Relations and Functions Test

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Relations and Functions Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

The solution of $$8x\equiv 6(mod \ 14) $$ is

A
$$\{8, 6\}$$

B
$$\{6,14\}$$

C
$$\{6,13\}$$

D
$$\{8,14,6\}$$

Solution

Since,$$8\mathrm{x}\equiv 6(mod \ 14)$$
i.e., $$8\mathrm{x}-6=14\mathrm{k}$$ for $$\mathrm{k}\in \mathrm{I}$$.
$$ \Rightarrow 8x = 14k+6$$
$$ \Rightarrow 4x = 7k+3$$
The values $$6$$ and $$13$$ satisfy this equation (when $$k =3$$ and $$k=7$$),
while $$8, 14$$ and $$16$$ do not.
Question 2
1 / -0

If $$f(x)=\dfrac{e^{x}+e^{-x}}{2}$$, then the inverse of $$f(x)$$ is

A
$$\log_{e}(x+\sqrt{x^{2}+1})$$

B
$$\log_{e}\sqrt{x^{2}-1}$$

C
$$\log_{e}\left(\dfrac {x+\sqrt{x^{2}-1}}{2}\right)$$

D
$$\log_{e}(x+\sqrt{x^{2}-1})$$

Solution

Let $$f\left ( x \right )=\dfrac{e^{x}+e^{-x}}{2}=y$$
$$\therefore x=f^{-1}\left ( y \right )$$
$$e^{x}+e^{-x}=2y$$
$$e^{2x}-2y e^{x}+1=0$$
$$\Rightarrow e^{x}=\dfrac{2y \pm \sqrt{4y^{2}-4}}{2}$$
$$e^{x}=\dfrac{y\pm \sqrt{y^{2}-1}}{1}$$
Range of $$f$$ is $$\left ( -\infty ,\infty \right )\Rightarrow e^{x}=\dfrac{y+\sqrt{y^{2}-1}}{1}$$
As if we take $$e^{x}=\dfrac{y-\sqrt{y^{2}-1}}{1}$$, which is always small
$$\therefore x=\log _{e}\left ( \dfrac{y+\sqrt{y^{2}-1}}{1} \right )$$$$\therefore f^{-1}\left ( x \right )=\log_{e}\left ( \dfrac{x+\sqrt{x^{2}-1}}{1} \right )$$
Question 3
1 / -0

If $$f(x)=(1-x)^{1/2}$$ and $$g(x)= \ln(x)$$ then the domain of $$(gof)(x)$$ is

A
$$(-\infty ,2)$$

B
$$(-1,1)$$

C
$$(-\infty ,1]$$

D
$$(-\infty ,1)$$

Solution

Given $$f(x)=(1-x)^{\frac{1}{2}}$$ and $$g(x)=ln(x)$$

$$gof(x)$$
$$=g(f(x))$$
$$=\ln\left ( 1-x \right )^{1/2}$$
$$=\dfrac{1}{2}\ln\left ( 1-x \right )$$
$$\therefore$$ For the composite function to be defined $$1-x>0$$
$$x<1$$
$$\therefore$$ Domain is $$\left ( -\infty ,1 \right )$$
Question 4
1 / -0

If $$f:R^{+}\rightarrow R$$ such that $$f(x)=\log_{5} x$$ then $$f^{-1}(x)=$$

A
$$\log_{x}10$$

B
$$5^{x}$$

C
$$3^{-x}$$

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

Solution

$$f\left ( x \right )=y$$

$$\Rightarrow \log _{5}x=y$$

$$\Rightarrow \dfrac{\log x}{\log 5}=y$$

$$\Rightarrow \log x=y\log 5$$

$$\Rightarrow e^{\log x}=e^{y\log 5}$$

$$\Rightarrow e^{\log x} =e^{\log 5^{y}}$$

$$[\because a \log x = \log x^{a}$$]

