Let A = { x ∈ N: x 2 < 121 and x is a prime }
A = { 2, 3, 5, 7 }
(ii) The set of positive roots of the equation (x – 1) ( x + 1) (x – 1 ) = 0
The set of positive roots of the equations
(x – 1) (x + 1) (x 2 – 1) = 0
(x – 1 ) (x + 1 ) (x + 1) (x – 1) = 0
(x + 1 ) 2 (x – 1) 2 = 0
(x + 1) 2 = 0 or (x – 1) 2 = 0
x + 1 = 0 or x – 1 = 0
x = -1 or x = 1
A = { 1 }
(iii) {x ∈ N: 4x + 9 < 52}
4x + 9 < 52
4x + 9 – 9 < 52 – 9
4x < 43
x < \(\frac{43}{4}\) (i.e.) x < 10.75 4
But x ∈ N
∴ A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(iv)
Let A =
⇒ \(\frac{x-4}{x+2}\) = 3
⇒ x – 4 = 3(x + 2)
⇒ x – 4 = 3x + 6
⇒ 3x – x = – 4 – 6
2x = – 10
⇒ x = \(-\frac{10}{2}\) = -5
A = { -5 }
A = {x: x 2 – 1 = 0, x ∈ R}
Let A = { x ∈ N: x is an even prime number )
A = {2}
A is a finite set.
(ii) {x ∈ N: x is an odd prime number }
Let B = {x ∈ N: x is an odd prime number}
B = {1, 3, 5, 7, 11, …………….. }
B is an infinite set.
(iii) {x ∈ Z: x is even and < 10 }
C = {x ∈ Z: x is even and< 10}
C = { ……….. -8, -6, -4, -2, 0, 2, 4, 6, 8}
C is an infinite set.
(iv) {x ∈ R: x is a rational number }
D = { x ∈ R: x is a rational number }
D is an infinite set.
(v) {x ∈ N: x is a rational number }
E = { x ∈ N: x is a rational number )
E = {1, 2, 3, 4, 5, 6, …………..)
Every integer is a rational number.
∴ E is an infinite set.
To prove: A × (B ∩ C) = (A × B) ∩ (A × C)
B ∩ C = {8}; A = {1, 2, 5, 7}
So A × (B ∩ C) = {1, 2, 5, 7} × {8}
= {(1, 8), (2. 8), (5, 8), (7, 8)}
Now A x B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9), (7, 2), (7, 7), (7, 8), (7, 9)} …. ( 1)
A × C = {(1, 1), (1, 5),(1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)}
(A × B) ∩ (A × C) = {(1, 8), (2, 8), (5, 8), (7, 8)} ……… (2)
(1) = (2)
⇒ A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) A × (B ∪ C) = (A × B) ∪ (A × C)
Let A = {1, 2}, B = {3, 4}, C = {4, 5}
B ∪ C = {3, 4} ∪ {4, 5}
B ∪ C = {3, 4, 5)
A × (B ∪ C) = {1, 2} × {3, 4, 5}
A × (B ∪ C) = { (1, 3),( 1, 4),(1, 5),(2, 3), (2, 4),(2,5)} ——– (1)
A × B = {1, 2} × {3, 4}
A × B = { (1, 3), (1, 4), (2, 3), (2, 4) }
A × C = {1, 2} × {4, 5}
A × C = { (1, 4), (1, 5), (2, 4), (2, 5 )}
(A × B) ∪ (A × C) = {(1, 3), (1, 4), (2, 3), (2, 4)} ∪ {(1, 4 ), (1, 5 ), ( 2, 4 ), (2, 5)}
(A × B) ∪ (A × C) = { (1, 3) (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)} —— (2)
From equations (1) and (2)
A × (B U C) = (A × B) U (A × C)
(iii) (A × B) ∩ (B × A) = (A ∩ B) × ( B ∩ A)
A × B = {(1, 2), (1, 7), (1, 8), (1, 9) (2, 2), (2, 7), (2, 8), (2, 9) (5, 2), (5, 7), (5, 8), (5, 9) (7, 2), (7, 7), (7, 8), (7, 9)}
B × A = {(2, 1), (2, 2), (2, 5), (2, 7) (7, 1), (7, 2), (7, 5), (7, 7) (8, 1), (8, 2), (8, 5), (8, 7) (9,1), (9, 2), (9, 5), (9, 7)}
L.H.S. (A × B) ∩ (B × A) = {(2, 2), (2, 7), (7, 2), (7, 7)} …. (1)
R.H.S. A ∩ B = {2, 7}
B ∩ A = {2, 7}
(A ∩ B) × (B ∩ A) = {2, 7} × {2, 7}
= {(2, 2), (2, 7), (7, 2), (7, 7)} ……… (2)
(1) = (2) ⇒ LHS = RHS
(iv) C – (B – A) = (C ∩ A) ∪ (C ∩ B’)
Let A = {1, 2, 3), B = {2, 3, 4), C = {3, 4, 5}
B – A = { 2, 3, 4 ) – {1, 2, 3}
B – A = {4}
C – (B – A) = {3, 4, 5} – {4}
C – (B – A) = {3, 5} —- (1)
C ∩ A = {3, 4, 5} ∩ { 1, 2, 3)
C ∩ A = {3}
B’ = {1, 5}
C ∩ B’ = {3, 4, 5} ∩ {1, 5}
C ∩ B’ = {5}
(C ∩ A) ∪ (C∩B’) = {3} ∪ {5}
(C∩A) ∪ (C ∩ B’) = {3, 5} —— (2)
From equations (1) and (2)
C – (B – A) = (C ∩ A) u (C ∩ B’)
(v) (B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A)
To prove (B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A)
A= {1, 2, 5, 7}, B = {2, 7, 8, 9}, C = {1, 5, 8, 10}
Now B – A = {8, 9}
(B – A) ∩ C = {8} ……. (1)
B ∩ C = {8}
A = {1, 2, 5, 7}
So (B ∩ C) – A = {8} …… (2)
C – A = {8, 10}
B = {2, 7, 8, 9}
B ∩ (C – A) = {8} …. (3)
(1) = (2) = (3)
(vi) (B – A) ∪ C = (B ∪ C) – (A – C)
Let A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8}
B – A = {3, 4, 5, 6} – {1, 2, 3, 4}
B – A = {5, 6}
(B – A) ∪ C = {5, 6} ∪ {5, 6, 7, 8}
(B – A) ∪ C = { 5, 6, 7, 8 } ——- (1)
B ∪ C = { 3, 4, 5, 6 } ∪ { 5, 6, 7,8 }
B ∪ C = { 3, 4, 5, 6, 7, 8 }
A – C = { 1, 2, 3, 4 } – { 5, 6, 7, 8 }
A – C = { 1, 2, 3, 4 }
(B ∪ C) – (A – C) = {3, 4, 5, 6, 7, 8} – {1, 2, 3, 4}
(B ∪ C) – (A – C) = { 5, 6, 7, 8 } —-(2)
From equations (1) and (2)
(B – A) ∪ C = (B ∪ C) – (A – C)
A set itself can be a subset of itself (i.e.) A ⊆ A.
