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NCERT Exemplar Class 11 Maths Chapter 2 Relations and Functions Exercise Solutions Fill in the Blank Questions
Table of Contents
Solved Fill in the Blank Type Questions of NCERT Exemplar Relations and Functions Exercise
Question 36
Let f and g be two real functions given by
f = { (0, 1), (2, 0), (3, −4), (4, 2), (5, 1) }
g = { (1, 0), (2, 2), (3, −1), (4, 4), (5, 3) }
Then the domain of f · g is given by __________.
Answer 36
Given that:
f(x) = { (0, 1), (2, 0), (3, −4), (4, 2), (5, 1) }
g(x) = { (1, 0), (2, 2), (3, −1), (4, 4), (5, 3) }
Domain of f = { 0, 2, 3, 4, 5 }
Domain of g = { 1, 2, 3, 4, 5 }
So,
Domain of f · g
= Domain of f ∩ Domain of g
= { 2, 3, 4, 5 }
Hence, the answer is:
{ 2, 3, 4, 5 }
Question 37. Let
f = { (2, 4), (5, 6), (8, −1), (10, −3) }
and
g = { (2, 5), (7, 1), (8, 4), (10, 13), (11, 5) }
be two real functions.
Then match the following:
(a) f − g
(b) f + g
(c) f · g
(d) f / g
(i) { (2, 4/5), (8, −1/4), (10, −3/13) }
(ii) { (2, 20), (8, −4), (10, −39) }
(iii) { (2, −1), (8, −5), (10, −16) }
(iv) { (2, 9), (8, 3), (10, 10) }
Answer 37.
Given that:
f = { (2, 4), (5, 6), (8, −1), (10, −3) }
g = { (2, 5), (7, 1), (8, 4), (10, 13), (11, 5) }
Domain
The functions f − g, f + g, f · g, and f / g are defined on
Domain of f ∩ Domain of g
= { 2, 5, 8, 10 } ∩ { 2, 7, 8, 10, 11 }
= { 2, 8, 10 }
(i) f − g
(f − g)(2) = f(2) − g(2) = 4 − 5 = −1
(f − g)(8) = f(8) − g(8) = −1 − 4 = −5
(f − g)(10) = f(10) − g(10) = −3 − 13 = −16
∴ f − g = { (2, −1), (8, −5), (10, −16) }
(ii) f + g
(f + g)(2) = f(2) + g(2) = 4 + 5 = 9
(f + g)(8) = f(8) + g(8) = −1 + 4 = 3
(f + g)(10) = f(10) + g(10) = −3 + 13 = 10
∴ f + g = { (2, 9), (8, 3), (10, 10) }
(iii) f · g
(f · g)(2) = f(2) · g(2) = 4 · 5 = 20
(f · g)(8) = f(8) · g(8) = (−1) · 4 = −4
(f · g)(10) = f(10) · g(10) = (−3) · 13 = −39
∴ f · g = { (2, 20), (8, −4), (10, −39) }
(iv) f / g
(f / g)(2) = f(2) / g(2) = 4 / 5
(f / g)(8) = f(8) / g(8) = −1 / 4
(f / g)(10) = f(10) / g(10) = −3 / 13
∴ f / g = { (2, 4/5), (8, −1/4), (10, −3/13) }
Final Matching
(a) f − g ↔ (iii)
(b) f + g ↔ (iv)
(c) f · g ↔ (ii)
(d) f / g ↔ (i)