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VOOZH | about |
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VOOZH | about |
Relation and Function are two ways of establishing links between two sets in mathematics. Relation and Function in maths are analogous to the relation that we see in our daily lives i.e., two persons are related by the relation of father-son, mother-daughter, brother-sister, and many more. On a similar pattern, two numbers can be related to each other as one number is the square of another number and many more. A function is a special kind of relation that is defined as a unique relation between two mathematical entities.
The chapter Relation and Function consists of 8 Marks Weightage in Board Exams.
It is predicted according to previous year's papers the chapter Relation and Function will be asked in 2 Marks which will be seen in MCQ.
And rest the marks can be distributed in any section of the Paper.
Solution:
A={1 ,2 ,3,...,13, 14}
R={(x,y): 3x-y=0}
Therefore R={(1,3),(2,6),(3,9),(4,12)}
R is not reflexive since (1,1),(2,2),(3,3),...,(14,14)∉R
Also, R is not reflexive since (1,3)∈R, but (3,1)∉R.[since 3(3)-1≠0]
Also, R is not transitive as (1,3), (3,9)∈R, but (1,9)∉R.[since 3(1)-9≠0]
Hence, R is not reflexive, nor symmetric nor transitive.
Solution:
R={(x, y): y=x+5 and x<4}={(1,6), (2, 7), (3,8)}
It is seen that (1, 1)∉R. Therefore, R is not reflexive.
(1, 6)∈R. But, (6, 1)∉R so, R is not symmetric.
Now, since their is no pair in R such that (x, y) and (y, z)∈R, so (x, z) can not belong to R. Therefore, R is not transitive.
So, we can conclude that R is neither reflexive, nor symmetric, nor transitive.
Solution:
A={1, 2, 3, 4, 5, 6}
R={(x, y): y is divisible by x}
We know that any number is always divisible by itself.⇒ (x, x)∈ R. Therefore, R is reflexive.
We see, (2, 4)∈R [4 is divisible by 2]. But, (4, 2)∉R [2 is not divisible by 4]. Therefore, R is not symmetric.
Let's assume (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y. Therefore, z is divisible by x.⇒ (x, z) ∈ R. Therefore, R is transitive.
Solution:
R={(x, y): x-y is an integer}
For every x ∈ Z, (x, x) ∈ R [x-x=0 which is an integer]. Therefore, R is reflexive.
For every x, y ∈ Z if (x, y) ∈ R, then x-y is an integer. ⇒-(x-y) is also an integer. ⇒ (y-x) is also an integer. Therefore, R is symmetric.
Let's assume, (x, y) and (y, z) ∈ R, where x, y and z ∈ Z. ⇒ (x-y) and (y-z) are integers. ⇒ x-z=(x-y)+(y-z) is an integer. ⇒ (x, z) ∈ R. Therefore, R is transitive.
(a)R={(x,y) : x and y work at the same place.
Solution:
We can see (x,x) ∈ R .Therefore, R is reflexive.
If (x,y) ∈ R, then x and y work at the same place. So, (y,x) ∈ R. Therefore, R is symmetric.
Let, (x,y), (y,z) ∈ R.
⇒ x and y work mat the same place and y and z work at the same place.
⇒ x and z work at the same place.
Therefore, R is transitive.
(b) R={(x,y): x and y live in the same locality}.
Solution:
We can see (x,x) ∈ R. Therefore, R is reflexive.
If (x,y) ∈ R, then x and y live in same locality. So, (y,x) ∈ R. Therefore, R is symmetric.
Let (x,y) ∈ R and (y,z) ∈ R. So, x, y and z live in the same locality. So, (x,z) ∈ R. Therefore, R is transitive.
(c) R={(x,y): x is exactly 7 cm taller than y}.
Solution:
(x,x)∉R since, human being can not be taller than himself. So, R is not reflexive.
Let (x,y) ∈ R ,then x is exactly 7 cm taller than y. Then, y is not taller than x. Therefore, R is not symmetric.
Let (x,y), (y,z) ∈ R, then x is exactly 7 cm taller than y and y is exactly 7 cm taller than z which means x is 14 cm taller than z. So, (x,z)∉R . Therefore, R is not transitive.
(d) R={(x,y): x is wife of y}.
Solution:
(x,x) ∉ R. Since, x can not be the wife of herself. Therefore, R is not reflexive.
