MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
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1. Determine whether each of the following relations are reflexive, symmetric and
transitive:
(i) Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as
R = {(x, y) : 3x – y = 0}
(ii) Relation R in the set N of natural numbers defined as
R = {(x, y) : y = x + 5 and x < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y) : y is divisible by x}
(iv) Relation R in the set Z of all integers defined as
R = {(x, y) : x – y is an integer}
(v) Relation R in the set A of human beings in a town at a particular time given by
(a) R = {(x, y) : x and y work at the same place}
(b) R = {(x, y) : x and y live in the same locality}
(c) R = {(x, y) : x is exactly 7 cm taller than y}
(d) R = {(x, y) : x is wife of y}.
•
Solution:
(i)R={(x,y): 3x-y =0}, where x,y ε A ={1, 2, 3, …, 13, 14}
∴R ={(1, 3), (2, 6), (3, 9), (4, 12)}
Reflexivity: Since (1,1)
R. So, R is not reflexive.
Symmetry: We observe that (1, 3) ε R but (3, 1)
R. So, R is not symmetric.
Transitivity: We observe that (1, 3) ε R and (3,9) ε R but (1, 9)
transitive.
R. So, R is not
(ii) R={(x,y): y = x + 5 and x < 4}, where x, y ε N
∴R ={(1, 6), (2, 7), (3, 8)}
Reflexivity: Clearly (1,1), (2,2) etc. are not in R. So, R is not reflexive.
R. So, R is not symmetric.
Transitivity: Since (1, 6) ε R and there is no order pair in R which has 6 as the
first element. Same is the case for(2, 7) and (3, 8). So, R is transitive.
in
Symmetry: We find that (1, 6) ε R but (6, 1)
(iii) R={(x,y): y is divisible by x}, where x, y ε A ={1, 2, 3, 4, 5, 6}
Reflexivity: We know that:
x is divisible by x for all x ε A
∴(x, x) ε R for all x ε A
R is reflexive.
Symmetry: We observe that 6 is divisible by 2 but not divisible by 6.
1
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:1 [email protected] web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
This means that (2, 6) ε R but (6, 2)
So R is not symmetric.
R
Transitivity: Let (x, y) ε R and (y, z) ε R. Then,
(x, y) ε R=> y is divisible by x and (y, z) ε R => z is divisible by y
thus z is divisible by x.
=>(x, z) ε R. Thus R, is transitive relation.
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(iv) R={(x,y):x- y is an integer}, where x, y ε Z
Reflexivity: We have
x –x = 0 which is an integer for all x ε Z
∴(x, x) ε R for all x ε Z
R is reflexive on Z .
Symmetry: Let (x, y) ε R, Then,
(x,y) ε R =>x – y is an integer, say, λ
=>y – x = - λ => y – x is an integer
=>(y, x) ε R
Thus (x, y) ε R=> (y, x) ε R for all x, yε Z
Transitivity: Let (x, y) ε R and (y, z) ε R. Then,
(x, y) ε R=> x - y is an integer and (y, z) ε R => y - z is an integer
=>(x – y) + (y – z) is an integer (Since sum of two integers is also an
integer)
x – z is an integer
=>(x, z) ε R. Thus R, is transitive relation on Z.
(v) a) Reflexive, symmetric & transitive
b) Reflexive, symmetric & transitive
c) neither reflexive, nor symmetric but transitive
d) neither reflexive, nor symmetric nor transitive.
2. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a
+ 1} is reflexive, symmetric or transitive.
•
in
Solution:
We observe the following properties
Reflexivity: Let a be an arbitrary element of R, Then A Σ R
⇒a+1≠a
∴R is not reflexive (a, a) ∉ R1
Symmetry Let (a,b) Σ R then (a, b) Σ R1
⇒b=a+1
a = b – 1 thus (b, a) ∉ R1
Hence not symmetry
Transitive : (1, 2) Σ R, & (2, 3) Σ R, but (1, 3) ∉ R, Thus it is not transitive.
3. Check whether the relation R in R defined by R = {(a, b) : a ≤ b3} is reflexive,
symmetric or transitive.
2
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
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EMAIL:2 [email protected] web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
•
Solution:
We have,
R = {(a, b); a ≤ b3} where a, b Σ R
Reflexivity:
≤
is not true. Therefore
∉ R. So R is not reflexive.
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We observe that
Symmetry:
We observe that –1 ≤ 33 but 3 ≤ 33 (ie) (-1, 3) ∈ R but (3, -1) ∉ R. So, R is not symmetric.
