Exercise 2.1
Q.1: If (a3+1,b–23) = (53,13), what is the value of a and b?
Sol:
(a3+1,b–23) = (53,13)
As the ordered pairs are equal, the corresponding elements will also be equal.
Therefore, a3+1=53 a3=53–1 a3=23
Therefore, a = 2
Now, b–23=13 b=13+23
Therefore, b = 1
Hence, a = 2 and b = 1
Q:2. If the set X has 4 elements and the set Y = {2, 3, 4, 5}, then find the number of elements in X × Y
Sol:
There are 4 elements in set X and the elements of set X are 2, 3, 4, and 5.
No. of elements in X × Y = (No. of elements in X) × (No. of elements in Y)
= 4 × 4
= 16
Therefore, the no. of elements in (X×Y) is 16.
Q.3: If A = {8, 9} and B = {4, 5, 2}, what is the value of A × B and B × A?
Sol:
A = {8, 9}
B = {4, 5, 2}
As we know that Cartesian product ‘P × Q’ of two non-empty sets ‘P’ and ‘Q’ is defined as P × Q = {(p, q): p ∈ P, q ∈ Q}
Therefore,
A × B = {(8, 4), (8, 5), (8, 2), (9, 4), (9, 5), (9, 2)}
B × A = {(4, 8), (4, 9), (5, 8), (5, 9), (2, 8), (2, 9)}
Q.4: State whether the given statements are True or False. If the statement is false, write that statement correctly.
(i). If X = {a, b} and Y = {b, a}, then X × Y = {(a, b), (b, a)}
(ii). If P and Q are non – empty sets, then P × Q is a non – empty set of ordered pairs (a, b) such that x ∈ P and b ∈ Q.
(iii). If M = {2, 3}, N = {4, 5}, then M × (N ∩Ø ) = Ø.
Sol:
(i). If X = {a, b} and Y = {b, a}, then X × Y = {(a, b), (b, a)}
The given statement is False.
X = {a, b}
Y = {b, a}
Therefore, X × Y = {(a, b), (a, a), (b, b), (b, a)}
(ii). If P and Q are non – empty sets, then P × Q is a non – empty set of ordered pairs (a, b) such that x ∈ P and b ∈ Q.
The given statement is True.
(iii). If M = {2, 3}, N = {4, 5}, then M × (N ∩Ø ) = Ø.
The given statement is True.
Q.5: If M = {-2, 2}, then find M × M × M.
Sol:
For any non – empty set ‘M’, M × M × M is defined as:
M × M × M = {(x, y, z): x, y, z ∈ M}
Since, M = {-2, 2} [ Given]
Therefore, M × M × M = {(–2, –2, –2), (–2, –2, 2), (–2, 2, –2), (–2, 2, 2), (2, –2, –2), (2, –2, 2), (2, 2, –2), (2, 2, 2)}
Q.6: If X × Y = {(a, m), (a, n), (b, m), (b, n)}. Find X and Y.
Sol:
X × Y = {(a, m), (a, n), (b, m), (b, n)}
As we know, that Cartesian product P × Q of two non-empty sets P and Q is defined as P × Q = {(p, q): p ∈ P, q ∈ Q}
Therefore, ‘X’ is the set of all the first elements and ‘Y’ is the set of all the second elements.
Therefore, X = {a, b} and Y = {m, n}
Q.7: Let P = {2, 3}, Q = {2, 3, 4, 5}, R = {6, 7} and S = {6, 7, 8, 9}. Verify the following:
(i). P×(Q∩R) = (P×Q)∩(P×R)
(ii). P × R is a subset of Q × S
Answer:
(i). P×(Q∩R) = (P×Q)∩(P×R)
Taking LHS:
(Q∩R) = {2, 3, 4, 5} ∩ {6, 7}
= Ø
P×(Q∩R) = P × Ø = Ø
Now Taking RHS:
P×Q = {(2, 2), (2, 3), (2, 4), (2, 5), (3, 2), (3, 3), (3, 4), (3, 5)}
P×R = {(2, 6), (2, 7), (3, 6), (3, 7)}
(P×Q)∩(P×R) = Ø
Therefore, LHS = RHS
P×(Q∩R) = (P×Q)∩(P×R)
(ii). P × R is a subset of Q × S
P × R = {(2, 6), (2, 7), (3, 6), (3, 7)}
Q × S = {(2, 6), (2, 7), (2, 8), (2, 9), (3, 6), (3, 7), (3, 8), (3, 9), (4, 6), (4, 7), (4, 8), (4, 9), (5, 6), (5, 7), (5, 8), (5, 9)}
We can see that all the elements of set P × R are the elements of the set Q × S.
