ML Aggarwal Class 12 Maths Solutions Section A Chapter 1 Relations and Functions Chapter Test

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 1 Relations and Functions Chapter Test

Question 1.
If a relation R on Z (set of all integers) is defined by R = {(a, b) : | a – b | ≤ 3}, then show that R is reflexive and symmetric but not transitive.
Solution:
Given relation R on Z (set of all integers) be defined by R = {(a, b) : | a – b | ≤ 3}
Reflexive :
∀ a ∈ Z, |a – a| = 0 ≤ 3
⇒ (a, a) ∈ R
∴ R is reflexive on Z.

Symmetric :
∀ a, b ∈ Z s.t (a, b) ∈ R
⇒ | a – b | ≤ 3
⇒ | – (b – a) | ≤ 3
⇒ | b – a | ≤ 3
⇒ (b, a) ∈ R
∴ R is symmetric on Z.

Transitive :
∀ a, b, c ∀ Z s.t (a, b), (b, c) ∈ R
(a, b) ∈ R ⇒ | a – b | ≤ 3
(b, c) ∈ R ⇒ | b – c | ≤ 3
Now
| a – c | = | a – b + b – c | ≤ | a – b | + | b – c | ≤ 3 + 3 = 6
∴ (a, c) ∉ R
Thus, R is not transitive on Z.
Hence R is reflexive, symmetric but not transitive on Z.

Question 2.
Show that the number of equivalence relations in the set {1, 2, 3} containing (1, 2) and (2, 1) is two.
Solution:
Let A= {1,2, 3}
R₁ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
which is clearly reflexive, symmetric and transitive
∴ R₁ is equivalence relation on Z.
R₂ = {(1, 1), (2, 2), (3, 3), (1,2), (2, 1). (1, 3), (3, 1), (2, 3), (3, 2)}
Since (1, 1), (2, 2), (3, 3) ∈ R
∴ R₂ is reflexive on Z.
Now (1, 2) ∈ R and (2, 1) ∈ R
Now (2, 3), (3, 2) ∈ R and (1,3), (3, 1} ∈ R
∴ R₂ is symmetric on Z.
Also R₂ is transitive on Z.
Hence, R₂ be reflexive, symmetric and transitive on Z.
Thus, R₂ be an equivalence relation on Z.

Question 3.
Consider the function f (x) = x + \(\frac{1}{x}\) ∈ R, x ≠ 0. Is f one-one ?
Solution:
∀ x, y ∈ R – {0} s.t.f (x) = f(y)
⇒ x + \(\frac{1}{x}\) = y + \(\frac{1}{y}\)
⇒ x – y + \(\frac{1}{x}\) – \(\frac{1}{y}\) = 0
⇒ (x – y) (1 – \(\frac{1}{xy}\)) = 0
⇒ x – y = 0 or 1 – \(\frac{1}{xy}\) = 0 i.e. xy = 1
f(x) = f(y) ≠ x = y
Thus, f is not one-one.
e.g. f(2) = 2 + \(\frac{1}{2}\)
and f(\(\frac{1}{2}\)) = \(\frac{1}{2}\) + 2
∴ f(2) = f (\(\frac{1}{2}\))
Thus, 2 and \(\frac{1}{2}\) have same images in R.
∴ f is not one-one.

Question 4.
Show that the function f : N → N defined by f (x) = 3x – 2 is one-one but not onto.
Solution:
Given function f : N → N defined by f(x) = 3x – 2 ∀ x ∈ N
∀ x, y ∈ N s.t f (x) = f(y)
⇒ 3x – 2 = 3y – 2
⇒ x = y
∴ f is one-one.
Since 2 ∈ N (codomain of f)
Let x ∈ N (D_f) s.t f (x) = 2
⇒ 3x – 2 = 2
⇒ x = \(\frac{4}{3}\) ∉ N
Hence 2 has no pre image in N (domain of f)
∴ f is not onto.
Hence f is one-one but not onto.