But it cannot be a proper subset.
Given n(P(A)) = 1024, n(A ∪ B) = 15, n(P(B)) = 32
n(P(A)) = 1024 = 2 10 n(A) = 10
n(P(B)) = 32 = 2 5 = n(B) = 5
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
15 = 10 + 5 – n(A ∩ B)
15 = 15 – n (A ∩ B)
n(A ∩ B) = 0
n(A ∪ B) = 10; n(A ∩ B) = 3
n(A ∆ B) = 10 – 3 = 7
and n(P(A ∆ B)) = 27 = 128
Given A × A contains 16 elements.
∴ A contains 4 elements.
Also, (1, 3) and (0, 2) are two elements of A × A.
∴ A = { 0, 1, 2, 3 }
n(A) = 3 ⇒ set A contains 3 elements
n(B) = 2 ⇒ set B contains 2 elements –
we are given (x, 1), (y, 2), (z, 1) are elements in A × B ⇒ A = {x, y, z} and B = {1, 2}
Given A × A has 16 elements.
∴ A has 4 elements.
Also S = {(a, b) ∈ A × A; a < b}
Given (-1, 2) and (0, 1) ∈ S
A = {-1, 0, 1, 2 }
The elements of S are
S = { (-1, 0), (-1, 1),(-1, 2),(0, 1), (0, 2), (1, 2)}
∴ The other elements of the sets are
(-1, 0), (-1, 1), (0, 2), (1, 2)
S = {set of all positive integers}
(a) mRm ⇒ ‘m’ divides’m’ ⇒ reflexive
(b) mRn ⇒ m divides n but
nRm ⇒ n does not divide m
(i.e.,) mRn ≠ nRm
It is not symmetric
(c) mRn ⇒ nRr as n divides r
It is transitive
(ii) Let P denote the set of all straight lines in a plane. The relation R defined by“ l R m if l is perpendicular to m”.
Let P denote the set of all straight lines in a plane. The relation R is defined by l R m if l is perpendicular to m.
R = {(l, m): l is perpendicular to m}
(a) Reflexive:
Let l be any line in the plane P. Then line l is not perpendicular to itself.
{1, 1) ∉ R
∴ R is not reflexive.
(b) Symmetric:
Let (1, m) ∉ R ⇒ l is perpendicular to m
∴ m is perpendicular to l.
Hence (m, l) ∈ R
∴ R is symmetric.
(c) Transitive;
Let (l, m), (m, n) ∈ R
⇒ l is perpendicular to m.
∴ l is parallel to n. (l, n) ∉ R
Hence R is not transitive.
(iii) Let A be the set consisting of all the members of a family. The relation R defined by “a R b if a is not a sister of b”.
A = {set of all members of the family}
aRb is a is not a sister of b
(a) aRa ⇒ a is not a sister of a It is reflexive
(b) aRb ⇒ a is not a sister of b.
bRa ⇒ b is not a sister of a.
It is symmetric
(c) aRb ⇒ a is not a sister of b.
bRc ⇒ b is not a sister of c.
⇒ aRc ⇒ a can be a sister of c
It is not transitive.
(iv) Let A be the set consisting of all the female members of a family. The relation R defined by “a R b if a is not a sister of b”.
Given A is the set containing female members of the family.
Let M = Mother
H = Female child
A = { M, H }
The relation R on A is defined by aRb if a is not a sister of b.
R = {(M, M), (M, H), (H, H), (H, M)}
(a) Reflexive:
Clearly (M, M) and (H, H) ∈ R.
∴ R is reflexive.
(b) Symmetric:
For (M, H) ∈ R, we have (H, M ) ∈ R
∴ R is symmetric.
(c) Transitive:
For (M,M),(M,H) ∈ R ⇒ (M, H) ∈ R
(M,H),(H,H) ∈ R ⇒ (M, H) ∈ R
(H,H),(H,M) ∈ R ⇒ (H, M) ∈ R
(H,M),(M,M) ∈ R ⇒ (H, M) ∈ R
(H,M),(M,H) ∈ R (H, H) ∈ R
∴ R is transitive.
(v) On the set of natural numbers the relation R defined by “x R y if x + 2y = 1”
N= {1, 2, 3, 4, 5,….}
xRy if x + 2y = 1 R is an empty set
(a) xRx ⇒ x + 2x = 1 ⇒ x = \(\frac{1}{3}\) ∉ N. It is not reflexive
xRy = yRx ⇒ x + 2y = 1 It does not imply that y + 2x = 1 as y = \(\frac{1-x}{2}\) It is not symmetric.
