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NCERT Exemplar Class 11 Maths Chapter 2 Relations and Functions Exercise Solutions Short Answer Type
Table of Contents
Solved Short Answer Type Questions of NCERT Exemplar Chapter 2 Relations and Functions Exercise
Question 1 Let A = {−1, 2, 3} and B = {1, 3}. Determine:
(i) A × B
(ii) B × A
(iii) B × B
(iv) A × A
Answer 1 Given:
A = {−1, 2, 3}
B = {1, 3}
(i) A × B = { (−1, 1), (−1, 3), (2, 1), (2, 3), (3, 1), (3, 3) }
(ii) B × A = { (1, −1), (1, 2), (1, 3), (3, −1), (3, 2), (3, 3) }
(iii) B × B = { (1, 1), (1, 3), (3, 1), (3, 3) }
(iv) A × A = { (−1, −1), (−1, 2), (−1, 3), (2, −1), (2, 2), (2, 3), (3, −1), (3, 2), (3, 3) }
Question 2 If
P = {x : x < 3, x ∈ N}
Q = {x : x ≤ 2, x ∈ W}
Find (P ∪ Q) × (P ∩ Q),
where W is the set of whole numbers.
Answer 2.
P = {1, 2}
Q = {0, 1, 2}
P ∪ Q = {0, 1, 2}
P ∩ Q = {1, 2}
Therefore,
(P ∪ Q) × (P ∩ Q)
= { (0,1), (0,2), (1,1), (1,2), (2,1), (2,2) }
Question 3. If
A = {x : x ∈ W, x < 2}
B = {x : x ∈ N, 1 < x < 5}
C = {3, 5}
Find:
(i) A × (B ∩ C)
(ii) A × (B ∪ C)
Answer 3.
A = {0, 1}
B = {2, 3, 4}
C = {3, 5}
B ∩ C = {3}
B ∪ C = {2, 3, 4, 5}
(i) A × (B ∩ C) = { (0,3), (1,3) }
(ii) A × (B ∪ C) = { (0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5) }
Question 4. In each of the following cases, find a and b:
(i) (2a + b, a − b) = (8, 3)
(ii) (a/4, a − 2b) = (0, 6 + b)
Answer 4. (i) 2a + b = 8
a − b = 3
Solving,
a = 11/3, b = 2/3
(ii)
a/4 = 0 ⟹ a = 0
a − 2b = 6 + b
Substituting a = 0:
−2b = 6 + b
b = −2
Question 5. Given
A = {1, 2, 3, 4, 5}
S = { (x, y) : x ∈ A, y ∈ A }
Find the ordered pairs which satisfy the condition given below:
(i) x + y = 5
(ii) x + y < 5
(iii) x + y > 8
Answer 5.
Given:
A = {1, 2, 3, 4, 5}
S = { (x, y) : x ∈ A, y ∈ A }
(i) x + y = 5
The ordered pairs satisfying the condition are:
{ (1,4), (4,1), (2,3), (3,2) }
(ii) x + y < 5
The ordered pairs satisfying the condition are:
{ (1,1), (1,2), (2,1), (1,3), (3,1), (2,2) }
(iii) x + y > 8
The ordered pairs satisfying the condition are:
{ (4,5), (5,4), (5,5) }
Question 6. Given R = {(x, y) : x, y ∈ W, x² + y² = 25}, find the domain and range of R.
Answer 6. Since x, y ∈ W (whole numbers),
W = {0, 1, 2, 3, …}
Given equation:
x² + y² = 25
Checking whole number solutions:
0² + 5² = 25 → (0, 5)
3² + 4² = 25 → (3, 4)
4² + 3² = 25 → (4, 3)
5² + 0² = 25 → (5, 0)
So,
R = {(0,5), (3,4), (4,3), (5,0)}
Domain of R:
{0, 3, 4, 5}
Range of R:
{0, 3, 4, 5}
Final Answer:
Domain = {0, 3, 4, 5}
Range = {0, 3, 4, 5}
In set notation, the order of elements does not matter. So:
{0,3,4,5} = {5,4,3,0}
Both represent the same set
Question 7. If R₁ = {(x, y) | y = 2x + 7, where x ∈ R and −5 ≤ x ≤ 5}
is a relation, find the domain and range.
Answer 7. Given
y = 2x + 7
Domain:
Since x ∈ R and −5 ≤ x ≤ 5,
Domain = [−5, 5]
Range:
For y = 2x + 7:
When x = −5,
y = 2(−5) + 7 = −10 + 7 = −3
When x = 5,
y = 2(5) + 7 = 10 + 7 = 17
So,
Range = [−3, 17]
Final Answer:
Domain = [−5, 5]
Range = [−3, 17]
Question 8. If R₂ = { (x, y) | x and y are integers and x² + y² = 64 }
is a relation, then find R₂.
Answer 8.
Given:
x² + y² = 64, where x, y ∈ Z
Since the sum of squares of two integers is 64,
For x = 0,
y² = 64 ⟹ y = ±8
For x = ±8,
y² = 0 ⟹ y = 0
Hence,
R₂ = { (0, 8), (0, −8), (8, 0), (−8, 0) }
Question 9. If R₃ = { (x, |x|) | x is a real number } is a relation, then find the domain and range of R₃.