Question 5.
Prove that f : N → N, defined by f(m) = m² + m + 1 for all m ∈ N, is one – one but not onto.
Solution:
Given f : N → N defined by
f (x) = x² + x + 1 ∀ x, y ∈ N
such that f (x) = f (y)
⇒ x² + x + 1 = y² + y + 1
⇒ x² + x – y² – y = 0
⇒ (x – y) (x + y) + x – y = 0
⇒ (x – y) (x + y + 1) = 0
⇒ x – y = 0 [x + y + 1 ≠ 0 ∀ x, y ∈ N]
⇒ x = y
Thus f is one-one.
Further f (x) = x² + x + 1 ≥ 3 ∀ x ∈ N
Thus f (x) = 1 and 2 and nave no pre-images in N (domain).
∴ f is not onto.
Thus f is 1 – 1 but not onto.

Question 6.
Show that the function f : R → R given by f (x) = 3x³ + 5 is bijective.
Solution:
Given f : R -> R be a function defined by
f (x) = 3x³ + 5 ∀ x ∈ R
∀ x, y ∈ R such that f (x) = f (y)
⇒ 3x³ + 5 = 3y³ + 5
⇒ x³ = y³
⇒ x = y
Thus f is one-one.
Let y ∈ R be any arbitrary element.
Then f (x) = y
∴ 3x³ + 5 = 3y³ + 5
⇒ x = \(\left(\frac{y-5}{3}\right)^{1 / 3}\)
Since y ∈ R
∴ \(\left(\frac{y-5}{3}\right)^{1 / 3}\) ∈ R
∀ y ∈ R ∃ x = \(\left(\frac{y-5}{3}\right)^{1 / 3}\) ∈ R
such that
f(x) = f \(\left[\left(\frac{y-5}{3}\right)^{1 / 3}\right]\)
= \(3\left(\frac{y-5}{3}\right)\) + 5
= y – 5 + 5 = y
Thus f is onto.
Hence f is bijective.

Question 7.
The functions f and g are given by f = {(1, 4), (2, 6), (3, 9)} and g = {(4, 2), (6, 1), (9, 1), (10, 3)}
(i) Find the domain of f and g.
(ii) Find the range of f and g.
(iii) Write gof and fog as sets of ordered pairs.
Solution:
Given f = {(1,4), (2, 6), (3, 9)}
and g = {(4, 2), (6, 1), (9, 1), (10, 3)}
(i) D_f = {1,2,3} ; R_f = {4, 6, 9}
(ii) D_g = {4, 6, 9, 10} ; R_g = {1, 2, 3}
(iii) Clearly R_f ⊂ D_g
∴ gof exists and be a function from {1, 2, 3} to {1, 2, 3}.
Since f (1) = 4 ;
f (2) = 6 ;
f (3) = 9
g (4) = 2;
g (6) = 1 ;
g (9) = 1 ;
g (10) = 3
(gof) (1) = g (f(1)) = g (4) = 2
(gof) (2) = g (f(2)) = g (6) = 1
(gof) (3) = g (f (3)) = g (9) = 1
gof = {(1, 2), (2, 1), (3, 1)}
Also, R_g ⊂ D_f
∴ fog exists and be a function from {4, 6, 9, 10} to {4, 6, 9}.
Thus, (fog) (4) = f (g (4)) = f (2) = 6
(fog) (6) = f (g(6)) = f (1) = 4
(fog) (9) = f (g (9)) = f (1) = 4
(fog) (10) = f (g (10)) = f (3) = 9
∴ fog = {(4, 6), (6, 4), (9, 4), (10, 9)}

Question 8.
If A = {α, β, γ, δ} and f corresponds to the subset {(α, γ), (β, α), (γ, δ), (δ, β)} of the cartesian product, show that f is bijective and hence find f^-1.
Solution:
Given A = {α, β, γ, δ} and f = {(α, γ), (β, α), (γ, δ), (δ, β)}
D_f = {α, β, γ, δ}
and R_f = {γ, α, δ, β}
Clearly, f (α) = γ ; f (β) = α ; f (γ) = δ ; f (δ) = β
So different elements of A have different images in A.
∴ f is one-one.
Also codomain of f = Range of f
= R_f = A
∴ f is onto.
Thus f is one-one and onto and hence f is bijective.
Thus f is invertible and f^-1 exists and be a function from A to A.
∴ f^-1 = {(γ, α), (α, β), (δ, γ), (β, δ)}.

