2 Marks - Maths STD 12 Science Questions - Vidyadip

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Question 12 Marks

Examine whether the operation ^* defined on R by $\text{a}^*\text{b}=\text{ab}+1$ is (i) a binary or not. (ii) if a binary operation, is it associative or not?

Answer

The given operation is $\text{a}^*\text{b}=\text{ab}+1$
If any operation is a binary operation then it must follow the closure property.
Let $\text{a}\in\text{R},\text{b}\in\text{R}$
then $\text{a}^*\text{b}\in\text{R}$
also $\text{ab}+1\in\text{R}$
i.e. $\text{a}^*\text{b}\in\text{R}$
So ^* on R satisfies the closure property
Now if this binary operation satisfies associative law then
$(\text{a}^*\text{b})^*\text{c}=\text{a}^*(\text{b}^*\text{c})$
$(\text{a}^*\text{b})^*\text{c}=(\text{ab}+1)^*\text{c}$
$=(\text{ab}+1)\text{c}+1$
$=\text{abc}+\text{c}+1$
$\text{a}^*(\text{b}^*\text{c})=\text{a}^*(\text{bc}+1)$
$=\text{a}(\text{bc}+1)+1$
$=\text{abc}+\text{a}+1$
$\therefore(\text{a}^*\text{b})^*\text{c}\neq\text{a}^*(\text{b}^*\text{c})$
i.e., ^* operation does not follow associative law.

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Question 22 Marks

Let $\text{f(x)=}\begin{cases}1+\text{x,}&0\leq\text{x}\leq2\\3-\text{x,}&2<\text{x}\leq3\end{cases}.$ Find fof.

Answer

$\text{f(x)=}\begin{cases}1+\text{x,}&0\leq\text{x}\leq2\\3-\text{x,}&2<\text{x}\leq3\end{cases}$
$\therefore$ Range of $\text{f}=[0,3]\subseteq$ Domain of f.
$\therefore$ fof(x) = f(f(x))
$=\text{f}\begin{cases}1+\text{x,}&0\leq\text{x}\leq2\\3-\text{x,}&2<\text{x}\leq3\end{cases}$
$\text{fof(x)}=\begin{cases}2+\text{x},&0\leq\text{x}\leq1\\2-\text{x},&1<\text{x}\leq2\\4-\text{x},&2<\text{x}\leq3\end{cases}$

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Question 32 Marks

If f : R → R is defined by f(x) = x², find f^-1(-25).

Answer

f : R → R defined by f(x) = x²

$\therefore$ f^-1(x²) = x

$\Rightarrow\ \text{f}^{-1}(-25)=\phi$ $[\because\ \sqrt{-25}\notin\text{R}]$

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Question 42 Marks

The binary operation *: R × R → R is defined as a * b = 2a + b. Find (2 * 3) * 4.

Answer

It is given that, a * b = 2a + b
Now,
(2 * 3) = 2 × 2 + 3
= 4 + 3
= 7
(2 * 3) * 4 = 7 * 4 = 2 × 7 + 4
= 14 + 4
= 18

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Question 52 Marks

Symmetric and transitive but not reflexive.

Answer

“is brother of” R = {( x, y) : x is a brother of y}

It is clear that $\text{x}\geq\text{x}$	$\therefore$	R is reflexive.
It is clear that x is not the brother of x.	$\therefore$	R is not symmetric.
Also if x is brother of y and y is brother of z then
x can be brother of z	$\therefore$	R is transitive.

Therefore, R is symmetric and transitive but not reflexive.

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Question 62 Marks

Define a symmetric relation.