(b) -x = y ⇒ (-1, 1) ∉ N
It is not transitive.
Given X = { a, b, c, d }
R = { (a, a), (b, b), (a, c) }
(i) The minimum ordered pairs to be included to R in order to make R to be reflexive is (c, c) and (d, d)
(ii) The minimum ordered pairs to be included to R in order to make R to be symmetric is (c, a).
(iii) R is transitive. We need not add any pair.
(iv) After adding the ordered pairs (c, c),(d, d), (c, a) the new relation becomes
R, = {(a, a), (b, b), (c, c), (d, d), (a, c), (c, a)}
The new relation satisfies, reflexive, symmetric and transitive property.
∴ R 1 is an equivalence relation.
Given A = {a, b, c }
R = { (a, a), (b, b),(a, c) }
(i) The minimum ordered pair to be included to R in order to make it reflexive is (c, c).
(ii) The minimum ordered pair to be included to R in order to make it symmetrical is (c, a).
(iii) R is transitive. We need not add any pair.
(iv) After including the ordered pairs (c, c),(c, a) to R the new relation becomes
R 1 = { (a, a), (b, b), (c, c), (a, c), (c, a) }
R 1 is reflexive symmetric and transitive.
∴ R 1 is an equivalence relation.
Given P = the set of all triangles in a plane.
R is the relation defined by a R b if a is similar to b.
R = {(a, b): a is similar to b for a, b ∈ p }
(a) Reflexive:
(a, a) ⇒ a is similar to a for all a ∈ P
∴ R is reflexive.
(b) Symmetric:.
Let (a,b) ∈ R ⇒ a is similar to b
⇒ b is similar to a
∴ (b, a) ∈ R
Hence R is symmetric.
c) Transitive:
Let (a, b) and ( b, c) ∈ R
(a, b) ∈ R ⇒ a is similar to b
(b, c) ∈ R ⇒ b is similar to c
∴ a is similar to c.
Hence R is transitive.
∴ R is an equivalence relation on P.
Given N = set of natural numbers
R is the relation defined by a R b if 2a + 3b = 30
When a > 15, b negative and does not belong to N.
∴ R = { (3,8),(6,6), (9,4), (12,2)}.
(i) R is not reflexive since (a, a) ∉ R for all a ∈ N.
(ii) R is not symmetric since for (3, 8) ∈ R, (8, 3) ∉ R
(iii) Clearly R is transitive since we cannot find elements (a, b), (b, c) in R such that (a, c) ∉ R
∴ R is not an equivalence relation.
(a) S = aRa (i.e. ) a person can be a friend to himself or herself.
So it is reflective.
(b) aRb ⇒ bRa so it is symmetric
(c) aRb, bRc does not ⇒ aRc so it is not transitive
⇒ It is not an equivalence relation
N = the set of natural numbers.
R is the relation defined on N by
a R b if a + b ≤ 6
R = { (a, b), a, b ∈ N / a + b ≤ 6}
a + b ≤ 6 ⇒ b ≤ 6 – a
a = 1,
b ≤ 6 – 1 = 5
b is 1, 2, 3, 4, 5
∴ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5) ∈ R
a = 2,
b ≤ 6 – 2 = 4
b is 1, 2, 3, 4
∴ (2, 1), (2, 2),(2, 3), (2, 4) ∈ R
a = 3,
b < 6 – 3 = 3
b is 1, 2, 3
∴ (3, 1), (3, 2), (3, 3) ∈ R
a = 4,
b < 6 – 4 = 2
b is 1, 2
∴ (4, 1), (4, 2) ∈ R
a = 5,
b < 6 – 5 = 1
b is 1
∴ (5, 1) ∈ R
∴ R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)}
(i) Reflexive:
R is not reflexive since (4, 4), (5, 5) ∈ R
(ii) Symmetric:
Cleary R is symmetric forever (x, y) ∈ R, we have (y, x) ∈ R.
(iii) Transitive:
(3, 1), (1, 5) ∈ R ⇒ (3,5) ∉ R
∴ R is not transitive.
(iv) R is not an equivalence relation.
R = {{a, a), (b, b), (c, c)} is this smallest cardinality of A to make it equivalence relation n(R) = 3
(i) R = {(a, a), {a, b), (a, c), (b, c), (b, b), {b, c), (c, a), (c, b), (c, c)}
n(R) = 9 is the largest cardinality of R to make it equivalence.
Z = set of all integers
Relation R is defined on Z by m R n if m – n is divisible by 7.
R = {(m, n), m, n ∈ Z/m – n divisible by 7}
m – n divisible by 7
∴ m – n = 7k where k is an integer.
a) Reflexive:
m – m = 0 = 0 × 7
m – m is divisible by 7
∴ (m, m ) ∈ R for all m ∈ Z
Hence R is reflexive.
b) Symmetric:
Let (m, n ) ∈ R ⇒ m – n is divisible by 7
m – n = 7k
n – m = – 7k
n – m = (-k)7
∴ n – m is divisible by 7
∴ (n, m) ∈ R.
c) Transitive:
Let (m, n) and (n, r) ∈ R
m – n is divisible by 7
m – n = 7k ——— (1)
n – r is divisible by 7
n – r = 7k 1 ——— (2)
(m – n) + (n – r) = 7k + 7k 1
m – r = ( k + k 1 ) 7
m – r is divisible by 7.
∴ (m, r) ∈ R
Hence R is transitive.
R is an equivalence relation.
Given A denotes the set of students and B denotes the set of sections. Aslo given there 120 students and 4 sections.
Let f be a relation from A to B as “x related to y if the student x belongs to the section y”
Two are more students in A may belong to same section in B. But one student in A cannot belong to two or more sections in B. Every student in A can belong to any one of the section in B. Therefore / is a function.