Therefore, P × R is a subset of Q × S.
Q.8: Let P = {2, 3} and Q = {4, 5}. Find P × Q and then find how many subsets will P × Q have? List them.
Sol:
P = {2, 3}
Q = {4, 5}
P × Q = {(2, 4), (2, 5), (3, 4), (3, 5)}
n (P × Q) = 4
As we know, that If ‘A’ is a set with n(A) = m,
Then, n[P(A)] = 2m
Therefore,
For the set P × Q = 24
= 16 subsets
The subsets are as following:
Ø, {(2, 4)}, {(2, 5)}, {(3, 4)}, {(3, 5)}, {(2, 4), (2, 5)}, {(2, 4), (3, 4)}, {(2, 4), (3, 5)}, {(2, 5), (3, 4)}, {(2, 5), (3, 5)}, {(3, 4), (3, 5)}, {(2, 4), (2, 5), (3, 4)}, {(2, 4), (2, 5), (3, 5)}, {(2, 4), (3, 4), (3, 5)}, {(2, 5), (3, 4), (3, 5)}, {(2, 4), (2, 5), (3, 4), (3, 5)}
Q.9: Let M and N be two sets where n (M) = 3 and n (N) = 2. If (a, 1), (b, 2), (c, 1) are in M × N, find M and N, where a, b and c are different elements.
Sol:
n(M) = 3
n(N) = 2
Since, (a, 1), (b, 2), (c, 1) are in M × N
[M = Set of first elements of the ordered pair elements of M × N] [N = Set of second elements of the ordered pair elements of M × N]
Therefore, a, b and c are the elements of M
And, 1 and 2 are the elements of N
As n(M) = 3 and n(N) = 2, hence M = {a, b, c} and N = {1, 2}
Q.10: The Cartesian product Z × Z has 9 elements among which are found (-2, 0) and (0, 2). Find the set Z and also the remaining elements of Z × Z.
Sol:
As we know that, If n(M) = p and n(N) = q, then n(M × N) = pq.
Now,
n (Z × Z) = n(Z) × n(Z)
But, it is given that, n(Z × Z) = 9
Therefore, n(Z) × n(Z) = 9
n(Z) = 3
The pairs (-2, 0) and (0, 2) are two of the nine elements of Z × Z
As we know, Z × Z = {(x, x): x ∈ Z}
Therefore, –2, 0, and 2 are elements of Z
Since, n(Z) = 3, we can see that Z = {–2, 0, 2}
Therefore, the remaining elements of the set Z × Z are (–2, –2), (–2, 2), (0, –2), (0, 0), (2, –2), (2, 0), and (2, 2).
Exercise 2.2
Q.1: Let X = {1, 2, 3, 4, . . . . . 14}. Define a relation Z from X to X by Z= {(a, b): 3a – b = 0, where a, b ∈ X}. Find its co – domain, domain and range.
Sol:
The relation ‘Z’ from ‘X to X’ is:
Z = {(a, b): 3a – b = 0, where a, b ∈ X}
Z = {(a, b): 3a = b, where a, b ∈ X}
Z = {(1, 3), (2, 6), (3, 9), (4, 12)}
The domain of Z is the set of all the first elements of the ordered pairs in the relation.
Domain of Z = {1, 2, 3, 4}
The set X is the co – domain of the relation Z.
Therefore, co – domain of Z = X = {1, 2, 3, 4, . . . . . . 14}
The range of Z is the set of the second elements of the ordered pairs in the relation.
Therefore, Range of Z = {3, 6, 9, 12}
Q.2: Define a relation Z on the set N of natural no. by Z = {(a, b): b = a + 5, a is a natural no less than 4; a, b ∈ N}. Give this relationship in the roaster form. Find the domain and the range.
