Download App

Maths 2 Marks

RELATIONS AND FUNCTIONS · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 181 questions.

Questions · Page 3 of 4

2 Marks

Question 1012 Marks
Determine whether the following operations define a binary operation on the given set or not:
'+6' on S = {0, 1, 2, 3, 4, 5} defined by, $\text{a}+_6\text{b}=\begin{cases}\text{a}+\text{b},&\text{if a}+\text{b}<6\\\text{a}+\text{b}-6,&\text{if a}+\text{b}\geq6\end{cases}$
Answer
We have,

S = {0, 1, 2, 3, 4, 5}

and, $\text{a}+_6\text{b}=\begin{cases}\text{a}+\text{b},&\text{if a}+\text{b}<6\\\text{a}+\text{b}-6,&\text{if a}+\text{b}\geq6\end{cases}$

Let $\text{a}\in\text{S}$ and $\text{b}\in\text{S}$ such that a + b < 6

Then $\text{a}+_6\text{b}=\text{a}+\text{b}\in\text{S}$ $\big[\because$ a + b < 6 = 0, 1, 2, 3, 4, 5$\big]$

Let $\text{a}\in\text{S}$ and $\text{b}\in\text{S}$ such that a + b > 6

Then $\text{a}+_6\text{b}=\text{a}+\text{b}-6\in\text{S}$ $\big[\because\ \text{if a}+\text{b}\geq6$ then $\text{a}+\text{b}-6\geq6$ = 0, 1, 2, 3, 4, 5$\big]$

$\therefore\ \text{a}+_6\text{b}\in\text{S}$ for $\text{a, b}\in\text{S}$

$\therefore$ +6 defined a binary operation on S.

View full question & answer→
Question 1022 Marks
A fair die is rolled. Consider events E = $\{1,\ 3,\ 5\},\ \text{F}=\{2,\ 3\}\ \text{and}\ \text{G}=\{2,\ 3,\ 4,\ 5\}.\ \text{Find}:$
$\text{P}(\text{E}|\text{G})\ \text{and}\ \text{P}(\text{G}|\text{E})$
Answer
$\text{P}\left(\text{E}\right)=\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}=\frac{3}{6}\ \ \ \ \ \ \ \ \text{P}\left(\text{G}\right)=\frac{\text{n}\left(\text{G}\right)}{\text{n}\left(\text{S}\right)}=\frac{4}{6}$
$\text{E}\ \cap\ \text{G}=(3,\ 5)\ \Rightarrow\ \ \ \ \ \text{n}\left(\text{E}\cap\text{G}\right)=2$
$\text{P}\left(\text{E}\cap\text{G}\right)=\frac{\text{n}\left(\text{E}\ \cap\ \text{G}\right)}{\text{n}\left(\text{S}\right)}=\frac{2}{6}$
$\text{P}\left(\text{E}|\text{G}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{G}\right)}{\text{P}\left(\text{G}\right)}=\frac{\frac{2}{6}}{\frac{4}{6}}=\frac{2}{4}=\frac{1}{2}\ \ \ \\ \text{and}\ \ \ \text{P}\left(\text{G}|\text{E}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{G}\right)}{\text{P}\left(\text{E}\right)}=\frac{\frac{2}{6}}{\frac{3}{6}}=\frac{2}{3}$
View full question & answer→
Question 1032 Marks
Let * be a binary operation on N given by a * b = LCM (a, b) for all $\text{a, b}\in\text{N.}$ Find 5 * 7.
Answer
As, a * b = LCM (a, b)
So, 5 * 7 = LCM (5, 7) = 35
View full question & answer→
Question 1042 Marks
If R = {(x, y): x2 + y2 ≤ 4; x, y ∈ Z} is a relation on Z, write the domain of R.
Answer
Domain of R is the set of values of x that satisfies the relation R.
Because x must be an integer, the provided values of x are:
$0,\pm1,\pm2$
Thus, Domain of R is $0,\pm1,\pm2$
View full question & answer→
Question 1052 Marks
Write the inverse of 5 under multiplication modulo 11 on the set {1, 2, ... ,10}.
Answer
As, e = 1 : 5 × 9 ≡ 1 (mod 11)
So, the inverse of 5 i.e. 5-1 = 9
View full question & answer→
Question 1062 Marks
If f(x) = 2x + 5 and g(x) = x2 + 1 be two real functions, then describe the following functions:
gof
Also, show that fof ≠ f2
Answer
f(x) and g(x) are polynomials.
⇒ f : R → R and g : R → R.
So, fog : R → R and gof : R → R.
(gof)(x) = g(f(x))
= g(2x + 5)
= (2x + 5)2 + 1
= 4x2 + 20x + 26
View full question & answer→
Question 1072 Marks
If f : C → C is defined by f(x) = x2, write f-1(-4). Here, C denotes the set of all complex numbers.
Answer
f : C → C defined by f(x) = x2