Let (x,y) ∈ R, then x is the wife of y. So, y is not the wife of x ,i.e., (y,x) ∉ R. Therefore, R is not symmetric.
Let (x,y), (y,z) ∈ R, then x is the wife of y and y is the wife of z which is not possible. So, R can not be transitive.
(e) R={(x,y): x is father of y}.
Solution:
(x,x) ∉ R. Since, x can not be the father of himself. Therefore, R is not reflexive.
Let (x,y) ∈ R, then x is the father of y. So, y can not be the father of x. So, (y,x) ∉ R. Therefore, R is not symmetric.
Let (x,y), (y,z) ∈ R, then x is the father of y and y is the father of z which means x is the grandfather of z. So, So, (x, z) ∉ R. Therefore, R is not transitive.
Solution:
It can be observed that (½, ½) ∉ R, since ½>(½)2 =¼. Therefore, R is not reflexive.
(1,4) ∈ R as 1<42 .But, (4,1) ∉ R. Therefore, R is not symmetric.
(3,2), (2,1.5) ∈ R. But, 3> (1.5)2 =2.25 . So, (3,1.5) ∉ R. Therefore, R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
Solution:
Let the set {1, 2, 3, 4, 5, 6} be named A.
R={(1,2), (2, 3), (3, 4), (4, 5
), (5, 6)}We can see (x, x) ∉ R. Since, x ≠ x+1. Therefore, R is not reflexive.
It is observed that (1,2) ∈ R but, (2,1) ∉ R. Therefore, R is not symmetric.
We can see, (1,2), (2, 3) ∈ R, but (1,3) ∉ R. Therefore, R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
Solution:
Clearly, (a,a) ∈ R as a=a. Therefore, R is reflexive.
(2,4) ∈ R (as 2<4) but (4,2) ∉ R as 4 is greater than 2. Therefore, R is not symmetric.
Let (a,b), (b,c) ∈ R. Then, a≤ b and b≤ c.
⇒a ≤ c.
(a, c) ∈ R. Therefore, R is transitive.
Hence, R is reflexive and transitive but not symmetric.
Solution:
It is observed that (½, ½) ∉ R as ½ > (½)3 =(1/8). Therefore, R is not reflexive.
(1,2) ∈ R(as 1<8) but, (2,1) ∉ R. Therefore, R is not symmetric.
We have, (3, 3/2), (3/2, 6/5) ∈ R but, (3, 6/5) ∉ R. Therefore, R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
Solution:
Let the set {1, 2, 3} be named A.
It is seen that, (1, 1), (2,2), (3,3)∉ R. Therefore, R is not reflexive.
As (1, 2) ∈ R and (2, 1) ∈ R. Therefore, R is symmetric.
However, (1, 1)∉ R. Therefore, R is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.
Solution:
Set A is the set of all books in the library of a college.
R={(x,y):x and y have the same number of pages}
R is reflexive since (x,x) ∈ R as x and x have the same number of pages.
Let (x,y) ∈ R
⇒x and y have the same number of pages
⇒y and x have the same number of pages.
⇒(y,x)∈ R
Therefore , R is symmetric.
Let (x,y) ∈ R and (y,z)∈ R.
⇒x and y have the same number of pages and y and z have the same number of pages.
⇒x and z have the same number of pages.
⇒(x,z) ∈ R
Therefore, R is transitive.
Hence, R is an equivalence relation.
Solution:
A={ 1,2,3,4,5}
R={(a,b):|a-b| is even }
It is clear that for any element a∈ A, we have |a-a|=0 (which is even).
Therefore, R is reflexive.
Let (a,b) ∈ R.
⇒|a-b| is even.
⇒|-(a-b)|=|b-a| is also even.
⇒(b,a)∈ R
Therefore, R is symmetric.
Now , let (a,b)∈ R and (b,c)∈ R.
⇒|a-b| is even and |b-c| is even.
⇒(a-b) is even and (b-c) is even.
⇒(a-c)=(a-b)+(b-c) is even. [Sum of two even integers is even]
⇒|a-c| is even.
Therefore, R is transitive.
Hence, R is an equivalence relation.
All elements of the set {1,2,3} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.
Similarly, all elements of the set {2,4} are related to each other as all the elements of this subset are even.
Also, no element of the subset {1,3,5} can be related to any element of {2,4} as all elements of {1,3,5} are odd and all elements of {2,4} are even . Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.
