**Miscellaneous Exercise**

Relations and Functions

**Question and Answers**

**Class 11 – Maths**

Relations and Functions

Class | Class 11 |

Subject | Mathematics |

Chapter Name | Relations and Functions |

Chapter No. | Chapter 2 |

Exercise | Miscellaneous Exercise |

Category | Class 11 Maths NCERT Solutions |

**Question 1 The relation f is defined by **

**f (x) = x ^{2}, 0 ≤ x ≤3**

**3x, 3 ≤ x ≤ 10**

**The relation g is defined by
g (x) = x ^{2}, 0 ≤ x ≤ 2**

**3x, 2 ≤ x ≤ 10 **

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

**Answer** In the given relation

f (x) = x^{2}, 0 ≤ x ≤3

3x, 3 ≤ x ≤ 10

It is seen that for 0 ≤ *x* < 3,

*f*(*x*) = *x*^{2 }and for 3 < *x* ≤ 10,

*f*(*x*) = 3*x*

Also, at *x* = 3

*f*(*x*) = 3^{2} = 9 or *f*(*x*) = 3 × 3 = 9

i.e., at *x* = 3, *f*(*x*) = 9

Hence, for 0 ≤ *x* ≤ 10, the images of *f*(*x*) are unique.

Therefore, the given relation is a function.

Now,

In the given relation,* g* is defined as

g (x) = x^{2}, 0 ≤ x ≤ 2

3x, 2 ≤ x ≤ 10

It is seen that, for *x* = 2

*g*(*x*) = 2^{2} = 4 and *g*(*x*) = 3 × 2 = 6

Thus, element 2 of the domain of the relation *g* corresponds to two different images, i.e., 4 and 6.

Therefore, this relation is not a function.

**Question 2 If f (x) = x ^{ 2} , find **

**[ f (1.1) – f(1) ]
****____________**

** (1.1 – 1)**

**Answer** We have the function, f ( x ) = x^{2}

f (1.1) – f(1)

__________

(1.1 – 1)

(1.1)^{2} – 1^{2}

________

(1.1 – 1)

[1.21 – 1 ]
__________

0.1

=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}−8x+12

=x^{2} + 2x + 1 /( x^{2}−6x−2x+12)

=x^{2} + 2x + 1 /[ x(x−6)−2(x−6)]

=(x−2)(x−6)

So, we see that the function is defined for every real numbers except 6,2

Thus, the domain of the function will be, R − {2,6}

**Question 4 Find the domain and the range of the real function f defined by f (x) = √(x-1) **

**Answer **Given real function,

*f*(x) = √(x – 1)

Clearly, √(x – 1) is defined for (*x* – 1) ≥ 0

So, the function *f*(x) = √(x – 1) is defined for *x* ≥ 1

Thus, the domain of *f* is the set of all real numbers greater than or equal to 1.

Domain of *f* = [1, ∞)

Now,

As *x* ≥ 1 ⇒ (*x* – 1) ≥ 0 ⇒ √(x – 1) ≥ 0

Thus, the range of *f* is the set of all real numbers greater than or equal to 0.

Range of *f* = [0, ∞ ]

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

**Answer **The function which is given is *f* (*x*) = |*x* – 1|

We can clearly see that, the function is well defined for all the real numbers.

Thus, it can be concluded that, the domain of the function is R

And for every x ∈ R , the function gives all non-negative real numbers.

So, the range of the function is the set of all non-negative real numbers. i.e, [0,∞]

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

**Answer** f = { (x , x^{2}/ 1+ x^{2} ) : x ∈ R }

Expressing it by term to term, we are getting,

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

The range of *f* is the set of all second elements. It can be observed that all these elements are greater than or equal to 0 but less than 1.

[As the denominator is greater than the numerator.]