Answer

A relation R on a set A is said to be symmetric if $\text{a, b}\in\text{R}$
Implies that, $\text{b, a}\in\text{R}$ for all $\text{a, b}\in\text{A}$
That is, aRb implies that bRa for all $\text{a, b}\in\text{A}$

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Question 72 Marks

f: R → R defined by f(x) = 1 + x²

Answer

f: R → R is defined as
f(x) = 1 + x²
Let $\text{x}_1,\text{x}_2\in\text{R}$ such that f(x₁) = f(x₂)
$\Rightarrow1+\text{x}_{1}^{2}=1+\text{x}_{2}^{2}$
$\Rightarrow\text{x}_{1}^{2}=\text{x}_{2}^{2}$
$\Rightarrow\text{x}_{1}=\pm\text{x}_{2}$
$\therefore$ f(x₁) = f(x₂) does not imply that x₁ = x₂.
For instance,
f(1) = f(-1) = 2
$\therefore$ f is not one-one.
Consider an element -2 in co-domain R.
It is seen that f(x) = 1 + x² is positive for all $\text{x}\in\text{R}.$
Thus, there does not exist any x in domain R such that f(x) = -2.
$\therefore$ f is not onto.
Hence, f is neither one-one nor onto.

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Question 82 Marks

Let R = {(x, y): |x² - y²| < 1} be a relation on set A = {1, 2, 3, 4, 5}. Write R as a set of ordered pairs.

Answer

Given: R = {(x, y): |x² - y²| < 1} be a relation on A = {1, 2, 3, 4, 5}
Then, R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}

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Question 92 Marks

Write the domain of the relation R defined on the set Z of integers as follows:
(a, b) ∈ R ⇔ a² + b² = 25

Answer

We have,
R = {(a, b) ∈ R ⇔ a² + b² = 25} be a relation on Z.
The domain of R is the value of 'a' ∈ Z, that satisfies a² + b² = 25
a² + b² = 25
$\Rightarrow\ \text{a}=\pm\sqrt{25-\text{b}^2}$
$\therefore$ Domain of $\text{R}=\{0,\pm3,\pm4,\pm5\}$

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Question 102 Marks

Let f : R → R, g : R → R be two functions defined by f(x) = x² + x + 1 and g(x) = 1 - x². Write fog (-2).

Answer

(fog)(-2) = f(g(-2))
= f(1 - (-2)²)
= f(-3)
= (-3)² + (-3) + 1
= 9 - 3 + 1
= 7

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Question 112 Marks

Relation R in the set Z of all integers defined as R = {(x, y) : x – y is an integer}

Answer

R = {( x, y) : x − y is an integer} in set Z of all integers.

Now (x, x) i.e., (1, 1) = 1 - 1 = $0\in\text{Z},$	$\therefore$	R is reflexive.
Again $(\text{x},\ \text{y})\in\text{R},\ \text{and}(\text{y},\ \text{x})\in\text{R},$ i.e., x - y and y - x are an integer	$\therefore$	R is symmetric.
$\text{Also}\ (\text{x}_1,\ \text{y}_1)=\text{x}_1-\text{y}_1\in\text{Z},\ \text{and}(\text{y}_1,\ \text{z}_1)=\text{y}_1-\text{z}_1\in\text{Z}\ \text{and}\\(\text{x}_1,\ \text{z}_1)\in\text{R}, $	$\therefore$	R is transitive.

Therefore, R is reflexive, symmetric and transitive.

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Question 122 Marks

If f : R → R, g : R → R are given by f(x) = (x + 1)² and g(x) = x² + 1, then write the value of fog(-3).

Answer

(fog)(-3) = f(g(-3))
= f((-3)² + 1)
= f(10)
= (10 + 1)²
= 121

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Question 132 Marks

Let * be a binary operation on the set Q of rational numbers as follows:
a * b = ab²

Answer

a * b = ab² and b * a = ba² $\neq\text{a}*\text{b}$
$\therefore$ operation * is not commutative.
(a * b) * c = (ab²) * c = (ab²)c² = ab²c²
And a * (b * c) = a * (bc²) = a(bc²)² = ab²c⁴
Here, $(\text{a}*\text{b})*\text{c}=\text{a}*(\text{b}*\text{c})$
$\therefore$ operation * not is associative.