Transitarty we find that
2 ≤ (-3)2 & -3 ≤ 1 but 2 ≤ 1
(ie) (2, -3) ∈ R & (-3, 1) ∈ R but (2, 1) ∉ R
So R is not transitive.
4. Show that the relation R in the set A of all the books in a library of a college, given
by R = {(x, y): x and y have same number of pages} is an equivalence relation.
•
Solution:
(ii) A = {x ∈ ≥ 0 ≤ x ≤ 12}
We have,
R = {(a, b), a = b }
Reflexivity : For any a ∈ A, we have
a = a, ⇒ (a, a) ∈ R, a ∈ A
So, R is reflexive.
Symmetry: Let (a, b) ∈ R, then (a, b) ∈ R
⇒ a = b & b = a ⇒ (b, a) ∈ R .
Thus R is symmetry.
Transitive Let (a, b) ∈ R & (b, c) ∈ R then
A = b & b = c ⇒ a = c Thus (a, c) ∈ R.
Hence R is transitive
It is an equivalence relation.
5. Show that the relation R defined in the set A of all polygons as R = {(P1, P2): P1 and
P2 have same number of sides}, is an equivalence relation. What is the set of all elements in
A related to the right angle triangle T with sides 3, 4 and 5?
Solution:
Reflaxive:
We know that every polygons have same number of sides.
There(P, P) ∈ R for all P ∈ A ⇒ R is reflaxive.
in
•
Symmetry
Let (P1, P2) ∈ R, then (P1, P2) ∈ R ⇒ P1 have same number of sides as P2
⇒ P2 have same number of sides of P1 ⇒ (P2, P1) ∈ R.
So, R is symmetric.
Transitivity: Let P1, P2, P3 ∈ A such that (P1, P2) ∈ R & (P2, P3) ∈ R.
P1 & P2 have same number of sides also P2 & P3 have same number of sides
3
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MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
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P1 & P3 have same number of sides
(P1, P2) ∈ R
So R is transitive.
Hence R is an equivalence relation.
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6. Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1,
L2) : L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines
related to the line y = 2x + 4.
•
Solution:
For each line l ∈ L we have L || L ⇒ (l, l) ∈ R such that l ∈ L (R is reflexive).
Symmetric: Let l1, l2 ∈ L such that (l1, l2) ∈ R. Then (l2, l1) ∈ R. So R is symmetric on R.
Transitive:
Let l1, l2, l3 ∈ L such that (l1, l2) ∈ R & (l2, l3) ∈ R. Then, (l1, l2) ∈ R & (l2, l3) ∈ R
⇒ l1 || l2 ⇒ l2 || l3 ⇒ l1 || l3 ⇒ (l1, l3) ∈ R.
∴R is transitive on L. Hence R being reflexive, symmetric l transitive it is an equivalence relation on
L.
7. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4,4),
(1, 3), (3, 3), (3, 2)}. Choose the correct answer.
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
•
Solution:
(B) R is reflexive and transitive but not symmetric.
8. Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose the
correct answer.
(A) (2, 4) ∈ R (B) (3, 8) ∈ R (C) (6, 8) ∈ R (D) (8, 7) ∈ R.
•
Solution:
(C) (6, 8) ∈ R.
(ii) f : Z → Z given by f(x) = x2
in
9. Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f(x) = x2
(iii) f : R → R given by f(x) = x2
(iv) f : N → N given by f(x) = x3
(v) f : Z → Z given by f(x) = x3.
4
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
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MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
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•
Solution:
(i) Let x, y be true arbitrary elements of R (domain of f) such that = f(y) then f(x) = f(y) ⇒ x2 = y2.
⇒ x = y. Hence f is injective.
Let y be an arbitrary element of N, then f(x) = y ⇒ x2 = y ⇒ x =
This is not subjective such that f(x) – f(c)
clearly for all y ∈ N,
∈ N.
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ii) Let x, y be two arbitrary element of z, then
f(x) = f(y) ⇒ x2 = y2 ⇒ x = ± y. But
∉ Z . Thus not subjectivity.
iii) Let x, y be two arbitrary element of R, such that f(x) = f(y). Here only x = y hence f is injective.
Let y be any arbitrary element of R then f(x) = y ⇒ x2 = y ⇒ x =
surgective.
. But
∈ R. Thus f is
iv) Let x, y be two arbitrary element of N, such that f(x) = f(y) then f(x) = f(y) ⇒ x3 = y3 ⇒ x = y.