Sol:
Z = {(a, b): b = a + 5, a is a natural number less than 4; a, b ∈ N}.
Natural numbers less than 4 are 1, 2 and 3.
Z = {(1, 6), (2, 7), (3, 8)}
The domain of Z is the set of all the first elements of the ordered pairs in the relation.
Domain of Z = {1, 2, 3}
The range of Z is the set of the second elements of the ordered pairs in the relation.
Therefore, Range of Z = {6, 7, 8}
Q.3: M = {1, 2, 3, 5} and N = {4, 6, 9}. Define a relation Z from M to N by Z = {(a, b): the difference between a and b is odd; a ∈ M, b ∈ N}. Find Z in roster form.
Sol:
M = {1, 2, 3, 5}
N = {4, 6, 9}
Z = {(a, b): the difference between a and b is odd; a ∈ M, b ∈ N}
Therefore, Z = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}
Q.4: The figure given below shows a relationship between the sets A and B. Find the following relation:
sets A and B
(i) In set-builder form
(ii) In roster form.
What is its range and domain?
Sol:
According to the information given in the figure:
A = {5, 6, 7}
B = {3, 4, 5}
(i). Z = {(a, b): b = a – 2; a ∈ A}
(or), Z = {(a, b): b = a – 2 for a = 5, 6, 7}
(ii). Z = {(5, 3), (6, 4), (7, 5)}
Domain of Z = {5, 6, 7}
Range of Z = {3, 4, 5}
Q.5: Let X = {1, 2, 3, 4, 6}. Let Z be the relation on X defined by {(p, q): p, q ∈ X, q is divisible by p}.
(i) Write Z in the roster form
(ii) Find domain of Z
(iii) Find range of Z
Sol:
X = {1, 2, 3, 4, 6}
Z = {(p, q): p, q ∈ X, q is divisible by p}
(i) Z = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}
(ii) Domain of Z = {1, 2, 3, 4, 6}
(iii) Range of Z = {1, 2, 3, 4, 6}
Q.6: Find the range and domain of the relation Z defined by Z = {(a, a + 5): a ∈ {0, 1, 2, 3, 4, 5}}.
Sol:
Z = {(a, a + 5): a ∈ {0, 1, 2, 3, 4, 5}}
Therefore, Z = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
Domain = {0, 1, 2, 3, 4, 5}
Range = {5, 6, 7, 8, 9, 10}
Q.7: Find the relation Z = {(a, a3): a is a prime number less than 10} in the roster form.
Sol:
Z = {(a, a3): a is a prime no. less than 10}
The prime number less than 10 are 2, 3, 5, and 7.
Therefore, Z = {(2, 8), (3, 27), (5, 125), (7, 343)}
Q.8: Let X = {a, b, c} and Y = {11, 12}. Find the no. of relations from X to Y.
Sol:
It is given that X = {a, b, c} and Y = {11, 12}.
X × Y = {(a, 11), (a, 12), (b, 11), (b, 12), (c, 11), (c, 12)}
As n(X × Y) = 6, the no of subsets of X × Y = 26.
Therefore, the number of relations from X to Y is 26.
Q.9: Let Z be the relation on P defined by Z = {(x, y): x, y ∈ P, x – y is an integer}. Find the range and domain of Z.
Sol:
Z = {(x, y): x, y ∈ P, x – y is an integer}
As we know that the difference between any two integers is always an integer.
Domain of Z = P
Range of Z = P
Exercise 2.3
Q.1: Which of the given relations are functions? Answer with reason. If it is a function, find its range and domain.
(i) {(12, 11), (15, 11), (18, 11), (1, 11), (4, 11), (7, 11)}
(ii) {(12, 11), (14, 12), (16, 13), (18, 14), (0, 15), (2, 16), (4, 17)}
(iii) {(11, 13), (11, 15), (12, 15)}
Sol:
(i). {(12, 11), (15, 11), (18, 11), (1, 11), (4, 11), (7, 11)}
Since, 12, 15, 18, 1, 4 and 7 are the elements of the domain of the given relation which have unique images.
Therefore, the given relation is a function.
Domain = {12, 15, 18, 1, 4, 7}
Range = {11}
(ii). {(12, 11), (14, 12), (16, 13), (18, 14), (0, 15), (2, 16), (4, 17)}
Since, 12, 14, 16, 18, 0, 2 and 4 are the elements of the domain of the given relation which have unique images.