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Question 142 Marks

Let A and B be two sets, each with a finite number of elements. Assume that there is an injective map from A to B and that there is an injective map from B to A. Prove that there is a bijection from A to B.

Answer

A and B are two non empty sets.
Let f be a function from A to B. It is given that there is injective map from A to B. That means f is one-one function. It is also given that there is injective map from B to A. That means every element of set B has its image in set A.
f is onto function or surjective.
$\therefore$ f is bijective.
If a function is both injective and surjective, then the function is bijective.

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Question 152 Marks

Let A = [-1, 1]. Then, discuss whether the following functions defined on A are one-one, onto or bijective:
g(x) = |x|

Answer

Let g(x₁) = g(x₂)
⇒ |x₁| = |x₂|
$\Rightarrow\ \text{x}_1=\pm\text{x}_2$
So, g(x) is not one-one.
Now, $\text{y}|\text{x}|\Rightarrow\ \text{x}=\pm\text{y}\notin\text{A},\ \forall\ \text{y}\in\text{A}$
So, g(x) is not onto, also, g(x) is not bijective.

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Question 162 Marks

If f: R → R is defined by f(x) = x² – 3x + 2, find f(f(x)).

Answer

It is given that f: R → R is defined by f(x) = x² - 3x + 2.
f(f(x)) = f(x² - 3x + 2)
= (x² - 3x + 2)² - 3(x² - 3x + 2) + 2
= x⁴ + 9x² + 4 - 6x³ - 12x + 4x² - 3x² + 9x - 6 + 2
= x⁴ - 6x³ + 10x² - 3x

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Question 172 Marks

Write the composition table for the binary operation ×₅ (multiplication modulo 5) on the set S = {0, 1, 2, 3, 4}.

Answer

Here,
1 ×₅1 = Remainder obtained by dividing 1 × 1 by 5 = 1
3 ×₅4 = Remainder obtained by dividing 3 × 4 by 5 = 2
4 ×₅4 = Remainder obtained by dividing 4 × 4 by 5 = 1
So, the composition table is as follows:

×₅	0	1	2	3	4
0	0	0	0	0	0
1	0	1	2	3	4
2	0	2	4	1	3
3	0	3	1	4	2
4	0	4	3	2	1

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Question 182 Marks

Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective:
{(a, b): a is a person, b is an ancestor of a}

Answer

g = {(a, b): a is a person, h is an ancestor of a}
Since,the ordered map (a, b) does not map 'a' - a person to a living person.
Therefore, g is not a function.

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Question 192 Marks

Write the domain of the real function $\text{f(x)}=\sqrt{\text{x}-[\text{x}]}.$

Answer

We have, $\text{f(x)}=\sqrt{\text{x}-[\text{x}]}$
The domain of will be real only if
$\text{x}-[\text{x}]\geq0$
⇒ Domain of f = x for all $\text{x}\in\text{R}$
$\therefore$ Domain of f = R
$[\because\ \text{f(x)}=\text{x}-[\text{x}]=\text{x}\ \forall\ \text{x}\in\text{R}]$

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Question 202 Marks

If A = {1, 2, 3, 4} define relations on A which have properties of being:
Symmetric but neither reflexive nor transitive.

Answer

The relation on A having properties of being symmetric, but neither reflexive nor transitive is,
R = {(1, 2), (2, 1)}
The relation R on A is neither reflexive nor transitive, but symmetric.

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Question 212 Marks

Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? If g is described by $\text{g}(\text{x})=\alpha\text{x}+\beta,$ then what value should be assigned to $\alpha$ and $\beta.$

Answer

Yes, g is a function since every element in domain has a unique image.
Now, Let g(x) = ax + b then given,
g(1) = a + b = 1 and,
g(2) = 2a + b = 3
Subtracting g(1) from g(2) gives
(2a + b) - (a + b) = a = 2 and Substituting it into g(1)
We have b = -1.

