Relations and Functions Test

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Relations and Functions Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

If $$D$$ be subset of the set of all rational numbers which can be expressed as terminating decimals, then $$D$$ is closed under the binary operations of:

A
addition, subtraction and division

B
addition, multiplication and division

C
addition, subtraction and multiplication

D
subtraction, multiplication and division

Solution

Division of rational number that can be expressed as terminating decimal may not be rational. $$\therefore$$ $$D$$ is closed under addition, subtraction, and multiplication.
Question 2
1 / -0

If $$a\ast b={ a }^{ 3 }+{ b }^{ 3 }$$ on $$z$$, then $$\left( 1\ast 2 \right) \ast 0=........$$

A
$$0$$

B
$$729$$

C
$$81$$

D
$$27$$

Solution

$$a*b=a^{3}+b{3}$$
So $$1*2=1^{3}+2{3}=1+8=9$$
$$(1*2)*0=9*0=9^{3}+0^{3}=729+0=729$$
Question 3
1 / -0

If $$f : R - \left \{\dfrac {3}{5}\right \}\rightarrow R - \left \{\dfrac {3}{5}\right \}; f(x) = \dfrac {3x + 1}{5x - 3}$$, then ___________.

A
$$f^{-1} (x) = 2f(x)$$

B
$$f^{-1} (x) = f(x)$$

C
$$f^{-1} (x) = -f(x)$$

D
$$f^{-1} (x)$$ does not exists

Solution

$$f(x) = \dfrac {3x + 1}{5x - 3}$$ .... $$(i)$$
We know that if $$f(x) = \dfrac {ax + b}{cx + d}$$ and $$a + d = 0$$, then
$$f(x) = f^{-1}(x)$$
From $$(i)$$, we get
$$a=3,b=1,c=5,d=-3$$
$$a+d=3-3=0$$
Hence, $$f^{-1}(x)=f(x)$$
Question 4
1 / -0

If a language of natural numbers has a binary regularly of $$0$$ and $$1$$, then which one of the following strings represents the natural number $$7$$?

A
$$1$$

B
$$101$$

C
$$110$$

D
$$111$$

Solution

Converting decimal 7 to binary number
Question 5
1 / -0

The number of binary operations on $$\left\{ 1,2,3,4 \right\} $$ is ______.

A
$${ 4 }^{ 2 }$$

B
$${ 4 }^{ 8 }$$

C
$${ 4 }^{ 3 }$$

D
$${ 4 }^{ 16 }$$

Solution

In $$\{1,2,3,4\}$$ there are 4 elements.
The number of binary operations in a set with n elements$$=n^{(n\times n)}$$
Here $$n=4$$
So the number of binary operations in a set with n elements$$=4^{(4\times 4)}=4^{16}$$
Question 6
1 / -0

Suppose that $$g\left( x \right) =1+\sqrt { x }$$ and $$f\left( g\left( x \right) \right) =3+2\sqrt { x } +x$$, then $$f\left( x \right)$$ is

A
$$1+2{ x }^{ 2 }$$

B
$$2+{ x }^{ 2 }$$

C
$$1+x$$

D
$$2+x$$

Solution

$$g(x)=1+\sqrt{x}$$
$$(g(x))^{2}=1+2\sqrt{x}+x$$
$$2+(g(x))^{2}=3+2\sqrt{x}+x$$
$$\therefore f(x)=2+x^{2}$$
Question 7
1 / -0

$$f(x)=x^{3}+4x+5$$
$$\int_{0}^{5} f^{-1}(x)dx=$$

A
$$\dfrac{11} {4} $$

B
$$\dfrac{-11} {4} $$

C
$$\dfrac{9} {4} $$

D
$$\dfrac{-13} {4} $$

Solution

$$\int_a^b f(x) dx$$ + $$\int_{f(a)} ^{f(b)} f^{-1}(x)dx = bf(b) - af(a) $$

For the given problem f(a)=0, f(b) = 5 which gives a=-1 and b=0
Thus putting the values we get $$\int_0^5f^{-1}dx=-\int_{-1}^0f(x)dx = \dfrac{-11} {4} $$
Question 8
1 / -0

Let $$f:N\times N\rightarrow N-\left\{ 1 \right\} $$ be defined as $$f(m,n)=m+n$$, then function $$f$$ is ______.