Question 9.
Find the number of all onto functions from the set {1, 2, 3, ……………., n} to itself.
Solution:
Given A = {1, 2, 3, ……………, n]
First of all, we prove that Every onto function f : A → A is one-one
Let if possible f is not one-one.
Then ∃ two elements (say 1 and 2) in domain (A) which has same image in codomain.
Also the last element 3 of set A has image in codomain A under f contains one element.
Thus the range set can have at most two element of codomain A = {1, 2, 3}
∴ f is not onto which is a contradiction to given fact that f is onto, thus our supposition is wrong .
∴ f is one-one.
∴ f is 1 – 1 and onto and hence f is bijective.
Thus the required number of onto functions from A to itself = number of bijections from A to A
= Total number of permutations of n symbols 1, 2, ………………., n = n!.

Question 10.
Let f : R → R be defined as f(x) = 7x + 3. Find the function g : R → R such that gof = fog = I_R.
Solution:
Given a function f : R → R defined by
f (x) = 7x + 3 ∀ x ∈ R
Let y ∈ R be any arbitrary element as y ∈ R
⇒ \(\frac{y-3}{7}\) ∈ R
Lrt us define a function g : R → R by
g (y) = \(\frac{y-3}{7}\) ∀ x ∈ R
Since f : R → R and g : R → R.
Thus both gof and fog exists as
D_f = R_f = D_g = R_g = R
Now (fog) (y) = f(g (y))
= \(f\left(\frac{y-3}{7}\right)=7\left(\frac{y-3}{7}\right)\) + 3
= y ∀ y ∈ R
∴ fog = I_R
and (gof) (x) = g (f (x)) = g (7x + 3)
= \(\frac{7 x+3-3}{7}\)
= x ∀ x ∈ R
∴ gof = I_R
Thus, gof = I_R = fog.

Question 11.
Let R⁺ be the set of all non-negative real numbers. If f : R⁺ → R⁺ and g : R⁺ → R⁺ are the functions defined by f (x) = x² and g (x) = √x. Find gof and fog. Are these equal functions ?
Solution:
Given,
f : R⁺ → R⁺ defined by
f (x) = x² ∀ x ∈ R⁺
and g : R⁺ → R⁺ defined by
g (x) = √x ∀ x ∈ R⁺
Since gof exists as R_f ⊂ D_g and be a function from R⁺ to R⁺
∀ x ∈ R⁺ ,
(gof) (x) = g (f(x)) = g (x²)
= \(\sqrt{x^2}\) = | x | = x
[∵ x ∈ R⁺
⇒ | x | = x Since x > 0]
∴ (gof) (x) = x ∀ x ∈ R⁺
⇒ gof = I_R⁺
Further, R_g ⊂ D_f
∴ fog exists and be a function from R⁺ to R⁺ .
∀ x ∈ R⁺ , (fog) (x) = f (g (x))
= f (√x) = (√x)² = x
⇒ fog = I_R⁺
^ Further fog = gof = I_R⁺.

Question 12.
Consider f : R₊ → [- 9, ∞) given by f (x) = 5x² + 6x – 9, where R₊ is the set of all non-negative real numbers. Prove that f is invertible with f^-1 (y) = \(\frac{\sqrt{5 y+54}-3}{5}\).
Solution:
Given f : R₊ → [- 9, ∞) defined by
f (x) = 5x² + 6x – 9

one-one :
∀ x, y ∈ R₊ s.t f (x) = f (y)
⇒ 5x² + 6x – 9 = 5y² + 6y – 9
⇒ 5 (x² – y²) + 6 (x – y) = 0
⇒ (x – y) (5x + 5y + 6) = 0
⇒ x – y = 0 [∵ x, y ∈ R₊
⇒ 5x + 5y + 6 ≠ 0]
⇒ x = y
∴ f is one-one.

onto:
As x ∈ R₊
⇒ x ≥ 0
⇒ 5x² + 6x ≥ 0
⇒ 5x² + 6x – 9 ≥ – 9
⇒ f (x) ≥ – 9
⇒ R_f = [- 9, ∞) = codomain of f
Thus f is onto.
∴ f is one-one and onto and hence f is bijective and invertible.

