Relations and Functions Test

Result

Relations and Functions Test - 53

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

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

B
does not exist

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

D
None of these

Solution

Since $$R\subseteq A\times B$$ and $$S\subseteq B\times C$$, we have
So $$R\subseteq A\times C$$.
$$\therefore$$ So R is a relation from $$A$$ to $$C.$$
Question 2
1 / -0

If f = {(1,3) (2,1) (3,4) (4,2)} and g = {(1,2) (2,3) (3,4) (4,1)} then find n(fog)

A
12

B
16

C
4

D
5

Solution

we can find $$ n(fog) $$ by finding the range of the domain of $$f $$
In $$ (1,3) $$ as the range $$ 3 $$ is mapped to $$ 4$$ in $$g $$, the ordered pair here for $$fog $$ is $$ (1,4) $$
In $$ (2,1) $$ as the range $$ 1 $$ is mapped to $$2 $$ in $$g $$, the ordered pair here for $$fog $$ is $$ (2,2) $$
In $$ (3,4) $$ as the range $$ 4 $$ is mapped to $$ 1 $$ in $$g $$, the ordered pair here for $$fog $$ is $$ (3,1) $$
In $$ (4,2) $$ as the range $$ 2 $$ is mapped to $$ 3 $$ in $$g $$, the ordered pair here for $$fog $$ is $$ (4,3) $$

As we have $$ 4 $$ ordered pairs for $$ fog $$ , we have $$ n(fog) = 4 $$
Question 3
1 / -0

Consider $$f: R\rightarrow R$$ given by $$f(x) = 4x + 3$$. Show that $$f$$ is invertible. Find the inverse of $$f$$

A
$$y=\dfrac{(x+3)}{4}$$

B
$$y=\dfrac{(x-3)}{4}$$

C
$$y=\dfrac{(-x-3)}{4}$$

D
$$y=\dfrac{(x-3)}{-4}$$

Solution

A function is a one-to-one function if and only if each second element corresponds to one and only one first element. (each x and y value is used only once)

Use the horizontal line test to determine if a function is a one-to-one function.
If ANY horizontal line intersects your original function in ONLY ONE location, your function will be a one-to-one function and its inverse will also be a function.
Now,
$$f(x)=4x+3$$ is one-to-one function , Therefore it has inverse function
Now express the given function in following way
$$x=4y+3$$ , Now express y in terms of x, y will be inverse
$$y=\frac{(x-3)}{4}$$ is $$f^{-1}$$
Question 4
1 / -0

What is the relation for the following diagram?

A
$$R =\{(2, 5), (2, 6), (3, 7)\}$$

B
$$R = \{(1, 5), (2, 6), (3, 7)\}$$

C
$$R =\{(1, 5), (2, 6), (1, 7)\}$$

D
$$R = \{(1, 5), (2, 6), (2, 7)\}$$

Solution

$$R:A\to B$$
$$A\times B = \{(1,5),(2,6),(3,7)\}$$
Question 5
1 / -0

Which of the following do(es) not belong to $$A \times B$$ for the sets $$A = \{1, 2\}$$ and $$B =\{0, 2\}$$?

A
$$R = \{(1, 0), (2, 2)\}$$

B
$$R = \{(1, 1), (2, 1)\}$$

C
$$R = \{(1, 0), (1, 2)\}$$

D
$$R = \{(1, 2), (2, 2)\}$$

Solution

A relation for the sets $$A$$ and $$B$$ from $$A$$ to $$B$$ will be a subset of the $$A\times B.$$
Now the ordered pair belonging to $$A\times B$$ are
$$\{1,2\}\times\{0,2\}$$ $$=\{(1,0),(1,2),(2,0),(2,2)\}$$
Hence, $$\{(1,1)\}$$ and $$\{(2,1)\}$$ do not belong to $$A\times B$$.
Question 6
1 / -0

If $$f: R\rightarrow R$$ be given by $$f(x) = (3 - x^{3})^{ {1}/{3}}$$, then find $$f(f (x))$$ is