(i) R={(a,b):|a-b| is a multiple of 4}
(ii) R={(a,b):a=b}
is an equivalence relation. Find the set of all elements related to 1 in each case.
A={x∈ Z:0<=x<=12}={0,1,2,3,4,5,6,7,8,9,10,11,12}
(i) R={(a,b):|a-b| is a multiple of 4}
For any element a ∈A , we have (a,a)∈R as |a-a|=0 is a multiple of 4.
Therefore, R is reflexive.
Now , let (a,b)∈R ⇒|a-b| is a multiple of 4.
⇒|-(a-b)|=|b-a| is a multiple of 4.
⇒(b,a)∈R
Therefore, R is symmetric.
Let (a,b) ,(b,c) ∈ R.
⇒|a-b| is a multiple of 4 and |b-c| is a multiple of 4.
⇒(a-b) is a multiple of 4 and (b-c) is a multiple of 4.
⇒(a-c)=(a-b)+(b-c) is a multiple of 4.
⇒|a-c| is a multiple of 4.
⇒(a,c)∈R
Therefore, R is transitive.
Hence, R is an equivalence relation .
The set of elements related to 1 is {1,5,9} since |1-1|=0 is a multiple of 4,
|5-1|=4 is a multiple of 4,and
|9-1|=8 is a multiple of 4.
(ii) R={(a,b):a=b}
For any element a∈A, we have (a,a) ∈ R , since a=a.
Therefore, R is reflexive.
Now , let (a,b)∈R.
⇒a=b
⇒b=a
⇒(b,a)∈R
Therefore, R is symmetric.
Now, let (a,b)∈R and (b,c)∈R.
⇒a=b and b=c
⇒a=c
⇒(a,c)∈R
Therefore, R is transitive.
Hence, R is an equivalence relation.
The elements in R that are related to 1 will be those elements from set A which are equal to 1.
Hence, the set of elements related to 1 is {1].
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii)Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.
Solution:
(i) Let A ={5,6,7}.
Define a relation R on A as R ={(5,6),(6,5)}.
Relation R is not reflexive as (5,5) , (6,6),(7,7)∉R.
Now, as (5,6)∈R and also (6,5)∈R, R is symmetric.
⇒(5,6),(6,5)∈R , but (5,5)∉R
Therefore , R is not transitive.
Hence, relation R is symmetric but not reflexive or transitive.
(ii) Consider a relation R in R defined as:
R= {(a,b): a<b}
For any a∈R, we have (a,a)∉R since a cannot be strictly less than itself . In fact a=a.
Therefore, R is not reflexive. Now,
(1,2)∈R (as 1<2)
But, 2 is not less than 1.
Therefore, (2,1)∉R
Therefore, R is not symmetric.
Now, let (a,b),(b,c)∈R.
⇒a<b and b<c
⇒a<c
⇒ (a,c)∈R
Therefore, R is transitive.
Hence, relation R is transitive but not reflexive and symmetric.
(iii) Let A={4,6,8}.
Define a relation R on A as:
A={(4,4),(6,6),(8,8),(4,6),(6,4),(6,8),(8,6)}
Relation R is reflexive since for every a∈A , (a,a)∈R i.e.,(4,4),(6,6),(8,8)∈R.
Relation R is symmetric since (a,b)∈R⇒(b,a)∈R for all a,b∈R.
Relation R is not transitive since (4,6),(6,8)∈R , but (4,8)∉R.
Hence, relation R is reflexive and symmetric but not transitive.
(iv) Define a relation R in R as:
R={ (a,b): a3 ≥ b3 }
Clearly (a,a)∈R as a^3=a^3
Therefore, R is reflexive.
Now, (2,1)∈R(as 23>=13)
But, (1,2)∉ R (as 13< 23)
Therefore, R is not symmetric
Let (a,b), (b,c) ∈ R.
⇒a3 >= b3 and b3 >=c3
⇒a3>=c3
⇒(a, c)∈R
Therefore, R is transitive.
Hence, relation R is reflexive and transitive but not symmetric.
(v) Let A={-5,-6},
Define a relation R on A as:
R={(-5,-6), (-6,-5), (-5,-5)}
Relation R is not reflexive as (-6,-6)∉ R.
Relation R is symmetric as (-5,-6)∈ R and (-6,-5)∈ R.