Answer 9. Given:
R₃ = { (x, |x|) | x is a real number }
Clearly,
Domain of R₃ = R
Range of R₃ = [0, ∞)
Question 10. Is the given relation a function? Give reason for your answer.
(i) h = { (4, 6), (3, 9), (−11, 6), (3, 11) }
(ii) f = { (x, x) | x is a real number }
(iii) g = { (n, 1/n) | n is a positive integer }
(iv) s = { (n, n²) | n is a positive integer }
(v) t = { (x, 3) | x is a real number }
Answer 10.
(i) h is not a function
Since 3 has two images, 9 and 11.
(ii) f is a function
Every element of the domain has a unique image.
(iii) g is a function
For every positive integer n, there is a unique image 1/n.
(iv) s is a function
Square of any integer is unique.
(v) t is a function
Every real number is mapped to the constant value 3.
Question 11. If f and g are real functions defined by
f(x) = x² + 7
g(x) = 3x + 5
find each of the following:
(i) f(3) + g(−5)
(ii) f(1/2) × g(14)
(iii) f(−2) + g(−1)
(iv) f(t) − f(t − 2)
(v) [ f(t) − f(5) ] / (t − 5), t ≠ 5
Answer 11.
Given:
f(x) = x² + 7
g(x) = 3x + 5
(i)
f(3) + g(−5)
= (3² + 7) + [3(−5) + 5]
= (9 + 7) + (−15 + 5)
= 16 − 10
= 6
(ii)
f(1/2) × g(14)
= [(1/2)² + 7] × [3(14) + 5]
= (1/4 + 7) × (42 + 5)
= (29/4) × 47
= 1363/4
(iii)
f(−2) + g(−1)
= [ (−2)² + 7 ] + [ 3(−1) + 5 ]
= (4 + 7) + (−3 + 5)
= 11 + 2
= 13
(iv)
f(t) − f(t − 2)
= (t² + 7) − [ (t − 2)² + 7 ]
= t² − (t² − 4t + 4)
= 4t − 4
(v)
[ f(t) − f(5) ] / (t − 5)
= [ (t² + 7) − (25 + 7) ] / (t − 5)
= (t² − 25) / (t − 5)
= t + 5, t ≠ 5
Question 12. Let f and g be real functions defined by
f(x) = 2x + 1
g(x) = 4x − 7
(i) For what real numbers, f(x) = g(x)?
(ii) For what real numbers, f(x) < g(x)?
Answer 12. (x) = 2x + 1
g(x) = 4x − 7
(i) f(x) = g(x)
2x + 1 = 4x − 7
⇒ −2x = −8
⇒ x = 4
Hence, the required real number is x = 4.
(ii) f(x) < g(x)
2x + 1 < 4x − 7
⇒ −2x < −8
⇒ x > 4
Hence, the required real numbers are x > 4.
Question 13. If f and g are two real valued functions defined as
f(x) = 2x + 1
g(x) = x² + 1
find:
(i) f + g
(ii) f − g
(iii) f·g
(iv) f / g
Answer 13. Given:
f(x) = 2x + 1
g(x) = x² + 1
(i) (f + g)(x)
= f(x) + g(x)
= (2x + 1) + (x² + 1)
= x² + 2x + 2
(ii) (f − g)(x)
= f(x) − g(x)
= (2x + 1) − (x² + 1)
= 2x − x²
(iii) (f·g)(x)
= f(x)g(x)
= (2x + 1)(x² + 1)
= 2x³ + x² + 2x + 1
(iv) (f / g)(x)
= f(x) / g(x)
= (2x + 1) / (x² + 1)
Question 14. Express the following function as a set of ordered pairs and determine their range:
f : X → R,
f(x) = x³ + 1, where
X = {−1, 0, 3, 9, 7}
Answer 14. Given:
f(x) = x³ + 1
X = {−1, 0, 3, 9, 7}
Evaluate f(x) for each x ∈ X:
For x = −1
f(−1) = (−1)³ + 1 = −1 + 1 = 0
For x = 0
f(0) = 0³ + 1 = 1
For x = 3
f(3) = 27 + 1 = 28
For x = 9
f(9) = 729 + 1 = 730
For x = 7
f(7) = 343 + 1 = 344
Set of ordered pairs:
{ (−1, 0), (0, 1), (3, 28), (7, 344), (9, 730) }
Range:
{ 0, 1, 28, 344, 730 }
Question 15. Find the values of x for which the functions
f(x) = 3x² − 1
and g(x) = 3 + x are equal.
Answer 15. Given:
f(x) = 3x² − 1
g(x) = 3 + x
Since f(x) = g(x),
3x² − 1 = 3 + x
⇒ 3x² − x − 4 = 0
Factorising:
(3x − 4)(x + 1) = 0
⇒ 3x − 4 = 0 or x + 1 = 0
⇒ x = 4/3 or x = −1
Hence, the required values of x are −1 and 4/3.