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Maths 3 Marks

RELATIONS AND FUNCTIONS · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 200 questions.

Questions · Page 2 of 4

3 Marks

Question 513 Marks
Let f : R → R be the function defined by f(x) = 2x – 3 ∀ x ∈ R. write f-1.
Answer
Given f(x) = 2x - 3 ∀ x ∈ R

Now, Let a, b ∈ R such that

f(a) = f(b)

⇒ 2a - 3 = 2b - 3

⇒ a = b

⇒ f(x) is One-One.

Also, If x, y ∈ R such that

f(x) = y ⇒ 2x - 3 = y

$\Rightarrow\ \text{x}=\frac{\text{y}+3}{2}=\text{g}(\text{y})\ \forall\ \text{y}\in\text{R}$

⇒ f(x) is Onto and therefore is bijective implies f(x) has an inverse

Let f-1 denote the inverse of f(x) then,

$\text{f}^{-1}\text{x}=\text{g}(\text{x})=\frac{\text{x}+3}{2}\ \forall\ \text{x}\in\text{R}$

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Question 523 Marks
If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.
Answer
A = {1, 2, 3}
Possible onto function from A to A can be the following:
  1. {(1, 1), (2, 2), (3, 3)}
  2. {(1, 1), (2, 3), (3, 2)}
  3. {(1, 2), (2, 2), (3, 3)}
  4. {(1, 2), (2, 1), (3, 3)}
  5. {(1, 3), (2, 2), (3, 1)}
  6. {(1, 3), (2, 1), (3, 2)}
Here, in each function, different elements of the domain have different images.
Therefore, all the function are one-one.
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Question 533 Marks
On the set Z of integers a binary operation * is defined by a * b = ab + 1 for all a, b ∈ Z. Prove that * is not associative on Z.
Answer
Let $\text{a, b, c}\in\text{Z}$
a * (b * c) = a * (bc + 1)
= a(bc + 1) + 1
= abc + a + 1
(a * b) * c = (ab + 1) * c
= (ab + 1)c + 1
= abc + c + 1
Thus, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Z.
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Question 543 Marks
Let f: X → Y be an invertible function. Show that the inverse of f –1 is f, i.e., (f –1)–1 = f.
Answer
Let f: X → Y be an invertible function.
Then f is one-one and onto ⇒ g: y → X where g is also one-one and onto such that
gof(x) = Ix and fog(y) = Iy ⇒ g = f-1
Now, f-1o(f-1)-1 = I and fo[f-1o(f-1)-1] = foI
⇒ [fof-1]0(f-1)-1 = f  ⇒ Io(f-1)-1 = f ⇒ (f-1)-1 = f
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Question 553 Marks
Let f be a function from R to R, such that f(x) = cos(x + 2). Is f invertible? Justify your answer.
Answer
Given: A and B are two sets with finite elements.
f : A → B and g : B → A are injective map.
To prove: f is bijective.
Proof: Since, f : A → B is injective we need to show f in surjective only.
Now,
g : B → A is injective.
⇒ Each element of B has image in A.
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Question 563 Marks
Give an example of a relation which is,
Reflexive and symmetric but not transitive.
Answer
Let A = {4, 6, 8}

Define a relation R on A as:

A = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}

Relation R is reflexive since for every $\text{a}\in\text{A},\ (\text{a, a})\in\text{R}$ i.e., (4, 4), (6, 6), (8, 8) $\in\text{R}$

Relation R is symmetric since $(\text{a, b})\in\text{R}\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a, b}\in\text{R.}$

Relation R is not transitive since (4, 6), (6, 8) $\in\text{R,}$ but $(4,8)\notin\text{R.}$

Hence, relation R is reflexive and symmetric but not transitive.

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Question 573 Marks
Let f : R → R and g : R → R be defined by f(x) = x + 1 and g(x) = x - 1. Show that fog = gof = IR.
Answer
Given, f : R → R and g : R → R

fog : R → R and gof : R → R (Also, we know that IR : R → R)

Therefore, the domains of all fog, and IR are the same.