⇒ f-1(x2) = x

⇒ f-1(-4) = f-1[(2i, -2i)2] = (2i, -2i)

$\therefore$ f-1(-4) = (2i, -2i)

View full question & answer→
Question 1082 Marks
Write the identity relation on set A = {a, b, c}.
Answer
Identity set of A is:
I = {(a, a), (b, b), (c, c)}
Every element of this relation is related to itself.
View full question & answer→
Question 1092 Marks
Let * be a binary operation on the set I of integers, defined by a * b = 2a + b - 3. Find the value of 3 * 4.
Answer
It is given that, a * b = 2a + b - 3

Now, 3 * 4 = 2 × 3 + 4 - 3

= 10 - 3

= 7

View full question & answer→
Question 1102 Marks
For the set A = {1, 2, 3}, define a relation R on the set A as follows:
R = {(1, 1), (2, 2), (3, 3), (1, 3)}
Write the ordered pairs to be added to R to make the smallest equivalence relation.
Answer
(3, 1) is the single ordered pair which needs to be added to R to make it the smallest equivalence relation.
View full question & answer→
Question 1112 Marks
Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. What minimum number of ordered pairs may be added to R so that it may become a transitive relation on A.
Answer
We have, A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)}
To make R transitive we shall add (1, 3) only.
$$ $\therefore \text{R}' = \big\{(1, 2), (1, 1), (2, 3), (1, 3)\big\}$
View full question & answer→
Question 1122 Marks
Write whether f : R → R, given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2},$ is one-one, many-one, onto or into.
Answer
f : R → R given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2}$
$\because\ \text{f(x)=}\begin{cases}2\text{x};&\text{for x}>0\\0;&\text{for x}<0\end{cases}$
$\therefore$ f is many-one function.
View full question & answer→
Question 1132 Marks
Let D be the domain of the real valued function f defined by $\text{f}(\text{x})=\sqrt{25-\text{x}^2}.$ Then, write D.
Answer
Consider the given function, $\text{f}(\text{x})=\sqrt{25-\text{x}^2}$
For f(x) to be real, the term inside the square root can’t be negative
i.e., $25-\text{x}^2\geq0$
$\Rightarrow\ \text{x}^2\leq25$
$\Rightarrow\ 5\leq\text{x}\leq-5$
Therefore, the domain of the function, f(x) is given by D = [-5, 5]
View full question & answer→
Question 1142 Marks
For each binary operation * defined below, determine whether * is commutative or associative.
On Q, define a * b = ab + 1
Answer
For commutativity: a * b = ab + 1 and b * a = ba + 1 = ab + 1 = a * b
For associativity: a * (b * c) = a * (bc + 1) = a(bc + 1) + 1 = abc + a + 1
Also, (a * b) * c = (ab + 1)c + 1 = abc + c + 1
$\therefore\ \ \text{a }*(\text{b }*\text{c})\neq(\text{a }*\text{b })*\text{c}$
Therefore, the operation * is commutative but not associative.
View full question & answer→
Question 1152 Marks
Is it true that every relation which is symmetric and transitive is also reflexive? Give reasons.
Answer
No, it is not necessary that a relation which is symmetric and transitive is reflexive as well.
For Example,
Let A = {a, b, c} be a set and
R2 = {(a, a)} is a relation defined on A.
Clearly,
R2 is symmetric and transitive but not reflexive.
View full question & answer→
Question 1162 Marks
Let f be a function from C (set of all complex numbers) to itself given by f(x) = x3. Write f-1(-1).
Answer
Let f-1(-1) = x .....(1)
⇒ f(x) = -1
⇒ x3 = -1
⇒ x3 + 1 = 0
⇒ (x + 1)(x2 - x + 1) = 0
[Using the identity: a3 + b3 = (a + b)(a2 - ab + b2)]
$\Rightarrow\ (\text{x}+1)(\text{x}+\omega)(\text{x}+\omega^2)=0,$ where $\omega=\frac{1\pm\text{i}\sqrt{3}}{2}$
$\Rightarrow\ \text{x}=-1,-\omega,-\omega^2$ $(\text{as x}\in\text{C})$
$\Rightarrow\ \text{f}^{-1}(-1)=\{-1,-\omega,-\omega^2\}$ [from 1]
View full question & answer→
Question 1172 Marks
Find gof and fog, if:
f(x) = |x| and g(x) = |5x – 2|
Answer
To find: gof and fog
f(x) = x and g(x) = |5x − 2|
gof = g[f(x)] = g[|x|] and fog = f[g(x)] = f[(5x - 2)] = |5x - 2| = |5|x|-2|
View full question & answer→
Question 1182 Marks
Show that the binary operation * on Z defined by a * b = 3a + 7b is not commutative.
Answer
The binary operator * defined on Z and is given by a * b = 3a + 7b
Commutativity: Let $\text{a, b}\in\text{Z},$ Then,
a * b = 1a + 7b and
b * a = 3b + 7a
$\therefore\ \text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Hence, '*' is not commutative on Z.
View full question & answer→
Question 1192 Marks
Which of the following functions from A to B are one-one and onto?
f3 = {(a, x), (b, x), (c, z), (d, z)}; A = {a, b, c, d,}, B = {x, y, z}
Answer
f3 = {(a, x), (b, x), (c, z), (d, z)}
A = {a, b, c, d,}, B = {x, y, z}
Since, f3(a) = x = f3(b) and f3(c) = z = f3(d)
$\therefore$ f3 in not one-one.
Again, $\text{y}\in\text{B}$ in not the image of any of the element of A.
$\therefore$ f3 in not on to.
View full question & answer→
Question 1202 Marks
For each binary operation * defined below, determine whether * is commutative or associative.
On Z, define a * b = a – b
Answer
For Commutativity: a * b = a - b and b * a = b - a = -(a - b) $\neq\text{a}*\text{b}$