Hence if is injective.
Let y be any arbitrary element of N, such that f(x) = y ⇒ x3 = y ⇒ x = y1/3. But y1/3 ∈ R. This f is
surjective.
v) Let x, y be two arbitrary element of Z, such that f(x) = f(y) then f(x) = f(y) ⇒ x3 = y3 ⇒ x = y.
Thus f is injective and let y be any arbitrary element of Z, then f(x) = y ⇒ x3 = y ⇒ x3 = y ⇒ x =
y1/3 but y1/3 ∈ Z.
Hence not surjective.
10. Show that the Signum Function f : R → R, given by
f(x) =
is neither one-one nor onto.
•
Solution:
Since f(1) = 4, f(2) = 5 & f(3) = 6 different element in A have different images1 under function f.
Thus f is one – one.
11. In each of the following cases, state whether the function is one-one, onto or bijective.
Justify your answer.
(i) f : R → R defined by f (x) = 3 – 4x
(ii) f : R → R defined by f (x) = 1 + x2.
in
•
Solution:
(i) Let x, y be two arbitrary element of R such that f(x) = f(y) then 3 – 4x = 3 – 4y ⇒ x = y thus f is
one – one.
Let y be any arbitrary element of R, then f(x) = y ⇒ 3 – 4x = y ⇒ 4x = 3 – y ⇒ x =
But
∈ R thus it is onto. Thus it is bijective.
(ii) Let x, y be two arbitrary elements of R such that f(x) = f(y) then 1 + x2 = 1 + y2 ⇒ x2 = y2 ⇒ x
= ± y, thus since x takes two values among which –ve is not possible. Thus f is one – one.
5
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:5 [email protected] web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
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Let y be arbitrary element of R, then
f(x) = y ⇒ 1 + x2 = y ⇒ x2 = y – 1 ⇒ x
bijective as it is one-one and onto.
which is a real number. Thus f is onto. Hence f is
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12. Let f : N → N be defined by f (n) =
•
Solution:
Let n, m be two even natural numbers then
f’(n) = f(m) ⇒
=
⇒n=m
If n; m are two odd natural numbers
f(n) = f(m) =
=
⇒n+1=m+1⇒n=m
In both the cases f(n) = f(m) ⇒ n = m
If n is even and m is odd then so f is an injective map.
Surjectivity: Let n ∈ N, if n is odd, then
is even such that f
=
If n is even then there exists,
is even such that f
=
.
Thus every n ∈ N has its pre image in N. Hence f: N → N is bijective.
13. Let A = R – {3} and B = R – {1}. Consider the function f: A → B defined by
f(x) =
•
. Is f one-one and onto? Justify your answer.
Solution:
Injectivity: Let x, y be any two elements of A. Then, f(x) = f(y)
Surjective: Let y be an arbitrary element of B, Then, f(x) = y ⇒
⇒ x – 2 = y(x – 3)
⇒ x – 2 = xy – 3y
⇒ x – xy = 2 – 3y
⇒ x(1 – y) = 2 – 3y
in
⇒
=
⇒ (x – 2) (y – 3) = (y – 2) (x – 3)
⇒ xy – 3x - 2y + 6 = xy – 3y – 2x + 6
⇒ -x = -y ⇒ x = y
Thus f(x) = f(y) ⇒ x = y & x, y ∈ A so, f is injective.
=y
6
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:6 [email protected] web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
⇒x=
clearly x =
is a real number for all y ≠ 1
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Also
≠ 3 for if we take
=2
We get 2 – 3y = 2 – 2y
⇒ 3 = 2 which is wrong. Thus every element y in B, has its pre-image in A given by x =
. So, if is a surjective.
Hence f is bijective map.
14. Let f : R → R be defined as f(x) = x4. Choose the correct answer.
(A) f is one-one onto (B) f is many-one onto
(C) f is one-one but not onto (D) f is neither one-one nor onto.
•
Solution:
(A) f is one-one onto.
15. Let f : R → R be defined as f (x) = 3x. Choose the correct answer.
(A) f is one-one onto (B) f is many-one onto
(C) f is one-one but not onto (D) f is neither one-one nor onto.
•
Solution:
(C) f is one-one but not onto.