(fog)(x) = f(g(x)) = f(x - 1) = x - 1 + 1 = x = IR(x) ..... (1)

(gof)(x) = g(f(x)) = g(x + 1) = x + 1 -1 = x = IR(x) ..... (2)

From (1) and (2),

(fog)(x) = (gof)(x) = IR(x), $\forall\text{ x}\in\text{R}$

Hence, fog = gof = IR

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Question 583 Marks
Let A = [-1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:

h(x) = x2

Answer
h(x) = x2

Injection test: Let x and y be any two elements in the domain (A), such that f(x) = f(y).

f(x) = f(y)

x2 = y2

$\text{x}=\pm\text{y}$

So, f is not one-one.

Surjection test: For y = -1, there is no value of x in A.

So, f is not onto.

So, f is not bijective.

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Question 593 Marks
State with reasons whether the following functions have inverse:
f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
Answer
f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
We have,
f(1) = f(2) = f(3) = f(4) = 10
⇒ f is not one-one.
⇒ f is not a bijection.
So, f does not have an inverse.
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Question 603 Marks
Find which of the binary operations are commutative and which are associative.
Let A = N × N and * be the binary operation on A defined by:
(a, b) * (c, d) = (a + c, b + d)
Answer
A = N × N and * is a binary operation defined on A.
(a, b) * (c, d) = (a + b, c + d) = (c + a, d + b) = (c, d) * (a, b)
$\therefore$ The operation is commutative
Again, [(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) = (a + c + e, b + d + f)
And (a, b)[(c, d) * (e, f)] = (a, b) * (c + e, e + f) = (a + c + e, b + d + f)
Here, [(a, b) * (c, d)] * (e, f) = (a, b)[(c, d) * (e, f)]
$\therefore$ The operation is associative.
Let the identity function be (e, f), then (a, b) * (e, f) = (a + e, b + f)
For identity function a = a + e $\Rightarrow\ \ \text{e}=0$
And for b + f = b $\Rightarrow\ \ \text{f}=0$
As $0\neq\text{N},$ therefore identity element does not exist.
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Question 613 Marks
Let A = [-1, 1]. Then, discuss whether the following functions defined on A are one-one, onto or bijective:
$\text{f}(\text{x})=\frac{\text{x}}{2}$
Answer
Let f(x1) = f(x2)

$\Rightarrow\ \frac{\text{x}_1}{2}=\frac{\text{x}_2}{2}\Rightarrow\ \text{x}_1=\text{x}_2$

So, f(x) is one-one.

Now, let $\text{y}=\frac{\text{x}}{2}$

$\Rightarrow\ \text{x}=2\text{y}\notin\text{A},\ \forall\ \text{y}\in\text{A}$

As for $\text{y}=1\in\text{A},\ \text{x}=2\notin\text{A}$

So, f(x) is not onto.

Also, f(x) is not bijective as it is not onto.

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Question 623 Marks
Let f be an invertible real function. Write (f-1of)(1) + (f-1of)(2) + ..... + (f-1of)(100).
Answer
Given that f is an invertible real function.

f-1of = I, where I is an identity function.

Therefore, (f-1of)(1) + (f-1of)(2) + ....... + (f-1of)(100)

= I(1) + I(2) + .... + I(100)

= 1 + 2 + .... + 100 $(\text{As Ix}=\text{x},\ \forall\ \text{x}\in\text{R})$

$=\frac{1001(1001+1)}{2}$ $\Big[$ Sum of first n natural numbers $=\frac{\text{n}(\text{n}+1)}{2}\Big]$

