Referring to the AP Inter 2nd Year Maths Study Material Chapter 1 Relations and Functions Exercise 1C Solutions makes it easier to understand complex problems.
AP Inter 2nd Year Maths Relations and Functions Solutions Exercise 1C
I.
Question 1.
Show that the function f : R →R given by f(x) = x3 is injective.
Solution:
f: R → R is given by f(x) = x3. Let x1; x2 ∈ R (domain)
Now f (x1) = f(x2) ⇒ x13 = x23⇒ x1 = x2.
∴ f is injective
Question 2.
Find the number of all onto functions from the set {1, 2, 3, ……….., n} to itself.
Solution:
The total number of onto function from {1, 2, 3, ………. n} to itself
= total number of permutations on {1, 2, 3,………..n} = nPn = n!
Note: In f: A → B if n(A) = n(B) then only we claim that no.of onto functions = no.of one-one functions.
Question 3.
Find the number of all one-one functions from set A = (1, 2, 3} to itself.
Solution:
Total number of one-one function from (1, 2, 3} to itself is
= Total number of permutations on {1, 2, 3} = 3P3 = 3! = 3 × 2 × 1 = 6.
II.
Question 1.
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 IfA c B. Is R an equivalence relation on Justify your answer.
Solution:
(i) ARA for all A e P(X) [ every set is a subset of itself]. So, R is reflexive.
(ii) Let ARB ⇒ A ⊂ B . This does not imply that B ⊂ A. So, R is not symmetric.
(iii) If ARB and BRC, then A ⊂ B and B ⊂ C ⇒ A ⊂ C ⇒ ARC . So, R is transitive.
R is not an equivalence relation as it is not symmetric.
Question 2.
Let A = {- 1, 0, 1, 2), B = {- 4, – 2, 0, 2) and f, g : A → B be functions defined by f(x) = x2 – x, x ∈ A and g(x) = 2|x – \(\frac{1}{2}\)|, x ∈ A. Are f and g equal?
Justify your answer. (Hint: One may note that two functions f : A → B and g : A B such that f (a) = g (a), ∀ a ∈ A, are called equal functions).
Solution:
Given that f(x) = x2 – x and g(x)= 2|x – \(\frac{1}{2}\)| – 1, x ∈ A
and A = {-1, 0, 1, 2}, B = {-4, -2, 0, 2}
Now f(-1) = (-1)2 – (-1) = 1 + 1 = 2 and
g(-1) = 2|(-1) – \(\frac{1}{2}\)| – 1 = 2(\(\frac{3}{2}\)) – 1 = 3 – 1 = 2 ⇒ f(-1) = g(-1);
f(0) = (0)2 – 0 = 0 and g(0) = 2|0 – \(\frac{1}{2}\)| – 1 = 2(\(\frac{1}{2}\)) – 1 = 1 – 1 = 0 ⇒ f(0) = g(0)
f(1) = (1)2 – 1 = 0 and g(0) = 2|1 – \(\frac{1}{2}\)| – 1 = 2(\(\frac{1}{2}\)) – 1 = 1 – 1 = 0 ⇒ f(1) = g(1)
f(2) = (2)2 – 2 = 2 and g(2) = 2|2 – \(\frac{1}{2}\)| – 1 = 2(\(\frac{3}{2}\)) – 1 = 3 – 1 = 2 ⇒ f(2) = g(2)
∴ f(a)= g (a), ∀a ∈ A Hence, the functions f and g are equal.
III.
Question 1.
Show that the function f: R → b {x ∈ R : — 1 < x < 1) defined by f(x) = \(\frac{x}{1+|x|}\) x ∈ R is one one and onto function.
Solution:
(i) To show f is one-one:
Case (i): When x1, x2 ≥ 0. Then |x1| = x1, |x2| = x2 [∵ From the def. of |x|]
Now f(x1) = f(x2) ⇒ \(\frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}\) ⇒ x1 (1 + x2) = x2(1 + x1)
⇒ x1 + x1x2 = x2 + x1x2 ⇒ x1 = x2
∴ f is one-one.
Similarly, the result can be verified when x1 and x2 both are negative.
Case (ii): When one is non-negative and other is negative.
Let x1 ≥ 0 and x2 < 0. Then f(x1) ≥ 0, f(x2) < 0
Here, f(x1) ≠ f(x2) ⇒ x1 ≠ x2.
This implies that f(x1) = f(x2) ⇒ x1 = x2
∴ f is one-one.
(ii) To show f is onto:
Case (i): When x is positive then |x| = x. Put f(x) = y
∴ y = \(\frac{x}{1+x}\) ⇒ y(1 + x) = x ⇒ y + yx = xy = x(1 – y) ⇒ x = \(\frac{y}{1+y}\) exists ∀ y ≠ 1
∴ f is onto.
Verification: f(x) = \(f\left(\frac{y}{1-y}\right)=\frac{\left(\frac{y}{1-y}\right)}{1+\left|\frac{y}{1-y}\right|}=\frac{\frac{y}{1-y}}{1+\left(\frac{y}{1-y}\right)}=\frac{y}{1-y+y}=y\)
Case (ii): When x is negative then |x|= – x. Put f(x) = y
∴ y = \(\frac{x}{1+x}\) ⇒ y(1 – x) = x ⇒ y – yx = x ⇒ y = x(1 + y) ⇒ x = \(\frac{y}{1+y}\) exists ∀ y ≠ 1
∴ f is onto.
Verification: f(x) = \(\left(\frac{y}{1+y}\right)=\frac{\left(\frac{y}{1+y}\right)}{1+\left|\frac{y}{1+y}\right|}=\frac{\frac{y}{1+y}}{1+\left(\frac{-y}{1+y}\right)}=\frac{y}{1+y-y}=y\)
Hence, f is one-one and onto in both the cases. Hence proved.