For associativity: a * (b * c) = a * (b - c) = a - (b - c) = (a - b + c)

Also, (a * b) * c = (a - b) * c = (a - b - c)

$\therefore\ \ \ \text{a}*(\text{b}*\text{c})\neq(\text{a}*\text{b})*\text{c}$

Therefore, the operation * is neither commutative nor associative.

View full question & answer→
Question 1212 Marks
Write the domain of the real function f defined by $\text{f(x)}=\sqrt{25-\text{x}^2}.$
Answer
We have, $\text{f(x)}=\sqrt{25-\text{x}^2}$

The function is defined only when $25-\text{x}^2\geq0$

$\text{x}^2-25\leq0$

$(\text{x}+5)(\text{x}-5)\leq0$

$\text{x}\in[-5,5]$

Therefore, the domain of the given function is [-5, 5].

View full question & answer→
Question 1222 Marks
For each binary operation * defined below, determine whether * is commutative or associative.
On Z+, define a * b = 2ab
Answer
For commutativity: a * b = 2ab and b * a = 2ba = 2ab = a * b
For associativity: a * (b * c) = a * 2bc = (2)
Also, (a * b) * c = (2ab) * 2 = 2ab × c
$\therefore\ \ \text{a} * \text{(b} * \text{c)}\neq\text{(a} * \text{b)}* \text{c}$
Therefore, the operation * is commutative but not associative.
View full question & answer→
Question 1232 Marks
Give an example of a relation which is,
Symmetric and transitive but not reflexive.
Answer
Let R be the relation on A such that,
R = {(1, 2), (2, 1), (1, 3), (3, 1), (2, 3)}
We see that the relation R on A is symmetric and transitive, but not reflexive.
View full question & answer→
Question 1242 Marks
Let A = {x ∈ R : -4 ≤ x ≤ 4 and x ≠ 0} and f : A → R be defined by $\text{f(x)}=\frac{|\text{x}|}{\text{x}}.$ Write the range of f.
Answer
We have, $\text{A}=\{\text{x}\in\text{R}:-4\leq\text{x}\leq4\text{ and x }\neq0\}$
f : A → R defined by $\text{f(x)}=\frac{|\text{x}|}{\text{x}}$
Clearly, $\text{f(x)}=\begin{cases}1;&\text{x}>0\\-1;&\text{x}<0\end{cases}$
$\therefore$ Range of f = {-1, 1}
View full question & answer→
Question 1252 Marks
Determine which of the following binary operations are associative and which are commutative:
'*' on N defined by a * b = 1 for all $\text{a, b}\in\text{N}.$
Answer
Clearly, by defination a * b = 1 = b * a, $\forall\ \text{a, b}\in\text{N}$
Also, (a * b) * c = (1 * c) = 1
and a * (b * c) = (a * 1) = 1 $\forall\ \text{a, b, c}\in\text{N}$
Hence, N is both associative and commutative.
View full question & answer→
Question 1262 Marks
Which one the following relations on A = {1, 2, 3} is a function?
f = {(1, 3), (2, 3), (3, 2)}, g = {(1, 2), (1, 3), (3, 1)}
Answer
As, each element of the domain set has unique image in the relation f = {(1, 3), (2, 3), (3, 2)}
So, f is a function.
Also, the element 1 of the domain set has two images 2 and 3 of the range set in the relation g = {(1, 2), (1, 3), (3, 1)}
So, g is not a function.
View full question & answer→
Question 1272 Marks
For the binary operation multiplication modulo 5 (×5) defined on the set S = {1, 2, 3, 4}. Write the value of (3 ×5 4-1)−1
Answer
The composition table for ×5 on the set S = {1, 2, 3, 4} is
×5 1 2 3 4
1 1 2 3 4
2 2 4 1 3
3 3 1 4 2
4 4 3 2 1
Now,
(3 ×5 4-1)-1 = (3 ×5 4)-1 [$\because$ 4-1 = 4]
= 2-1 [3 ×5 4 = 2]
= 3 [$\because$ 2-1 = 3]
View full question & answer→
Question 1282 Marks
If f : A → A, g : A → A are two bijections, then prove that:
fog is a surjection.
Answer
Given: A → A, g : A → A are two bijections.