= 5050

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Question 633 Marks
If R and S are transitive relations on a set A, then prove that $\text{R}\cup\text{S}$ may not be a transitive relation on A.
Answer
Let A = {a, b, c} and R and S be two relations on a, given by R = {(a, a), (a, b), (b, a), (b, b)}
And S = {(b, b), (b, c), (c, b), (c, c)}
Here, the relations R and S are transitive on A.
$\text{a, b}\in\text{R}\cup\text{S}$ and $\text{b, c}\in\text{R}\cup\text{S}$ But $\text{a, c}\notin\text{R}\cup\text{S}$
Hence, $\text{R}\cup\text{S}$ is not a transitive relation on A.
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Question 643 Marks
Give examples of two functions f : N → N and g : N → N, such that gof is onto but f is not onto.
Answer
Let us consider a function f : N → N given by f(x) = x + 1, which is not onto.
[This not onto because if we take 0 in N (co-domain), then, 0 = x + 1, x = -1 $\notin\text{N}$]
Let us consider,
$\text{g(x)}:\begin{Bmatrix}\text{x}=\text{x}-1&\text{if x}>1\\1,&\text{if x}=1\end{Bmatrix}$
Now, let us find (gof)(x)
Case 1: x > 1
(gof)(x) = g(f(x)) = g(x + 1) = x + 1 - 1 = x
Case 2: x = 1
(gof)(x) = g(f(x)) = g(x + 1) = 1
From case 1 and case 2, x = x, $\forall\ \text{x}\in\text{N,}$
which is an identity function and, hence, it is onto.
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Question 653 Marks
The cartesian equation of a line is $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}.$ Write its vector form.
Answer
Given: The Cartesian equation of the line is
$\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}=\lambda\ (\text{say})$
$\Rightarrow\ \ \text{x}-5=3\lambda,\ \text{y}+4=7\lambda,\ \text{z}-6=2\lambda$
$\Rightarrow\ \ \text{x}=5+3\lambda,\ \text{y}=-4+7\lambda,\ \text{z}=6+2\lambda$
General equation for the required line is
$\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
Putting the values of x, y, z in this equation,
$\vec{\text{r}}=(5+3\lambda)\hat{\text{i}}+(-4+7\lambda)\hat{\text{j}}+(6+2\lambda)\hat{\text{k}}$
$=5\hat{\text{i}}+3\lambda\hat{\text{i}}-4\hat{\text{j}}+7\lambda\hat{\text{j}}+6\hat{\text{k}}+2\lambda\hat{\text{k}}$
$\Rightarrow\ \ \ \ \vec{\text{r}}=\Big(5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}\Big)+\lambda\Big(3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}\Big)\ \ \ \ \Big(\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\Big)$
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Question 663 Marks
Show that the Modulus Function f: R → R, given by f (x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is –x, if x is negative.
Answer
f: R → R is given by,

$\text{f(x)}=|\text{x}|=\begin{cases}\text{x},&\text{if }\text{x }\geq0\\-\text{x},&\text{if }\text{x }<0\end{cases}$

It is seen that f(-1) = |-1| = 1, f(1) = |1| = 1

$\therefore$ f(-1) = f(1), but $-1\neq1.$

$\therefore$ f is not one-one.

Now, consider $-1\in\text{R}.$It is known that f(x) = |

Hence, the greatest integer functio

x| is always non-negative integer. Thus, there does not exist any element x in domain R such that f(x) = |x| = -1.

$\therefore$ f is not onto.

n is neither one-one nor onto.

