Relations and Functions Test

Result

Relations and Functions Test - 24

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 the function $$f(x) = \log\left(\dfrac{1+x}{1-x}\right)$$, then the value of $$f\left(\dfrac{2x}{1+x^2}\right)$$ is equal to

A
$$f(x)$$

B
$$2f(x)$$

C
$$3f(x)$$

D
$$4f(x)$$

Solution

Given:
$$f\left(x\right) = \log\left(\dfrac{1+x}{1-x}\right)$$

Substitute $$x$$ as $$ \dfrac{2x}{1+x^2}$$, in the above function, we get

$$ f\left(\dfrac{2x}{1+x^2}\right)=\log\left(\dfrac{1+\dfrac{2x}{1+x^2}}{1-\dfrac{2x}{1+x^2}}\right)$$

$$\Rightarrow f\left(\dfrac{2x}{1+x^2}\right)=\log\left(\dfrac{x^2+2x+1}{x^2-2x+1}\right)$$

$$\Rightarrow f\left(\dfrac{2x}{1+x^2}\right)=\log\left(\dfrac{\left(x+1\right)^2}{\left(x-1\right)^2}\right)$$

$$\Rightarrow f\left(\dfrac{2x}{1+x^2}\right)=\log\left(\dfrac{x+1}{x-1}\right)^2$$

Using $$\log x^m = m\cdot\log x$$, we get

$$ f\left(\dfrac{2x}{1+x^2}\right)=2\cdot \log\left(\dfrac{x+1}{x-1}\right)=2f\left(x\right)$$

Hence, option B.
Question 2
1 / -0

R is a relation in A and (a, b) $$\notin$$ r, implies (b, a) $$\notin$$ R then R is said to be ____ relation

A
symmetric

B
contradictory

C
skew symmetric

D
none of these

Solution

$$R$$ is a relation in $$A$$ and $$(a,b)$$.
$$\Rightarrow (b,a)\notin R$$
Hence $$'R'$$ is a skew symmetric relation.
Answer:-$$C$$
Question 3
1 / -0

A $$\times$$ (B - C) =

A
$$(A \times B) - (A \times C)$$

B
$$(A \times C) - (A \times B)$$

C
$$(A \times B) \bigcup (A \times C)$$

D
$$(A \times B) \bigcap (A \times C)$$

Solution
Question 4
1 / -0

Directions For Questions

Based on this information answer the question
(i) If $$A=\left \{ 1, 2, 3, 4 \right \}$$ the $$p(A)=\left \{ \phi , \left \{ 1 \right \},
\left \{ 2 \right \}, \left \{ 3 \right \}, \left \{ 1, 2 \right \}, \left \{ 2, 3 \right \},
\left \{ 1, 3 \right \}, \left \{ 1, 2, 3 \right \} \right \}$$
Power set of an empty set, $$p(\phi )=\left \{ \left ( \phi \right ) \right \}$$ and $$n\{p(\phi
)\}=1.$$
(ii) Any subset of $$A\times A $$ is called a relation on A. Number of relations on $$A=2^{n^{2}}$$

...view full instructions

Find the number of relations from $$\left \{ m, o, t, h, e, r \right \}$$ to $$\left \{ c, h, i, l, d \right \}$$

A
$$30$$

B
$$30^{2}$$

C
$$2^{30}$$

D
$$60$$

Solution

$$n \{m.o,t,b,e,r\} = 6 = m$$ (let)
$$n \{c,h,i,l,d\} = 5 = n$$ (let)

$$\therefore $$ numbers of Relations
$$= { 2 }^{ mn }$$
$$= { 2 }^{ 6\times 5 }$$
$$= { 2 }^{ 30 }$$
Question 5
1 / -0

If $$R$$ is a relation from a set $$A$$ to a set $$B$$ and $$S$$ is a relation from $$B$$ to a set $$C$$, then the relation $$SOR$$

A
is from $$A$$ to $$C$$

B
is from $$C$$ to $$A$$

C
does not exist

D
none of these

Solution

Domain of $$SOR$$ is the domain of $$R$$ and the range is the range of $$S$$. Hence the mapping is from set $$A$$ to set $$C$$.
Question 6
1 / -0

Let $$A=\left \{ 1,2,3 \right \}$$. The total number of distinct relations that can be defined over $$A$$ is:

A
$$2^{9}$$

B
$$6$$

C
$$8$$

D
None of the above

Solution

Total number of distinct binary relations over the set $$ A$$ will be
$$2^{n^2}$$
$$=2^{3^{2}}$$
$$=2^{9}$$
$$=512$$
Question 7
1 / -0

If $$\displaystyle f\left ( x \right ) = x^{2} + bx + c$$ and $$\displaystyle f\left ( 2 + t \right ) = f\left ( 2 - t \right )$$ for all real numbers $$t$$, then which of the following is true?

