# Class 11 NCERT Solutions- Chapter 2 Relation And Functions – Miscellaneous Exercise on Chapter 2

• Last Updated : 19 Jan, 2021

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

Solution:

The given relation f is defined as:

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12.

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

f(x) = {x2, 0≤x≤3

3x, 3≤x≤10}

It is given that, for the condition 0 ≤ x < 3,

Solution of f(x) = x2 and

For the condition 3 < x ≤ 10, solution of f(x) = 3x.

Now for the value of x = 3, solution of f(x) by putting the value of x, f(x) = 32 = 9

or, f(x) = 3 × 3 = 9.

That means, at x = 3, f(x) = 9 [Single image]

So that, 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) = {x2, 0≤x≤2

3x, 2≤x≤10}

It is seen that, in case of both the condition, for x = 2,

The value of g(x), by putting the value of x, g(x) = 22= 4 and g(x) = 3 × 2 = 6.

So that, 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) = x2, find .

Solution:

Given:

f(x) = x2 .

Hence, by putting the condition of f(x) in f(1.1) and f(1),

we can find the result of the given equation

((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)

= 2.1

## Question 3. Find the domain of the function  f(x) = ((x2+2x+1)/(x2-8x+12)).

Solution:

Given function:

f(x) = ((x2+2x+1)/(x2-8x+12))

= (((x2+2x+1)/((x-6)(x-2)))

It is clearly notified that, the function f is defined for all real numbers except

at x = 6 and x = 2 as the denominator becomes zero otherwise.

### 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 it is notified, √(x – 1) is defined for (x – 1) ≥ 0.

Hence, the function f(x) = √(x – 1) is defined for x ≥ 1.

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

Domain of f = [1, ∞).

Now,

According to the condition, x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒ √(x – 1) ≥ 0

That’s why, the range of f is the set of all real numbers greater than or equal to 0.

Range of f = [0, ∞).

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

### Question 5. 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 it is notified that, the function |x – 1| is defined for all real numbers.

Hence, Domain of f = R

Also, according to the condition , for x ∈ R, |x – 1| assumes all real numbers.

So that, the range of f is the set of all non-negative real numbers.

### Question 6. Let f={(x, )}: x ∈ R} be a function from R into R. Determine the range of f.

Solution:

Given function:

f = {(x, x2/1+x2): x ∈ R}

Substituting values and determining the images, we have

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

From the above equation, the range of f is the set of all second elements.

It can be notified 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,

x2≥ 0

Then,

x2 + 1 ≥ x2

1 ≥ (x2 / (x2 + 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:

According to the question, let us assume, the functions f, g: R → R is defined as

given conditions f(x) = x + 1, g(x) = 2x – 3.

Now,

We find that (f + g) (x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2

So that, (f + g) (x) = 3x – 2

Now, we find that, (f – g) (x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = – x + 4

So that, (f – g) (x) = -x + 4

(f/g(x)) = f(x)/g(x), g(x) ≠ 0, x ∈ R

(f/g(x)) = x + 1/ 2x – 3, 2x – 3 ≠ 0

So that, (f/g(x)) = 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, for some integers a, b. Determine a, b.

Solution:

Given the values, f = {(1, 1), (2, 3), (0, –1), (–1, –3)}

And according to the question, the function is 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

So, b = -1

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

Putting the value here, a + (–1) = 1 ⇒ a = 1 + 1 = 2.

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

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

Solution:

According to the question, Given relation R = {(a, b): a, b ∈ N and a = b2}

(i) It can be notified that 2 ∈ N; however, 2 ≠ 22 = 4.

Hence, it is notified that the statement “(a, a) ∈ R, for all a ∈ N” is not true.

(ii) Its clearly notified that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 32.

Now, it’s clear that 3 ≠ 92 = 81; therefore, (3, 9) ∉ N

Hence, it is notified that the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.

(iii) Its clearly notified that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 42 and 4 = 22.

Now, it is clear that 16 ≠ 22 = 4; therefore, (16, 2) ∉ N

Hence, it is notified that the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.

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

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 in the question that, f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

(i) The verified statement is : 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 notified that f is a subset of A × B.

Hence, it’s clear that f is a relation from A to B.

(ii) From the given condition, 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:

According to the question, Given relation f is defined as f = {(ab, a + b): a, b ∈ Z}.

Now, 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 notified 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

Hence, it can be expressed as:

Here,

f(9) means the highest prime factor of 9 = 3

f(10) means the highest prime factor of 10 = 5

f(11) means the highest prime factor of 11 = 11

f(12) means the highest prime factor of 12 = 3

f(13) means the highest prime factor of 13 = 13

So that, the range of f is the set of all f(n), where n ∈ A.

Therefore,

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

My Personal Notes arrow_drop_up