Sets Test

Result

Sets Test - 41

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

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 $$4$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

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

$$100$$

$$150$$

$$200$$

None of these

Solution

Question 3

3

12

108

none of these

Solution

Question 4

$$\phi$$

$${4}$$

$$\mu$$

$${2,4}$$

Solution

Question 5

16

64

256

1024

Solution

Question 6

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

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

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

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

Solution

Question 7

$$10$$

$$1$$

$$9$$

$$0$$

Solution

Question 8

$$\dfrac{1}{2}$$

$$1$$

$$\dfrac{7}{4}$$

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

Solution

Question 9

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

$$\{-1, 3\}$$

$$\{-3\}$$

$$\{-1\}$$

Solution

Question 10

$$16$$

$$4$$

$$8$$

$$24$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.