Therefore, the given relation is a function.
Domain = {12, 14, 16, 18, 0, 2, 4}
Range = {11, 12, 13, 14, 15, 16, 17}
(iii). {(11, 13), (11, 15), (12, 15)}
Since, the same first element that is 11 corresponds to two images that is 13 and 15.
Therefore, the given relation is not a function.
Q.2: Find the range and domain of the given real function:
(i) f(y) = -|y|
(ii) f(y) = 9–y2−−−−√
Sol:
(i) f(y) = -|y|, y ∈ R
As we know:
|y| = {y,ify≥0−y,ify<0
Therefore,
f(y) = -|y| = {−y,ify≥0y,ify<0
Since, f(y) is defined for y ∈ R, the domain of ‘f’ is R.
The range of f(y) = -|y|is all the real number except positive real number.
Therefore, range of ‘f’ is (−∞,0].
(ii) f(y) = 9–y2−−−−√ 9–y2−−−−√ is defined for the real number which are greater than or equal to -3 and less than or equal to 3.
Therefore, Domain of f(y) = {y: -3 ≤ y ≤ 3} or [-3, 3].
For any value of ‘y’ such that −3≤y≤3, the value of f(y) will lie between 0 and 3.
Therefore, the Range of f(y) = {y: 0 ≤ y ≤ 3} or [0, 3]
Q.3: A function f is f(y) = 3y – 6. Find the values of the following:
(i) f(1)
(ii) f(8)
(iii) f(-2)
Sol:
The function of ‘f’ is:
f(y) = 3y – 6
(i) f(1) = (3 × 1) – 6
= 3 – 6 = -3
(ii) f(8) = (3 × 8) – 6
= 24 – 6= 18
(iii) f(-2) = (3 × -2) – 6
= -6 – 6= -12
Q.4: The function ‘f’ which shows temperature in degree Celsius into temperature in degree Fahrenheit is expressed as: 9C5+32.
Find for the following values:
(i) f(0)
(ii) f(28)
(iii) f(-10)
(iv) The value of C, when f(C) = 212
Sol:
F = 9C5+32
(i) f(0):
= 9×05+32
= 0 + 32= 32
(ii) f(28):
= 9×285+32
= 252+1605
= 4125
(iii) f(-10):
= 9×−105+32
= 9×(−2)+32
= -18 + 32= 14
(iv) The value of C, when f(C) = 212:
⇒ 212=9×C5+32 ⇒ 9×C5=212−32 ⇒ 9×C5=180 ⇒ 9×C=180×5 ⇒ C=180×59
i.e. C=100
Therefore, The value of f, when f(C) = 212 is 100.
Q.5: Calculate range of the given functions:
(i) f(y) = 2 – 3y, y ∈ R, y > 0.
(ii) f(y) = y2+2, is a real no.
(iii) f(y) = y, y is a real no.
Sol:
(i) f(y)=2–3y,y∈R,y>0
We can write the value of f(y) for different real numbers x > 0 in tabular form as:
y 0.01 0.1 0.9 1 2 2.5 4 5 …
f(y) 1.97 1.7 –
0.7
–
1
–
4
–
5.5
–
10
–
13
…
Thus, we can clearly observe that the range of ‘f’ forms the set for all real numbers which are less than 2.
i.e. Range of f = (−∞,2)
Alternative:
Let, y > 0
3y > 0
2 – 3y < 2
i.e f(y) < 2
Therefore, Range of f = (−∞,2)
(ii) f(y)=y2+2, y, is a real number.
We can write the value of f(y) for different real numbers x, in tabular form as:
y 0 ±0.3 ±0.8 ±1 ±2 ±3 . . .
f(y) 2 2.09 2.64 3 6 11 . . .
Thus, we can clearly observe that the range of f forms the set for all real numbers which are less than 2.
i.e. Range of f = (−∞,2)
Alternative:
Let ‘y’ be any real number. Then,
y2≥0=>y2+2≥0+2=>y2+2≥2f(y)≥2
Therefore, Range of f = [2,∞)
(iii) f(y) = y, where y is a real number
Here, we can see that the range of f is the set of all the real numbers.
