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Published on: 19/09/2019
Relations and Functions
Download Tamil Nadu 10th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Maths Test1.
Let A = {0, 1, 2, 3} and B = {1, 3, 5, 7, 9} be two sets. Let f: A \(\rightarrow\)B be a function given by f(x) = 2x + 1. Represent this function as a graph.
2.
Let A = {0, 1, 2, 3} and B = {1, 3, 5, 7, 9} be two sets. Let f : A \(\rightarrow\)B be a function given by f(x) = 2x + 1. Represent this function as an arrow .
3.
State whether the graph represent a function. Use vertical line test.
4.
If f(x) = 2x + 3, g(x) = 1 - 2x and h(x) = 3x. Prove that f o(g o h) = (f o g) o h.
5.
Find k if f o f(k) = 5 where f(k) = 2k - 1.
6.
If f(x) = 3x - 2, g(x) = 2x + k and if f o g = f o f, then find the value of k..
7.
A relation ‘f’ \(X \rightarrow Y\) is defined by f(x) = x2 - 2 where x \(\in \) {-2, -1, 0, 3} and Y = R
(i) List the elements of f
(ii) Is f a function?
8.
Let X = {1, 2, 3, 4} and Y = {2, 4, 6, 8,10} and R = {(1, 2),(2, 4),(3, 6),(4, 8)} Show that R is a function and find its domain, co-domain and range?
9.
If A x B = {(3,2), (3, 4), (5,2), (5, 4)} then find A and B.
10.
Let A = {x \(\in \) N| 1 < x < 4}, B = {x \(\in \) W| 0 ≤ x < 2) and C = {x \(\in \) N| x < 3} Then verify that
(i) A x (B U C) = (A x B) U (A x C)
(ii) A x (B ∩ C) = (A x B) ∩ (A x C)
1.
A Graph f = {(x, f(x) / x \(\epsilon \) A}
{(0, 1), (1, 3), (2, 5), (3, 7)}
2.
An arrow diagram
3.
It is not a function as the vertical line PQ cuts the graph at two points
4.
f(x) = 2x + 3, g(x) = 1 - 2x, h(x) = 3x
Now, (f o g)(x) = f(g(x)) = f(1 - 2x) = 2(1 - 2x) + 3 = 5 - 4x
Then, (f o g) o h(x) = (f o g)(3x) = 5 - 4(3x) = 5 - 12x..(1)
(g o h)(x) = g(h(x)) = g(3x) = 1 - 2(3x) = 1 - 6x
So, f o (g o h)(x) = f(1 - 6x) = 2(1 - 6x) + 3 = 5 - 12x...(2)
From (1) and (2), we get (f o g) oh = f o (g o h)
5.
f o f(k) = f(f(k))
= 2(2k - 1) -1 = 4k - 3
Thus, f o f(k) = 4k - 3
But, it is given that f o f(k) = 5
Therefore 4k - 3 = 5 ⇒ k = 2
6.
f(x) = 3 x -2, g(x) = 2x + k
f o g = f(g(x)) = f(2x + k) = 3(2x + k) - 2 = 6x + k - 2
Thus, f o g(x) = 6x + 3k - 2
g o f(x) = g(3x - 2) = 2(3x - 2) + k
Thus, g o f(x) = 6x - 4 + k
Given that f o g = g o f
Therefore, 6x + 3k-2 = 6z - 4 + k
6x - 6x + 3k - k = -4 + 2 ⇒ -1
7.
f(x) = x2 - 2 where x \(\in \){ -2, -1, 0, 3}
(i) f( -2) = ( -2)2 - 2 = 2; f( -1) = ( -1)2 - 2 = -1
f(0) = (0)2 - 2 = - 2 ; f(3) = (3)2 - 2 = 7
Therefore, f = {(-2, 2), (-1, -1), (0, -2), (3, 7)}
(ii) We note that each element in the domain of f has a unique image. Therefore f is a function.
8.
Pictorial representation of R . From the diagram, we see that for each x \(\in \) X, there exists only one y \(\in \) Y. Thus all elements in X have only one image in Y. Therefore R is a function Domain X = {1, 2, 3, 4}; Co-domain Y = {2, 3, 6, 8,10}; Range of f = {2, 4, 6, 8}.
9.
A x B = {(3,2), (3,4), (5,2), (5,4)}
We have A = {set of all first coordinates of elements of A x B}. Therefore, A = {3,5}
B = {set of all second coordinates of elements of A x B}. Therefore, B = {2,4}
Thus A = {3,5} and B = {2,4}.
10.
A = {x \(\in \) N| 1 < x < 4} = {2,3), B = {x \(\in \) W| 0 ≤ x < 2) = (0,1), C = {x \(\in \) N| x < 3} = (1,2)
(i) A x (B U C) = (A x B) U (A x C)
B U C = (0,1) U (1,2) = {0,1,2}
A x (B U C) = {2,3) x {0,1,2} = {(2,0),(2,1)(2,2)(3,0)(3,1),(3,2) ..(1)
A x B = {2,3} x {0,1} = {(2,0),(2,1),(3,0),(3,1)}
A x C = {2,3} x {1,2} = {(2,1),(2,2),(3,1)(3,2)}
(A x B) U (A x C) = {(2,0),(2,1),(3,0),(3,1)} U {(2,1),(2,2),(3,1),(3,2)}
= {(2,0),(2,1),(2,2),(3,0),(3,1),(3,2)} ...(2)
From (1) and (2), A x (B U C) = (A x B) U (A x C) is verified.
(ii) A x (B ∩ C) = (A x B) ∩ (A x C)
(B ∩ C) = {0,1} ∩ {1,2} = {1}
A x (B ∩ C) = {2,3} x {1} = {(2,1),(3,1)} .... (3)
A x B = {2,3} x {0,1} = {(2,0),(2,1),(3,0),(3,1)}
A x C = {2,3} x {1,2} = {(2,1),(2,2),(3,1),(3,2)
(A x B) ∩ (A x C) = {(2,0),(2,1),(3,0),(3,1)} ∩ {(2,1),(2,2),(3,1),(3,2)}
= {(2,1),(3,1)} .... (4)
From (3) and (4), A x (B ∩ C) = (A x B) ∩ (A x C) is verified.
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