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Question 673 Marks
Let f be any real function and let g be a function given by g(x) = 2x. Prove that gof = f + f.
Answer
Given, f : R → R
Since g(x) = 2x is a polynomial, g : R → R
Clearly, gof : R → R and f + f : R → R
So, domain of gof and f + f are the same.
(gof)(x) = g(f(x)) = 2f(x)
(f + f)(x) = f(x) + f(x) = 2f(x)
⇒ (gof)(x) = (f + f)(x), $\forall\ \text{x}\in\text{R}$
Hence, gof = f + f
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Question 683 Marks
Find f-1 if it exists: f : A → B, where, A = {0, -1, -3, 2}; B = {-9, -3, 0, 6} and f(x) = 3x.
Answer
A = {0, -1, -3, 2}; B = {-9, -3, 0, 6}
f : A → B is defined by f(x) = 3x
Since different elements of A have different images in B.
$\therefore$ f is one-one.
Again, each element in B has a preim-age in A.
$\because$ f in one-one bijective.
⇒ f-1 : B → A exists and is given by
$\text{f}^{-1}(\text{x})=\frac{\text{x}}{3}$
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Question 693 Marks
If a function g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by $\text{g(x)}=\alpha\text{x}+\beta\alpha\text{x}+\beta,$ then find the values of $\alpha$ and $\beta.$
Answer
We have,
A function g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by $\text{g(x)}=\alpha\text{x}+\beta$
As, g(1) = 1 and g(2) = 3
Therefore, $\alpha(1)+\beta=1$
$=\alpha+\beta=1\ ....(\text{i})$
and $\alpha(2)-\beta=3$
$2\alpha-\beta=3\ ....(\text{ii})$
(ii) - (i) we get
$2\alpha-\alpha=2$
$\alpha=2$
Substituting $\alpha=2$ in (i), we get
$2+\beta=1$
$\beta=1$
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Question 703 Marks
Each of the following defines a relation on N:
x is greater than $\text{y},\ \text{x},\ \text{y}\in\text{N}$
Determine which of the above relations are reflexive, symmetric and transitive.
Answer
x is greater than $\text{y},\ \text{x},\ \text{y}\in\text{N}$

If $(\text{x},\text{x})\in\text{R},$ then x > x, which is not true for any $\text{x}\in\text{N}$

Therefore, R is not reflexive.

Let $(\text{x},\text{y})\in\text{R}$

⇒ xRy

⇒ x > y

⇒ y > x, which is not true for any $\text{x},\text{y}\in\text{N}$

Thus, R is not symmetric.

Let xRy and yRz

⇒ x > y and y > z

⇒ x > z

⇒ xRz

So, R is transitive.

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Question 713 Marks
Three relation R2 is defined in set A = {a, b, c} as follows:
R2 = {(a, a)}
Find whether or not the relation R2 on A is:
  1. Reflexive.
  2. Symmetric.
  3. Transitive.
Answer
R2 is Reflexive: Clearly $\text{a, a}\in\text{R}_2$
Therefore, R2 is reflexive.
Symmetric: Clearly, $\text{a, a}\in\text{R}\Rightarrow\ \text{a, a}\in\text{R}.$
Therefore, R2 is symmetric.
Transitive: R2 is clearly a transitive relation, since there is only one element in it.
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Question 723 Marks
Show that f: [–1, 1] →R, given by $f(\text{x})=\frac{\text{x}}{(\text{x}+2)}$ is one-one. Find the inverse of the function f:[-1, 1] → Range f.
(Hint: For $\text{y}\in\text{Range f},\text{y}=f(\text{x})=\frac{\text{x}}{\text{x}+2},$ for some x in [-1, 1], i.e., $\text{x}=\frac{2\text{y}}{(1-\text{y})}$)
Answer
Part I: f: [-1, 1] → R given by $f(\text{x})=\frac{\text{x}}{(\text{x}+2)},\text{x}\neq-2$

$\text{Let }\text{x}_1,\text{x}_2\in[-1,1],\text{then }f(\text{x}_1)=\frac{\text{x}_1}{\text{x}_1+2}\text{ and }f(\text{x}_2)=\frac{\text{x}_2}{\text{x}_2+2}$

When f(x1) = f(x2), then $\frac{\text{x}_1}{\text{x}_1+2}=\frac{\text{x}_2}{\text{x}_2+2}$

⇒ x1x2 + 2x1 = x1x2 + 2x2 ⇒ x1 = x2

$\therefore$ f is one-one.

Part II: Let $\text{y}\in\text{Range of f}\Rightarrow\text{y}=f(\text{x})=\frac{\text{x}}{\text{x}+2}$ for some x in [-1, 1] 

As $\text{y}=\frac{\text{x}}{\text{x}+2}$ ⇒ yx + 2y = x ⇒ x(1 - y) = 2y
⇒ $\text{x}=\frac{2\text{y}}{1-\text{y}}$ ⇒ $f^{-1}\text{y}=\frac{2\text{y}}{1-\text{y}}$ $\therefore$ f is onto.

