tn 10th maths ex 1.2

 

Q1.
Let A = {1, 2, 3, 7} and B = {3, 0, -1, 7}, which of the following are relation from A to B ?
(i) R1 = {(2, 1), (7, 1)}
(ii) R2 = {(-1, 1)}
(iii) R3 = {(2, -1), (7, 7), (1, 3)}
(iv) R4 = {(7,-1), (0, 3), (3, 3), (0, 7)}
(i) A = {1, 2, 3, 7}, B = {3, 0,-1, 7}
Solution:

R1 = {(2,1), (7,1)}

It is not a relation there is no element as 1 in B.

(ii) R2 = {(-1, 1)}

It is not [∵ -1 ∉ A, 1 ∉ B]
(iii) R3 = {(2, -1), (7, 7), (1, 3)}
It is a relation.
R4 = {(7,-1), (0, 3), (3, 3), (0, 7)}
It is also not a relation. [∵ 0 ∉ A]

 Q 2.

Let A = {1, 2, 3, 4, ….., 45} and R be the relation defined as “is square of ” on A. Write R as a subset of A × A. Also, find the domain and range of R.
Answer:
A = {1,2, 3, 4 . . . . 45}
The relation is defined as “is square of’
R = {(1,1) (2, 4) (3, 9)
(4, 16) (5,25) (6, 36)}
Domain of R = {1, 2, 3, 4, 5, 6}
Range of R = {1, 4, 9, 16, 25, 36}

Q 3.
A Relation R is given by the set {(x, y) /y = x + 3, x ∈ {0, 1, 2, 3, 4, 5}}. Determine its domain and range.
answer:
x = {0, 1, 2, 3, 4, 5}
y = x + 3

Y=

(0+3)=3

(1+3)=4

(2+3)=5

(3+3)=6

(4+3)=7

(5+3)=8






⇒ y = {3, 4, 5, 6, 7, 8}
R = {(x, y)}
= {(0, 3),(1, 4),(2, 5),(3, 6), (4, 7), (5, 8)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {3, 4, 5, 6, 7, 8}

Q 4.
Represent each of the given relation by (a) an arrow diagram, (b) a graph and (c) a set in roster form, wherever possible.
(i) {(x, y)|x = 2y,x ∈ {2, 3, 4, 5},y ∈ {1, 2, 3, 4)
(ii) {(x, y)|y = x + 3, x, y are natural numbers <10}
answer:




q5
A company has four categories of employees given by Assistants (A), Clerks (C), Managers (M) and an Executive Officer (E). The company provide ₹10,000, ₹25,000, ₹50,000 and ₹1,00,000 as salaries to the people who work in the categories A, C, M and E respectively. If A1, A2, A3, A4 and As were Assistants; C1, C2, C3, C4 were Clerks; M1, M2, M3 were managers and E1,E2 were Executive officers and if the relation R is defined by xRy, where x is the salary given to person y, express the relation R through an ordered pair and an arrow diagram.

answer






Comments

Post a Comment

Popular posts from this blog

tn10th maths ex:1.5

( v) f(x) = 4x 2  – 1, g(x) = 1 + x fog(x) = f(g(x)) = f(1 + x) = 4(1 + x) 2  – 1 = 4(1 + x 2  + 2x) – 1 = 4 + 4x 2  + 8x – 1 = 4x 2  + 8x + 3 ……………. (1) gof(x) = g(f(x)) = g(4x 2  – 1) = 1 + 4x 2  – 1 = 4x 2  …………….. (2) (1) ≠ (2) ∴ fog(x) ≠ gof(x) Question 2. Find the value of k, such that f o g = g o f (i) f(x) = 3x + 2, g(x) = 6x – k Answer: f(x) = 3x + 2 ;g(x) = 6x – k fog = f[g(x)] = f (6x – k) = 3(6x – k) + 2 = 18x – 3K + 2 g0f= g [f(x)] = g (3x + 2) = 6(3x + 2) – k = 18x + 12 – k But given fog = gof. 18x – 3x + 2 = 18x + 12 – k -3k + 2 = 12 – k -3 k + k = 12-2 -2k = 10 k =  − 10 2  = -5 The value of k = -5 (ii) f(x) = 2x – k, g(x) = 4x + 5 Answer: f(x) = 2x – k ; g(x) = 4x + 5 fog = f[g(x)] = f(4x + 5) = 2(4x + 5) – k = 8x + 10 – k gof = g [f(x)] = g(2x – k) = 4(2x – k) + 5 = 8x – 4k + 5 But fog = gof 8x + 10 – k = 8x – 4k + 5 -k + 4k = 5 – 10 3k = -5 k =  − 5 3 The value of k =  − 5 3 Q 4. (i) If f (x) = x 2  – 1,...
Read more

ex:1.4

Image
  Q1. Determine whether the graph given below represent functions. Give reason for your answers concerning each graph. answer:(i) (ii)&(iii) (iv) (i) It is not a function. The graph meets the vertical line at more than one points. (ii) It is a function as the curve meets the vertical line at only one point. (iii) It is not a function as it meets the vertical line at more than one points. (iv) It is a function as it meets the vertical line at only one point. Q 2. Let f :A → B be a function defined by f(x) =  x 2  – 1, Where A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9}. Represent f by (i) set of ordered pairs; (ii) a table; (iii) an arrow diagram; (iv) a graph answer : f: A → B A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9} f(x)=x/2-1 f(2)=2/2-1=0 f(4)=4/2-1=1 f(6)=6/2-1=2 f(10)=10/2-1=4 f(12)=12/2-1=5 (i) Set of ordered pairs = {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)} (ii) a table (iii) an arrow diagram;
Read more