Relations and Functions Test

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Relations and Functions Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

If $$ f(2x+3y,2x-7y)=20x$$,then $$ f(x,y)$$ equal to

A
$$7x-3y$$

B
$$7x+3y$$

C
$$3x-7y$$

D
$$x-y$$

Solution

In $$f\left( 2x+3y,2x-7y \right) =20x$$
Let $$2x+3y=X$$ ...(1)
and $$2x-7y=Y$$ ...(2)
Multiplying (1) by $$7$$ and (2) by $$(3)$$ we get
$$14x+21y=7X$$ and $$6x-21y=3Y$$
Adding them
$$20x=7X+3Y$$
Hence
$$f\left( X,Y \right)=7X+3Y$$
or $$f\left( x,y \right)=7x+3y$$
Question 2
1 / -0

If $$f(x) = \cos (\log x)$$, then $$f(x)\, f(y)\, -\, \displaystyle \frac{1}{2}\, [f\left(\dfrac{x}{y}\right)\, +\, f(xy)]$$ is equal to :

A
$$- 1$$

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

C
$$- 2$$

D
$$0$$

Solution

Substituting the values we get $$\cos\left(\log\left(x\right)\right).\cos\left(\log\left(y\right)\right)-\dfrac{1}{2}\left[\cos\left(\log\left(\dfrac{x}{y}\right)\right)+\cos\left(\log\left(xy\right)\right)\right]$$

$$=\cos\left(\log\left(x\right)\right).\cos\left(\log\left(y\right)\right)-\dfrac{1}{2}\left[\cos\left(\log\left(x\right)-\log\left(y\right)\right)+\cos\left(\log\left(x\right)+\log\left(y\right)\right)\right]$$

$$=\cos\left(\log\left(x\right)\right).\cos\left(\log\left(y\right)\right)-\dfrac{1}{2}\left[2\cos\left(\log\left(x\right)\right).\cos\left(\log\left(y\right)\right)\right]$$$$=0$$
Question 3
1 / -0

Let $$A$$ and $$B$$ be two finite sets having $$m$$ and $$n$$ elements respectively. Then the total number of mapping from $$A$$ to $$B$$ is:

A
$$mn$$

B
$$2^{mn}$$

C
$$m^{n}$$

D
$$n^{m}$$

Solution

Consider an element $$a\in $$A, it can be assigned to any of the $$n$$ elements of B i.e. it has $$n$$ images. Similarly each of the $$m$$ elements of A can have $$n$$ images in B. Hence, the number of mappings is $$\displaystyle

n\times n\times n\times $$...m times$$=$$ $$n^{m}$$
Question 4
1 / -0

$$A$$ and $$B$$ are two sets having $$3$$ and $$4$$ elements respectively and having $$2$$ elements in common. The number of relations which can be defined from $$A$$ to $$B$$ is

A
$${2}^{5}$$

B
$${2}^{10}-1$$

C
$${2}^{12}-1$$

D
None of these

Solution

Given $$n(A)=3$$, $$n(B)=4$$

Now, number of elements (ordered pairs) in $$A\times B$$ is $$3 \times 4=12$$

Number of subsets of $$A\times B$$ is $$2^{12}$$ including empty set.

Since, every subset of $$A\times B$$ is a relation from $$A$$ to $$B.$$

Hence, number of relations from $$A$$ to $$B$$ is $$2^{12}-1$$
Question 5
1 / -0

If $$A=\left\{ 1,2,3 \right\} , B=\left\{ 1,4,6,9 \right\} $$ and $$R$$ is a relation from $$A$$ to $$B$$ defined by '$$x$$ is greater than $$y$$'. Then range of $$R$$ is

A
$$\left\{ 1,4,6,9 \right\} $$

B
$$\left\{ 4,6,9 \right\} $$

C
$$\left\{ 1 \right\} $$

D
None of these

Solution

Here $$R$$ is a relation $$A$$ to $$B$$ defined by '$$x$$ is greater then $$y$$'
$$R=\left\{ (2,1);(3,1) \right\} $$.Hence, range of $$R=\left\{ 1 \right\} $$
Question 6
1 / -0

Let $$g\left( x \right)=1+x-\left[ x \right] $$ and $$f\left( x \right)=\begin{cases} \begin{matrix} -1 & x<0 \end{matrix} \\ \begin{matrix} 0 & x=0 \end{matrix} \\ \begin{matrix} 1 & x>0 \end{matrix} \end{cases}$$. Then for all $$x,f\left[ g\left( x \right) \right] $$ is equal to

A
$$x$$

B
$$1$$

C
$$f(x)$$

D
$$g(x)$$

Solution

We have,
$$g\left( x \right)=\begin{cases} 1+n-n=1,x=n\in Z \\ 1+n+k-n=1+k,x=n+k \end{cases}$$
where $$n\in Z,0<k<1$$
Now $$f\left[ g\left( x \right) \right] =\begin{cases} \begin{matrix} -1 & g\left( x \right)<0 \end{matrix} \\ \begin{matrix} 0 & g\left( x \right)=0 \end{matrix} \\ \begin{matrix} 1 & g\left( x \right)>1 \end{matrix} \end{cases}$$
Clearly, $$g(x)>0$$ for all $$x$$. So $$f\left[ g\left( x \right) \right] =1$$, for all $$x$$.
Question 7
1 / -0

