A)=4 and n(B)=6. Then the least possible number of elements in the power set of $(A\cup B)$ is

A
16

B
64

C
256

D
1024

Solution

Given,
$n(A)=4,n(B)=6$

Then the least number of possible elements in
$n(A\cup B)=2^{n(A)}.2^{n(B)}=2^4.2^6=2^{10}=1024$
Question 6
1 / -0

Let $A , B$ and $C$ be pairwise independent events with $P ( C ) > 0$ and $P ( A \cap B \cap C ) = 0$ Then, $P \left( A ^ { C } \cap B ^ { C } / C \right)$ is equal to

A
$P \left( A ^ { C } \right) - P ( B )$

B
$P ( A ) - P \left( B ^ { C } \right)$

C
$P \left( A ^ { C } \right) + P \left( B ^ { C } \right)$

D
$P \left( A ^ { C } \right) - P \left( B ^ { C } \right)$

Solution

$\begin{array}{l} P\left( { \frac { { { A^{ C } }\cap { B^{ C } } } }{ C } } \right) =\frac { { P\left( { \left( { { A^{ C } }\cap { B^{ C } } } \right) \cap C } \right) } }{ { P\left( C \right) } } \\ =\frac { { P\left( { \left( { 1-A } \right) \left( { 1-B } \right) \cdot C } \right) } }{ { P\left( C \right) } } \\ =\frac { { P\left[ { \left( { 1-A-B+AB } \right) \cdot C } \right] } }{ { P\left( C \right) } } \\ =\frac { { P\left( C \right) -P\left( { CA } \right) -P\left( { CB } \right) +P\left( { ABC } \right) } }{ { P\left( C \right) } } \\ =\frac { { P\left( C \right) -P\left( C \right) \cdot P\left( A \right) -P\left( C \right) \cdot P\left( B \right) +P\left( A \right) P\left( B \right) P\left( C \right) } }{ { P\left( C \right) } } \\ =\frac { { P\left( C \right) -P\left( C \right) \cdot P\left( A \right) -P\left( C \right) \cdot P\left( B \right) +P\left( { A\cap B\cap C } \right) } }{ { P\left( C \right) } } \\ =1-P\left( A \right) -P\left( B \right) \\ =P\left( { { A^{ C } } } \right) -P\left( B \right) \\ Hence\, ,\, option\, A\, is\, correct\, answer. \end{array}$
Question 7
1 / -0

Let $\displaystyle S=\left\{a\in N,a\le100\right\}$ if the equation $\displaystyle[{\tan}^{2}x]-\tan x-a=0$ has real roots, then the number of elements $S$ is (when $[.]$ is greatest integer function)

A
$10$

B
$1$

C
$9$

D
$0$

Solution

Given equation is,
$[tan^2x]-tanx-a=0$
$\therefore D=b^2-4ac=(-1)^2-4(1)(-a)$

$=1+4a$
So D must be a odd perfect square

$\implies \sqrt{1+4a}=2\lambda+1$

$\implies 1+4a=4\lambda^2+1+4\lambda$

$\implies a=\lambda(\lambda+1)$

For different values og=f $\lambda$ we get
$a=0,2,6,12,20,30,42,56,72,90$

Hence there are $10$ values of $a$ .
Question 8
1 / -0

The function $f(x)$ satisfies the condition $(x-2)f(x)+2f\left(\dfrac{1}{x}\right)=2$ for all $x\neq 0$ . Then the value of $f(2)$ is

A
$\dfrac{1}{2}$

B
$1$

C
$\dfrac{7}{4}$

D
$\dfrac{-3}{2}$

Solution

Putting $x=2$ in $\left( x-2 \right) f\left( x \right) +2f\left( \dfrac { 1 }{ x } \right) =2$
$\left( 2-2 \right) f\left( 2 \right) +2f\left( \dfrac { 1 }{ 2 } \right) =2$
$2f\left( \dfrac { 1 }{ 2 } \right) =2$
$f\left( \dfrac { 1 }{ 2 } \right) =1$
Question 9
1 / -0

The value of c for which the set $\{(x, y)|x^2+y^2+2x\leq 1\}\cap \{(x, y)|x-y+c\geq 0\}$ contains only one point in common is?

A
$(-\infty, -1]\cup [3, \infty)$

B
$\{-1, 3\}$

C
$\{-3\}$

D
$\{-1\}$

Solution

$x^2+y^2+2x-1\le 0$ $x-y+c\ge 0$
To contain only one point in common the line should be a tangent to the circle.
$\Rightarrow x^2+y^2+2x+1-1-1 \le 0$
$\Rightarrow (x+1)^2+y^2\le 2$
Centre= $(-1,0)$
Radius= $\sqrt {2}$
$(-1,0)$ $x-y+c \ge 0$
$d=\left|\cfrac {-1+c}{\sqrt {2}}\right|=\sqrt {2}$
$\Rightarrow |c-1|=2$
$c-1=2$ $c-1=-2$
$c=3$ $c=-1$

$\therefore c={3,-1}$
Question 10
1 / -0

If $A=\left\{1, 2, 3, 4\right\}$ , then the number of subsets of $A$ that contain the element $2$ but not $3$ , is

A
$16$

B
$4$

C
$8$

D
$24$

Solution

The subsets are be $\left\{1, 2, 4\right\},\left\{1, 2\right\}, \left\{2, 4\right\}, \left\{2\right\}$
Number of subsets of $A$ that contain the element $2$ but not $3$ is $4$

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Re-Attempt Weekly Quiz Competition

Sets Test - 41

Result

Sets Test - 41

Score

-

Rank

-

SHARING IS CARING

Question 1

336

8

36

320

Solution

Question 2

100100100

150150150

200200200

None of these

Solution

Question 3

3

12

108

none of these

Solution

Question 4

ϕ\phiϕ

4{4}4

μ\muμ

2,4{2,4}2,4

Solution

Question 5

16

64

256

1024

Solution

Question 6

P(AC)−P(B)P \left( A ^ { C } \right) - P ( B )P(AC)−P(B)

P(A)−P(BC)P ( A ) - P \left( B ^ { C } \right)P(A)−P(BC)

P(AC)+P(BC)P \left( A ^ { C } \right) + P \left( B ^ { C } \right)P(AC)+P(BC)

P(AC)−P(BC)P \left( A ^ { C } \right) - P \left( B ^ { C } \right)P(AC)−P(BC)

Solution

Question 7

101010

111

999

000

Solution

Question 8

12\dfrac{1}{2}21​

111

74\dfrac{7}{4}47​

−32\dfrac{-3}{2}2−3​

Solution

Question 9

(−∞,−1]∪[3,∞)(-\infty, -1]\cup [3, \infty)(−∞,−1]∪[3,∞)

{−1,3}\{-1, 3\}{−1,3}

{−3}\{-3\}{−3}

{−1}\{-1\}{−1}

Solution

Question 10

161616

444

888

242424

Solution

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Question Analysis

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$100$

$150$

$200$

$\phi$

${4}$

$\mu$

${2,4}$

$P \left( A ^ { C } \right) - P ( B )$

$P ( A ) - P \left( B ^ { C } \right)$

$P \left( A ^ { C } \right) + P \left( B ^ { C } \right)$

$P \left( A ^ { C } \right) - P \left( B ^ { C } \right)$

$10$

$1$

$9$

$0$

$\dfrac{1}{2}$

$1$

$\dfrac{7}{4}$

$\dfrac{-3}{2}$

$(-\infty, -1]\cup [3, \infty)$

$\{-1, 3\}$

$\{-3\}$

$\{-1\}$

$16$

$4$

$8$

$24$