Or, We know that, for x ∈ R,

x^{2 }≥ 0

Then, x^{2} + 1 ≥ x^{2}

1 ≥ x^{2 }/ (x^{2 }+ 1)

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

**Question 7 Let f, g : R→R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, ****f – g and f / g .**

**Answer** *f*(*x*) = *x *+ 1, *g*(*x*) = 2*x* – 3

Now,

(*f* + *g*) (*x*) = *f*(*x*) + *g*(*x*) = (*x* + 1) + (2*x* – 3) = 3*x* – 2

Thus, (*f + g*) (*x*) = 3*x* – 2

(*f – g*) (*x*) = *f*(*x*) – *g*(*x*) = (*x* + 1) – (2*x* – 3) = *x* + 1 – 2*x* + 3 = – *x* + 4

Thus, (*f – g*) (*x*) = –*x* + 4

*f/g*(x) = *f*(x)*/g*(x), g(x) ≠ 0, x ∈ R

*f/g*(x) = *x *+ 1/ 2*x* – 3, 2*x* – 3 ≠ 0

Thus, *f/g*(x) = *x *+ 1/ 2*x* – 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, for some integers a, b. Determine a, b.**

**Answer** Given, *f *= {(1, 1), (2, 3), (0, –1), (–1, –3)}

And the function defined as, *f*(*x*) = *ax* + *b*

For (1, 1) ∈ *f*

We have, *f*(1) = 1

So, *a* × 1 + *b* = 1

*a* + *b* = 1 …. (i)

And for (0, –1) ∈ *f*

We have *f*(0) = –1

*a* × 0 + *b* = –1

*b* = –1

On substituting *b* = –1 in (i), we get

*a* + (–1) = 1 ⇒ *a* = 1 + 1 = 2.

Therefore, the values of *a* and *b* are 2 and –1, respectively.

**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?Justify your answer in each case.**

** (i) (a,a) ∈ R, for all a ∈ N**

**(i)** Given relation R = {(*a*, *b*): *a*, *b* ∈ N and *a* = *b*^{2}}

It can be seen that 2 ∈ N; however, 2 ≠ 2^{2} = 4.

Thus, the statement “(*a*, *a*) ∈ R, for all* a *∈ N” is not true.

**(ii) (a,b) ∈ R, implies (b,a) ∈ R**

**(ii)** Given relation R = {(*a*, *b*): *a*, *b* ∈ N and *a* = *b*^{2}}

Its clearly seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3^{2}.

Now, 3 ≠ 9^{2} = 81; therefore, (3, 9) ∉ N

Thus, the statement “(*a*, *b*) ∈ R, implies (*b*, *a*) ∈ R” is not true.

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

**(iii)** Given relation R = {(*a*, *b*): *a*, *b* ∈ N and *a* = *b*^{2}}

It’s clearly seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 4^{2} and 4 = 2^{2}.

Now, 16 ≠ 2^{2} = 4; therefore, (16, 2) ∉ N

Thus, the statement “(*a*, *b*) ∈ R, (*b*, *c*) ∈ R implies (*a*, *c*) ∈ R” is not true.

**Question 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? Justify your answer in each case.**

** (i) f is a relation from A to B**

(i) A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}

Thus, the Cartesian product of these two sets will be,

A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}

Also, given that, *f *= {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.

It’s clearly seen that *f* is a subset of A × B.

Therefore, *f* is a relation from A to B.

**(ii) f is a function from A to B. **

**(ii)** A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}

Thus, the Cartesian product of these two sets will be,

A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}

Also, given that, *f *= {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

We see that the first element 2 is providing us two different value of the image 9,11

So, it can be concluded that f is not a function from A to B.

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

**Answer** Given relation, *f* is defined as

*f *= {(*ab*, *a* + *b*): *a*, *b* ∈ Z}

We know that a relation *f* from a set A to a set B is said to be a function if every element of set A has unique images in set B.

Let us take 4 elements 2, 6, –2, –6 ∈ Z

(2 × 6, 2 + 6) ∈ *f*

(12, 8) ∈ *f*

, (–2 × –6, –2 + (–6)) ∈ *f*

(12, –8) ∈ *f*

It’s clearly seen that the same first element, 12, corresponds to two different images (8 and –8).

Therefore, the relation *f* is not a function.

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

**Answer** A = {9, 10, 11, 12, 13}

Now, *f*: A → **N** is defined as

*f*(*n*) = The highest prime factor of *n*

Prime factor of 9 = 3

Prime factors of 10 = 2, 5

Prime factor of 11 = 11

Prime factors of 12 = 2, 3

Prime factor of 13 = 13

Thus, it can be expressed as

*f*(9) = The highest prime factor of 9 = 3

*f*(10) = The highest prime factor of 10 = 5

*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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