Functions Test 1

Result

Functions Test 1

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

Let $$N$$ be the set of natural numbers and two functions $$f$$ and $$g$$ be defined as $$f,g : N\to N$$ such that :
$$f (n)= \begin{cases}\dfrac{n+1}{2}& \text{if n is odd}\\ \dfrac{n}{2} & \text{in n is even} \end{cases}$$
and $$g(n) = n - (-1)^n$$. The fog is:

A
Both one-one and onto

B
One-one but not onto

C
Neither one-one nor onto

D
onto but not one-one

Solution

$$fx = \begin{cases} \dfrac{n+1}{2}& \text{n is odd} \\\dfrac{n}{2}& \text{n is even}\end{cases}$$

$$g(x) = n -(-1)^n \begin{cases}n+1; \text{n is odd} \\n-1; \text{n is even} \end{cases}$$
$$f(g(n)) = \begin{cases}\dfrac{n}{2}; & \text{n is even}\\ \dfrac{n+1}{2}; &\text{n is odd} \end{cases}$$
$$\therefore$$ onto but not one-one
Question 2
1 / -0

If $$g(x)=x^2+x-1$$ and
$$(gof)(x)=4x^2-10x+5$$, then
$$f\left(\dfrac{5}{4}\right)$$ is equal to:

A
$$\dfrac{3}{2}$$

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

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

D
$$-\dfrac{1}{2}$$

Solution

$$g(x)=x^2+x-1$$

$$g\left(f\left(\dfrac{5}{4}\right)\right)=4\left(\dfrac{5}{4}\right)^2-10\dfrac{5}{4}+5=-\dfrac{5}{4}$$

$$g\left(f\left(\dfrac{5}{4}\right)\right)=f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)-1$$

$$-\dfrac{5}{4}=f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)-1$$

$$f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)+\dfrac{1}{4}=0$$

$$\left(f\left(\dfrac{5}{4}\right)+\dfrac{1}{2}\right)^2=0$$

$$\boxed{f\left(\dfrac{5}{4}\right)=\dfrac{-1}{2}}$$
Question 3
1 / -0

Domain of definition of the function $$\displaystyle{ f }({ x })=\sqrt { \sin ^{ -1 } (2{ x })+\frac { \pi }{ 6 } } $$ for real valued $$x$$, is

A
$$[-\displaystyle \frac{1}{4}, \displaystyle \frac{1}{2}]$$

B
$$[-\displaystyle \frac{1}{2}, \displaystyle \frac{1}{2}]$$

C
$$(-\displaystyle \frac{1}{2}, \displaystyle \frac{1}{9})$$

D
$$[-\displaystyle \frac{1}{4}, \displaystyle \frac{1}{4}]$$

Solution

Here $$\displaystyle f\left( x \right) =\sqrt { \sin ^{ -1 }{ \left( 2x \right) } +\frac { \pi }{ 6 } } $$ to find domain we must have

$$\displaystyle \sin ^{ -1 }{ \left( 2x \right) } +\frac { \pi }{ 6 } \ge 0$$ ( but $$\displaystyle -\frac { \pi }{ 2 } \le \sin ^{ -1 }{ \theta } \le \frac { \pi }{ 2 } $$)
As $$\displaystyle -\frac { \pi }{ 6 } \le \sin ^{ -1 }{ \left( 2x \right) } \le \frac { \pi }{ 2 } \\\Rightarrow \sin { \left( -\cfrac { \pi }{ 6 } \right) } \le 2x\le \sin { \left( \cfrac { \pi }{ 2 } \right) } $$
$$\displaystyle \Rightarrow -\cfrac { 1 }{ 2 } \le 2x\le 1\\\Rightarrow -\cfrac { 1 }{ 4 } \le x\le \cfrac { 1 }{ 2 } \\\Rightarrow x\in \left[ -\cfrac { 1 }{ 4 } ,\cfrac { 1 }{ 2 } \right] $$
Question 4
1 / -0

A constant function $$f:A\rightarrow B$$ will be one-one if

A
$$n (A) = n(B)$$

B
$$n(A) = 1$$

C
$$n (B) = 1$$

D
$$n (A) < n (B)$$

Solution

Given f is a constant functions.
$$\Rightarrow$$ range of f is $$\left \{ c \right \}(say)$$
Since f is one-one $$\Rightarrow$$ domain of A should also contain
one element.
$$\therefore n(A)=1.$$
Question 5
1 / -0

A constant function $$f:A\rightarrow B $$ will be onto if

A
$$n(A) = n(B)$$

B
$$n (A) =1$$

C
$$n (B) = 1$$

D
$$n(A) > n(B)$$

Solution

Given the $$f:A\rightarrow$$ B is onto & is constant.
Co domain=Range and so it should contain only one element.
$$\therefore n(B)=1$$
Question 6
1 / -0

If $$f:\mathbb{N} \rightarrow \mathbb{N}$$ and $$f(x) = x^{2}$$ then the function is

