### Question 1. Show that the function f:** R**_{* }⇢ R_{*} defined by f(x)=(1/x) is one-one and onto, where **R**_{*} is the set of all non-zero real numbers. Is the result true, if the domain R_{*} is replaced by **N** with co-domain same as **R**_{*}?

_{* }⇢ R

_{*}

_{*}is the set of all non-zero real numbers. Is the result true, if the domain R

_{*}

_{*}

**Solution:**

One-one:

f(x)=f(y)

⇒1/x =1/y

⇒x=y

Therefore, f is one-one.

Onto:

It is clear that for y∈

Rthere exists x=(1/y)∈ R_{*}_{*}(exists as y ≠ 0) such that f(x)=1/(1/y)=yTherefore, f is onto.

Thus, consider function g:

N⇢Rdefined by g(x)=1/x_{*}We have, f(x

_{1})=g(x_{2})⇒1/x_{1}=1/x_{2}⇒x_{1}=x_{2}Therefore, g is one-one.

Further, it is clear that g is not onto as for 1.2∈

Rthere does not exist any x in_{*}Nsuch that g(x)=1/(1.2)Hence, function g is one-one but not onto.

### Question 2. Check the injectivity and surjectivity of the following functions:

### (i) f: **N⇢N **given by f(x)=x^{2}

**Solution: **

It is seen that for x, y ∈

N,f(x)=f(y) ⇒x^{2}=y^{2}⇒x=yTherefore, f is injective.

Now, 2 ∈

Nbut there does not exist any x inNsuch that f(x)=x^{2}=2.Therefore, f is not surjective.

### (ii) f: **Z**⇢**Z given by f(x)=x**^{2}

^{2}

**Solution: **

It is seen that f(-1)=f(1), but -1 ≠1. Therefore, f is not injective.

-2 ∈

Z.But, there does not exist any x inZsuch that f(x)= x^{2}=-2.Therefore, f is not surjective

**(iii) f:** R⇢ **R **given by f(x)=x^{2}

**Solution: **

It is seen that f(-1)=f(1), but -1 ≠1. Therefore, f is not injective.

-2 ∈

R. But, there does not exist any x inRsuch that f(x)= x^{2}=-2.Therefore, f is not surjective.

#### (iv)f: **N**⇢N given by f(x)=x^{3}

**Solution: **

It is seen that for x, y ∈

N,f(x)=f(y)⇒x^{3}=y^{3}⇒x=y. Therefore, f is injective.2∈

N.But, there does not exist any element x in domainNsuch that f(x)=x^{3}=2. Therefore, f is not surjective.

#### (v) f: **Z**⇢**Z **given by f(x)=x^{3}

**Solution: **

It is seen that for x, y ∈

Z,f(x)=f(y)⇒x^{3}=y^{3}⇒x=y. Therefore, f is injective.2∈

Z.But, there does not exist any element x in domainZsuch that f(x)=x^{3}=2. Therefore, f is not surjective.

### Question 3. Prove that the Greatest Integer Function f: **R⇢R **given by f(x)=[x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

**Solution:**

It is seen that f(1.2)=[1.2]=1, f(1.9)=[1.9]=1.

f(1.2)=f(1.9), but 1.2≠1.9. Therefore, f is not one-one.

Consider 0.7∈

R. It is known that f(x)=[x] is always an integer. Thus, there does not exist any element x ∈Rsuch that f(x)=0.7. Therefore, f is not onto.Hence, the greatest integer function is neither one-one nor onto.

### Question 4. Show that the Modulus Function f:**R⇢R **given by f(x)=|x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is -x, if x is negative.

**Solution:**

It is seen that f(-1)=|-1|=1, f(1)=|1|=1.

f(-1)=f(1), but -1≠1. Therefore, f is not one-one.

Consider, -1∈

R.It is known that f(x)=|x| is always non-negative. Thus, there does not exist any element x in domainRsuch that f(x)=|x|=-1. Therefore, f is not onto.Hence, the modulus function is neither one-one nor onto.

### Question 5. Show that the signum function f: **R⇢R **given by, f(x)={ (1, if x>0), (0, if x=0), (-1, if x<0)} is neither one-one nor onto.

**Solution:**

It is seen that f(1)=f(2)=1, but 1≠2. Therefore, f is not one-one.

As f(x) takes only 3 values (1, 0, or -1) for the element -2 in co-domain

R,there does not exist any x in domainRsuch that f(x)=-2. Therefore, f is not onto.Hence, the signum function is neither one-one nor onto.

### Question 6. Let A={1, 2, 3}, B={4, 5, 6, 7} and let f={(1,4), (2,5), (3,6)} be a function from A to B. Show that f is one-one.

**Solution:**

It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.

f:A

⇢B is defined as f={(1,4), (2,5), (3,6)}Therefore, f(1)=4, f(2)=5, f(3)=6

It is seen that the images of distinct elements of A under f are distinct.

