Relations and Functions Test

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Relations and Functions Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)=\dfrac{x+1}{x-1}$$ and $$g(x)=2x-1, f[g(x)]=$$

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

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

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

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

E
$$\dfrac{2x-1}{2x+1}$$

Solution

Given, $$f(x) = \dfrac { x+1 }{ x-1 } $$ and $$g(x) = 2x-1$$
$$\therefore f(g(x)) = f(2x-1)$$
$$ = \dfrac { 2x-1+1 }{ 2x-1-1 } $$
$$=\dfrac { 2x }{ 2x-2 } $$
$$=\dfrac { x }{ x-1 } $$
Question 2
1 / -0

In the set of integers under the operation $$\ast $$ defined by $$a\ast b=a+b-1$$, the identity element is:

A
$$0$$

B
$$1$$

C
$$a$$

D
$$b$$

Solution

Let $$e$$ be the identity element, then
$$\Rightarrow a*e=a$$
or
$$a+e-1=a$$
$$e-1=0$$
$$e=1$$.
Hence the identity element for the above operation is $$1$$.
Question 3
1 / -0

The number of onto functions from the set $$\{1, 2, .........., 11\}$$ to set $$\{1, 2, ....., 10\}$$ is

A
$$5\times \underline{|11}$$

B
$$\underline{|10}$$

C
$$\dfrac{\underline{|11}}{2}$$

D
$$10\times \underline{|11}$$

Solution

Finding no. of such functions is same as putting 11 balls in 10 boxes with each box getting at least one ball. Then the only possible way is to put 2 balls in a box which can be done in $$^{11}C_2$$ ways.

$$^{11}C_{2}\times ^{9}C_{1} \times ^8C_1...................^1C_1$$

$$=\dfrac{11\times 10}{2}\times 9\times8\times7.................................$$

$$11!\times 5$$
Question 4
1 / -0

Which of the following is not a binary operation on $$R$$?

A
$$a \times b = ab$$

B
$$a \times b = a - b$$

C
$$a \times b = \sqrt{ab}$$

D
$$a \times b = \sqrt{a^2 + b^2}$$

Solution

When you will multiply, add , subtract, any two real number the resulting
number will be real number, so option $$A$$ and $$B$$ are binary operations.
Also if $$a$$ and $$b$$ are real number then $$\sqrt{a^2+b^2}$$ will be a real number, so option $$D$$ is also binary operation

Check option 'C', If sign of product of $$a$$ and $$b$$ will be negative, then $$\sqrt{ab}$$ will not be real number,
Hence it is not a binary operation

Therefore option $$C$$ is correct choice
Question 5
1 / -0

The number of real linear functions $$f(x)$$ satisfying $$f(f(x))=x+f(x)$$ is

A
$$0$$

B
$$4$$

C
$$5$$

D
$$2$$

Solution

Since $$f(x)$$ is a linear function, so take $$f(x)=ax+b$$
Given $$f(f(x))=x+f(x)$$
$$\Rightarrow$$ $$a(ax+b)+b=x+(ax+b)$$
Comparing the coefficients we get,
$$ { a }^{ 2 }=a+1$$ and $$ab+b=b$$
$$a^{ 2 }-a-1=0$$ and $$ab=0\Rightarrow a\ne 0,b=0$$
$$\therefore a=\cfrac { 1\pm \sqrt { 5 } }{ 2 } $$ and $$b=0$$
Thus $$f(x)=\left( \cfrac { 1\pm \sqrt { 5 } }{ 2 } \right) x+0$$
So there are two possible such functions exists
Question 6
1 / -0

Consider the function $$f(x)=\displaystyle\frac{x-1}{x+1}$$. What is $$f(f(x))$$ equal to?

A
$$x$$

B
$$-x$$

C
$$-\displaystyle\frac{1}{x}$$

D
None of the above

Solution

$$f(x)=\dfrac{x-1}{x+1}$$

$$f(f(x))=\dfrac{\dfrac{x-1}{x+1} - 1}{\dfrac{x-1}{x+1}+1}$$

$$f(f(x)) = \dfrac{-1}{x}$$

hence $$Ans$$ is $$option C$$
Question 7
1 / -0

Let N denote the set of all non-negative integers and Z denote the set of all integers. The function $$ f : Z \rightarrow N$$ given by f(x) = |x| is :

A
One-one but not onto

B
Onto but not one-one

C
Both one-one and onto

D
Neither one-one nor onto

Solution

$$f: X\rightarrow Y$$
For a function to be one one or injective, every element in the domain is the image of at most one element of it's co-domain.
In simple words, no value of $$y$$ must be same for $$2$$ or more different values of $$x$$.
For $$f(x)=\left| x \right| $$, we see that $$f(a)=f(-a)$$, for $$a\in Z$$
Hence, the function is not one one

