# Chapter 2 Miscellaneous sets ncert maths solution class 11

# Chapter 2 Miscellaneous(Sets) Class 11

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

** ****The relation g is defined by **

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

**Solution: **The given relation *f* is defined as:

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 [Single image]

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

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**

**Solution: **Given, *f*(*x*) = *x*^{2}

Hence,

**Question 3: Find the domain of the function **

**Solution: **Given function,

It’s clearly seen that the function *f* is defined for all real numbers except at *x* = 6 and *x* = 2, as the denominator becomes zero otherwise.

Therefore, the domain of *f* is R – {2, 6}.

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

**Solution: **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|.**

**Solution: **Given a real function,

*f* (*x*) = |*x* – 1|

Clearly, the function |*x* – 1| is defined for all real numbers.

Hence,

Domain of *f* = R

Also, for *x* ∈ R, |*x* – 1| assumes all real numbers.

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

**Question 6: Let ** **be a function from R into R. Determine the range of f.**

**Solution: **Given function,

Substituting values and determining the images, we have

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.**

**Solution: **Given the functions *f*, *g*: R → R is defined as

*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.**

**Solution: **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?**

**(i) ( a, a) ∈ R, for all a ∈ N**

(ii) (a, b) ∈ R, implies (b, a) ∈ R

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

**Justify your answer in each case.**

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

(i) 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) 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) 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?**

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

**Justify your answer in each case.**

**Solution: **Given,

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

So,

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)}

**(i)** 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)** As the same first element, i.e., 2 corresponds to two different images (9 and 11), relation *f *is not a function.

**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.**

**Solution: **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.

As 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ *f*

i.e., (12, 8), (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.**

**Solution: **Given,

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

Now, *f*: A → **N** is defined as

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

So,

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}