In B we can have sections without students. Every element in B need not have preimage in A.
∴ f need not be onto.
Thus, f is a function and inverse relation for f need not exist.
When x = -4
f(x) = – x + 4
f(-4) = – (-4) + 4
= 4 + 4 = 8
When x = 1
f(x) = x – x 2
f(1) = 1 – 1 2
= 1 – 1 = 0
When x = -2
f(x) = x 2 – x
f(-2) = (-2) 2 – (-2)
= 4 + 2 = 6
When x – 7
f(x) = 0
⇒ f(7) = 0
When x = 0
f(x) = x 2 – x
⇒ f(0) = 0 2 – 0 = 0
When x = – 3
f(x) = x 2 + x – 5
f(-3) = (-3) 2 + (-3) – 5
= 9 – 3 – 5
= 9 – 8 = 1
When x = 5
f(x) = x 2 + 3x – 2
f(5) = 5 2 + 3(5) – 2
= 25 + 15 – 2
= 40 – 2 = 38
When x = 2
f(x) = x 2 – 3
f(2) = 2 2 – 3 = 4 – 3 = 1
When x = – 1
f(x) = x 2 + x – 5
f(-1) = (-1) 2 – 1 – 5 = 1 – 1 – 5 = – 5
When x = 0
f(x) = x 2 – 3
f(0) = 0 2 – 3
A = { a, b, c }
f = {(a, c), (b, c), (c, b)}; f: A → A
f is a function since every element in the domain has a unique image in the codomain.
f is not one-one.
a, b belonging to the domain A has the same image in the codomain A. f is not onto since belonging to the codomain A does not have preimage in the domain A Thus the relation / is a function from A to A and it is neither one-one nor onto.
(ii) If X = { x, y, z } and f = { (x, y), (x, z), (z, x) }; (f: X → X)
X = { x, y, z }
f = {(x, y), (x, z), (z, x) } f: X → X
The relation f: X → X is not a function since the element x in the domain has two images in the co-domain.
f = { (1, b), (2, c), (3, d), (4, d)
f is a function, it not one to one and not onto.
(ii) not one – to – one but onto
Does not exists
(iii) one – to – one but not onto
Does not exist
(iv) one – to – one and onto
f = { (1, a), (2, b), (3, c), (4, d) }
f is a function which is one – to – one and onto.
Let f(x) = \(\frac{1}{1-2 \sin x}\)
When 1 – 2 sin x = 0
⇒ 1 = 2 sin x
sin x = \(\frac{1}{2}\)
⇒ sin x = sin \(\left(\frac{\pi}{6}\right)\)
x = nπ + (- 1) n \(\frac{\pi}{6}\), n ∈ Z
sin x = sin α ⇒ x = nπ + (-1) n d, n ∈ Z
∴ Domain of f(x) is
∴ For no real values of x, f (x) is defined.
∴ Domain of f(x) = { }
Let f(x) = \(\frac{1}{2 \cos x-1}\)
Range of cosine function is
– 1 ≤ cos x ≤ 1
– 2 ≤ 2 cos x ≤ 2
– 1 ≤ 2 cos x – 1 ≤ 2 – 1
– 3 ≤ 2 cos x – 1 ≤ 1
\(-\frac{1}{3}\) ≥ \(\frac{1}{2 \cos x-1}\) ≥ 1
xy = – 2 ⇒ y = -2/x
which is a function
The domain is (-∞, 0) ∪ (0, ∞) and range is R – {0}
Given
Given f: R → R, g: R → R and h: R → R (f + g) oh: R → R and (f o h + g o h): R → R for any x ∈ R.
[(f + g)oh] (x) = (f + g) h(x)
= f(h(x)) + g(h(x))
= foh(x) + goh(x)
∴ (f + g)oh = foh + goh
Also fo(g + h)(x) = f((g + h)(x)) for any x ∈ R
= f(g(x) + h(x))
= f(g(x)) + f(h(x))
= fog (x) + foh(x)
∴ fo(g + h) = fog +foh
Given f(x) = 3x – 5
Let y = 3x – 5
y + 5 = 3x
⇒ \(\frac{y+5}{3}\) = x
Let g(y) = \(\frac{y+5}{3}\)
gof (x) = g(f(x))
= g(3x – 5)
= \(\frac{3 x-5+5}{3}\) = \(\frac{3 x}{3}\) = x
gof (x) = x
fog (y) = f(g(y))
= f\(\left(\frac{y+5}{3}\right)\)
= 3\(\left(\frac{y+5}{3}\right)\) – 5
= y + 5 – 5
fog(y) = y
∴ gof = I x and fog = I Y
Hence f and g are bijections and inverses to each ot1er.
Hence f is a bijection and f -1 (y) = \(\frac{y+5}{3}\)
Replacing y by x we get f -1 (x) = \(\frac{x+5}{3}\)
W(x) = 0.35x
Since bodyweight x is positive and if it increases then W(x) also increases.
Domain is (0, ∞) i.e.,x > 0
Given s (t) = – 16t 2
s (t 1 ) = s (t 2 ) ⇒ – 16t 1 2 = – 16t 2 2
⇒ t 1 2 = t 2 2
⇒ ± t 1 = ± t 2
Since s (t 1 ) = s (t 1 )
14 t 1 = t 2
∴ The function s(t) is not one-one
Graph of s(t) = – 16t 2
Take the time along x – axis and distance along y – axis.