A
$$x^{{1}/{3}}$$

B
$$x^{3}$$

C
$$x$$

D
$$3 - x^{3}$$

Solution

$$f(x)= (3-x^3)^{\dfrac{1}{3}}\\
f(f(x))=f\left((3-x^3)^{\frac{1}{3}}\right)$$
$$ = (3-((3-x^3)^{\frac{1}{3}})^3)^{\frac{1}{3}}=x$$
Question 7
1 / -0

If $$f:R\rightarrow R$$ and $$g:R\rightarrow R$$ are defined by $$f\left( x \right) =\left| x \right| $$ and $$g\left( x \right) =\left[ x-3 \right] $$ for $$x\in R$$, then
$$g\left( f\left( x \right) \right) :\left\{ -\dfrac { 8 }{ 5 } < x < \dfrac { 8 }{ 5 } \right\} $$ is equal to
[.] is Greatest integer function

A
$$\left\{ 0,1 \right\} $$

B
$$\left\{ 1,2 \right\} $$

C
$$\left\{ -3,-2 \right\} $$

D
$$\left\{ 2,3 \right\} $$

Solution

Given that, $$f\left( x \right) =\left| x \right| $$ and $$g\left( x \right) =\left[ x-3 \right] $$
For $$-\dfrac { 8 }{ 5 } < x < \dfrac { 8 }{ 5 } , 0\le f\left( x \right) < \dfrac { 8 }{ 5 } $$
Now, for $$0 < f \left( x \right) < 1$$,
$$g\left( f\left( x \right) \right) = \left[ f\left( x \right) -3 \right] $$
$$=-3\quad \because -3\le f\left( x \right) -3 < -2$$
Again, for $$1 < f\left( x \right) < 1.6$$
$$g\left( f\left( x \right) \right) =-2$$
$$\because -2\le f\left( x \right) -3 < -1.4$$
Hence, required set is $$\left\{ -3,-2 \right\} $$
Question 8
1 / -0

If $$R$$ be the set of all real numbers and $$f:R\rightarrow R$$ is given by $$f\left( x \right)=3{ x }^{ 2 }+1$$. Then, the set $$f^{ -1 }\left( \left[ 1,6 \right] \right)$$ is

A
$$\left\{ -\sqrt { \dfrac { 5 }{ 3 } } ,0,\sqrt { \dfrac { 5 }{ 3 } } \right\} $$

B
$$\left[ -\sqrt { \dfrac { 5 }{ 3 } } ,\sqrt { \dfrac { 5 }{ 3 } } \right] $$

C
$$\left[ -\sqrt { \dfrac { 1 }{ 3 } } ,\sqrt { \dfrac { 1 }{ 3 } } \right] $$

D
$$\left( -\sqrt { \dfrac { 5 }{ 3 } } ,\sqrt { \dfrac { 5 }{ 3 } } \right) $$

Solution

Given, $$f\left( x \right) =3{ x }^{ 2 }+1$$
Let $$y=3{ x }^{ 2 }+1$$
$$\Rightarrow 3{ x }^{ 2 }=y-1\Rightarrow { x }^{ 2 }=\dfrac { y-1 }{ 3 } $$
$$\Rightarrow x=\pm \sqrt { \dfrac { y-1 }{ 3 } } $$
$$\therefore f^{ -1 }\left( x \right) =\pm \sqrt { \dfrac { x-1 }{ 3 } } $$
When $$x\in \left[ 1,6 \right] $$, then, $$f^{ -1 }\left( x \right) \in \left[ -\sqrt { \dfrac { 5 }{ 3 } } ,\sqrt { \dfrac { 5 }{ 3 } } \right] $$
Question 9
1 / -0

Let $$R$$ be the set of real numbers and the functions $$f: R \rightarrow R$$ and $$g: R\rightarrow R$$ be defined by $$f(x) = x^{2} + 2x - 3$$ and $$g(x) = x + 1$$. Then the value of $$x$$ for which $$f(g(x)) = g(f(x))$$ is

