Miscellaneous Exercise
Relations and Functions
Question and Answers
Class 11 – Maths
| 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) = x2, 0 ≤ x ≤3
3x, 3 ≤ x ≤ 10
The relation g is defined by
g (x) = x2, 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) = x2, 0 ≤ x ≤3
3x, 3 ≤ x ≤ 10
It is seen that for 0 ≤ x < 3,
f(x) = x2 and for 3 < x ≤ 10,
f(x) = 3x
Also, at x = 3
f(x) = 32 = 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) = x2, 0 ≤ x ≤ 2
3x, 2 ≤ x ≤ 10
It is seen that, for x = 2
g(x) = 22 = 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 ) = x2
f (1.1) – f(1)
__________
(1.1 – 1)
(1.1)2 – 12
________
(1.1 – 1)
[1.21 – 1 ]
__________
0.1
=2.1
Question 3 Find the domain of the function f (x) = [ x2 + 2x + 1 ] / [x2 -8x + 12 ]
Answer
f (x) = [ x2 + 2x + 1 ]
_______________
[x2 -8x + 12 ]
x2 + 2x + 1 / x2−8x+12
=x2 + 2x + 1 /( x2−6x−2x+12)
=x2 + 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 , x2/ 1+ x2 ) : x ∈ R } be a function from R into R. Determine the range of f.
Answer f = { (x , x2/ 1+ x2 ) : 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,
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 .
Answer f(x) = x + 1, g(x) = 2x – 3
Now,
(f + g) (x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2
Thus, (f + g) (x) = 3x – 2
(f – g) (x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 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/ 2x – 3, 2x – 3 ≠ 0
Thus, 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.
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 = b2 }. 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 = b2}
It can be seen that 2 ∈ N; however, 2 ≠ 22 = 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 = b2}
Its clearly seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 32.
Now, 3 ≠ 92 = 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 = b2}
It’s clearly seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 42 and 4 = 22.
Now, 16 ≠ 22 = 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}