To find f^-1:

Let f(x) = y = 5x² + 6x – 9
⇒ 5x² + 6x – 9 – y = 0
⇒ x = \(\frac{-6 \pm \sqrt{6^2-4 \times 5(-9-y)}}{10}\)
x = \(\frac{-6 \pm \sqrt{20 y+216}}{10}\)
x = – \(\frac{-3 \pm \sqrt{5 y+34}}{5}\)
since x ∈ R₊
∴ x = \(\frac{-3+\sqrt{5 y+54}}{5}\)
Now f(x) = y and f is invertible
⇒ f^-1 (y) = x
⇒ x = f^-1 (y) = \(\frac{\sqrt{5 y+54}-3}{5}\)
⇒ f^-1 (x) = \(\frac{\sqrt{5 y+54}-3}{5}\).

Question 12 (old).
Let X be a non-empty set. P (X) be its power set. Let be an operation defined on elements of P (X) by A * B = A ∩ B for all A, B ∈ P (X). Then
(i) Prove that * is a binary operation in P (X).
(ii) Is * commutative ?
(iii) Is * associative ?
(iv) Find the identity element in P (X) w.r.t. *.
(v) Find all the invertible elements of P (X).
(vi) If o is another binary operation defined on P (X) as AoB = A ∪ B, then verify that a distributes itself over *.
Solution:
(i) Given operation * defined on P (X) by
A * B = A ∩ B ∀ A, B ∈ P (X)
Since A, B ∈ P(X)
∴ A ∩ B ∈ P(X)
∴ A * B – A ∩ B ∈ P (X)
∴ * is a binary operation on P (X).

(ii) ∀ A, B ∈ P (X),
A * B = A ∩ B = B ∩ A = B * A
∴ * be commutative on P (X).

(iii) ∀ A, B, C ∈ P (X), we have
(A ∩ B) ∩ C = A ∩ (B ∩ C)
Thus * is associative on P (X).

(iv) Let E be the identity element in P (X) under *.
Then, E * A = A = A * E ∀ A ∈ P(X)
E ∩ A = A = A ∩ E ∀ A ∈ P(X)
⇒ E = X [X = universal set and A ⊆ X]
Thus, X be the identity element in P (X) with respect to *.

(v) Let A be any arbitrary invertible element of P (X) and let P be its inverse.
Then A * P = X = P * A ∀ A ∈ P(X)
⇒ A ∩ P = X = P ∩ A
⇒ P = A = X
Thus X be the only invertible element in P (X) under *.

(vi) Given o be the binary operation on P (X) defined by
A o B = A ∪ B ∀ A, B, C ∈ P (X),
A o (B * C) = A o (B ∩ C) = A ∪ (B ∩ C)
= (A ∪ B) ∩ (A ∪ C) [Distributive law]
∴ A o (B * C) = (A o B) * (A o C) ∀ A, B, C ∈ P (X)
Thus o be distributes itself over *.

Question 13.
Examine whether the operation * defined on R by a * b = a + ab is binary or not. If it is a binary operation, whether it is associative or not?
Solution:
Given operation * defined on R by a * b = a + ab
since a, b ∈ R
⇒ a + ab ∈ R
[closure property holds under multiplication and addition on R]
Therefore * is a binary operation on R.
∀ a, b, c ∈ R, (a * b) * c = (a + ab) * c
= a + ab + (a + ab) c
= a + ab + ac + abc
and a * (b * c) = a * (b + bc)
= a + a (b + bc)
= a + ab + abc
Thus (a * b) * c ≠ a * (b * c)
Now 2, 3, 4 ∈ R,
(2 * 3) * 4 = (2 + 2 × 3) * 4
= 8 * 4
= 8 + (8*4)
= 40
and 2 * (3*4)
= 2* (3 + 3 × 4)
= 2 * 15
= 2 + 2 × 15 = 32
Thus, 2 * (3 * 4) ≠ 2 * (3 * 4)
Thus * is not associative on R.

