Question 1: The function of f is defined by:

The relation g is defined by

Show that f is a function and g is not a function.

**Solution:** Given,

The function of f is defined by:

Therefore,

f (0) = 0

f (1) = 1

f (2) = 4

f (3) = 9

f (4) = 12

f (5) = 15

f (6) = 18

f (7) = 21

f (8) = 24

f (9) = 27

f (10) = 30

Here, each element of domain which correspond exactly one element of co-domain. Therefore, it is clear that f is a function.

Now,

The relation g is defined by

Therefore,

g (0) = 0,

g (1) = 1

g (2) = 4

g (3) = 6

Here ‘g’ is not a function as (2,4) and (2,6) have the same first component.

Clearly, g is Clearly, g is not a function.

Question 2: If `f(x)=x^2`, find `(f(1.1)-f(1))/(1.1-1)`

**Answer:** Given, `f(x)=x^2`

Hence, `(f(1.1)-f(1))/(1.1-1)`

`=((1.1)^2-1^2)/(1.1-1)`

`=(1.21-1)/(0.1)=(0.21)/(0.1)`

`=(21)/(10)=2.1`

Question 3: Find the domain of the function `f(x)=(x^2+2x+1)/(x^2-8x+12)`

**Answer:** `f(x)=(x^2+2x+1)/(x^2-8x+12)`

`=(x^2+2x+1)/(x^2-6x-2x+12)`

`=(x^2+2x+1)/(x(x-6)-2(x-6))`

`=(x^2+2x+1)/((x-6)(x-2))`

Here, the function f is defined as all real numbers except `x=6` and `x=2`

Therefore, domain of f is R-{2, 6}

Question 4: Find the domain and the range of the real function f defined by `f(x)=sqrt((x-1))`

**Answer:** Given, `f(x)=sqrt((x-1))`

It is clear that `sqrt((x-1))` is defined if `(x-1)≥0`

This means `x≥1`

Hence, the domain of g is the set of all real numbers greater than or equal to 1.

That is the domain of `f=(1, ∞)`

As `x≥1`

Hence, `(x-1)≥0`

Or, `sqrt(x-1)≥0`

Hence, the range of f is the set of all real numbers which are greater than or equal to 0.

Hence, range of `f=(0, ∞)`

Question 5: Find the domain and the range of the real function *f *defined by; *f (x)*= | *x*-1 |

**Solution:**

Given, the real function is * f* (*x*)= | *x*-1 |

From the above function it is clear that | *x *- 1| is defined for all real numbers.

Therefore, Domain of *f = R*

Now, For x ∈ R, | x-1 | assumes all real numbers.

Therefore, the range of *f * is the set of all non-negative real numbers.

Hence, Domain = *R*

Range = All non-negative real numbers

Question 6: Let `f={(x, (x^2)/(1+x^2)):x∈R}` be a function from R into R. Determine the range of f.

**Solution:** Here,

`f={(x, (x^2)/(1+x^2)):x∈R}`

`=(0,0), (±0.5,1/5), (±1,1/2)`,`(±1.5,(9)/(13)), (±2,4/5)`, `(3, (9)/(10)), (4, (16)/(17)`…..

It is clear that the range of f is the set of all second elements. SIt is clear that the range of *f* is the set of all second elements. So, it can be observed that all these elements are greater than or equal to 0 but less than 1(Because all denominators are greater than numerator).

Therefore, the range of *f *= [ 0, 1)

Or, Range = Any positive real number x is such that 0 ≤ x < 1

Question 7: Let *f,,* *g*∶*R* → *R *be defined, respectively by *f *(*x*)= *x *+1, *g* (*x*) = 2*x*-3.

Find `f+g`, `f-g` and `f/g`

**Solution:** Let *f *(*x*) =* x* + 1

g (x) = 2x – 3

Therefore,

f + g = f (x) + g (x)– 3

= 3x – 2

f – g = f (x) – g (x)

= (x + 1) – (2x – 3)

= x + 1 – 2x + 3

= – x + 4

`f/g=(f(x))/(g(x))`

`=(x+1)/(2x-3)`, `x≠3/2`

Hence, `f+g=3x-2`

`f-g=-x+4`

`f/g=(x+1)/(2x-3)`, `x≠3/2`

Question 8: Let *f* = { (1, 1), (2, 3), (0, - 1), (-1, -3 ) } be a function from Z to Z defined by *f* (*x*) = *ax* + *b*, from some integers *a, b*. Determine *a, b*.

**Solution:** Given, *f *= {( 1, 1 ), ( 2, 3 ), ( 0, - 1 ), ( -1, -3)}

f (x) = ax + b ⇒ f (1) = 1

Therefore, a ×1 + b = 1

⇒ a + b = 1

As (1,1)∈ f

⇒ f (0) = -1

⇒ b= -1

Now, after substituting b = - 1 in a + b = 1, we get

a + ( - 1) = 1

⇒ a-1 =1

Therefore, the values of

a = 2 and b = - 1

Question 9: Let R be a relation from N to N defined by

R = { ( a,b) ∶ a,b ∈ N and a = b^{2}

Are the following true?

(i) (a,a) ∈ R, f or all a∈ N

(ii) (a,a) ∈ R,implies (b,a)∈ R

(iii) (a,b) ∈ R,(b,c) ∈ R implies (a,c)∈(iii) (a,b) ∈ R,(b,c) ∈ R implies (a,c)∈ R

**Solution:**

(i) (a,a) ∈ R, for all a ∈ N

It is not true, because

When a = 2, ^{>2}

⇒ 2 = 4

Which is not true.

(ii) (a,a)∈R,implies (b,a)∈R

Which is not true.

(ii) (a,a)∈R,implies (b,a)∈R

It is not true

As,(a,b)∈R implies b= aich is not true,

Therefore,

(b,a) ∉ R

(iii) (a,b)∈R,(b,c)∈R implies (a,c)∈R

No, it is (b,a) ∉ R

(iii) (a,b)∈R,(b,c)∈R implies (a,c)∈R

No, it is also not true

Because,(a,b) ∈ R, (b,c)∈R implies a = b

Qeustion 10: Let A = { 1, 2, 3, 4 }, B = {1, 5, 9, 11, 15, 16} and f = { (1,5), (2,9), (3,1), (4,5), (2,11) } Are the following true?

(i) f is a relation from A to B

(ii) f is a function from A to B

Justify your answer

**Solution:**

(i) f is a relation from A to B

It is true.in f have the same first components.

Question 11: Let f be the subset of Z x Z defined by f={(ab, a+b): a, b ∈ Z}. Is f a function from Z to Z? Justify your answer.

**Solution:** Here,1 x 6 = 6 and 2 x 3 = 6nd let f : A → N be defined by f(n) = highest prime factor of n. Find the range of f.

HereSo, f is not a function from Z to Z.

Question 12: Let A = {9, 10, 11, 12, 13} and let f : A → N be defined by f(n) = highest prime factor of n. Find the range of f.

**Solution:** Here, A = {9, 10, 11, 12, 13}

* f : A → N *is defined as

Prime factor of 9 = 3

f (11) = The highest prime factor of 11 = 11

f (12) = The highest prime factor of 12 = 3

f (13) = The highest prime factor of 13 = 13

The range of f is the set of all f (n), where n ∈ A

Therefore,

Range of f = {3, 5, 11, 13}

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