Relations and Functions Test

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Relations and Functions Test - 68

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Question 1
1 / -0

The numbers system which uses alphabets as well as numbers is-

A
Binary numbers system

B
Octal numbers system

C
Decimal numbers system

D
Hexadecimal numbers system

Solution

HEXADECIMAL NUMBERS SYSTEM
Question 2
1 / -0

Number of one-one functions from A to B where $$n(A)=4, n(B)=5$$.

A
$$4$$

B
$$5$$

C
$$120$$

D
$$90$$

Solution

$$n(A)=4$$ and $$n(B)=5$$

For one-one mapping
4 elements can be selected out of 5 elements of set B in $${}^5C_4$$ ways
and then those 4 selected elements can be mapped with 4 elements of set A in $$4!$$ ways.

Number of one-one mapping from $$A$$ to $$B$$ $$={}^5C_4\times4!={}^5P_4=\dfrac{5!}{(5-4)!}=5!=120$$
Question 3
1 / -0

If is a binary operation such that $$a * b = a^2 + b^2$$ then $$3 * 5$$ is

A
34

B
9

C
8

D
25

Solution

$$a \star b = {a}^{2} + {b}^{2} \quad \left( \text{Given} \right)$$
Therefore,
$$3 \star 5 = {3}^{2} + {5}^{2} = 9 + 25 = 34$$
Hence the correct answer is $$34$$.
Question 4
1 / -0

If a binary operation is defined $$a\star b=a^b$$ then 2$$\star 2$$ is equal to:

A
$$4$$

B
$$2$$

C
$$9$$

D
$$8$$

Solution

$$a \star b = {a}^{b} \quad \left( \text{Given} \right)$$
Therefore,
$$2 \star 2 = {2}^{2} = 4$$
Hence the correct answer is $$4$$.
Question 5
1 / -0

Let $$f(x)=x^ {135}+x^ {125}-x^ {115}+x^ {5}+1$$. If $$f(x)$$ divided by $$x^ {3}-x$$, then the remainder is some function of $$x$$ say $$g(x)$$. Then $$g(x)$$ is an:-

A
one-one function

B
many one function

C
into function

D
onto function

Solution
Question 6
1 / -0

If $$ f : R \rightarrow R $$ be given by $$ f(x)=\left(3-x^{3}\right)^{\dfrac{1}{3}}, $$ then $$fof(x)$$ is

A
$$
x^{\dfrac{1}{3}}
$$

B
$$
1^{3}
$$

C
x

D
$$
\left(3-x^{3}\right)
$$

Solution

Given, $$ f(x)=\left(3-x^{3}\right)^{\dfrac{1}{3}} $$.
Now,
$$fof(x)$$
$$=f[f(x)]$$

$$ =\left(3-[f(x)]^{3}\right)^{\dfrac{1}{3}}, $$

$$ =\left(3-[(3-x^3)^{\dfrac{1}{3}}]^{3}\right)^{\dfrac{1}{3}}, $$

$$ =\left(3-[(3-x^3)]\right)^{\dfrac{1}{3}}, $$

$$=[x^3]^{\dfrac{1}{3}}$$

$$=x$$.
Question 7
1 / -0

Let : $$R\rightarrow R$$ defined as $$f\left( x \right) =\dfrac { x\left( x+1 \right) \left( { x }^{ 4 }+1 \right) +{ 2x }^{ 4 }+{ x }^{ 2 }+2 }{ { x }^{ 2 }+x+1 } $$

A
odd and one-one

B
even and one-one

C
many to one and even

D
many to one and neither even nor odd
Question 8
1 / -0

Let f : $$R\rightarrow R$$ be a function defined by f(x) = $${ x }^{ 3 }+{ x }^{ 2 }+3x+sin\times .$$ Then f is.

A
one-one & onto

B
one-one & into

C
many one & onto

D
many one & into

Solution

$$f(x)=x^{3}+x^{2}+3 x+\sin x$$
$$Suppose$$
$$f(x_{1})=f(x_{2})$$
$$x_{1}^{3}+x_{1}^{2}+3 x_{1}+\sin x_{1}=x_{2}+x_{2}+\sin x-\sin x_{2}$$
$$Equating\;above\;equations\;to \;0$$
$$x_{1}^{3}+x_{1}^{2}+3 x_{1}+\sin x_{1}=x_{2}+x_{2}+\sin x-\sin x_{2}=0$$
$$Solving\;them$$
$$It\; will\; not\; prove\; that$$
$$x_{1}=x_{2}$$
$$So,\;f(x)\;is\;one-one$$
$$Range\;f(x)=Real$$
$$Co-domain\;f(x)=Real$$
$$Range=Co-domain$$
$$Hence,\; f(x) \;is \;onto$$
$$So,\;option\; C \;is\;correct.$$
Question 9
1 / -0