C – base cost,
S = fuel surcharge,
m = mileage
C(m) = 0.4 m + 50
S(m) = 0.03 m
Total cost = C(m) + S(m)
= 0.4 m + 50 + 0.03 m
= 0.43 m + 50
for 1600 miles
T(c) = 0.43 (1600) + 50 = 688 + 50 = ₹ 738
Given A (x) = 30,000 + 0.04 x
S (x) = 25,000 + 0.05x
A(x) + S(x) = 30,000 + 0.04 x + 25,000 + 0.05x
(A + S)(x) = 55,000 + 0.09 x
Given x = 1,50,00,000
Then (A + S)(x) = 55000 + 0.09 × 1,50,00,000
= 55000 + 1,35000000
Total family income = Rs. 14,05,000
Given f(x) = 1.23x
where x represents the number of American dollars
g(y) = 50.50y
where y represents the number of Singapore dollars.
To convert American dollars to Indian rupees, we must find
gof (x) = g(f(x))
= g (1.23x)
= 50.50 (1.23x)
= 62.115x
∴ The function for the exchange rate of American can dollars in terms of Indian rupees is
gof (x) = 62.1 15x
cost of one meal = ₹ 100
Total cost = ₹ 100 (200 – x)
Number of customers = 200 – x
Day revenue = ₹ (200 – x) x
Total profit = day revenue – total cost
= (200 – x) x – (100) (200 – x)
Given f(x) = 3x – 4
Let y = 3x – 4
⇒ y + 4 = 3x
⇒ x = \(\frac{y+4}{3}\)
Let g(y) = \(\frac{y+4}{3}\)
gof (x) = g (f(x) )
= g(3x – 4)
= \(\frac{3 x-4+4}{3}=\frac{3 x}{3}\)
gof(x) = x
and fog(y) = f(g(y))
= f\(\left(\frac{y+4}{3}\right)\)
= 3\(\left(\frac{y+4}{3}\right)\)
= y + 4 – 4 = y
fog (y) = y
Hence g of = I x and fog = I y
This shows that f and g are bijections and inverses of each other.
Hence f is bijection and f -1 (y) = \(\frac{y+4}{3}\)
Replacing y by x we get f -1 (x) = \(\frac{x+4}{3}\)
The line y = x
f(x) =
The line y =3x-4
When x = 0 ⇒ y = 3 × 0 – 4 = -4
When x = 1 ⇒ y = 3 × 1 – 4 = -1
When x = -1 ⇒ y = 3 × -1 – 4 = -7
When x = 2 ⇒ y = 3 × 2 – 4 = 2
When x = -2 ⇒ y = 3 × -2 – 4 = -10
When x = 3 ⇒ y = 3 × 3 – 4 = 5
When x = -3 ⇒ y = 3 × -3 – 4 = -13
The line y = \(\frac{x+4}{3}\)
(i) y = – x 1/3
-y = x 1/3
(-y) 3 = x
-y 3 = x
When y = 0 ⇒ – 0 3 ⇒ x = 0
y = 1 ⇒ – 1 3 = x ⇒ x = – 1
y = 2 ⇒ – 2 3 = x ⇒ x = – 8
y = 3 ⇒ – 3 3 = x ⇒ x = – 27
y = -1 ⇒ – (-1) 3 = x ⇒ x = 1
y = -2 ⇒ – (-2) 3 = x ⇒ x = 8
y = -3 ⇒ – (- 3) 3 = x ⇒ x = 27
The graph of y = – x 1/3 is the reflection of the graph of y = x 1/3 about the x-axis.
The graph of y = – f(x) is the reflection of the graph of y = f(x) about x – axis.
(ii) y = x 1/3 + 1
y – 1 = x 1/3
⇒ (y – 1) 3 = x
When
y = 0 ⇒ (0 – 1 ) 3 = x ⇒ x = – 1
y = 1 ⇒ (1 – 1) 3 = x ⇒ x = 0
y = 2 ⇒ ( 2 – 1 ) 3 = x ⇒ x = 1
y = 3 ⇒ (3 – 1) 3 = x ⇒ x = 8
y = -1 ⇒ (-1 – 1) 3 = x ⇒ x = – 8
y = -2 ⇒ (-2 – 1) 3 = x ⇒ x = – 27
The graph of y = x 1/3 + 1 causes the graph y = x 1/3 a shift to the upward by 1 unit.
The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.
(iii) y = x 1/3 – 1
y + 1 = x 1/3
( y + 1) 3 = x
When
y = 0 ⇒ (0 + 1) 3 = x ⇒ x = 1
y = 1 ⇒ ( 1 + 1) 3 = x ⇒ x = 8
y = 2 ⇒ (2 + 1) 3 = x ⇒ x = 27
y = – 1 ⇒ (-1 + 1) 3 = x ⇒ x = 0
y = – 2 ⇒ (-2 + 1) 3 = x ⇒ x = – 1
y = – 3 ⇒ (-3 + 1) 3 = x ⇒ x = – 8
The graph of y = x 1/3 – 1 causes the graph y = x 1/3 a shift to the downward by 1 unit.
The graph of y = f(x) – d, d > 0 causes the graph y = f(x) a shift to the downward by d units.
(iv) (x + 1) 1/3
y 3 = x + 1
When
y = 0 ⇒ 0 3 = x + 1 ⇒ x = -1
y = 1 ⇒ 1 3 = x + 1 ⇒ x = 0
y = 2 ⇒ 2 3 = x + 1 ⇒ x = 8 – 1 = 7
y = 3 ⇒ 3 3 = x + 1 ⇒ x = 27 – 1 = 26
y = – 1 ⇒ (-1) 3 = x + 1 ⇒ x = – 1 – 1 = – 2
y = -2 ⇒ (-2) 3 = x + 1 ⇒ x = – 8 – 1 = – 9
The graph of y = (x + 1) 3 causes the graph y = x 1/3 a shift to the left by 1 unit.
The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.
Given functions are f(x) = x3 and g(x) = x 1/3
fog (x) = f(g(x))
= f\(\left(x^{\frac{1}{3}}\right)\)
= \(\left(x^{\frac{1}{3}}\right)^{3}\) = x
f(x) = x 3
g(x) = x 1/3
Graph of fog(x) = x
Since fog(x) = x is symmetric about the line y = x, g(x) is the inverse image of f(x).