Question 13 (old).
Define a binary operation * on the set {0, 1, 2, 3, 4, 5} as a * b = (a + b) (mod 6). Show that zero is the identity for this operation and each element a (except 0) of the set is invertible with 6 – a being the inverse of a. Also write the operation table for the given operation.
Solution:
Given binary operation * on A defined by
a * b = (a + b) (mod 6) where A = {0, 1, 2, 3, 4, 5}
Thus a* b = \(\left\{\begin{array}{cc}
a+b ; & a+b<6 \\
a+b-6 ; & a+b \geq 6
\end{array}\right.\)
Since 0 ∈ A and let a ∈ A be any arbitrary element
Now, 0 * a = 0 + a = a [∵ a ∈ A ⇒ a < 6]
and a * 0 = a + 0 = a
∴ 0 * a = a = a * 0
Thus 0 behaves as the identity element for this operation.
Let a ≠ 0 ∈ A be any arbitary element.
Then 6 – a ∈ A
Now, a * (6 – a) = a + (6 – a) – 6 = 0
and (6 – a) * a = (6 – a) + o – 6 = 0
∴ (6 – a) * a = 0 = a * (6 – a) ∀ a ≠ 0 ∈ A
Thus 6 – a be the inverse of a and a ≠ 0 ∈ A be any arbitrary element. Hence all elements of the set A (except 0) are invertible.
The operation table for the binary operation * on A is given as under :

Question 14.
Examine whether the operation * defined on R by a * b = \(\sqrt{a^2+b^2}\) is a binary operation or not. It is a binary operation, whether it is associative or not ?
Solution:
Given operation * defined on R by
a * b = \(\sqrt{a^2+b^2}\) ∀ a, b ∈ R
[∵ closure property holds under multiplication on R]
⇒ \(\sqrt{a^2+b^2}\) ∈ R
[∀ a, b ∈ R ⇒ a + b ∈ R]
Thus * be a binary operation on R.
2, 3, 4 ∈ R, (2 * 3) * 4 = \(\sqrt{2^2+3^2}\) * 4
= \(\sqrt{13}\) * 4
= \(\sqrt{(\sqrt{13})^2+4^2}=\sqrt{29}\)

2 * (3 * 4) = 2 * \(\sqrt{3^2+4^2}\)
= 2 * 5
= \(\sqrt{2^2+5^2}=\sqrt{29}\)

Now a * (b * c)
= a * \(\sqrt{b^2+c^2}\)
= \(\sqrt{a^2+\left(\sqrt{b^2+c^2}\right)^2}\)
= \(\sqrt{a^2+b^2+c^2}\)

and (a * b) * c = \(\sqrt{a^2+b^2}\) * c
= \(\sqrt{\left(\sqrt{a^2+b^2}\right)^2+c^2}\)
= \(\sqrt{a^2+b^2+c^2}\)
Thus, a * (b * c) = (a * b) * c ∀ a, b, c ∈ R
Hence * be associative on R.

Question 14 (old).
Let S be the set of all rational numbers except 1 and * be defined on S by a * b = a + b – ab for all a, b ∈ S. Prove that
(i) * is a binary operation on S.
(ii) the operation is commutative as well as associative.
Find the identity element. Also find the inverse of an element a e A.
Solution:
Given operation * on R – {1} defined as
a * b = a + b – ab ∀ a, b ∈ R – { 1}
if a + b – ab = 1
⇒ a + b – ab – 1 = 0
⇒ (a – 1) – b (a – 1) = 0
⇒ (a – 1) (1 – b) = 0
⇒ a = 1 or b = 1 which is false as a, b ∈ R – {1}
Thus a * b = a + b – ab ∈ R – {1}

Commutativity :
∀ a, b ∈ R – {1}, we have
a * b = a + b – ab
b * a = b + a – ba = a + b – ab
[Since commutativity holds under addition and multiplication i.e. a + b = b + a; ab = ba]
∴ a * b = b * a ∀ a, b ∈ R – {1}
Thus * is commutative on R – {1}.

