Relations and Functions Test

Result

Relations and Functions Test - 29

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 $$g(x)=x^2+x-1$$ 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 2
1 / -0

Let $$f:N\rightarrow Y$$ be a function defined as $$f(x)=4x+3$$, where $$Y=\{y\in N:y=4x+3,$$ for some $$x\in N\}$$. Show that $$f$$ is invertible and its inverse is

A
$$g(y)=\displaystyle \frac{y+3}{4}$$

B
$$g(y)=\displaystyle \frac{y-3}{4}$$

C
$$g(y)=\displaystyle \frac{3y+4}{3}$$

D
$$g(y)=4+\displaystyle \frac{y+3}{4}$$

Solution

Given $$f:N\rightarrow Y$$ defined by
$$f(x)=4x+3$$
For $$f$$ to be invertible, $$f$$ should be one-one and onto.

Let $$x_1,x_2 \in N$$ such that
$$f(x_1)=f(x_2)$$
$$4x_1+3=4x_2+3$$
$$\Rightarrow x_1=x_2$$
Hence, f is one-one.

$$Y=\{y\in N:y=4x+3; \text{for some} x\in N\}$$
By the definition, it follows that for every $$y\in Y$$, there is some $$x\in X$$.
Hence, f is onto.
Hence, f is invertible.

Let $$y=4x+3$$
$$\Rightarrow x=\dfrac{y-3}{4}$$
$$\Rightarrow f^{-1}(y)=\dfrac{y-3}{4}=g(y)$$
Question 3
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 4
1 / -0

Let $$f : A \rightarrow B$$ be a function defined as $$f(x) = \dfrac {x - 1}{x - 2}$$, where $$A = R - \left \{2\right \}$$ and $$B = R - \left \{1\right \}$$. Then $$f$$ is :

A
invertible and $$f^{-1}(y) = \dfrac {2y + 1}{y - 1}$$

B
invertible and $$f^{-1}(y) = \dfrac {3y - 1}{y - 1}$$

C
not invertible

D
invertible and $$f^{-1}(y) = \dfrac {2y - 1}{y - 1}$$

Solution

Let $$y = f(x)$$
$$\Rightarrow y = \dfrac{x-1}{x-2}$$
$$\Rightarrow yx - 2y = x - 1$$
$$\Rightarrow (y-1)x = 2y-1 $$
$$\Rightarrow x = f^{-1} (y) = \dfrac{2y-1}{y-1}$$
So on the given domain the function is invertible and its inverse can be computed as shown above.
So, option D is the correct answer.
Question 5
1 / -0

If $$f(x) =x+\tan x$$ and $$f$$ is inverse of $$g$$, then $$g'(x)$$ is equal to

A
$$\dfrac{1}{1+[g(x)-x]^2}$$

B
$$\dfrac{1}{2-[g(x)+x]^2}$$

C
$$\dfrac{1}{2+(x-g(x))^2}$$

D
$$None\ of\ these$$

Solution

Since,
$$f=g^{-1}$$$$\implies f(g(x))=x$$
$$\Rightarrow f(g(x))=g(x)+\tan(g(x))=x$$
$$\tan (g(x))=x-g(x)$$

Differentiating both sides with respect to $$x$$
$$g'(x)+\sec ^2(g(x))g'(x)=1$$
$$g'(x)=\dfrac{1}{1+\sec ^2(g(x))}=\dfrac{1}{2+\tan ^2(g(x))}$$
$$g'(x)=\dfrac{1}{2+(x-g(x))^2}$$

Hence, option $$C$$ is correct answer.
Question 6
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 7
1 / -0

$$f:R\rightarrow R$$ is a function defined by $$f(x)=10x-7$$. If $$g=f^{-1}$$ then $$g(x)=$$

A
$$\dfrac{1}{10x-7}$$

B
$$\dfrac{1}{10x+7}$$

C
$$\dfrac{x+7}{10}$$

D
$$\dfrac{x-7}{10}$$

Solution

Given $$f(x)=10x-7$$ and $$g=f^{-1}$$
Let $$f\left ( x \right )=10x-7=y$$
$$\Rightarrow x=\dfrac{y+7}{10}=f^{-1}\left ( y \right )=g\left ( y \right )$$
$$\therefore g\left ( x \right )=\dfrac{x+7}{10}$$
Question 8
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 9
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 10
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.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 29

Result

Relations and Functions Test - 29

Score

-

Rank

-

SHARING IS CARING

Question 1

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

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

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

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

Solution

Question 2

$$g(y)=\displaystyle \frac{y+3}{4}$$

$$g(y)=\displaystyle \frac{y-3}{4}$$

$$g(y)=\displaystyle \frac{3y+4}{3}$$

$$g(y)=4+\displaystyle \frac{y+3}{4}$$

Solution

Question 3

Both one-one and onto

One-one but not onto

Neither one-one nor onto

onto but not one-one

Solution

Question 4

invertible and $$f^{-1}(y) = \dfrac {2y + 1}{y - 1}$$

invertible and $$f^{-1}(y) = \dfrac {3y - 1}{y - 1}$$

not invertible

invertible and $$f^{-1}(y) = \dfrac {2y - 1}{y - 1}$$

Solution

Question 5

$$\dfrac{1}{1+[g(x)-x]^2}$$

$$\dfrac{1}{2-[g(x)+x]^2}$$

$$\dfrac{1}{2+(x-g(x))^2}$$

$$None\ of\ these$$

Solution

Question 6

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

$$n (A) =1$$

$$n (B) = 1$$

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

Solution

Question 7

$$\dfrac{1}{10x-7}$$

$$\dfrac{1}{10x+7}$$

$$\dfrac{x+7}{10}$$

$$\dfrac{x-7}{10}$$

Solution

Question 8

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

$$n(A) = 1$$

$$n (B) = 1$$

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

Solution

Question 9

$$0$$

$$1$$

$$\sqrt{5}$$

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

Solution

Question 10

not one to one function

one to one function

into function

none of these

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.