16. Let f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} be given by
f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
•
Solution:
f : {1, 3, 4} → {1, 2, 5} & g : {1, 2, 5} → {1, 3}
f = {(1, 2), (3, 5), (4, 1)}, g = {(1, 3), (2, 3), (5, 1)}
f(1) = 2, f(3) = 5, f(4) = 1 & g(1) = 3, g(2) = 3, g(5) = 1
go f(x) must be obtained for x = 1, 3, 4.
gof(1) = g(f(i) = g(2) = 3
gof(3) = g(f(3)= g(5) = 1
gof(4) = g(f(4)= g(1) = 3.
•
in
17. Find gof and fog, if
(i) f (x) = | x | and g(x) = | 5x – 2 |
(ii) f(x) = 8x3 and g(x) = x1/3 .
Solution:
7
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:7 [email protected] web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
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(i) f(x) =
& g(x0=)
=
Gof = gof(x) = go[f(x)]
=
(ii) f(x) = 8x3 and g(x) = x1/3
gof = gof(x)
= go[8x3] = (8x)1/3
= 2x
fog = fog(x)
= fo(x1/3) = 8(x1/3)3
= 8x.
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•
18. If f(x) =
•
x≠
, show that fof(x) = x, for all x ≠
. What is the inverse of f?
Solution:
fof(x) = fo
=
=
=
=
=
=
= x ∴ fof(x) = x for x ≠
f-1(x)
(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
in
19. State with reason whether following functions have inverse
(i) f : {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}.
•
Solution:
i) f is not one to one, since f(1) = f(2) =f(3) = f(4) = 10, so f is not invertible. Hence f-1 doesn’t
exist.
ii) since g(5) = g(7) = 4 if is not one to one, so f is not invertible, this f-1 doesn’t exist.
8
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iii) f is 1 – 1 and onto thus f is invertible, hence f-1 exists.
20. Show that f : [–1, 1] → R, given by f (x) =
function f:[-1, 1] → range f.
is one-one. Find the inverse of the
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(Hint: For y ∈ Range, y = f(x) =
•
, for some x in [-1, 1], i.e., x =
is one.
Solution:
For y ∈ Range f, y = f(x) but f(x) =
(i.e.) xy – 2y = x
2y = x – xy ⇒ x(1 – y) = 2y
⇒x=
where x ∈ [-1, 1]
.
21. Consider f : R → R given by f (x) = 4x + 3. Show that f is invertible. Find the inverse of
f.
•
Solution:
f : R → R, f(x) = 4x + 3
An arbitrary element y in Y in the form y = 4x + 3 for some x ∈ R. This implies x =
Define g: R → R by g(y) =
. Now gof(x) = g(f(x) = g(4x + 3) =
This shows that gof = IR and fog = IR. Hence f is invertible with f-1 = g.
=x
22. Consider f : R+ → [4, ∞) given by f (x) = x2 + 4. Show that f is invertible with the
inverse f–1 of f given by f–1(y) =
numbers.
•
, where R+ is the set of all non-negative real
Solution:
An arbitrary element y in Yin the form y= x2 + 4 for some x ∈ R.This implies x2 = y –4
⇒x=
23. f : R+ → [-5, ∞ ] given by f(x) = 9x2 + 6x –5.
•
=
in
Define g: R → (4, ∞ ) by g(y) =
. Now gof(x) = g(x2 + 4) =
This shows that gof = IR+ and fog = [4, ∞ ]. Hence f is invertible with f-1 = g.
Solution:
An arbitrary element y in Y in the form y = 9x2 + 6x –5 for some x ∈ R ⇒ y = (3x + 1)2 – 6 this year
x=
.
Define g: R+ → [-5, ∞ ] by g(y) =
9
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Now gof(x) = g(9x2 + 6x +5) =
=
=
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09
=
=2
This shows gof = IR and fog = [-5, ∞ ]. Hence f is invertible & f-1 = g.
24. Let f : X → Y be an invertible function. Show that f has unique inverse. (Hint: suppose
g1 and g2 are two inverses of f. Then for all y ∈ Y, fog1(y) = 1Y(y) = fog2(y). Use one-one
ness of f).
•
Solution:
f: X → y is a invertible for suppose g1 & g2 are two inverse of f. Then for all y ∈ Y, fog1(y).
1
•
25. Consider f : {1, 2, 3} → {a, b, c} given by f (1) = a, f (2) = b and f (3) = c. Find f–
and show that (f–1)–1 = f.
Solution:
f-1 = [(a, 1), (b, 2), (c, 3)}. But (f-1)-1 = {(1, a), (2, b), (3, c)}
Thus (f-1)-1 = f.