Associativity :
∀ a, b, c ∈ R – {1}
(a * b) * c = (a + b – ab) * c
= a + b – ab + c – (a + b – ab) c
= a + b + c – ab – ac – bc + abc …………… (1)
and a * (b * c) = a * (b + c – bc)
= a + b + c – bc – a (b + c – bc)
= a + b + c – bc – ab – ac + abc ………………. (2)
[Since commutativity, associativity holds under addition and multiplication]
∴ from (1) and (2) ; we have
(a * b) * c = a * (b * c) ∀ a, b, c ∈ R – {1}
∴ * is associative on R – {1}.
Let e be the identity element in R – {1}.
Then a * e = a = e * a ∀ a ∈ R – {1}
⇒ a * e = a and e * a = a ∀ a ∈ R – { 1}
⇒ a + e – ae = a and e + a – ca = a ∀ a ∈ R – { 1}
⇒ e (1 – a) = 0 and e (1 – a) = 0 ∀ a ∈ R – { 1}
⇒ e = 0 [∵ a ∈ R – {1} ⇒ a ≠ 1 ⇒ a – 1 ≠ 0]
Hence 0 be the identity element in R – {1}.
Let a be the arbitrary element of R – { 1} and
let x be the inverse of a.
Then a * x = e = x * a
⇒ a * x = 0 = x * a
⇒ a * x = 0 and x * a = 0
⇒ a + x – ax = 0 and x + a – xa = 0
⇒ a + x (1 – a) = a and a + x (1 – a) = 0
⇒ x = – \(\frac{a}{1-a}\) ∈ R – {1}
^ [∵ a ∈ R – {1} ∴- \(\frac{a}{1-a}\) ∈ R – {1}]
[For x = 1 ⇒ 1 – a = – a ⇒ 1 = 0, which is false]
Thus every element of R – {1} is invertible and — be the inverse of a in R – {1}.

Question 14.
Let S be the set of all rational numbers except 1 and * be defined on S by & a * b = a + b – ab for all a, b ∈ S. Prove that
(i) * is a binary operation on S.
(ii) the operation is commutative as well as associative.
Find the identity element. Also find the inverse of an element a ∈ A.
Solution:
Given operation * on R – {1} defined as
a * b = a + b – ab ∀ a, b ∈ R – {1}
if a + b – ab = 1
⇒ a + b – ab – 1 = 0
⇒ (a – 1) – b (a – 1) = 0
⇒ (a – 1) (1 – b) = 0
⇒ a = 1 or b = 1 which is false as a, b ∈ R – {1}
Thus a * b = a + b – ab ∈ R – {1}

Commutativity: ∀ a, b ∈ R – {1}, we have
a * b = a + b – ab
b * a = b + a – ba = a + b – ab
[Since commutativity holds under addition and multiplication i.e. a + b = b + a; ab = ba]
∴ a * b = b * a ∀ a, b ∈ R- {1}
Thus * is commutative on R – {1}.

Associativity: ∀ a, b ∈ R – {1}
(a * b ) * c = {a + b – ab) * c
= a + b – ab + c – (a + b – ab) c
= a + b + c – ab – ac – be + abc …………….(1)
and a * (b * c) = a * (b + c – bc)
= a + b + c – bc – a (b + c – bc)
= a + b + c – be – ab – ac + abc …………….. (2)
[Since commutativity, associativity holds under addition and multiplication]
∴ from (1) and (2) ; we have
(a * b) * c = a * (b * c) ∀ a, b, c ∈ R – {1}
∴ * is associative on R – {1}.
Let e be the identity element in R-{1}.
Then a * e = a = e * a ∀ a, b ∈ R – {1}
⇒ a * e = a and e * a = a ∀ a, b ∈ R – {1}
⇒ a + e – ae = a and e + a – ea = a ∀ a, b ∈ R -{ 1}
⇒ e (1 – a) = 0 and e (1 – a) = 0 ∀ a, b ∈ R – {1}
⇒ e = 0 [∵ a ∈ R – {1} ⇒ a ^ 1 ⇒ a – 1 ≠ 0]
Hence o be the identity element in R- (1}.
Let a be the arbitrary element of R – {1}
and let x be the inverse of a.
Then a * x = e = x * a
⇒ a * x = 0 = x * a
⇒ a * x = 0 and x * a = 0
⇒ a + x – ax = 0 and x + a – xa = 0
⇒ a + x (1 – a) = a and a + x (1 – a) = 0
⇒ x = – \(\frac{a}{1-a}\) ∈ R – {1}
[∵ a ∈ R – {1}
∴ – \(\frac{a}{1-a}\) ∈ R – {1}]
[For x = 1 ⇒ 1 – a = – a ⇒ 1 = 0, which is false]
Thus every element of R – {1} is invertible and – \(\frac{a}{1-a}\) be the inverse of a in R – {1}.