∴ g(x) = f -1 (x)
y = sin x
(i) y = sin(-x)
y = – sin x
The graph of y = sin (- x) is the reflection of the graph of y = sin x about y-axis.
The graph of y = f(- x) is the reflection of the graph of y = f(x) about y – axis.
(ii) y = – sin (-x)
y = sin x
y = – sin (-x) is the reflection of y = sin (-x) about the x – axis.
The graph of y = – f( x) is the reflection of the graph of y = f( x) about x – axis.
(iii) y = sin\(\left(\frac{\pi}{2}+x\right)\)
The graph of y = sin \(\left(\frac{\pi}{2}+x\right)\) causes y = sin x a shift to the left by \(\frac{\pi}{2}\) units.
The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.
(iv) y = sin\(\left(\frac{\pi}{2}-x\right)\)
The graph of sin \(\left(\frac{\pi}{2}-x\right)\) causes the graph y = sin x a shift to the right by \(\frac{\pi}{2}\) unit.
The graph of y = f(x – c), c > 0 causes the graph y = f(x) a shift to the right by c units.
(i) y = -x
Graph of y = – x is the reflection of the graph of y = x about the x – axis.
The graph of y = – f(x) is the reflection of the graph of y = f(x) about x – axis.
(ii) y = 2x
The graph of y = 2x compresses the graph y = x towards the y-axis that is moving away from the x-axis since the multiplying factor is 2 which is greater than 1.
The graph of y = k f(x), k > 0 moves away from the x-axis if k is greater than 1.
(iii) y = x + 1
The graph of y = x + 1 causes the graph y = x shift to upward by 1 unit.
The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.
(iv) y = \(\frac{1}{2}\)x + 1
When
x = 0 ⇒ y = \(\frac{1}{2}\) × 0 + 1 = 1
x = 2 ⇒ y = \(\frac{1}{2}\) × 2 + 1 = 2
x = 4 ⇒ y = \(\frac{1}{2}\) × 4 + 1 = 2 + 1 = 3
x = 6 ⇒ y = \(\frac{1}{2}\) × 6 + 1 = 3 + 1 = 4
x = – 2 ⇒ y = \(\frac{1}{2}\) × – 2 +1= – 1 + 1 = 0
x = – 4 ⇒ y = \(\frac{1}{2}\) × – 4 + 1 = – 2 + 1 = – 1
x = – 6 ⇒ y = \(\frac{1}{2}\) × – 6 + 1 = – 3 + 1 = – 2
The graph of y = \(\frac{1}{2}\)x + 1 stretches the graph y = x towards the x – axis since the multiplying factor is \(\frac{1}{2}\) which is less than 1 and shifts to the upward by 1 unit.
The graph of y = kf(x), k > 0 moves towards the x-axis if k is less than 1.
The graph of y = f(x) + d, d >0 causes the graph y = f(x) a shift to the upward by d units.
(v) 2x + y + 3 = 0
y = -2x – 3
When
x = 0 ⇒ y = -2 × 0 – 3 = -3
x = 1 ⇒ y = -2 × 1 – 3 = -5
x = \(\frac{1}{2}\) ⇒ y = – 2 × \(\frac{1}{2}\) – 3 = – 1 – 3 = – 4
x = 2 ⇒ y = -2 × 2 – 3 = – 4 – 3 = – 7
x = – 1 ⇒ y = -2 × – 1 – 3 = 2 – 3 = – 1
x = – 2 ⇒ y = 2 × – 2 – 3 = 4 – 3 = 1
x = – 3 ⇒ y = -2 × -3 – 3 = 6 – 3 = 3
The graph of y = – 2x – 3 stretches the graph y = x towards the x-axis since the multiplying factor is – 2 which is less than 1 and causes the shift to the downward by 3 units.5
The graph of y = kf(x), k > 0 moves towards the x-axis if k is less than 1.
The graph of y = f(x) – d, d >0 causes the graph y = f(x) a shift to the downward by d units.
y = |x|
(i) y = |x – 1| + 1
(a) Consider y = |x – 1|
x = 0 ⇒ y = – x + 1 ⇒ y = 1
x = 1 ⇒ y = x – 1 ⇒ y = 0
x = 2 ⇒ y = x – 1 ⇒ y = 1
x = 3 ⇒ y = x – 1 ⇒ y = 2
x = – 1 ⇒ y = – x + 1 ⇒ y = 2
x = – 2 ⇒ y = – x + 1 ⇒ y = 3
The graph of y = |x – 1| causes the graph y = |x| a shift to the right by 1 unit.
The graph of y = f(x – c), c > 0 causes the graph y = f(x) a shift to the right by c units.
(b) Consider y = |x – 1| + 1
x = 0 ⇒ y = – x + 2 ⇒ y = 2
x = 1 ⇒ y = x ⇒ y = 1
x = 2 ⇒ y = x ⇒ y = 2
x = 3 ⇒ y = x ⇒ y = 3
x = – 1 ⇒ y = – x + 2 ⇒ y = 3
x = – 2 ⇒ y = – x + 2 ⇒ y = 4
The graph of y = |x – 1| + 1 shift the graph y = |x| to the right by 1 unit and causes a shift to the upward by 1 unit.
The graph of y = f( x – c), c > 0 causes the graph y = f(x) a shift to the right by c units.
The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.
(ii) y = |x + 1| – 1
(a) Consider y = |x + 1|
x = 0 ⇒ y = x + 1 ⇒ y = 1
x = 1 ⇒ y = x + 1 ⇒ y = 2
x = 2 ⇒ y = x + 1 ⇒ y = 3
x = 3 ⇒ y = x + 1 ⇒ y = 4
x = – 1 ⇒ y = x + 1 ⇒ y = 0
x = – 2 ⇒ y = – (x + 1) ⇒ y = 1
x = – 3 ⇒ y = – (x + 1) ⇒ y = 2
The graph of y = |x + 1| shifts the graph y = |x| to the left by 1 unit.