Then, fog : A → A

Surjectivity of fog: let z be an element in the co-domain of fog (A).

Now, $\text{z}\in\text{A}$ (co-domain of f) and f is a surjection.

So, z = f(y), where $\text{y}\in\text{A}$ (domain of f) .....(1)

Now, $\text{y}\in\text{A}$ (co-domain of g) and g is a surjection.

So, y = g(x), where $\text{x}\in\text{A}$ (domain of g) .....(2)

From (1) and (2),

z = f(y) = f(g(x)) = (fog)(x), where $\text{x}\in\text{A}$ (domain of fog)

So, fog is a surjection.

View full question & answer→
Question 1292 Marks
Find which of the binary operations are commutative and which are associative.
State whether the following statements are true or false. Justify
If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a
Answer
* being a binary operation on N.
$\therefore$ c * b = b * c
$\therefore$ (c * b) * a = (b * c) * a = a * (b * c)
Thus, a * (b * b) = (c * b) * a, therefore, the given statement is true.
View full question & answer→
Question 1302 Marks
f(x) = 8x3 and g(x) $=\text{x}^{\frac{1}{3}}.$
Answer
f(x) = 8x3 and g(x) $\text{x}^{\frac{1}{3}}$
gof = g[f(x)] = g[8x3] $(8\text{x}^3)^{\frac{1}{3}}=2\text{x}$
and fog = f[g(x)] $=f\Big[\Big(\text{x}^{\frac{1}{3}}\Big)\Big]=8\Big(\text{x}^{\frac{1}{3}}\Big)^3=8\text{x}$
View full question & answer→
Question 1312 Marks
Define a reflexive relation.
Answer
A relation R on a set A is said to be reflexive if every element of A is related to itself.
Mathematically, reflexive relation is written as R = {(a, a): for all $\text{a}\in\text{A}$}
For example if A = {1, 2, 3}, then a reflexive relation on A will be R = {(1, 1), (2, 2), (3, 3)}
View full question & answer→
Question 1322 Marks
Determine whether the following operations define a binary operation on the given set or not:
'×6' on S = {1, 2, 3, 4, 5} defined by, a ×6 b = Remainder when ab is divided by 6.
Answer
Consider the composition table,
×6
1
2
3
4
5
1
1
2
3
4
5
2
2
4
0
2
4
3
3
0
3
0
3
4
4
2
0
4
2
5
5
4
3
2
1
Here all the elements of the table are not in S.
For a = 2 and b = 3,
$\text{a}\times_6\text{b}= 2\times_63$ = remainder when 6 divided by $6=0\neq\text{S}$
Thus, ×6 is not a binary operation on S.
View full question & answer→
Question 1332 Marks
F = {(a, 2), (b, 1), (c, 1)}.
Answer
F: S → T is defined as:
F ={(a, 2), (b, 1), (c, 1)}
Since F(b) = F(c) = 1, F is not one-one.
Hence, F is not invertible i.e., F-1 does not exist.
View full question & answer→
Question 1342 Marks
A = {1, 2, 3, 4, 5, 6, 7, 8} and if R = {(x, y): y is one half of x; x, y ∈ A} is a relation on A, then write R as a set of ordered pairs.
Answer
Since R = {(x, y): y is one half of x; x, y ∈ A}
So, R = {(2, 1), (4, 2), (6, 3), (8, 4)}
View full question & answer→
Question 1352 Marks
Let f, g : R → R be defined by f(x) = 2x + 1 and g(x) = x2 - 2 for all x ∈ R, respectively. Then, find gof.
Answer
We have,
f, g : R → R are defined by f(x) = 2x + 1 and g(x) = x2 - 2 for all x ∈ R, respectively
Now,
gof(x) = g(f(x))
= g(2x + 1)
= (2x + 1)2 - 2
= 4x2 + 4x + 1 - 2
= 4x2 + 4x - 1
View full question & answer→
Question 1362 Marks
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 *′ same as the operation * defined in Exercise 4 above? Justify your answer.
Answer
Let A = {1, 2, 3, 4, 5} and a *' b = H.C.F. of a and b.
*'
1
2
3
4
5
1
1
1
1
1
1
2
1
2
1
2
1
3
1
1
3
1
1
4
1
2
1
4
1
5
1
1
1
1
5
We observe that the operation *' is the same as the operation * in ex. 4.
View full question & answer→
Question 1372 Marks
Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.
On Z+ define * by a * b = |a - b|
Here, Z+ denotes the set of all non-negative integers.
Answer
On Z+, * is defined by a * b = |a - b|.