It is seen that (-5,-6),(-6,-5)∈R . Also , (-5,-5)∈R.
Therefore, the relation R is transitive .
Hence, relation R is symmetric and transitive but not reflexive.
Solution:
We can see (P, P) ∈ R since, the distance of point P from the origin is always the same as the distance of the same point P from the origin. Therefore, R is reflexive.
Let (P,Q)∈ R.
⇒The distance of point P from the origin is the same as the distance of point Q from the origin.
⇒The distance of point Q from the origin is the same as the distance of point P from the origin.
So, (Q,P) ∈ R. Therefore, R is symmetric.
Let (P,Q), (Q,S) ∈ R.
⇒The distance of point P from the origin is the same as the distance of point Q from the origin and also, the distance of point P from the origin is the same.
⇒ The distance of points P and S from the origin is the same.
⇒(P, S) ∈ R. Therefore, R is transitive.
Therefore, R is an equivalence relation.
The set of all points related to P ≠ (0, 0) will be those points whose distance from the origin is the same as the distance of point P from the origin.
In other words, if O(0 0) is the origin and OP = k, then the set of all points related to P is at the same as k from the origin. Hence, this set of points forms a circle with the centre as the origin and the circle passes through point P.
Solution:
R is reflexive since, every triangle is similar to itself.
If (T1, T2) ∈ R, then T1 is similar to T2. So, (T2, T1) ∈ R. Therefore, R is symmetric.
Let (T1, T2), (T2, T3) ∈ R, then T1 is similar to T2 and T2 is similar to T3. So, T1 is also similar to T3. Therefore, (T1, T3) ∈ R so, R is transitive.
Thus, R is an equivalence relation.
We observe,
(3/6)=(4/8)=(5/10)=1/2
Therefore, the corresponding sides of triangles T1 and T3 are in the same ratio. Then, triangle T1 is similar to triangle T3.
Hence, T1 is related to T3.
Solution:
R is reflexive since(P1, P2) ∈ R as the same polygon has the same number of sides with itself.
Let (P1, P2) ∈ R, then P1 and P2 have the same number of sides. So, (P2, P1) ∈ R. Therefore, R is symmetric.
Let (P1, P2), (P2, P3) ∈ R, then P1 and P2 have the same number of sides. Also, P2 and P3 have the same number of sides. So, P1 and P3 have the same number of sides ,i.e., (P1, P3) ∈ R. Therefore, R is transitive.
Hence, R is an equivalence relation.
The elements in A related to the right-angled triangle (T) with sides 3, 4 and 5 are those polygons which have 3 sides(since, T is a polygon with 3 sides).
Hence, the set of all elements in A related to triangle T is the set of all triangles.
Solution:
R is reflexive as any line L1 is parallel to itself i.e., (L1, L2) ∈ R.
Let (L1, L2) ∈ R, then L1 is parallel to L2. So, (L2, L1) ∈ R. Therefore, R is symmetric.
Let (L1, L2), (L2, L3) ∈ R, then L1 is parallel to L2 and L2 is parallel to L3. So, L1 is parallel to L3. Therefore, R is transitive.
Hence, R is an equivalence relation.
The set of all lines related to the line y=2x +4 is the set of all lines that are parallel to the line y=2x+4.Slope of the line is m=2.
It is known that parallel lines have the same slopes. The line parallel to the given line is of the form y=2x +c, where c ∈ R.
Hence, the set of all lines related to the given line is given by y=2x +c, where c ∈ R.
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is equivalence relation.
Solution:
It is seen that (a,a) ∈ R, for every a ∈ {1, 2, 3, 4}. Therefore, R is reflexive.
It is seen that (1,2) ∈ R but (2,1) ∉ R. Therefore, R is not symmetric.
Also, it is observed that (a,b), (b,c) ∈ R⇒ (a,c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}. Therefore, R is transitive.
Hence, R is reflexive and transitive but not symmetric. The correct answer is B.
(A) (2,4) ∈ R
(B) (3,8) ∈ R
(C) (6,8)
(D) (8,7) ∈ R
Solution:
Since b>a, (2,4) ∉ R also, as 3≠8-2, (3,8)∉R and as 8≠7-2. Therefore, (8,7)∉R
Consider (6,8). We have 8>6 and also, 6=8-2. Therefore, (6,8) ∈ R.
The correct answer is C.