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Question 222 Marks

f: R → R given by f(x) = x²

Answer

f: R → R is given by,
f(x) = x²
It is seen that f(-1) = f(1) = 1, but $-1\neq1.$
$\therefore$ f is not injective.
Now, $-2\in\text{R}.$ But, there does not exist any element $\text{x}\in\text{R}$ such that f(x) = x² = -2.
$\therefore$ f is not surjective.
Hence, function f is neither injective nor surjective.

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Question 232 Marks

Let the relation R be defined on N by aRb if 2a + 3b = 30. Then write R as a set of ordered pairs.

Answer

If $\text{a, b}\in\text{N}$ then b must be an even integer so that $\text{a}\in\text{N}$
Hence only possible values for b are 2, 4, 6, 8.
if b = 2, it gives a = 12
if b = 4, it gives a = 9
if b = 6, it gives a = 6
if b = 8, it gives a = 3
Hence $(\text{a, b})\in\big\{(3, 8), (6, 6), (9, 4), (12, 2)\big\}$ $$

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Question 242 Marks

Define an associative binary operation on a set.

Answer

An operation * on a set A is called associative binary operation if and only if it is a binary operation as well as associative, i.e. it must satisfy the following two conditions:

$\text{a}\times\text{b}\in\text{A},\forall\text{ a},\text{b}\in\text{A}$ (Binary operation)
$\text{a}\times\text{b}\times\text{c}=\text{a}\times\text{b}\times\text{c},\forall\text{ a, b, c}\in\text{A}$ (Associative)

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Question 252 Marks

Transitive but neither reflexive nor symmetric.

Answer

Relation R = {( x, y) : x > y}
We know that x > x is false. Also x > y but y > x is false and if x > y , y > z this implies x > z.
Therefore, R is transitive, but neither reflexive nor symmetric.

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Question 262 Marks

For each binary operation * defined below, determine whether * is commutative or associative.
On R - {-1}, define $\text{a}*\text{b}=\frac{\text{a}}{\text{b}+1}$

Answer

For commutativity: $\text{a}*\text{b}=\frac{\text{a}}{\text{b}+1}\ \text{and}\ \text{b}*\text{a}=\frac{\text{b}}{\text{a}+1}\Rightarrow\ \ \text{a}*\text{b}\neq\text{b}*\text{a}$
For associativity: $\text{a}*(\text{b}*\text{c})=\text{a}*\Big(\frac{\text{b}}{\text{c}+1}\Big)=\frac{\text{a}}{\frac{\text{a}}{\text{c}+1}+1}=\frac{\text{a(c + a)}}{\text{b + c}+1}$
Also, $(\text{a}*\text{b})*\text{c}=\Big(\frac{\text{a}}{\text{b}+1}\Big)*\text{c}=\frac{\text{a}/\text{b}+1}{\text{c}+1/\text{c}}=\frac{\text{a}}{(\text{b}+1)(\text{c}+1)}$
$\therefore\ \ \text{a} * \text{(b} * \text{c)}\neq\text{(a} * \text{b)}* \text{c}$
Therefore, the operation * is neither commutative nor associative.

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Question 272 Marks

Let 'o' be a binary operation on the set Q₀ of all non-zero rational numbers defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{2} $ for all $\text{a},\text{b}\in\text{Q}_0.$
Find the identity element in Q₀.

Answer

We have,
$\text{a }^*\text{ b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0$
Let $\text{e}\in\text{Q}_0$ be the identity element with respect to *.
By identity property, we have,
a * e = e * a = a for all $\text{a}\in\text{Q}_0$
$\Rightarrow\frac{\text{ae}}{2}=\text{a}\Rightarrow\text{e}=2$
Thus the required identity element is 2.

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Question 282 Marks

Define identity element for a binary operation defined on a set.