Therefore, Range of f = R
Miscellaneous Exercise
Q-1: The relation ‘m’ is defined by:
m (y) = y2, 0≤y≤5
= 5y, 5≤y≤30
The relation ‘n’ is defined by
n (y) = y2, 0≤y≤4
= 5y, 4≤y≤30
Now, prove that ‘m’ is a function and ‘n’ is not a function.
Sol:
Here,
m (y) = y2, 0≤y≤5
= 5y, 5≤y≤30
Now, 0≤y≤5, m(y) = y2
And, 5≤y≤30, m(y) = 5y
Now, at y = 5, m(y) = 52 = 25 or m(y) = 5 × 5 = 25
i.e., at y = 5, m(y) = 25
Therefore, for 0≤y≤30 , the images of m(y) are unique. Thus, the given relation is a function.
Now, n(y) = y2, 0≤y≤4
= 5y, 4≤y≤30
Now, at y = 4, m (y) = 42 = 16 or m(y) = 5 × 4 = 20
Thus, element 4 of the domain 0≤y≤30 of relation ‘n’ has 2 different images i.e., 16 and 20
Therefore, this relation is not a function.
Q-2: If g(y) = y2 then, Find g(1.2)–g(1)(1.2–1)
Sol:
Here, g(y) = y2
Therefore, g(1.2)–g(1)(1.2–1) =(1.2)2–(1)2(1.2–1)=1.44–10.2 =0.440.2=2.2
Q-3: Find the domain for the function given below:
g(y)=y2–2y+1y2–9y+20
Sol:
Here, g(y)=y2–2y+1y2–9y+20 =y2–2y+1(y–5)(y–4)
Now, it clear from above equation that the function ‘g’ is defined for all real numbers except ‘y = 4’ and ‘y = 5’.
Therefore, the required domain is R: {4, 5}
Q-4: Find the range and domain of the function given below:
g(y)=(y–5)−−−−−√
Sol:
Here, g(y)=(y–5)−−−−−√ is the given function.
So, it is clear that the function is defined for y≥5.
So, the domain will be the set of all real numbers greater than or equal to 5. i.e., The domain for g(y) is [5,∞)
Now, for range of the given function, we have:
y≥5 ⇒(y–5)≥0 ⇒(y–5)−−−−−√≥0
Therefore, the range of g(y) is the set of all real numbers greater than or equal to 0.
i.e, the range of g(y) is [0,∞).
Q-5: Find the range and domain of the function: g(y) = |y – 4|
Sol:
Here |y – 4| is the given function.
So, it is clear that the function is defined for all the real numbers.
The domain for g(y) is R
Now, for range of the given function, we have:
yϵR,, |y – 4| assumes for all real numbers.
Therefore, the range of g(y) is the set of all non- negative real numbers.
Q-6: A function from ‘R into R’ is given below. Find the range of ‘g’.
g=[(y,y21+y2);yϵR]
Sol:
g=[(y,y21+y2);yϵR]
=[(0,0),(±0.5,15),(±1,12),(±1.5,913),(±2,45),(3,910),…….]
Thus, the range of ‘g’ is the set of all 2nd elements. It can be seen that all these elements are: ≥0but<1.
Therefore, the range of ‘g’ is = [0,1)
Q-7: Assume that function ‘m’ and ‘n’ is defined from: R→R.
m (y) = y + 2, n(y) = 3y – 2
Find m + n, m – n and mn
Sol:
Here, m(y) = y + 2 and n(y) = 3y – 2 are defined from R→R.
Now, (m + n) (y) = m(y) + n(y) = (y + 2) + (3y – 2) = 4y
Therefore, (m + n) (y) = 4y
(m – n) (y) = m(y) – n(y) = (y + 2) – (3y – 2) = -2y + 4
Therefore, (m – n) (y) = -2y + 4
(mn)(y)=m(y)n(y),n(y)≠0,yϵR
Therefore, (mn)(y)=y+23y–2,3y–2≠0,3y≠2 (mn)(y)=y+23y–2,y≠23
Q-8: Let g = {(1, 1), (2, 3), (0, -1), (-1, -3)} be a function from ‘Z to Z’ defined by g(y) = uy + v, for some integers u,v. Find u,v.
