NCERT Solutions Class 12- Mathematics Part I β Chapter 1 Relations And Functions -Miscellaneous Exercise on Chapter 1
Question 1. Show that the function f : R β {x β R : β 1 < x < 1} defined by f(x) = [Tex]\frac{x}{1+|x|} [/Tex], x β R is one one and onto function.
Solution:
As, it is mentioned here
f : R β {x β R : β 1 < x < 1} defined by [Tex]f(x) = \frac{x}{1+|x|} [/Tex], x β R
As, we know f is invertible, if and only if f is one-one and onto.
ONE-ONE
For the pair of number, we will deal with three cases:
Case 1: When both numbers p and p are positive numbers.
The function f is defined as
Case 1: When both numbers p and q are positive numbers.
[Tex]f(p) = \frac{p}{1+|p|}[/Tex]
[Tex]f(q) = \frac{q}{1+|q|}[/Tex]
f(p) = f(q)
[Tex]\frac{p}{1+|p|} = \frac{q}{1+|q|}[/Tex]
[Tex]\frac{p}{1+p} = \frac{q}{1+q}[/Tex]
p(1+q) = q(1+p)
p = q
Case 2: When number p and q are negative numbers.
[Tex]f(p) = \frac{p}{1+|p|}[/Tex]
[Tex]f(q) = \frac{q}{1+|q|}[/Tex]
f(p) = f(q)
[Tex]\frac{p}{1+|p|} = \frac{q}{1+|q|}[/Tex]
[Tex]\frac{p}{1-p} = \frac{q}{1-q}[/Tex]
p(1-q) = q(1-p)
p = q
Case 3: When p is positive and q is negative
[Tex]f(p) = \frac{p}{1+|p|}[/Tex]
[Tex]f(q) = \frac{q}{1+|q|}[/Tex]
f(p) = f(q)
[Tex]\frac{p}{1+|p|} = \frac{q}{1+|q|}[/Tex]
[Tex]\frac{p}{1+p} = \frac{q}{1-q}[/Tex]
p(1-q) = q(1+p)
p + q = 2pq
Here, RHS will be negative and LHS will be positive. Hence, the case 3 result is impossible.
So, the function f is one-one, for case 1 and case 2.
ONTO
Case 1: When p>0.
[Tex]f(p) = \frac{p}{1+|p|}[/Tex]
[Tex]y = \frac{p}{1+p}[/Tex]
[Tex]p = \frac{y}{1-y} (yβ 1)[/Tex]
Case 2: When p <0
[Tex]f(p) = \frac{p}{1+|p|}[/Tex]
[Tex]y = \frac{p}{1-p}[/Tex]
[Tex]p = \frac{y}{1+y} (yβ -1)[/Tex]
Hence, p is defined for all the values of y, pβ R
Hence f is onto.
As, f is one-one and onto. This f is an invertible function.
Question 2. Show that the function f : R β R given by f(x) = x3 is injective.
Solution:
As, it is mentioned here
f : R β R defined by f(x) = x3, x β R
To prove f is injective (or one-one).
ONE-ONE
The function f is defined as
f(x) = x3
f(y) = y3
f(x) = f(y)
x3 = y3
x = y
The function f is one-one, so f is injective.
Question 3. Given a non empty set X, consider P(X) which is the set of all subsets of X.
Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A β B. Is R an equivalence relation on P(X)? Justify your answer.
Solution:
Given, A and B are the subsets of P(x), Aβ B
To check the equivalence relation on P(X), we have to check
- Reflexive
As, we know that every set is the subset of itself.
Hence, Aβ A and Bβ B
ARA and BRB is reflexive for all A,Bβ P(X)
- Symmetric
As, it is given that Aβ B. But it doesnβt make sure that Bβ A.
To be symmetric it has to be A = B
ARB is not symmetric.
- Transitive
When Aβ B and Bβ C
Then of course, Aβ C
Hence, R is transitive.
So, as R is not symmetric.
R is not an equivalence relation on P(X).
Question 4. Find the number of all onto functions from the set {1, 2, 3, β¦ , n} to itself.
Solution:
Onto function from the set {1,2,3,β¦..,n} to itself is just same as the permutations of n.
1Γ2Γ3Γ4Γβ¦β¦.Γn
Which is n!.
Question 5. Let A = {β 1, 0, 1, 2}, B = {β 4, β 2, 0, 2} and f, g : A β B be functions defined by f(x) = x2 β x, x β A and [Tex]g(x) = 2|x-\frac{1}{2}|-1 [/Tex] x β A. Are f and g equal? Justify your answer.
(Hint: One may note that two functions f : A β B and g : A β B such that f(a) = g (a) β a β A, are called equal functions).
Solution:
Given, f, g : A β B be functions defined by f(x) = x2 β x, x β A and g(x) = [Tex]2|x-\frac{1}{2}|-1 [/Tex] x β A
At x = -1
f(0) = (-1)2 β (-1) = 2
g(0) = [Tex]2|-1-\frac{1}{2}|-1 [/Tex] = 2
Here, f(-1) = g(-1) and 2=2
At x = 0
f(0) = 02 β 0 = 0
g(0) = [Tex]2|0-\frac{1}{2}|-1 [/Tex] = 0
Here, f(0) = g(0) and 0=0
At x = 1
f(1) = 12 β 1 = 0
g(1) = [Tex]2|1-\frac{1}{2}|-1 [/Tex] = 0
Here, f(1) = g(1) and 1=1
At x = 2
f(1) = 22 β 2 = 2
g(1) = [Tex]2|2-\frac{1}{2}|-1 [/Tex] = 2
Here, f(2) = g(2) and 2=2
For, every cβ A, f(c) = g(c)
Hence, f and g are equal functions.
Question 6. Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
(A) 1
(B) 2
(C) 3
(D) 4
Solution:
R = {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1),(3,3)}
Reflexive : (1,1), (2,2), (3,3) β R
Symmetric: (1,2), (2,1)β R and (1,3), (3,1) β R
R is not Transitive because, (1,2), (1,3) β R but (3,2) βR
So, if we will add (3,2) and (2,3) or both, then R will become transitive.
New, R = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)}
Hence, A is the correct option.
Question 7. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
(A) 1
(B) 2
(C) 3
(D) 4
Solution:
Smallest equivalence relations containing (1, 2):
R = {(1,1),(2,2),(1,2),(2,1),(3,3)}
or R = {(1,1),(2,2),(1,2),(2,1),(3,2),(2,3)(3,3)}
Hence, B is the correct option.