10th Maths Chapter 1
Relations and Functions
Unit Exercise 1
10th Maths Chapter 1 Unit Exercise 1, 10th Maths Chapter 1 Relations and Functions Exercise 1.6, 10th Maths Chapter 1 Relations and Functions Exercise 1.4, 10th Maths Chapter 1 Relations and Functions Exercise 1.5
Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.5
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10th Maths Chapter 1 video classes
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Class 10 Chapter 1 All Exercise Sums
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Class 10 Chapter 1 Example 1.1 to 1.5
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Class 10 Chapter 1 Example 1.6 to 1.10
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Class 10 Chapter 1 Example 1.11 to 1.15
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Class 10 Chapter 1 Example 1.16 to 1.20
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10th Maths Chapter 1 Exercise 1.6, 10th Maths Exercise 1.2 | 10th Maths Ex 1.2 | 10th Maths Chapter 1 Relations and Functions Exercise 1.2 | 10th Maths Chapter 1 Relations and Functions Exercise 1.1 | 10th Maths Chapter 1 Relations and Functions Exercise 1.3 | 10th Maths Chapter 1 Relations and Functions Exercise 1.4 | 10th Maths Chapter 1 Relations and Functions Exercise 1.5 | 10th Maths Chapter 1 Relations and Functions Exercise 1.6 | Class 10 Chapter 1 Relations and Functions Example 1.11 to 1.15 | Class 10 Chapter 1 Relations and Functions Example 1.6 to 1.10 | 10th Maths Example 1.6 | 10th Maths Example 1.7 | 10th Maths Example 1.8 | 10th Maths Example 1.9 | 10th Maths Example 1.10 | Class 10 Chapter 1 Relations and Functions Example 1.1 to 1.5 TN New Syllabus 10th Maths Chapter 1 Relations And Functions Example 1.1 1 Relations and Functions | 1.1 Introduction |1.2 Ordered Pair | 1.3 Cartesian Product | 1.4 Relations | 1.5 Functions | 1.6 Representation of Functions | 1.7 Types of functions | 1.8 Special cases of Functions | 1.9 Composition of Functions 1.10 Identifying the graphs of Linear, Quadratic, | Cubic and Reciprocal functions | TN (Samacheer) 10 Maths Relations and Functions New syllabus Ex 1.1 TN (Samacheer) 10 Maths New Syllabus Relations and Functions.
Answer:
(x^2 – 3x, y^2 + 4y) = (-2, 5)
x^2 – 3x = -2
x^2 – 3x + 2 = 0
(x – 2) (x – 1) = 0
x – 2 = 0 or x – 1 = 0
x = 2 or 1
y^2 + 4y = 5
y^2 + 4y – 5 = 0
(y + 5) (y – 1) = 0
y + 5 = 0 or y – 1 = 0
y = -5 or y = 1
The value of x = 2, 1
and 7 = -5, 1
2. The Cartesian product A × A has 9 elements among which (-1, 0) and (0, 1) are found. Find the set A and the remaining elements of A × A.
Solution:
A = {-1, 0, 1}, B = {1, 0, -1}
A × B = {(-1, 1), (-1, 0), (-1, -1), (0, 1), (0, 0), (0, -1), (1, 1), (1, 0), (1, -1)}.
4 x<1.
Find
(i) f(0) (ii)f (3) (iii) f(a + 1) in terms of a.(Given that a > 0)
Answer:
f(x) = √x−1 ; f(x) = 4
(i) f(0) = 4
(ii) f(3) = √3−1= √2
(iii) f(a + 1) = √a+1−1 = √a
A = {9, 10, 11, 12, 13, 14, 15, 16, 17}
f: A → N
f(n) = the highest prime factor of n ∈ A
f = {(9, 3), (10, 5), (11, 11), (12, 3), (13, 13), (14, 7), (15, 5), (16, 2), (17, 17)}
Range = {3, 5, 11, 13, 7, 2, 17}
= {2, 3, 5, 7, 11, 13, 17}
Answer:
Domain of f(x) = {-1, 0, 1}
Solution:
f(x) = x^2
g(x) = 3x
h(x) = x – 2
(fog)oh = x – 2
LHS = fo(goh)
fog = f(g(x)) = f(3x) = (3x)^2 = 9x^2
(fog)oh = (fog) h(x) = (fog) (x – 2)
= 9(x – 2)^2 = 9(x^2 – 4x + 4)
= 9x^2 – 36x + 36 ……………. (1)
RHS = fo(goh)
(goh) = g(h(x)) = g(x – 2)
= 3(x – 2) = 3x – 6
fo(goh) = f(3x – 6) = (3x – 6)2
= 9x^2 – 36x + 36 ………….. (2)
(1) = (2)
LHS = RHS
(fog)oh = fo(goh) is proved.
B = {1, 2, 3, 4}
C = {5,6}
D = {5,6, 7,8}
A × C = {1,2} × {5,6}
= {(1,5) (1,6) (2, 5) (2, 6)}
B × D = {1,2, 3, 4} × {5, 6, 7, 8}
= {(1,5) (1,6) (1,7) (1,8)
(2, 5) (2, 6) (2,7) (2, 8)
(3, 5) (3, 6) (3, 7) (3, 8)
(4, 5) (4, 6) (4, 7) (4, 8)}
∴ A × C ⊂ B × D
Hence it is verified
f(f(x)) = – 1/x, Provided x ≠ 0.
Answer:
(i) Calculate the value of gg [latex]\frac { 1 }{ 2 } [/latex]
(a) Write an expression for gf (x) in its simplest form.
f(x) = 6x + 8 ; g(x) = x−2/3
(i) f (x) = 2x+1/x−9
If the denominator vanishes when x = 9
So f(x) is not defined at x = 9
∴ Domain is x ∈ [R – {9}]
(ii) if p(x) = =−5/4x^2+1
p(x) is defined for all values of x. So domain is x ∈ R.
(iii) g(x) = √x−2
When x < 2 g(x) becomes complex. But given “g” is real valued function.
So x > 2
Domain x ∈ (2, α)
(iv) h (x) = x + 6
For all values of x, h(x) is defined. Hence domain is x ∈ R.
Class 10 Chapter 1 Relations and Functions Example 1.11 to 1.15 | Class 10 Chapter 1 Relations and Functions Example 1.6 to 1.10 | 10th Maths Example 1.6 | 10th Maths Example 1.7 | 10th Maths Example 1.8 | 10th Maths Example 1.9 | 10th Maths Example 1.10 | Class 10 Chapter 1 Relations and Functions Example 1.1 to 1.5 TN New Syllabus 10th Maths Chapter 1 Relations And Functions Example 1.1 1 Relations and Functions | 1.1 Introduction |1.2 Ordered Pair | 1.3 Cartesian Product | 1.4 Relations | 1.5 Functions | 1.6 Representation of Functions | 1.7 Types of functions | 1.8 Special cases of Functions | 1.9 Composition of Functions 1.10 Identifying the graphs of Linear, Quadratic, | Cubic and Reciprocal functions | TN (Samacheer) 10 Maths Relations and Functions New syllabus Ex 1.1 TN (Samacheer) 10 Maths New Syllabus Relations and Functions
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