If $$A = \left\{1,2,3\right\}$$, $$B = \left\{1,4,6,9\right\}$$ and $$R$$ is relation from $$A$$ to $$B$$ defined by $$'x'$$ is greater than $$'y'$$. Then range of $$R$$ is

A
$$\left\{1,4,6,9\right\}$$

B
$$\left\{4,6,9\right\}$$

C
$$\left\{1\right\}$$

D
None of these

Solution

Here $$R$$ is a relation $$A$$ to $$B$$ defined by $$x$$ is greater than $$y$$.
$$\therefore\quad R = \left\{(2,1); (3,1)\right\}$$. Hence, range of $$R = \left\{1\right\}$$
Question 8
1 / -0

Given the relation $$R = \left\{(1,2), (2,3)\right\}$$ on the set $$A = \left\{1,2,3\right\}$$, the minimum number of ordered pairs which when added to $$R$$ make it an equivalence relation is

A
$$5$$

B
$$6$$

C
$$7$$

D
$$8$$

Solution

$$R$$ is reflexive if it contains $$(1,1) (2,2) (3,3)$$

$$\because (1,2) \in R,\space (2,3) \in R$$

$$\therefore R$$ is symmetric if $$(2,1), (3,2) \in R$$

Now, $$R = \left\{(1,1), (2,2), (3,3), (2,1), (3,2), (2,3), (1,2)\right\}$$

$$R$$ will be transitive if $$(3,1); (1,3)\in R$$. Thus, $$R$$ becomes and

equivalence relation by adding $$(1,1) (2,2) (3,3) (2,1) (3,2) (1,3) (1,2)$$. Hence,

the total number of ordered pairs is $$7$$.
Question 9
1 / -0

If $$A = \left\{2,3\right\}$$ and $$B = \left\{1,2\right\}$$, then $$A \times B$$ is equal to

A
$$\left\{(2,1), (2,2), (3,1), (3,2)\right\}$$

B
$$\left\{(1,2), (1,3), (2,2), (2,3)\right\}$$

C
$$\left\{(2,1), (3,2)\right\}$$

D
$$\left\{(1,2), (2,3)\right\}$$

Solution

If $$A$$ and $$B$$ are any two non-empty sets.
then $$A\times B$$$$=\left\{(x,y):x\in A and y\in B\right\}$$
As $$A = \left\{2,3\right\}$$ and $$B = \left\{1,2\right\}$$
$$A \times B$$$$=\left\{(2,1), (2,2), (3,1), (3,2)\right\}$$
Hence, option A.
Question 10
1 / -0

Let $$n(A) = n$$. Then the number of all relations on $$A$$ is

A
$$\displaystyle 2^{n}$$

B
$$\displaystyle 2^{\left ( n \right )!}$$

C
$$\displaystyle 2^{n^{2}}$$

D
none

Solution

For any set $$A$$ such that $$n(A)=n$$
then number of all relations on $$A$$ is $${ 2 }^{ n^{2} }$$
As the total number of Relations that can be defined from a set $$A$$ to $$B$$ is the number of possible subsets of $$ A \times B$$. If $$n(A) = p$$ and $$n(B) = q$$ then $$n(A \times B) = pq$$ and the number of subsets of $$A \times B$$ = $$2^{pq}$$.

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Relations and Functions Test - 27

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Relations and Functions Test - 27

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Question 1

$$7x-3y$$

$$7x+3y$$

$$3x-7y$$

$$x-y$$

Solution

Question 2

$$- 1$$

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

$$- 2$$

$$0$$

Solution

Question 3

$$mn$$

$$2^{mn}$$

$$m^{n}$$

$$n^{m}$$

Solution

Question 4

$${2}^{5}$$

$${2}^{10}-1$$

$${2}^{12}-1$$

None of these

Solution

Question 5

$$\left\{ 1,4,6,9 \right\} $$

$$\left\{ 4,6,9 \right\} $$

$$\left\{ 1 \right\} $$

None of these

Solution

Question 6

$$x$$

$$1$$

$$f(x)$$

$$g(x)$$

Solution

Question 7

$$\left\{1,4,6,9\right\}$$

$$\left\{4,6,9\right\}$$

$$\left\{1\right\}$$

None of these

Solution

Question 8

$$5$$

$$6$$

$$7$$

$$8$$

Solution

Question 9

$$\left\{(2,1), (2,2), (3,1), (3,2)\right\}$$

$$\left\{(1,2), (1,3), (2,2), (2,3)\right\}$$

$$\left\{(2,1), (3,2)\right\}$$

$$\left\{(1,2), (2,3)\right\}$$

Solution

Question 10

$$\displaystyle 2^{n}$$

$$\displaystyle 2^{\left ( n \right )!}$$

$$\displaystyle 2^{n^{2}}$$

none

Solution

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