The graph of y = f( x + c), c > 0 causes the graph y = f(x) a shift to the left by e units.
(b) Consider y = |x + 1| – 1
x = 0 ⇒ y = x ⇒ y = 0
x = 1 ⇒ y = x ⇒ y = 1
x = 2 ⇒ y = x ⇒ y = 2
x = 3 ⇒ y = x ⇒ y = 3
x = – 1 ⇒ y = x ⇒ y = – 1
x = – 2 ⇒ y = – x – 2 ⇒ y = 0
x = – 3 ⇒ y = – x – 2 ⇒ y = 1
x = – 4 ⇒ y = – x – 5 ⇒ y = -1
The Graph of y = |x + 1 | – 1 shift the graph y = |x| to the left by 1 unit and causes a shift downward by 1 unit.
The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.
The graph of y = f(x) – d, d > 0 causes the graph y = f(x) a shift to the downward by d units.
(iii) y = |x + 2| – 3
(a) Consider the curve y = |x + 2|
x = 0 ⇒ y = x + 2 ⇒ y = 2
x = 1 ⇒ y = x + 2 ⇒ y = 3
x = 2 ⇒ y = x + 2 ⇒ y = 4
x = 3 ⇒ y = x + 2 ⇒ y = 5
x = – 1 ⇒ y = x + 2 ⇒ y = 1
x = – 2 ⇒ y = x + 2 ⇒ y = 0
x = – 3 ⇒ y = – x – 2 ⇒ y = 1
The graph of y = |x + 2| shifts the graph y = |x| to the left by 2 units.
The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.
(b) Consider the curve y = |x + 2| – 3
x = 0 ⇒ y = x – 1 ⇒ y = – 1
x = 1 ⇒ y = x – 1 ⇒ y = 0
x = 2 ⇒ y = x – 1 ⇒ y = 1
x = 3 ⇒ y = x – 1 ⇒ y = 2
x = – 1 ⇒ y = x – 1 ⇒ y = – 2
x = – 2 ⇒ y = x – 1 ⇒ y = – 3
x = – 3 ⇒ y = – x – 5 ⇒ y = – 2
The graph of y = |x + 2| – 3 shifts the graph y = |x| to the left by 2 units and causes a shift downward by 3 units.
The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.
The graph of y = f(x) – d, d > 0 causes the graph y = f(x) a shift to the down ward by d units.
y = sin |x|
(a) y = sin x
(b) Consider y = sin |x|
(3) 1
Explaination:
Given
A = {(x,y): y = e x, x ∈ R}
B = {(x,y): y = e -x, x ∈ R}
Consider the curve y = e x
When x = 0 ⇒ y = e -0 = 1
When x = ∞ ⇒ y = e -∞ = ∞
When x = -∞ ⇒ y = e ∞ = 0
Consider the curve y = e -x.
When x = 0 ⇒ y = e -0 = 1
When x = ∞ ⇒ y = e -∞ = 0
When x = -∞ ⇒ y = e ∞ = ∞
(0, 1) is the only point common to y = e<sup.x and y = e -x
∴ A ∩ B = {(0,1)} ⇒ n (A ∩ B ) = 1
(2) infinitely many elements
Explaination:
Given A = { (x, y): y = sin x, x ∈ R}
B = {(x, y): y = cos x, x ∈ R }
Consider the equations y = sin x and y = cos x
sin x = cos x ⇒ \(\frac{\sin x}{\cos x}\) = 1
tan x = 1 ⇒ x = nπ + \(\) for n ∈ z
There are infinite number of common points for the sets A and B
(4) Range of R is {0, -1, 1}
Explaination:
A= {0, -1, 1, 2}
|x 2 + y 2 | ≤ 2
The values of x and y can be 0, -1 or 1
So range = {0, -1, 1}
(1)
Explaination:
f(x) = |x – 2| + |x + 2|, x ∈ R
Divide the Real line into three intervals
In the interval (2, ∞), both the factors x – 2 and x + 2 are positive.
∴ f(x) = x – 2 + x + 2 = 2x
f(x) = 2x for all x ∈ (2, ∞)
In the interval (- ∞, – 2 ] both the factors x – 2 and x + 2 are negative.
∴ f(x) = – (x – 2) – (x + 2)
= – x + 2 – x – 2 = – 2x
∴ f(x) = – 2x for all x ∈ (- ∞,- 2]
In the interval (—2, 2], the factor x – 2 is negative and the factor x + 2 is positive.