26. Let f : X → Y be an invertible function. Show that the inverse of f–1 is f, i.e.,(f –1)–1 = f.
•
Solution:
If f:x → y is an invertible function then f g: y → X such that gof = Ix and fog = Iy. Then g is called
the inverse of f & it is denoted by f—1.
Inverse of
27. If f : R → R be given by f(x) = (3 – x3)1/3, then fof (x) is
(A) x1/3 (B) x3 (C) x (D) (3 – x3).
•
28. Let f : R –
map
g: Range f → R –
→ R be a function defined as f(x) =
in
Solution:
Fof(x) = f(3 – x3)1/3
=(3 – [(3 – x3)1/3]3)1/3
= (3 – [(3 - x3)]1/3
= (3 – 3 + x3)1/3 = x = (c).
. The inverse of f is the
given by
10
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b) g(y) =
c) g(y) =
d) g(y) =
.
Solution:
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•
a) g(y) =
f:R -
→ R defined by f(x) =
g: Range f → R -
Let y ∈ Y be any arbitrary in Y defined as y =
y(3x+4) = 4x ⇒ x =
thus g(y) =
⇒ x(4 – 3y) = 4y.
for some x in R, which implies that
(b)
29. Determine whether or not each of the definition of * given below gives a binary
operation. In the event that * is not a binary operation, give justification for this.
(i) On Z+, define * by a * b = a – b
(ii) On Z+, define * by a * b = ab
(iii) On Z+, define * by a * b = | a – b |.
•
Solution:
(i) Z+ - the set of +ve integers .
The * defined by a * b = a – b is not a binary operation because the inverse of +ve integers is –ve
integers which doesn’t exists.
(ii) The * defined by a * b = ab is not a binary operation because the inverse of a is
a element of z+.
(iii) The * defined by a * b =
–ve integers which doesn’t exists.
which is not
is not a binary operation because the inverse of +ve integers is
in
30. For each binary operation * defined below, determine whether * is commutative or
associative.
(i) On Z, define a * b = a – b
(ii) On Q, define a * b = ab + 1
(iii) On Q, define a * b =
(iv) On Z+, define a * b = 2ab
(v) On Z+, define a * b = ab
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(vi) On R – {– 1}, define a * b =
•
.
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Solution:
(i)a * B = a - b
since 3* -4 = 3-(-4) = 7; –4 × 3 = -4 – 3 = -7
This prove the operation * is not commutative
But (3 * -4) * 2 = (3 – (-4) * 2
= 7 * 2 = 14
3 *((-4) * 2)= 3 * [(-4) –2)]
= 3 × -6
= 3 –(-6)
=3+6=9
Thus * is not associative.
(ii) a * b = ab +1
⇒ Here the operation * is both commutative * associative
(iii) The binary operation * is both commutative * associative.
(iv) The binary operation * on Z+ is both commutative * associative.
v) The binary operation * on Z+ defined as a * b = ab is not commutative & not associative because
3 * 2 = 32 & 2 * 3 = 23
=9=8
Also (3 * 2) * 4
= 32 * 4
= 9 * 4 = 94
and
3 * (2 * 4)
= 3 * 24 = 3 * 16
= 316
So (3 * 2) * 4 ≠ 3 * (2 * 4)
vi) The binary operation * on R - {– 1} defined as a =
is not commutative & not associative.
31. Consider the binary operation ∧ on the set {1, 2, 3, 4, 5} defined by a ∧ b = min {a,
b}. Write the operation table of the operation ∧ .
•
∧
1
2
3
4
5
1
1
1
1
1
1
2
1
2
2
2
2
3
1
2
3
3
3
4
1
2
3
4
4
5
1
2
3
4
1
in
Solution:
a ∧ b = min {a, b}
32. Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following
multiplication table (Table 1.2).
(i) Compute (2 * 3) * 4 and 2 * (3 * 4)
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(ii) Is * commutative?
(iii) Compute (2 * 3) * (4 * 5).
(Hint: use the following table)
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•
Solution:
i) (2 * 3) * 4 = (1 * 4) = 1
2*(3 * 4) = (2 * 1) = 1
⇒ (2 * 3) * 4 = 2( 3 * 4)
(ii) Yes it is commutative
(iii) (2 * 3) * (4 * 5)
= 1 * (4 * 5)
= 1 * 1 = 1.
33. Let *′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a *′ b = H.C.F.
of a and b. Is the operation a′ same as the operation *’ defined in Exercise 4 above? Justify
your answer.