A
$$\displaystyle f \left ( 1 \right ) < f \left ( 2 \right ) < f\left ( 4 \right )$$

B
$$\displaystyle f \left ( 2 \right ) < f \left ( 1 \right ) < f\left ( 4 \right )$$

C
$$\displaystyle f \left ( 2 \right ) < f \left ( 4 \right ) < f\left ( 1 \right )$$

D
$$\displaystyle f \left ( 4 \right ) < f \left ( 2 \right ) < f\left ( 1 \right )$$

Solution

Given, $$f(x)=x^2+bx+c$$
and $$f(2+t)=f(2-t)$$
$$\therefore (2+t)^{2}+b(2+t)+c=(2-t)^{2}+b(2-t)+c$$
$$8t=-2bt$$
$$t=-4$$
Substituting in $$f(x)$$, we get
$$f(x)=x^2-4x+c$$
Hence $$f(4)=c$$
$$f(2)=c-4$$
$$f(1)=c-3$$
Hence $$f(2)<f(1)<f(4)$$.
Question 8
1 / -0

Let $$f:R\rightarrow R, g:R\rightarrow R$$ be the two given functions, then the function $$\displaystyle h(x)=2\times$$ min$$\{f(x)-g(x), 0\}$$ is same as

A
$$f(x)+g(x)-\left|g(x)-f'(x) \right|$$

B
$$f(x)+g(x)+\left|g(x)-f(x) \right|$$

C
$$f(x)-g(x)+\left|g(x)-f(x) \right|$$

D
$$f(x)-g(x)-\left|g(x)-f(x) \right|$$

Solution

if $$f\left( x \right) > g\left( x \right)$$, then $$\displaystyle h(x)=2\times min (f(x)-g(x), 0)$$ will be equal to 0.
if $$f\left( x \right) < g\left( x \right)$$ then $$\displaystyle h(x)=2\times min (f(x)-g(x), 0)$$ will be same as $$2\times\left( {f\left( x \right) - g\left( x \right)} \right)$$
Only option D satisfies both the the conditions. Hence the correct answer is option D.
Question 9
1 / -0

A real valued function $$f(x)$$ satisfies the functional equation $$\displaystyle f(x-y)=f(x)f(y)-f(a-x)f(a+y)$$, where $$a$$ is a given constant and $$f(0)=1$$, then $$f(2a-x)$$ equals

A
$$f(x)$$

B
$$-f(x)$$

C
$$f(-x)$$

D
$$f(a)+f(a-x)$$

Solution

$$\displaystyle f(x-y)=f(x)f(y)-f(a-x)f(a+y) ......(i)$$

Substitute $$x=a$$, and $$y = x-a$$

$$\Rightarrow f(a-(x-a))=f(a)f(x-a)-f(a-a)f(a+x-a)$$

$$\Rightarrow f(2a-x) = f(a).f(x-a)-f(0).f(x) = f(a).f(x-a)-f(x) ......(ii)$$

Now put $$x=y=0$$ in $$(i)$$,

$$\Rightarrow f(0) =\{f(0)\}^2-\{f(a)\}^2\Rightarrow f(a) =0$$ since $$f(0)=1$$ (given)

$$\therefore f(2a-x) =-f(x)$$
Question 10
1 / -0

Let $$R=\{(1, 3), (2, 2), (3, 2)\}$$ and $$S=\{(2, 1), (3, 2), (2, 3)\}$$ be two relations on set $$A=\{1, 2, 3\}$$.Then $$R$$o$$S$$ is equal

A
$$\{(2, 3), (3, 2), (2, 2)\}$$

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

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

D
$$\{(2, 3), (3, 2)\}$$

Solution

If $$f\epsilon X\times Y$$ and $$g\epsilon Y\times Z$$

Then
$$gof\epsilon X\times Z$$ i.e
$$ (x,z)\epsilon gof$$

Applying the above concept, we gives us
$$RoS=\{(2,1)|(1,3),(3,2)|(2,2),(2,3)|(3,2)\}$$

$$=\{(2,3),(3,2),(2,2)\}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 24

Result

Relations and Functions Test - 24

Score

-

Rank

-

SHARING IS CARING

Question 1

$$f(x)$$

$$2f(x)$$

$$3f(x)$$

$$4f(x)$$

Solution

Question 2

symmetric

contradictory

skew symmetric

none of these

Solution

Question 3

$$(A \times B) - (A \times C)$$

$$(A \times C) - (A \times B)$$

$$(A \times B) \bigcup (A \times C)$$

$$(A \times B) \bigcap (A \times C)$$

Solution

Question 4

$$30$$

$$30^{2}$$

$$2^{30}$$

$$60$$

Solution

Question 5

is from $$A$$ to $$C$$

is from $$C$$ to $$A$$

does not exist

none of these

Solution

Question 6

$$2^{9}$$

$$6$$

$$8$$

None of the above

Solution

Question 7

$$\displaystyle f \left ( 1 \right ) < f \left ( 2 \right ) < f\left ( 4 \right )$$

$$\displaystyle f \left ( 2 \right ) < f \left ( 1 \right ) < f\left ( 4 \right )$$

$$\displaystyle f \left ( 2 \right ) < f \left ( 4 \right ) < f\left ( 1 \right )$$

$$\displaystyle f \left ( 4 \right ) < f \left ( 2 \right ) < f\left ( 1 \right )$$

Solution

Question 8

$$f(x)+g(x)-\left|g(x)-f'(x) \right|$$

$$f(x)+g(x)+\left|g(x)-f(x) \right|$$

$$f(x)-g(x)+\left|g(x)-f(x) \right|$$

$$f(x)-g(x)-\left|g(x)-f(x) \right|$$

Solution

Question 9

$$f(x)$$

$$-f(x)$$

$$f(-x)$$

$$f(a)+f(a-x)$$

Solution

Question 10

$$\{(2, 3), (3, 2), (2, 2)\}$$

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

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

$$\{(2, 3), (3, 2)\}$$

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.