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution

According to the question,

$$f(g(x)) = g(f(x))$$
$$\Rightarrow f(x + 1) = g(x^{2} + 2x - 3)$$

$$\Rightarrow (x + 1)^{2} + 2 (x + 1) - 3 = x^{2} + 2x - 3 + 1$$

$$\Rightarrow x^{2} + 1 + 2x + 2x + 2 - 3 = x^{2} + 2x - 2$$

$$\Rightarrow x^{2} + 4x = x^{2} + 2x - 2$$
$$\Rightarrow x^{2} + 4x - x^{2} - 2x + 2 = 0$$
$$\Rightarrow 2x + 2 = 0$$

$$\Rightarrow 2x = -2$$

$$\Rightarrow x = -1$$
Question 10
1 / -0

Let $$f:R\rightarrow R$$ be such that $$f$$ is injective and $$f(x)f(y)=f(x+y)$$ for all $$x,y\in R$$, if $$f(x), f(y)$$ and $$f(z)$$ are in GP, then $$x,y$$ and $$z$$ are in

A
AP always

B
GP always

C
AP depending on the values of $$x,y$$ and $$z$$

D
GP depending on the values of $$x,y$$ and $$z$$

Solution

Let the funtion $$f(x)={a}^{kx}$$
which define in $$f:R\rightarrow R$$ and injective also.

Now, we have
$$f(x)f(y)=f(x+y)$$

$$\Rightarrow$$ $${a}^{kx}.{a}^{ky}={a}^{k(x+y)}$$

$$\Rightarrow$$ $${a}^{k(x+y)}={a}^{k(x+y)}$$

$$\because$$ $$f(x), f(y)$$ and $$f(z)$$ are in GP

$$\therefore$$ $$f({y}^{2})=f(x).f(z)$$

$$\Rightarrow$$ $${a}^{2ky}={a}^{kx}.{a}^{kz}$$

$$\Rightarrow$$ $${e}^{2ky}={e}^{k(x+z)}$$

On comparing, we get
$$2ky=k(x+z)$$ $$\Rightarrow$$ $$2y=x+z$$

$$\Rightarrow$$ $$x,y$$ and $$z$$ are in AP

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 53

Result

Relations and Functions Test - 53

Score

-

Rank

-

SHARING IS CARING

Question 1

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

does not exist

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

None of these

Solution

Question 2

12

16

4

5

Solution

Question 3

$$y=\dfrac{(x+3)}{4}$$

$$y=\dfrac{(x-3)}{4}$$

$$y=\dfrac{(-x-3)}{4}$$

$$y=\dfrac{(x-3)}{-4}$$

Solution

Question 4

$$R =\{(2, 5), (2, 6), (3, 7)\}$$

$$R = \{(1, 5), (2, 6), (3, 7)\}$$

$$R =\{(1, 5), (2, 6), (1, 7)\}$$

$$R = \{(1, 5), (2, 6), (2, 7)\}$$

Solution

Question 5

$$R = \{(1, 0), (2, 2)\}$$

$$R = \{(1, 1), (2, 1)\}$$

$$R = \{(1, 0), (1, 2)\}$$

$$R = \{(1, 2), (2, 2)\}$$

Solution

Question 6

$$x^{{1}/{3}}$$

$$x^{3}$$

$$x$$

$$3 - x^{3}$$

Solution

Question 7

$$\left\{ 0,1 \right\} $$

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

$$\left\{ -3,-2 \right\} $$

$$\left\{ 2,3 \right\} $$

Solution

Question 8

$$\left\{ -\sqrt { \dfrac { 5 }{ 3 } } ,0,\sqrt { \dfrac { 5 }{ 3 } } \right\} $$

$$\left[ -\sqrt { \dfrac { 5 }{ 3 } } ,\sqrt { \dfrac { 5 }{ 3 } } \right] $$

$$\left[ -\sqrt { \dfrac { 1 }{ 3 } } ,\sqrt { \dfrac { 1 }{ 3 } } \right] $$

$$\left( -\sqrt { \dfrac { 5 }{ 3 } } ,\sqrt { \dfrac { 5 }{ 3 } } \right) $$

Solution

Question 9

$$-1$$

$$0$$

$$1$$

$$2$$

Solution

Question 10

AP always

GP always

AP depending on the values of $$x,y$$ and $$z$$

GP depending on the values of $$x,y$$ and $$z$$

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.