$$\Rightarrow x=5^{y}=f^{-1}\left ( y \right )$$

$$\therefore f^{-1}\left ( x \right )=5^{x}$$
Question 5
1 / -0

If $$f(x)=\dfrac{x+1}{x-1}(x\neq 1)$$ then $$fofofof(x)=$$

A
$$f(x)$$

B
$$2\left ( \dfrac{x+1}{x-1} \right )$$

C
$$\dfrac{x-1}{x+1}$$

D
$$x$$

Solution

$$fofofof\left(x\right)=fofof\left(\dfrac{x+1}{x-1}\right)$$
$$=fof\left(\dfrac{\dfrac{x+1}{x-1}+1}{\dfrac{x+1}{x-1}-1}\right)=fof\left(\dfrac{2x}{2}\right)$$
$$=fof\left(x\right)$$
$$=f\left(\dfrac{x+1}{x-1}\right)=x$$
Question 6
1 / -0

If $$f:[1,\infty )\rightarrow B$$ defined by the function $$ f(x)=x^{2}-2x+6$$ is a surjection, then $$B$$ is equals to

A
$$[1,\infty )$$

B
$$[5,\infty )$$

C
$$[6,\infty )$$

D
$$[2,\infty )$$

Solution

$$f(x)=x^2-2x+6$$ is a surjection.
So the range of $$f(x)$$ will be equal to its codomain.
$$f(x)=x^2-2x+6$$
$$f^1(x)=2x-2$$
$$=2(x-1)$$
$$f(x)$$ will be increasing when $$x\geqslant 1$$.
$$\therefore f(1)=1-2+6$$
$$=5$$
$$\therefore B=[5, \infty)$$
Question 7
1 / -0

If $$f:R\rightarrow R^{+}$$ then $$\displaystyle f(x)=\left(\dfrac{1}{3}\right)^{x}$$, then $$f^{-1}(x)=$$

A
$$\displaystyle \left(\dfrac{1}{3}\right)^{-x}$$

B
$$3^{x}$$

C
$$\displaystyle \log_{1/3}$$$$ x$$

D
$$\displaystyle \log_{x}\left(\dfrac{1}{3}\right)$$

Solution

Let $$\displaystyle f(x)=\left(\frac{1}{3}\right)^{x}=y$$
Taking logarithm on both sides,
$$\displaystyle x\log \frac{1}{3}=\log y$$
$$\displaystyle \Rightarrow x=\frac{\log y}{\log \frac{1}{3}}$$
$$\displaystyle \Rightarrow x=\log _ { 1/3 } y\quad \quad \quad \left[ \because \frac { \log b }{ \log a } =\log _{ a }{ b } \right] $$
$$\therefore f^{-1}\left ( x \right )=\log _{1/3}x$$
Question 8
1 / -0

If $$ X =\{1, 2,3,4,5\} $$ and $$Y =\{1,3,5,7,9\}$$, determine which of the following sets represent a relation and also a mapping?

A
$$R_{1}= \{(x,y)$$:$$ y=x+2, x \in Y,y \in Y\}$$

B
$$R_{2}=\{(1,1), (1,3), (3,5), (4,7), (5,9)\}$$

C
$$R_{3}=\{(1,1), (2,3), (3,5), (3,7), (5,7)\}$$

D
$$R_{4}=\{(1,3), (2,5), (4,7), (5,9), (3,1)\}$$

Solution

Here we will check all options one by one:
Option A:
$$R_1 = \{(x,y): y= x+2, x\in X, y\in Y\}$$
$$\implies \mathrm{R}_{1}=\{(1,3),(2, 4),(3,5),(4,6),(5,7)\}$$
Since $$4$$ amd $$6$$ are the images of $$2$$ and $$4$$ respectively but $$4$$ and $$6$$ do not belong to $$Y$$
$$\therefore (2, 4),(4,6)\not\in X \times Y$$
Hence $$R_{1}$$ is not a relation as well as not a mapping.

Option B:
$$R_{2}$$: It is a relation but not a mapping because the element $$1$$ has two different images.