∴ f(x) = – (x – 2) + (x + 2)
f(x) = – x + 2 + x + 2 = 4
Thus f(x) = 4 for all x ∈ (- 2, – 2]
(1) T is an equivalence relation but S Is not an equivalence relation
Explanation:
(0, 1), (1, 2) it is not an equivalence relation
T is an equivalence relation
(4) N
Explaination:
Let N = {1, 2, 3, ……….. 10}
A = { 1, 2, 3, 4, 5 }
B = {6, 7, 8, 9, 10}
A’ = {6, 7, 8, 9, 10 }
B’ = { 1, 2, 3, 4, 5 }
A ∪ B = {1, 2, 3, 4, 5} ∪ {6, 7, 8, 9, 10}
A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(A ∪ B) ∩ B’ = {1,2, 3, 4, 5, 6,7, 8, 9,10} ∩ { 1, 2, 3, 4, 5 }
(A ∪ B) ∩ B’= {1,2, 3,4, 5}
A’ ∪ [(A ∪ B ) ∩ B’] = { 6, 7, 8, 9, 10 } ∪ {1, 2, 3, 4, 5 }
= {1, 2, 3, 4,5, 6, 7, 8, 9, 10}
A’ ∪ [(A ∪ B) ∩ B’] = N
(2) 1130
Explanation:
n(M ∪ C) = n(M) + n(C) – n(M ∩ C)
= 700 + 500 – 70
= 1130
(2) 4
Explaination:
Given n[(A × B) n (A × C)] = 8
n(B ∩ C) = 2
n[(A × B) ∩ (A × C)] = 8
A × (B ∩ C) = (A × B) ∩ (A × C) ]
⇒ n [A × (B ∩ C)] = 8
⇒ n(A). n (B ∩ C) = 8
⇒ n(A). 2 = 8
⇒ n(A) = \(\frac{8}{2}\) = 4
(3) 6
Explaination:
n[(A × B) ∪ (A × C)] = n[ A × (B ∪ C)] = n(A) × n(B ∪ C)
= 2 × 3
= 6
(2) 17 2
Explanation:
If two sets A and B have 17 elements in common, then the number of elements common to the set A × B
and B × A is 17 2
(2) A × A
Explanation:
When A ⊂ B, (A × B) ∩ (B × A) = A × A
(3) 512
Explanation:
The number of relations on a set containing n elements is 2 n 2. Here n = 3
∴ Required number = 2 3 2 = 2 9
= 512
(3) transitive
Explanation:
Given R is a universal relation on the set X.
The universal relation is always an equivalence relation.
R is reflexive, symmetric, and transitive.
(2) Symmetric
Explanation:
(4, 4} ∉ R ⇒ R is not reflexive
(1, 4), (4, 1) ∈ R ⇒ R is symmetric
(1, 4), (4, 1) ∈ R but (4, 4) ∉ R
So R is not transitive
(4) (- ∞, – 1] ∪ [\(\frac{1}{3}\), ∞)
Explaination:
(3) [0, 1)
Explaination:
f(x) = |[x] – x|
When x = 1,
we have [x] = [1] = 1
f(1) = |1 – 1| = o
When x = 1.5
we have [x} = [1.5] = 1
f(1.5) = |1 – 1.5| = |- 0.5| = 0.5
When x = 2.5
we have [x] = [2.5] = 2
f (2.5) = |2 – 2.5| = |- 0.5| = 0.5
When x = – 2.5
we have [x] = [- 2.5] = – 3
f (-2.5) = |- 3 – (-2.5| = |- 3 + 0.5| = |- 0.5| = 0.5
∴ Range of f(x) = |[x] – x| is [0, 1)
(4) [0, ∞), [0, ∞)
Explaination:
Let x ∈ R, then x can be negative or zero or positive.
Given f(x) = x 2
The image of x under f is always positive since x 2 is positive for x = 1 and x = – 1 ∈ R
f(1) = 1 2 = 1
f(-1) = (-1) 2 = 1
∴ 1, – 1 have the same image
∴ f is not one – one if the domain is R.
Suppose the domain is [0, ∞) then f is one – one and onto.
Domain = [0, ∞)
Co-domain = [0, ∞)
(3) n
Explanation:
Let A be a set having m elements and B be a set having n elements.
When all the elements of A mapped onto the first element of B we get the first constant function. When all the elements of A mapped onto the second element of B we get the second constant function.
When all the elements of A mapped on to the n th element of B, we get the n th constant function.
∴ The number of constant functions possible is n.
(2) onto
Explaination:
f: [0, 2π] → [- 1, 1]
Defined by f (x) = sin x
f(0) = sin 0 = 0
(4) [0, 9]
Explaination:
f: [-3, 3] → S defined by f(x) = x 2
f(-3) = (-3) 2 = 9
f(0) = 0 2 = o
f(3) = 3 2 = 9
∴ S = [0, 9]
(4) not a function
Explaination:
X = {1, 2, 3, 4}, Y = {a, b, c, d>
f = {(1, a), (4, b), (2, c), (3, d), (2, d)}
f is not a function since 2 ∈ X has two images c and d.
(1)
Explaination:
Let f(x) = x if x < 1 —— (1)
Put y = x then
(1) ⇒ f(x) = y
⇒ x = f -1 (y) if y < 1
⇒ y = f -1 (y) if y < 1
⇒ f -1 (x) = x if x < 1
Let f(x) = x 2 if 1 ≤ x ≤ 4 —– (2)
Put y = x 2 ⇒ x = √y, if 1 ≤ y ≤ 16
then (2) ⇒ f(x) = y
⇒ x = f -1 (y) if 1 ≤ y ≤ 16
⇒ √(y) = f -1 (y) if 1 ≤ y ≤ 16
⇒ √x = f -1 (y) if 1 ≤ x ≤ 16
Let f(x) = 8√x if x > 4 ———– (3)
Put y = 8√x ⇒ y 2 = 64 x
⇒ x = \(\frac{y^{2}}{64}\) if y>16
then (3) ⇒ f(x) = y
⇒ x = f -1 (y) if y > 16
⇒ √y = f -1 (y) if y > 16
⇒ \(\frac{y^{2}}{64}\) = f -1 (x) if x > 16
(4) (- ∞, 1)
Explaination:
f: R ➝ R defined by
f(x) = 1 – |x|
For example,
f(1) = 1 – 1 = 0
f(8) = 1 – 8 = -1
f(-9) = 1 – 9 = -8
f(-0.2) = 1 – 0.2 = 0.8
so range = (-∞, 1]
(2) neither an odd function nor an even function
Explaination:
f: R → R is defined by f(x) = sin x + cos x
f(-x) = sin (-x) + cos (-x) = -sin x + cos x ≠ f (x)
If f(-x) = -f(x) then f(x) is an odd function.
If f(-x) = f(x) then f(x) is an even function.
∴ f (x) is neither odd function nor an even function.