A
not one to one function

B
one to one function

C
into function

D
none of these

Solution

Given : $$f(x)=x^2$$
$$\because x^2>0 \implies f(x)>0$$ for every $$x\in \mathbb{N}$$
Let' find domain and range of $$f(x)$$
$$ x =1\longrightarrow f(x)=1$$
$$x =2\longrightarrow f(x)=4$$
$$x =3\longrightarrow f(x)=9$$
$$ x=4\longrightarrow f(x)=16$$
$$x =5\longrightarrow f(x)=25$$
Here we get that, no two elements of the domain has the same image and no element of co-domain is the image of more than one element in the domain.
$$\therefore$$ $$f$$ is one-one.
Function $$f:X\rightarrow Y$$ is onto if, for any $$y\in {Y}$$ there exist $$x\in {X}$$ such that $$f(x)=y$$
Let's prove that $$f(x)=x^2$$ is not onto.
Let's take an example $$y=3\in \mathbb{N}$$
$$\implies f(x)=y$$ ............ by definition of onto function
$$\implies x^2=3$$
$$\implies x=\sqrt{3}\ or\ -\sqrt3$$ which does not belong to $$\mathbb{N}$$
Hence, for $$y=3\in \mathbb{N}$$ there does not exist any $$x\in \mathbb{N}$$ such that $$f(x)=y$$.
$$\therefore$$ $$f$$ is not onto.
Hence, $$f$$ in only one-one.
Question 7
1 / -0

$$f(x)=1$$, if $$x$$ is rational and $$f(x)=0$$, if $$x$$ is irrational
then $$(fof) (\sqrt{5})=$$

A
$$0$$

B
$$1$$

C
$$\sqrt{5}$$

D
$$\dfrac{1}{\sqrt{5}}$$

Solution

Given,
$$f(x)=\begin{cases}1 & \text {; if x is rational} \\ 0 & \text{; if x is irrational}\end{cases}$$
Consider $$x=\sqrt{5}$$,
$$\text{(f o f)}(\sqrt{5})=f(f(\sqrt{5}))$$
$$=f(0)$$ .......... $$[\because \sqrt{5}$$ is irrational$$]$$
$$=1$$ .......... $$[\because 0$$ is rational$$]$$
$$\therefore \text{(f o f)}(\sqrt{5})=1$$
Hence, option B is correct.
Question 8
1 / -0

The domain of $$f(x) = x!$$ is

A
$$R$$

B
$$Z$$

C
$$Q$$

D
$${0,1,2,3,.....}$$

Solution

$$x!$$ is defined only for whole numbers.
$$\therefore$$ domain is $$\left \{ 0,1,2,........ \right \}$$
Question 9
1 / -0

If $$f(x) = 3x + 2, g(x) = x^2 + 1$$, then the value of $$(fog) (x^2 +1)$$ is

A
$$3x^4 + 6x^2 + 8$$

B
$$3x^4 + 3x + 4$$

C
$$6x^4 + 3x^2 + 2$$

D
$$3x^2 + 6x + 2$$

Solution

We have $$f(x) = 3x + 2, g(x) = x^2 + 1$$
$$fog(x) = f[g(x)] =3(x^2+1)+2= 3x^2 + 5$$
so $$(fog) (x^2 +1) = 3(x^2 + 1)^2 + 5 =3(x^4+2x^2+1)+5= 3x^4+ 6x^2 + 8$$
Question 10
1 / -0

If $$f:A\rightarrow B $$ is surjective then

A
no two elements of $$A$$ have the same image in $$B$$

B
every element of $$A$$ has an image in $$B$$

C
every element of $$B$$ has at least one pre-image in $$A$$

D
$$A$$ and $$B$$ are finite non empty sets

Solution

Surjective means onto function.
co domain $$=$$ Range
So every element of $$B$$ has at least one pre-image in $$A$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Functions Test 1

Result

Functions Test 1

Score

-

Rank

-

SHARING IS CARING

Question 1

Both one-one and onto

One-one but not onto

Neither one-one nor onto

onto but not one-one

Solution

Question 2

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

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

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

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

Solution

Question 3

$$[-\displaystyle \frac{1}{4}, \displaystyle \frac{1}{2}]$$

$$[-\displaystyle \frac{1}{2}, \displaystyle \frac{1}{2}]$$

$$(-\displaystyle \frac{1}{2}, \displaystyle \frac{1}{9})$$

$$[-\displaystyle \frac{1}{4}, \displaystyle \frac{1}{4}]$$

Solution

Question 4

$$n (A) = n(B)$$

$$n(A) = 1$$

$$n (B) = 1$$

$$n (A) < n (B)$$

Solution

Question 5

$$n(A) = n(B)$$

$$n (A) =1$$

$$n (B) = 1$$

$$n(A) > n(B)$$

Solution

Question 6

not one to one function

one to one function

into function

none of these

Solution

Question 7

$$0$$

$$1$$

$$\sqrt{5}$$

$$\dfrac{1}{\sqrt{5}}$$

Solution

Question 8

$$R$$

$$Z$$

$$Q$$

$${0,1,2,3,.....}$$

Solution

Question 9

$$3x^4 + 6x^2 + 8$$

$$3x^4 + 3x + 4$$

$$6x^4 + 3x^2 + 2$$

$$3x^2 + 6x + 2$$

Solution

Question 10

no two elements of $$A$$ have the same image in $$B$$

every element of $$A$$ has an image in $$B$$

every element of $$B$$ has at least one pre-image in $$A$$

$$A$$ and $$B$$ are finite non empty sets

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.