A
One-One and onto

B
Many- One and not onto

C
One- One and not onto

D
Many- One and onto

Solution

Given $$f(m,n)=m+n$$
Range of this function is set of all natural numbers except $$1$$
[since least value of $$m$$ and $$n$$ is $$1$$]
Therefore the range and co-domain are equal
Therefore it is onto
For every $$(m,n)$$, there exists only one $$m+n$$
Therefore it is one-one also
Hence, $$f$$ is one-one and onto.
Question 9
1 / -0

$$f:R \to R,f\left( x \right) = \dfrac {{x^2} + 2x + c}{{x^2} + 4x + c}$$ is onto only if

A
$$c < 2$$

B
$$c \ge 0$$

C
$$c < 0$$

D
$$|c| \le 1$$

Solution

$$f:R\rightarrow R,f\left( x \right) =\cfrac { { x }^{ 2 }+2x+c }{ { x }^{ 2 }+4x+c } \\ \Rightarrow y=\cfrac { { x }^{ 2 }+2x+c }{ { x }^{ 2 }+4x+c } \\ \Rightarrow { x }^{ 2 }y+4xy+cy={ x }^{ 2 }+2x+c\\ \Rightarrow { x }^{ 2 }(y-1)+2x(2y-1)+(cy-c)=0\\ \because x\epsilon R\quad \quad \therefore 4{ (2y-1) }^{ 2 }-4c{ (y-1) }^{ 2 }\ge 0\\ { (2y-1) }^{ 2 }-c{ (y-1) }^{ 2 }\ge 0\\ 4{ y }^{ 2 }+1-4y-c{ y }^{ 2 }+2cy-1\ge 0\\ { y }^{ 2 }(4-c)-y(4-2c)\ge 0\\ \because y\epsilon R\quad \quad { (4-2c) }^{ 2 }-4(4-c)\times 0<0\\ { (4-2c) }^{ 2 }<0\\ { (2c-4) }^{ 2 }<0\\ 2c-4<0\\ c<2$$
Question 10
1 / -0

$$f: R\rightarrow R$$ is defined by $$f(x)=x^2-5x$$. Then the inverse image set of $$\{-6\}$$ is?

A
$$\phi$$

B
$$\{3, 2\}$$

C
$$\{3\}$$

D
$$\{-3, -2\}$$

Solution

For inverse image set of $$\{-6\}$$
we have, $$x^2-5x = -6$$
$$x^2-5x+6 = 0 $$
$$x^2-3x-2x+6 = 0$$
$$(x-2)(x-3) = 0$$
Hence, the set is $$\{3,2\}$$

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Relations and Functions Test - 33

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Relations and Functions Test - 33

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Question 1

addition, subtraction and division

addition, multiplication and division

addition, subtraction and multiplication

subtraction, multiplication and division

Solution

Question 2

$$0$$

$$729$$

$$81$$

$$27$$

Solution

Question 3

$$f^{-1} (x) = 2f(x)$$

$$f^{-1} (x) = f(x)$$

$$f^{-1} (x) = -f(x)$$

$$f^{-1} (x)$$ does not exists

Solution

Question 4

$$1$$

$$101$$

$$110$$

$$111$$

Solution

Question 5

$${ 4 }^{ 2 }$$

$${ 4 }^{ 8 }$$

$${ 4 }^{ 3 }$$

$${ 4 }^{ 16 }$$

Solution

Question 6

$$1+2{ x }^{ 2 }$$

$$2+{ x }^{ 2 }$$

$$1+x$$

$$2+x$$

Solution

Question 7

$$\dfrac{11} {4} $$

$$\dfrac{-11} {4} $$

$$\dfrac{9} {4} $$

$$\dfrac{-13} {4} $$

Solution

Question 8

One-One and onto

Many- One and not onto

One- One and not onto

Many- One and onto

Solution

Question 9

$$c < 2$$

$$c \ge 0$$

$$c < 0$$

$$|c| \le 1$$

Solution

Question 10

$$\phi$$

$$\{3, 2\}$$

$$\{3\}$$

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

Solution

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