It is seen that for each $\text{a, b}\in\text{Z}^{+},$ there is a unique element |a - b| in Z+.

This means that * carries each pair (a, b) to a unique element a * b = |a - b| in Z+.

Therefore, * is a binary operation.

View full question & answer→
Question 1382 Marks
Determine whether the following operations define a binary operation on the given set or not:
'O' on Z defined by a O b = ab for all $\text{a, b}\in\text{Z.}$
Answer
We have,
a O b = ab for all $\text{a, b}\in\text{Z}$
Let $\text{a}\in\text{Z}$ and $\text{b}\in\text{Z}$
$\Rightarrow\ \text{a}^{\text{b}}\notin\text{Z}\ \Rightarrow\ \text{a O b}\notin\text{Z}$
For example, if a = 2, b = -2
$\Rightarrow\ \text{a}^{\text{b}}=2^{-2}=\frac{1}{4}\notin\text{Z}$
$\therefore$ The operation 'O' does not define a binary operation on Z.
View full question & answer→
Question 1392 Marks
Reflexive and symmetric but not transitive.
Answer
“is friend of” R = {( x, y) : x is a friend of y}
It is clear that x is friend of x. $\therefore$ R is reflexive.
Also x is friend of y and y is friend of x. $\therefore$ R is symmetric.
Also if x is friend of y and y is friend of z then    
x cannot be friend of z. $\therefore$ R is not transitive.
Therefore, R is reflexive and symmetric but not transitive.
View full question & answer→
Question 1402 Marks
Let A = {2, 3, 4, 5} and B = {1, 3, 4}. If R is the relation from A to B given by a R b if "a is a divisor of b". Write R as a set of ordered pairs.
Answer
We have, A = {2, 3, 4, 5}, B = {1, 3, 4} and relation from A to B is given by aRb if ''is divisor of'' B