Answer

Let S be a non-empty set and * be a binary operation on S.
If there exist an element $\text{e}\in\text{S}$ such that
a * e = e * a = a for all $\text{e}\in\text{S}$
Then e is called the identity element for the binary operation * on S.
'0' is the identity element for '+' on Z
1 is the identity element for '×' on Z.

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Question 292 Marks

Reflexive and transitive but not symmetric.

Answer

“is greater or equal to” $\text{R}=\{(\text{x},\text{y}):\text{x}\geq\text{y}\}$

It is clear that $\text{x}\geq\text{x}$	$\therefore$	R is reflexive.
And $\text{x}\geq\text{y}$ does not imply $\text{y}\geq\text{x}$	$\therefore$	R is not symmetric.
But $\text{x}\geq\text{y},\text{y}\geq\text{z}\Rightarrow\text{x}\geq\text{z}$	$\therefore$	R is transitive.

Therefore, R is reflexive and transitive but not symmetric.

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Question 302 Marks

Find the total number of binary operations on {a, b}.

Answer

We have,
S = {a, b}
The total number of binary operation on S = {a, b} in $2^{2^{2}}= 2^4=16$

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Question 312 Marks

Let * be a binary operation on Z defined by a * b = a + b - 4 for all a, b ∈ Z.
Find the identity element in Z.

Answer

Let e be the identity element in Z with respect to * such that
a * e = a = e * a, $\forall\ \text{a}\in\text{Z}$
a * e = a and e * a = a, $\forall\ \text{a}\in\text{Z}$
a + e - 4 = a and e + a - 4 = a, $\forall\ \text{a}\in\text{Z}$
e = 4, $\forall\ \text{a}\in\text{Z}$
Thus, 4 is the identity element in Z with respect to *.

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Question 322 Marks

Let the relation R be defined on the set A = {1, 2, 3, 4, 5} by R = {(a, b): |a² - b²| < 8}. Write R as a set of ordered pairs.

Answer

Given: A = {1, 2, 3, 4, 5}

R = {(a, b): |a² - b²| < 8}

R = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (3, 4), (4, 3), (4, 4), (5, 5)}

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Question 332 Marks

Determine whether the following operations define a binary operation on the given set or not:
'*' on Q defined by $\text{a}\ ^*\ \text{b}=\frac{\text{a}-1}{\text{b}+1}$ for all $\text{a, b}\in\text{Q.}$

Answer

If a = 2 and b = -1 in Q,
$\text{a}\ ^*\ \text{b}=\frac{\text{a}-1}{\text{b}+1}$
$=\frac{2-1}{-1+1}$
$=\frac{1}{0}$ [which is not defined]
For a = 2 and b = -1,
$\text{a}\ ^*\ \text{b}\notin\text{Q}$
Therefore,
* is a binary operation on Q.

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Question 342 Marks

If the binary operation o is defined by a o b = a + b - ab on the set Q - {-1} of all rational numbers other than 1, shown that o is commutative on Q - [1].

Answer

Let $\text{a, b}\in\text{Q}-1.$ Then,
a o b = a + b - ab
= b + a - ba
= b o a
Therefore,
a o b = b o a, $\forall\ \text{a, b}\in\text{Q}-1$
Thus, o is commutative on Q - {1}.

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Question 352 Marks

Let * be a binary operation on Q - {-1} defined by a * b = a + b + ab for all a, b ∈ Q - {-1}. Then,
Find the identity element in Q − {−1}.

Answer

We have,
a * b = a + b + ab for all a, b ∈ Q - {-1}
Let e be identity element with respect to *.
By identity property,
a * e = a = e * a for all a ∈ Q - {-1}
⇒ a + e + ae = a
⇒ e(1 + a) = 0 ⇒ e = 0 $[\because\ 1+\text{a}\neq0\text{ as }\text{a}\neq-1]$
e = 0 is the identity element with respect to *.