Sol:
g = {(1, 1), (2, 3), (0, -1), (-1, -3)} and g(y) = uy + v
(1,1)ϵg⇒g(1)=1⇒u×1+v=1 ⇒u+v=1 (0,−1)ϵg⇒g(0)=−1⇒u×0+v=−1 ⇒v=−1
By putting v = -1 in u + v = 1, we get
u = 2.
Therefore, u = 2 and v = -1.
Q-9: Let ‘f’ be a relation from ‘N to N’ defined by f = {(x,y): x,yϵN and x = y2}. Find out which of the following is true and which one is false.
1.(x,y)ϵf,(y,z)ϵf⇒(x,z)ϵf.
2.(x,x)ϵf,forallxϵN
3.(x,y)ϵf⇒(y,x)ϵf
Also justify your answer.
Sol:
f = {(x,y): x,yϵN and x = y2}
(1). Now, take (4,2)ϵf,(25,5)ϵfbecause4,2,25,5ϵN and 4 = 22 and 16 = 42.
Therefore, the given statement is true.
(2). Now, let 3ϵNbut3≠32=9
Therefore, the statement is false.
(3). (16,4)ϵNbecause16,4ϵN and 16 = 42.
Now, 4≠162=256; therefore (4, 16) does not belongs to N.
Therefore, the statement is false.
Q-10: Assume U = {1, 2, 3, 4}, V = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2,9), (3,1), (4, 5), (2, 11)}. Find out which of the following is true and which one is false.
(1). ‘f’ is a function from U to V.
(2). ‘f’ is a relation from U to V.
Justify your answer.
Sol:
Here, U = {1, 2, 3, 4} and V = {1, 5, 9, 11, 15, 16}
Therefore, U x V = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}
f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
(1). As, the 1st member i.e. ‘2’ corresponds to 2 different images i.e. ‘9’ and ‘11’, Therefore, ‘f’ is not a function.
(2). A relation of non-empty set ‘U’ to a non-empty set ‘V’ is a subset of the Cartesian product U x V.
Also it can be seen that ‘f’ is a subset of U x V.
Therefore, ‘f’ is a relation from U to V.
Q-11: Assume ‘g’ be the subset of ‘Z to Z’ defined by f = {(xy, x + y): x,yϵZ}. Is ‘g’ a function from ‘Z to Z’, also justify your answer.
Sol:
Here,
f = {(xy, x + y): x,yϵZ}
As we know that a relation g from set X to a set Y is said to be a function only if every element of set X is having a unique images in set Y.
Since, 4,12,−4,−12ϵZ [4×12,4+12],[(−4)×(−12),−4+(−12)]ϵg
i.e. [ (48, 16), (48, -16) ] ϵg
Here, the same 1st member ‘48’ is having 2 images ‘16’ and ‘-16’.
Therefore, relation ‘g’ is not a function from ‘Z to Z’.
Q-12: Assume ‘X’ = {5,7,9, 10, 11, 12, 13} and let ‘g’: X→N be defined by g(n) = The highest prime factor of ‘n’. Find the range of ‘g’.
Sol:
X = {5, 7, 9, 10, 11, 12, 13, 14, 15};
g: X→N be defined by g(n) = The highest prime factor of ‘n’
Prime factor of 5 = 5
Prime factor of 7 = 7
Prime factor of 9 = 3
Prime factor of 10 = 2,5
Prime factor of 11 = 11
Prime factor of 12 = 2,3
Prime factor of 13 = 13
Prime factor of 14 = 2, 7
Prime factor of 15 = 3, 5
Therefore,
g(5) = The highest prime factor of 5 = 5
g(7) = The highest prime factor of 7 = 7
g(9) = The highest prime factor of 9 = 3
g(10) = The highest prime factor of 10 = 5
g(11) = The highest prime factor of 11 = 11
g(12) = The highest prime factor of 12 = 3
g(13) = The highest prime factor of 13 = 13
g(14) = The highest prime factor of 14 = 7
g(15) = The highest prime factor of 15 = 5
Thus, the range of ‘g’ is the set of all ‘g(n)’, where nϵX.
Therefore, Range of g = {3, 5, 7, 11, 13}
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