$\therefore$ R can be written as ordered pair as R = {(2, 4), (3, 3), (4, 4)}

View full question & answer→
Question 1412 Marks
Let R = {(a, a3): a is a prime number less than 5} be a relation. Find the range of R.
Answer
We have,
R = {(a, a3): a is a prime number less than 5}
Or,
R = {(2, 8), (3, 27)}
So, the range of R is {8, 27}.
View full question & answer→
Question 1422 Marks
Write the smallest reflexive relation on set A = {1, 2, 3, 4}.
Answer
The smallest reflexive relation R on any set A is the identity relation IA on the set A.
We are given, A = {1, 2, 3, 4}
$\therefore$ R = {(1, 1), (2, 2), (3, 3), (4, 4)}
View full question & answer→
Question 1432 Marks
Let * be a binary operation defined by a * b = 3a + 4b − 2. Find 4 * 5.
Answer
Given: a * b = 3a + 4b - 2
Here,
4 * 5 = 3(4) + 4(5) - 2
= 12 + 20 - 2
= 30
View full question & answer→
Question 1442 Marks
If R is a symmetric relation on a set A, then write a relation between R and R-1.
Answer
Here, R is symmetric on the set A.
Let $(\text{a, b})\in\text{R}$
$\Rightarrow\ (\text{b, a})\in\text{R}$ [Since R is symmetric]
$\Rightarrow\ (\text{a, b})\in\text{R}^{-1}$ [By definition of inverse relation]
$\Rightarrow\ \text{R}\subset\text{R}^{-1}$
Let $(\text{x, y})\in\text{R}^{-1}$
$\Rightarrow\ (\text{y, x})\in\text{R}$ [By definition of inverse relation]
$\Rightarrow\ (\text{x, y})\in\text{R}$ [Since R is symmetric]
$\Rightarrow\ \text{R}^{-1}\subset\text{R}$
Thus, $\text{R}=\text{R}^{-1}$
View full question & answer→
Question 1452 Marks
Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.
Answer
(i) $\text{a}\leq\text{a}$ which is true, so $(\text{a},\text{a})\in\text{R},$  $\therefore$ R is reflexive.
(ii) $\text{a}\leq\text{b}\ \text{but}\ \text{b}\leq\text{a}$ $\therefore$ R is not symmetric.
(iii) $\text{a}\leq\text{b}\ \text{and}\ \text{b}\leq\text{c}\Rightarrow\text{a}\leq\text{c}$ which is true. $\therefore$ R is transitive.

Therefore, R is reflexive and transitive but not symmetric.

View full question & answer→
Question 1462 Marks
If f : R → R defined by f(x) = 3x - 4 is invertible, then write f-1(x).
Answer
Let f-1(x) = y .....(1)

⇒ f(y) = x

⇒ 3y - 4 = x

⇒ 3y = x + 4

$\Rightarrow\ \text{y}=\frac{\text{x}+4}{3}$

$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+4}{3}$ [from (1)]

View full question & answer→
Question 1472 Marks
Define a commutative binary operation on a set.
Answer
Commutativity: Let S be a non-empty set. A function F: S × S → S is said to be binary operation on S.

Mathematically: Let * be a binary operation. It is said to be commutative binary operation if it satisfies commutativity with respect to *.

That is, if $\text{a, b}\in\text{S}$, then

a * b = b * a for all $\text{a, b}\in\text{S}$.

View full question & answer→
Question 1482 Marks
The following defines a relation on N:
$\text{x} +\text{y} = 10,\text{x, y}\in\text{N}$ $$
Determine which of the above relations are reflexive, symmetric and transitive.
Answer
A relation R in A is said to be reflexive if aRa for all $\text{a}\in\text{A},$ R is symmetric if aRb ⇒ bRa, for all $\text{a, b}\in\text{A}$ and it is said to be transitive if aRb and bRc ⇒ aRc for all $\text{a, b, c}\in\text{A.}$

$\text{x} +\text{y} = 10,\text{x, y}\in\text{N}$

 

$(\text{x, y})\in\big\{(1, 9), (9, 1), (2, 8), (8, 2), (3, 7), (7, 3),\\ (4, 6), (6, 4), (5, 5)\big\}$ $$

This is not reflexive as (1, 1), (2, 2), .... are absent.

This only follows the condition of symmetric set as $(1,9)\in\text{R}$ also $(9,1)\in\text{R}$ Similarly other cases are also satisfy the condition.

This is not transitive because {(1, 9), (9, 1)} $\in\text{R}$ but (1, 1) is absent.

View full question & answer→
Question 1492 Marks
If f(x) = x + 7 and g(x) = x - 7, x ∈ R, write fog (7).
Answer
(fog)(7) = f(g(7))
= f(7 - 7)
= f(0)
= 0 + 7
= 7
View full question & answer→
Question 1502 Marks
Check the injectivity and surjectivity of the following functions:
f: N → N given by f(x) = x2
Answer
f: N → N is given by,
f(x) = x2
It is seen that for $\text{x},\text{y}\in\text{N},$ f(x) = f(y) ⇒ x2 = y2 ⇒ x = y.
$\therefore$ f is injective.
Now, $2\in\text{N}.$ But, there does not exist any x in N such that f(x) = x2 = 2.
$\therefore$ f is not surjective.
Hence, function f is injective but not surjective.
View full question & answer→
Generate Paper FreeAll RELATIONS AND FUNCTIONS topics
2 Marks - Page 3 - Maths STD 12 Science Questions - Vidyadip