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Question 362 Marks

Determine whether the following operations define a binary operation on the given set or not:
'*' on N defined by a * b = a + b - 2 for all $\text{a, b}\in\text{N.}$

Answer

If a = 1 and b = 1,

a * b = a + b - 2

= 1 + 1 - 2

$=0\notin\text{N}$

Thus, there exist a = 1 and b = 1 such that $\text{a}\ ^*\ \text{b}\notin\text{N}$

So, * is not a binary operation on N.

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Question 372 Marks

Which of the following functions from A to B are one-one and onto?
f₂ = {(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B = {a, b, c}

Answer

f₂ = {(2, a), (3, b), (4, c)}

A = {2, 3, 4}, B = {a, b, c}

It in not clear that different elements of A have different images in B.

$\therefore$ f₂ in not one-one.

Again, each element of B is the image of some element of A.

$\therefore$ f₂ in not on to.

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Question 382 Marks

Let R₀ denote the set of all non-zero real numbers and let A = R₀ × R₀. If '*' is a binary operation on adefined by,
(a, b) * (c, d) = (ac, bd) for all (a, b), (c, d) ∈ A
Find the identity element in A

Answer

Let (x, y) be the identity element in $\text{A}\forall\text{ x, y}\in\text{A}$. Then,
(a, b) * (x, y) = (a, b) = (x, y) * (a, b)
Implies that (a, b) * (x, y) = (a, b) and (x, y) * (a, b) = (a, b)
Implies that (ax, by) = (a, b) and (xa, yb) = (a, b)
Implies that x = 1 and y = 1
Thus, (1, 1) is the identity element of A.

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Question 392 Marks

Determine whether each of the following relations are reflexive, symmetric and transitive:
Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y): 3x – y = 0}

Answer

R = {( x, y): 3x − y = 0}, in A = {1, 2, 3, 4, 5, 6, ……13, 14}
Clearly R = {(1, 3), (2, 6), (3, 9), (4, 12)}

Since, $(\text{x},\text{x})\notin\text{R},$	$\therefore$	R is not reflexive.
Again $(\text{x},\text{y})\in\text{R}\ \text{but}\ (\text{y},\text{x})\notin\text{R}$	$\therefore$	R is not symmetric.
Also $(1,3)\in\text{R}\ \text{and}\ (3,9)\in\text{R}\ \text{but}\notin\text{R},$	$\therefore$	R is not transitive.

Therefore, R is neither reflexive, nor symmetric and nor transitive.

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Question 402 Marks

If f : R → R be defined by $\text{f(x)} = (3 - \text{x}^3)^\frac{1}{3},$ then find fof(x).

Answer

f : R → R defined by $\text{f(x)}=(3-\text{x}^3)^\frac{1}{3}$
$\therefore$ fof(x) = f(f(x))
$=\text{f}(3-\text{x}^3)^\frac{1}{3}$
$=\bigg\{3-\Big[(3-\text{x}^3)^\frac{1}{3}\Big]^3\bigg\}^\frac{1}{3}$
$=\{3-3+\text{x}^3\}^\frac{1}{3}$
$\therefore$ fof(x) = x

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Question 412 Marks

Find fog (2) and gof (1) when : f : R → R; f(x) = x² + 8 and g : R → R; g(x) = 3x³ + 1.

Answer

(fog)(2) = f(g(2)) = f(3 × 2³ + 1) = f(25) = 25² + 8 = 633
(gof)(1) = g(f(1)) = g(1² + 8) = g(9) = 3 × 9³ + 1 = 2188

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Question 422 Marks

Let * be a binary operation on Z defined by a * b = a + b - 4 for all a, b ∈ Z.
Find the invertible elements in Z.

Answer

Let $\text{a}\in\text{Z}$ and $\text{b}\in\text{Z}$ be the inverse of a. Then,
a * b = e = b * a
a * b = e and b * a = e
a + b - 4 = 4 and b + a - 4 = 4
$\text{b}=8-\text{a}\in\text{Z}$
Thus, 8 - a is the inverse of $\text{a}\in\text{Z.}$

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Question 432 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⁺, defined * 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 not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 - 2
$=-1\notin\text{Z}^{+}$

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Question 442 Marks

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.