Option C:
$$R_{2}$$: It is a relation but not a mapping because the element $$3$$ has two different images.

Option D:
$$R_{4}$$: It is both a relation and a mapping because every element in $$X$$ is mapped to the elements in $$Y$$. Also, every element of $$X$$ has a one and only one image in $$Y$$ and every element in $$Y$$ has its pre-image in $$X$$. Hence, it is also one-one and onto mapping and hence it is a bijection.
Question 9
1 / -0

If A $$=$${$$x : x^{2}-3x+2= 0$$}, and $$R$$ is a universal relation on $$A$$, then $$R$$ is

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

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

C
$$\phi $$

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

Solution

Consider, $$x^2-3x+2=0$$
$$\implies x^2-2x-x+2=0$$
$$\implies (x-2)(x-1)=0$$
$$\implies x=1,2$$
$$\therefore A=\{1,2\}$$
Also R is universal relation on set A, then every element of set A is related every other element of A
So $$R= \{(1,1), (1,2), (2,1), (2,2)\}$$.
Question 10
1 / -0

If $$f(x)=2+x^{3}$$, then $$f^{-1}(x)$$ is equal to

A
$$\sqrt[3]{x}+2$$

B
$$\sqrt[3]{x}-2$$

C
$$\sqrt[3]{x-2}$$

D
$$\sqrt[3]{x+2}$$

Solution

Let $$f\left ( x \right )=y$$
$$\Rightarrow 2+x^{3}=y$$
$$\Rightarrow x^{3}=y-2$$
$$\Rightarrow x=\left ( y-2 \right )^{1/3}$$
$$\therefore f^{-1}\left ( y \right )=x$$$$=\left ( y-2 \right )^{{1}/{3}}$$
$$\therefore f^{-1}\left ( x \right )=\left ( x-2 \right )^{{1}/{3}}$$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 38

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Relations and Functions Test - 38

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SHARING IS CARING

Question 1

$$\{8, 6\}$$

$$\{6,14\}$$

$$\{6,13\}$$

$$\{8,14,6\}$$

Solution

Question 2

$$\log_{e}(x+\sqrt{x^{2}+1})$$

$$\log_{e}\sqrt{x^{2}-1}$$

$$\log_{e}\left(\dfrac {x+\sqrt{x^{2}-1}}{2}\right)$$

$$\log_{e}(x+\sqrt{x^{2}-1})$$

Solution

Question 3

$$(-\infty ,2)$$

$$(-1,1)$$

$$(-\infty ,1]$$

$$(-\infty ,1)$$

Solution

Question 4

$$\log_{x}10$$

$$5^{x}$$

$$3^{-x}$$

$$3^{1/x}$$

Solution

Question 5

$$f(x)$$

$$2\left ( \dfrac{x+1}{x-1} \right )$$

$$\dfrac{x-1}{x+1}$$

$$x$$

Solution

Question 6

$$[1,\infty )$$

$$[5,\infty )$$

$$[6,\infty )$$

$$[2,\infty )$$

Solution

Question 7

$$\displaystyle \left(\dfrac{1}{3}\right)^{-x}$$

$$3^{x}$$

$$\displaystyle \log_{1/3}$$$$ x$$

$$\displaystyle \log_{x}\left(\dfrac{1}{3}\right)$$

Solution

Question 8

$$R_{1}= \{(x,y)$$:$$ y=x+2, x \in Y,y \in Y\}$$

$$R_{2}=\{(1,1), (1,3), (3,5), (4,7), (5,9)\}$$

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

$$R_{4}=\{(1,3), (2,5), (4,7), (5,9), (3,1)\}$$

Solution

Question 9

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

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

$$\phi $$

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

Solution

Question 10

$$\sqrt[3]{x}+2$$

$$\sqrt[3]{x}-2$$

$$\sqrt[3]{x-2}$$

$$\sqrt[3]{x+2}$$

Solution

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