Hence, function f is one-one.

### Question 7. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

### (i) f:**R⇢R **defined by f(x)=3-4x

**Solution: **

Let x

_{1}, x_{2}∈R such that f(x_{1})=f(x_{2})⇒3-4x

_{1}=3-4x_{2}⇒-4x

_{1}=-4x_{2}⇒x

_{1}=x_{2}Therefore, f is one-one.

For any real number (y) in

R,there exists {(3-y)/4} inRsuch that f((3-y)/4)=3-4((3-y)/4)=y.Therefore, f is onto

Hence, f is bijective.

### (ii) f:**R⇢R **defined b f(x)=1+x^{2}

**Solution:**

Let x

_{1}, x_{2}∈Rsuch that f(x_{1})=f(x_{2})⇒1+x

_{1}^{2}=1+x_{2}^{2}⇒x

_{1}^{2}=x_{2}^{2}⇒x

_{1}=±x_{2}Therefore, f(x

_{1})=f(x_{2}) does not imply that x_{1}=x_{2}For instance, f(1)=f(-1)=2

Therefore, f is not one-one.

Consider, an element -2 in co-domain

R.It is seen that f(x)=1+x

^{2}is positive for all x ∈R.Thus, there does not exist any x in domain

Rsuch that f(x)=-1.Therefore, f is not onto.

Hence, f is neither one-one nor onto.

### Question 8. Let A and B be sets. Show that f: A x B ⇢B x A such that (a, b)=(b, a) is bijective function.

**Solution:**

Let (a

_{1}, b_{1}), (a_{2}, b_{2}) ∈ A x b such that f(a_{1}, b_{1})=f(a_{2}, b_{2})⇒(b

_{1}, a_{1})=(b_{2}, a_{2})⇒b

_{1}=b_{2}and a_{1}=a_{2}⇒(a

_{1}, b_{1})=(a_{2}, b_{2})Therefore, f is one-one.

Let (b,a) ∈ B x A such that f(a, b)=(b,a).

Therefore, f is onto.

Hence, f is bijective.

### Question 9. Let f: **N**⇢ **N **defined by f(n)={((n+1)/2, if n is odd), (n/2, if n is even) for all n ∈ **N. **State whether the function f is bijective. Justify your answer.

**Solution:**

It can be observed that:

f(1)=(1+1)/2=1 and f(2)=2/2=1

So, f(1)=f(2), where, 1≠2

Therefore, f is not one-one.

Therefore, it is not bijective. (Since, it needs to be both one-one and onto to be bijective).

### Question 10. Let A=**R**-{1}. Consider the function f: A⇢B defined by f(x)=(x-2)/(x-3). Is f one-one and onto? Justify your answer.

**Solution:**

Let x, y ∈ A such that f(x)=f(y)

⇒ (x-2)/(x-3)=(y-2)/(y-3)

⇒(x-2)(y-3)=(y-2)(x-3)

⇒ xy-3x-2y+6=xy-3y-2x+6

⇒ -3x-2y=-3y-2x

⇒ 3x-2x=3y-2y

⇒ x=y

Therefore, f is one-one.

Let, y ∈ B=

R-{1}. Then y≠1.The function f is onto if there exists x ∈ A such that f(x)=y

Now,

f(x)=y

⇒ (x-2)/(x-3)=y

⇒ x-2=xy-3y

⇒ x(1-y)=-3y+2

⇒ x=(2-3y)/(1-y) ∈ A

Thus, for any y ∈ B, there exists (2-3y)/(1-y) ∈ A such that f((2-3y)/(1-y))={((2-3y)/(1-y))-2}/{((2-3y)/(1-y))-3}=(2-3y-2+2y)/(2-3y-3+3y)=(-y)/(-1)=y

Therefore, f is onto.

Hence, function f is one-one and onto.

### Question 11. Let f: **R⇢R **be defined as f(x)=x^{4}. Choose the correct answer:

### (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto

#### Solution:

Let x, y ∈

Rsuch that f(x)=f(y)⇒ x4=y4

⇒ x=±y

Therefore, f(x1)=f(x

^{2}) does not imply that x1=x^{2}For instance, f(1)=f(-1)=1

Therefore, f(1)=f(-1)=1

Therefore, f is not one-one

Consider an element 2 in co-domain

R.It is clear that there does not exist any x in domainRsuch that f(x)=2Therefore, f is not onto.

The correct answer is D.

### Question 12. Let f:**R**⇢**R **be defined as f(x)=3x. Choose the correct answer:

### (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto

**Solution:**

Let x, y ∈

Rsuch that f(x)=f(y)⇒ 3x = 3y

⇒ x=y

Therefore, f is one-one.

Also, for any real number (y) in co-domain

R,there exists y/3 inRsuch that f(y/3) = 3(y/3) = yTherefore, f is onto.

Hence, the

correct answer is A.

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