Answer

f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}
Now, f(1) = 2, f(3) = 5, f(4) =1 and g(1) = 3, g(2) = 3, g(5) =1
(gof)(n) = g[f(x)] = g[f(1)] = g(2) = 3
g[f(3)] = g(5) = 1 and g[f(4)] = g(1) = 3
Hence, gof = {(1, 3), (3, 1), (4, 3)}

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Question 452 Marks

Let * be a binary operation on the set Q of rational numbers as follows:
$\text{a} * \text{b} = \frac{\text{ab}}{4}$

Answer

$\text{a}*\text{b}=\frac{\text{ab}}{4}=\frac{\text{ba}}{4}=\text{b}*\text{a}$
$\therefore$ operation * is commutative.
$(\text{a}*\text{b})*\text{c}=\frac{\text{ab}}{4}*\text{c}=\frac{\frac{\text{ab}}{4}\text{c}}{4}=\frac{\text{abc}}{16}$
And $\text{a}*(\text{b}*\text{c})=\text{a}*\frac{\text{bc}}{4}=\frac{\text{a}\frac{\text{bc}}{4}}{4}=\frac{\text{abc}}{16}$
Here, $(\text{a}*\text{b})*\text{c}=\text{a}*(\text{b}*\text{c})$
$\therefore$ operation * is associative.

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Question 462 Marks

If the binary operation * on the set Z is defined by a * b = a + b - 5, the find the identity element with respect to *.

Answer

Let e be the identity element in Z with respect to * such that,
a * e = a = e * a, $\forall\ \text{a}\in\text{Z}$
a * e = a and e * a = a, $\forall\ \text{a}\in\text{Z}$
a + e - 5 = a and e + a - 5 = a, $\forall\ \text{a}\in\text{Z}$
e = 5, $\forall\ \text{a}\in\text{Z}$
Thus, 5 is the identity element in Z with respect to *.

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Question 472 Marks

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer:
f: R → R defined by f(x) = 3 – 4x

Answer

f: R → R is defined as f(x) = 3 - 4x.

Let $\text{x}_1,\text{x}_2\in\text{R}$ such that f(x₁) = f(x₂)

⇒ 3 - 4x₁ = 3 - 4x₂

⇒ -4x₁ = -4x₂

⇒ x₁ = x₂

$\therefore$ f is one-one.

For any real number (y) in R, there exists $\frac{3-\text{y}}{4}$ in R such that $f\Big(\frac{3-\text{y}}{4}\Big)=3-4\Big(\frac{3-\text{y}}{4}\Big)=\text{y}.$

$\therefore$ f is onto.

Hence, f is bijective.

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Question 482 Marks

If A = {3, 5, 7} and B = {2, 4, 9} and R is a relation given by "is less than", write R as a set ordered pairs.

Answer

Since, R = x, y: x, y $\in\text{N}$ and x < y,
Hence, R = {(3, 4), (3, 9), (5, 9), (7, 9)}

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Question 492 Marks

Give an example of a relation which is,

Reflexive and transitive but not symmetric.

Answer

Let R be the relation on A such that
R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 3)}
Clearly, the relation R on A is reflexive and transitive, but not symmetric.

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Question 502 Marks

Let f : R → R and g : R → R be defined by f(x) = x² and g(x) = x + 1. Show that fog ≠ gof.

Answer

Given, f : R → R and g : R → R.
So, the domains of f and g are the same.
(fog)(x) = f(g(x)) = f(x + 1) = (x + 1)² = x² + 1 + 2x
(gof)(x) = g(f(x)) = g(x²) = x² + 1
So, fog ≠ gof.

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2 Marks - Maths STD 12 Science Questions - Vidyadip