For a function $$f: X\rightarrow Y$$, to be surjective,
every element $$y$$ in the co-domain Y must be linked with at least one element $$x$$ in the domain.
Every element in the co-domain of $$f(x)=\left| x \right| $$ is linked to at-least one element in domain.
Thus, $$f(x)=\left| x \right| $$ is onto but not one one.
Hence, b is correct.
Question 8
1 / -0

If $$N$$ is a set of natural numbers, then under binary operation $$a\cdot b = a + b, (N, .)$$ is

A
Quasi-group

B
Semi-group

C
Monoid

D
Group

Solution

Let $$a,b$$ be any element in $$N$$ $$\implies$$ $$a+b\ \epsilon N$$

$$\therefore a \cdot b=a+b\ \epsilon N$$
$$\implies\ a \cdot b\ \epsilon N$$ ......... $$(1)$$
$$(a \cdot b) \cdot c=(a\cdot b)+c=a+b+c$$

Also, $$a\cdot (b\cdot c)=a+(b \cdot c)=a+b+c$$
$$\therefore\ a \cdot (b \cdot c)=(a \cdot b) \cdot c$$ ........ $$(2)$$
$$a\cdot b=a+b=b+a=b\cdot a$$ ......... $$(3)$$

From $$(1), (2)$$ and $$(3)$$,

The structure $$(N, .)$$ satisfies the closure property, associativity and commutativity but the identity element $$0$$ does not belong to $$N$$.

Hence, $$(N,.)$$ is a semi-group.
Question 9
1 / -0

If $$f(x) = \log_{e}\left (\dfrac {1 + x}{1 - x}\right ), g(x) = \dfrac {3x + x^{3}}{1 + 3x^{2}}$$ and $$go f(t) = g(f(t))$$, then what is $$go f\left (\dfrac {e - 1}{e + 1}\right )$$ equal to?

A
$$2$$

B
$$1$$

C
$$0$$

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

Solution

$$f\left(\dfrac{e-1}{e+1}\right)=\log_e\left\lbrace\dfrac{1+\left(\dfrac{e-1}{e+1}\right)}{1-\left(\dfrac{e-1}{e+1}\right)}\right\rbrace=\log_e\left(\dfrac{2e}{2}\right)=\log_ee=1$$
$$\therefore gof\left(\dfrac{e-1}{e+1}\right)=g\left(f\left(\dfrac{e-1}{e+1}\right)\right)=g(1)=\dfrac{3+1}{1+3}=1$$
Question 10
1 / -0

If $$g(x)=\dfrac{1}{f(x)}$$ and $$f(x)=x, x\ne 0,$$ then which one of the following is correct?

A
$$f(f(f(g(g(f(x))))))=g(g(f(g(f(x)))))$$

B
$$f(g(f(g(g(f(g(x)))))))=g(g(f(g(f(x)))))$$

C
$$f(g(f(g(g(f(g(x)))))))=f(g(f(g(f(x)))))$$

D
$$f(f(f(g(g(f(x))))))=f(f(f(g(f(x)))))$$

Solution

From the given data, it is evident that $$gof(x)=\dfrac{1}{x}=fog(x)$$
$$f(f(f(g(g(f(x))))))=x$$ and $$g(g(f(g(f(x)))))=\dfrac{1}{x}$$
$$f(g(f(g(g(f(g(x)))))))=x$$ and $$g(g(f(g(f(x)))))=\dfrac{1}{x}$$
$$f(g(f(g(g(f(g(x)))))))=x$$ and $$f(g(f(g(f(x)))))=x$$
$$f(f(f(g(g(f(x))))))=x$$ and $$f(f(f(g(f(x)))))=\dfrac{1}{x}$$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 58

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Relations and Functions Test - 58

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SHARING IS CARING

Question 1

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

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

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

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

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

Solution

Question 2

$$0$$

$$1$$

$$a$$

$$b$$

Solution

Question 3

$$5\times \underline{|11}$$

$$\underline{|10}$$

$$\dfrac{\underline{|11}}{2}$$

$$10\times \underline{|11}$$

Solution

Question 4

$$a \times b = ab$$

$$a \times b = a - b$$

$$a \times b = \sqrt{ab}$$

$$a \times b = \sqrt{a^2 + b^2}$$

Solution

Question 5

$$0$$

$$4$$

$$5$$

$$2$$

Solution

Question 6

$$x$$

$$-x$$

$$-\displaystyle\frac{1}{x}$$

None of the above

Solution

Question 7

One-one but not onto

Onto but not one-one

Both one-one and onto

Neither one-one nor onto

Solution

Question 8

Quasi-group

Semi-group

Monoid

Group

Solution

Question 9

$$2$$

$$1$$

$$0$$

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

Solution

Question 10

$$f(f(f(g(g(f(x))))))=g(g(f(g(f(x)))))$$

$$f(g(f(g(g(f(g(x)))))))=g(g(f(g(f(x)))))$$

$$f(g(f(g(g(f(g(x)))))))=f(g(f(g(f(x)))))$$

$$f(f(f(g(g(f(x))))))=f(f(f(g(f(x)))))$$

Solution

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