A function $$f$$ from the set of natural numbers to integers defined by $$f(n)=\begin{cases} \cfrac { n-1 }{ 2 } ,\quad \text{when n is odd} \\ -\cfrac { n }{ 2 } ,\quad \text{when n is even} \end{cases}$$ is

A
neither one-one nor onto

B
one-one but not onto

C
onto but not one-one

D
one-one and onto both

Solution

$$one-one$$ test of $$f:$$
Let $$x_1$$ and $$x_2$$ be any two elements in the domain $$(N).$$
$$Case\,I:$$ When both $$x_1$$ and $$x_2$$ are even.

Let $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$\dfrac{-x_1}{2}=\dfrac{x_2}{2}$$
$$\Rightarrow$$ $$-x_1=-x_2$$
$$\Rightarrow$$ $$x_1=x_2$$

$$Case\,II:$$ When both $$x_1$$ and $$x_2$$ are odd.

Let $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$\dfrac{x_1-1}{2}=\dfrac{x_2-1}{2}$$
$$\Rightarrow$$ $$x_1-1=x_2-1$$
$$\Rightarrow$$ $$x_1=x_2$$

$$Case\,III:$$ When $$x_1$$ be even and $$x_2$$ be odd.

Then, $$f(x_1)=\dfrac{-x_1}{2}$$ and $$f(x_2)=\dfrac{x_2-1}{2}$$
Then clearly,
$$\Rightarrow$$ $$x_1\ne x_2$$
$$\Rightarrow$$ $$f(x_1)\ne f(x_2)$$
From, all the cases, we can say that, $$f$$ is one-one.

$$onto$$ test of $$f:$$
Co-domain of $$f=Z=\{....,-3,-2,-1,0,1,2,3,...\}$$
Range of $$f=\left\{...,\dfrac{-2-1}{2},\dfrac{-(-2)}{2},\dfrac{-1-1}{2},\dfrac{0}{2},\dfrac{1-1}{2},\dfrac{-2}{2},\dfrac{3-1}{2},...\right\}$$
Range of $$f=\{...,-2,1,-1,0,0,-1,1,..\}$$
$$\Rightarrow$$ Co-domain of $$f=$$ Range of $$f$$
$$\therefore$$ $$f$$ is onto.
Question 10
1 / -0

Let $$f:[2,\infty )\rightarrow X$$ be defined by $$f(x)=4x-{x}^{2}$$. Then, $$f$$ is invertible, if $$X=$$

A
$$[2,\infty)$$

B
$$(-\infty,2]$$

C
$$(-\infty,4]$$

D
$$[4,\infty)$$

Solution

Since, $$f$$ is invertible, range of $$f=$$ Co-domain of $$f=X$$
So, we need to find the range of $$f$$ to find $$X.$$
Foe finding the range, let
$$f(x)=y$$
$$\Rightarrow$$ $$4x-x^2=y$$
$$\Rightarrow$$ $$x^2-4x=-y$$
$$\Rightarrow$$ $$x^2-4x+4=4-y$$ [ Adding $$4$$ on both sides ]
$$\Rightarrow$$ $$(x-2)^2=4-y$$
$$\Rightarrow$$ $$x-2=\pm\sqrt{4-y}$$
$$\Rightarrow$$ $$x=2\pm\sqrt{4-y}$$
This is defined only when,
$$4-y\ge 0$$
$$\Rightarrow$$ $$y\le 4$$
$$X=$$ Range of $$f=(-\infty , 4]$$

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Relations and Functions Test - 68

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Relations and Functions Test - 68

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SHARING IS CARING

Question 1

Binary numbers system

Octal numbers system

Decimal numbers system

Hexadecimal numbers system

Solution

Question 2

$$4$$

$$5$$

$$120$$

$$90$$

Solution

Question 3

34

9

8

25

Solution

Question 4

$$4$$

$$2$$

$$9$$

$$8$$

Solution

Question 5

one-one function

many one function

into function

onto function

Solution

Question 6

$$ x^{\dfrac{1}{3}} $$

$$ 1^{3} $$

x

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

Solution

Question 7

odd and one-one

even and one-one

many to one and even

many to one and neither even nor odd

Question 8

one-one & onto

one-one & into

many one & onto

many one & into

Solution

Question 9

neither one-one nor onto

one-one but not onto

onto but not one-one

one-one and onto both

Solution

Question 10

$$[2,\infty)$$

$$(-\infty,2]$$

$$(-\infty,4]$$

$$[4,\infty)$$

Solution

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$$
x^{\dfrac{1}{3}}
$$

$$
1^{3}
$$

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