•
Solution:
*’
1
2
3
4
5
•
1
1
1
1
1
1
2
1
1
1
2
1
3
1
1
3
1
1
4
1
2
1
4
1
5
1
1
1
1
5
Yes the operation *’ is same as the operation * defined above.
in
34. Let * be the binary operation on N given by a * b = L.C.M. of a and b. Find
(i) 5 * 7, 20 * 16 (ii) Is * commutative?
(iii) Is * associative? (iv) Find the identity of * in N
•
Solution:
(i) 5 * 7 = 35
20 * 16 = 80
(ii) Yes, * is commulative
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(iii) No, * is not associative
(iv) The identity of * in N is 1.
35. Is * defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binary operation?
Justify your answer.
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•
Solution:
*
1
2
3
4
5
•
1
1
2
3
4
5
2
2
2
6
4
10
3
3
6
3
12
15
4
4
4
12
4
20
5
5
10
15
20
25
It is not a binary operation because a × b = LCM of a * b contains elements other than 1 to 5.
36. Let * be the binary operation on N defined by a * b = H.C.F. of a and b. Is *
commutative? Is * associative? Does there exist identity for this binary operation on N?
•
Solution:
* is commutative, * is not associative, There exist an identity element for the binary operation.
37. Let * be a binary operation on the set Q of rational numbers as follows:
(i) a * b = a – b
(ii) a * b = a2 + b2
(iii) a * b = a + ab
(iv) a * b = (a – b)2
(v) a * b =
(vi) a * b = ab2
Find which of the binary operations are commutative and which are associative.
•
in
Solution:
(i) For any a, b ∈ Q we have a * b = a – b and b * a = b – a. Thus a – b ≠ b – a for all a, b ∈ Q. Thus
it is not commutative.
For any a, b, c ∈ Q we have (a * b)*c = (a – b) * c
=a–b–c
a *(b * c) = a *(b – c) = a – (b – c) = a –b +c
Thus it is not association.
(ii) For any a, b ∈ Q we have a * b = a2 + b2 & b * a = b2 + a2. Thus a2 + b2 = b2 + a2 for all a,
b ∈ Q.
Thus it is commutative
For any a, b, c ∈ Q we have (a * b)*c = (a2 + b2) * c
=(a2 + b2) + c2
and a * (b*c) = a * (b2 + c2)
= a2 + (b2 + c2)2
Here (a * b)*c ≠ a (b*c) ….. It is not associative.
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also b * a = b + ab. This implies a * b ≠ b * a thus * is not
(a + ab) * c
+ ab + (a+ab)
+ ab +ac +abc
+a(b+c)+abc
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(iii) a * b= a + ab for any a, b ∈ Q &
commulative.
For any a, b, c ∈ Q, here (a * b)*c =
=a
=a
=a
a*(b*c) = a*(b+bc)
= a + a(b+bc)
= a +ab + abc
iv) For a, b ∈ Q, a * b = (a – b)2. Also b * a = (b – a)2
= b2 – 2ab + a2
= a2 – 2ab + b2
Thus a * b = b * a. Hence * is commutative.
a * (b*c) = a * (b – c)2
=[a – (b-c)2] = a2 – 2a(b – c)2 + (b – c)4
(a * b)*c = [(a – b)2 * c]
= [(a – b)2 - c]2 = (a – b)4 – 2c(a – b)2 + c2
= a *(b*c) ≠ (a*b) * c
v) a *b =
for a, b ∈ Q, a * b =
&b*a=
thus * is commutative. Now(a*b)*c =
*c=
a*(b*c) = a *
=
Thus (a*b)* c = a (b*c).
vi) for a, b ∈ Q, a * b = ab2 & b * a = ba2. Here a *b ≠ b * a because ab2 ≠ ba2. Thus * is not
commutative. Now(a * b) * c = ab2 * c = ab2c2
Also a*(b*c) =a*bc2 = ab2c4
Thus * is not associative.
38. State whether the following statements are true or false. Justify.
(i) For an arbitrary binary operation * on a set N, a * a = a ∀ a ∈ N.
(ii) If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a.
•
Solution:
(i) False. Only if a is the identity element a * a = a is possible.
in
(ii) False.
39. Consider a binary operation * on N defined as a * b = a3 + b3. Choose the correct
answer.
(A) Is * both associative and commutative?
(B) Is * commutative but not associative?
(C) Is * associative but not commutative?
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(D) Is * neither commutative nor associative?
•
Solution:
b) Is * commutative but not associative.
in
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