Set Theory Test 28

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Set Theory Test 28

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Weekly Quiz Competition

Question 1
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 2
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 3
1 / -0

Left A = {1, 2, 3, 4, 5, 6} and B= {1, 2, 3, 4} be two sets, then the number of functions that can be defined from A to B such that the element '2' in B has exactly 3 pre - images i A, is equal ot

A
540

B
440

C
810

D
620
Question 4
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 5
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 6
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 7
1 / -0

If $$A=\left\{ 1,3,5,7,9,11,13,15,17 \right\} ,B=\left\{ 2,4,....,18 \right\} $$ and N is the universal set, then $$A'\cup \left( \left( A\cup B \right) \cap B' \right) $$

A
A

B
N

C
B

D
R

Solution

$$A=\left\{1,3,5,7,9,11,13,15,17\right\}$$

$${A}^{\prime}=N-A=N-\left\{1,3,5,7,9,11,13,15,17\right\}=\left\{2,4,6,8,10,12,14,16,18\right\}$$

$$A\cup B=\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\right\}$$

$${B}^{\prime}=N-B=N-\left\{2,4,6,8,10,12,14,16,18\right\}=\left\{1,3,5,7,9,11,13,15,17\right\}$$

$$\left(A\cup B\right)\cap {B}^{\prime}$$

$$=\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\right\}\cap\left\{1,3,5,7,9,11,13,15,17\right\}$$

$$=\left\{1,3,5,7,9,11,13,15,17\right\}$$

$${A}^{\prime}\cup \left(\left(A\cup B\right)\cap {B}^{\prime}\right)$$

$$=\left\{2,4,6,8,10,12,14,16,18\right\}\cup\left\{1,3,5,7,9,11,13,15,17\right\}$$

$$=\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\right\}=N$$
Question 8
1 / -0

If $$A=\left\{ 1,2,4 \right\} ,B=\left\{ 2,4,5 \right\} $$ and $$C=\left\{ 2,5 \right\} $$, then $$\left( A-B \right) \times \left( B-C \right) =$$

A
$$\left\{ \left( 1,2 \right) ,\left( 1,5 \right) ,\left( 2,5 \right) \right\} $$

B
$$\left\{ \left\{ 1,4 \right\} \right\} $$

C
$$\left( 1,4 \right) $$

D
$$\left\{ \left( 1,2 \right) \right\} $$

Solution

$$A=\left\{1,2,4\right\}$$ and $$B=\left\{2,4,5\right\}$$

$$A-B=\left\{1,2,4\right\}-\left\{2,4,5\right\}=\left\{1\right\}$$

$$B=\left\{2,4,5\right\}$$ and $$C=\left\{2,5\right\}$$

$$B-C=\left\{2,4,5\right\}-\left\{2,5\right\}=\left\{4\right\}$$

$$\left(A-B\right)\times\left(B-C\right)=\left\{1\right\}\times \left\{4\right\}=\left(1,4\right)$$
Question 9
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 $$4$$
Question 10
1 / -0

Let $$p$$ and $$q$$ be two statements. amongst the following, the statement is equivalent to $$p\rightarrow q$$ is

A
$$\displaystyle p\wedge \sim q$$

B
$$\displaystyle \sim p\vee q$$

C
$$\displaystyle \sim p\wedge q$$

D
$$\displaystyle p\vee \sim q$$

Solution

$$ \begin{array}{l} \text { Given, } \quad P \longrightarrow q \\ \text { ne trinow that, } \rightarrow \text { - if then } \\ \Rightarrow \text { if } P \text { then } q \end{array} $$
$$ \begin{array}{lll} p & q & p \rightarrow q \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} $$
$$ \begin{array}{c} \Rightarrow \text { Evaluating option } A(p \wedge \sim q) \\ P \text { iq } p \wedge \sim q \\ 111 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ \text { Incorrect assumption } \end{array} $$
$$ \begin{array}{c} \Rightarrow \text { Eualuating } B(\sim p \wedge q) \\ \sim p \quad q \quad \sim p \wedge q \\ 0 \\ 0 \\ 1 \\ 1 \quad 0 \\ \text { Incorrect assumption } \end{array} $$
$$ \begin{array}{rlrl} \Rightarrow \text { Evaluating } & c(\sim p \vee q) \\ & \sim p & q & \sim p v q \\ & 0 & 1 & & \\ 0 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ & \text { correct assumption } \end{array} $$

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Re-Attempt Weekly Quiz Competition

Set Theory Test 28

Result

Set Theory Test 28

Score

-

Rank

-

SHARING IS CARING

Question 1

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

$$1$$

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

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

Solution

Question 2

16

64

256

1024

Solution

Question 3

540

440

810

620

Question 4

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

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

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

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

Solution

Question 5

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

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

$$\{-3\}$$

$$\{-1\}$$

Solution

Question 6

$$10$$

$$1$$

$$9$$

$$0$$

Solution

Question 7

A

N

B

R

Solution

Question 8

$$\left\{ \left( 1,2 \right) ,\left( 1,5 \right) ,\left( 2,5 \right) \right\} $$

$$\left\{ \left\{ 1,4 \right\} \right\} $$

$$\left( 1,4 \right) $$

$$\left\{ \left( 1,2 \right) \right\} $$

Solution

Question 9

$$16$$

$$4$$

$$8$$

$$24$$

Solution

Question 10

$$\displaystyle p\wedge \sim q$$

$$\displaystyle \sim p\vee q$$

$$\displaystyle \sim p\wedge q$$

$$\displaystyle p\vee \sim q$$

Solution

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