Therefore, $f^{-1}\text{x}=\frac{2\text{x}}{1-\text{x}}$

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Question 733 Marks
Give examples of two functions f: N → N and g: N → N such that gof is onto but f is not onto.
(Hint: Consider f(x) = x + 1 and $\text{g(x)}=\begin{cases}\text{x}-1\ \text{if x}>1\\\ \ \ 1\ \ \ \ \ \text{if x}=1\end{cases}$
Answer
Define f: N → N by,

f(x) = x + 1

And, g: N → N by,

$\text{g(x)}=\begin{cases}\text{x}-1\ \text{if x}>1\\\ \ \ 1\ \ \ \ \ \text{if x}=1\end{cases}$

We first show that g is not onto.

For this, consider element 1 in co-domain N. It is clear that this element is not an image of any of the elements in domain N.

$\therefore$ f is not onto.

Now, gof: N → N is defined by,

gof(x) = g(f(x)) = g(x + 1) = (x + 1) - 1 = x $[\text{x}\in\text{N}\Rightarrow(\text{x}+1)>1]$

Then, it is clear that for $\text{y}\in\text{N},$ there exists $\text{x}=\text{y}\in\text{N}$ such that gof(x) = y.

Hence, gof is onto.

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Question 743 Marks
Given a non empty set X, consider P(X) which is the set of all subsets of X.

Define the relation R in P(X) as follows:

For subsets A, B in P(X), ARB if and only if $\text{A}\subset\text{B}.$ Is R an equivalence relation on P(X)? Justify your answer.

 

Answer
Since every set is a subset of itself, ARA for all $\text{A}\in\text{P(X)}.$
$\therefore$ R is reflexive.
Let $\text{ARB}\Rightarrow\text{A}\subset\text{B}.$
This cannot be implied to $\text{B}\subset\text{A}.$
For instance, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.
$\therefore$ R is not symmetric.
Further, if ARB and BRC, then $\text{A}\subset\text{B}\ \text{and}\ \text{B}\subset\text{C}.$
$\Rightarrow\text{A}\subset\text{C}$
$\Rightarrow\text{ARC}$
$\therefore$ R is transitive.
Hence, R is not an equivalence relation since it is not symmetric.
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Question 753 Marks
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
Answer
A = {1, 2, 3, 4, 5} and R = {(a, b) : a − b is even} , then R = {(1, 3), (1, 5), (3, 5), (2, 4)}

  1. For (a, a), |a − a| = 0 which is even. $\therefore$ R is reflexive
    If |a − b| is even, then |b − a| is also even. $\therefore$ R is symmetric
    Now, if |a − b| and |b − c| is even then |a − b +  b − c| ⇒ |a − c|    
    is also even. $\therefore$ R is transitive
    Therefore, R is an equivalence relation.

  2. Elements of {1, 3, 5} are related to each other.

Since |1− 3| = 2, |3− 5| = 2, |1− 5| = 4, all are even numbers

⇒ Elements of {1, 3, 5} are related to each other.

Similarly elements of (2, 4) are related to each other.

Since |2 − 4| = 2 an even number, then no element of the set {1, 3, 5} is related to any element of (2, 4).

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Question 763 Marks
Given a non-empty set X, consider the binary operation *: P(X) × P(X) → P(X) given by $\text{A}*\text{B}=\text{A}\cap\text{B}\ \ \forall\ \text{A},\text{ B}$ in P(X), where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation *.
Answer
It is given that *: P(X) × P(X) → P(X) is defined as $\text{A}*\text{B}=\text{A}\cap\text{B}\ \ \forall\ \text{A},\text{ B}\in\text{P(X)}$
We know that $\text{A}\cap\text{X}=\text{A}=\text{X}\cap\text{A}\forall\text{A}\in\text{P(X)}$
Thus, X is the identity element for the given binary operation *.
Now, an element $\text{A}\in\text{P(X)}$ is invertibleif there exists $\text{B}\in\text{P(X)}$ such that
A * B = X = B * A. (As X is the identity element)
i.e.,
$\text{A}\cap\text{B}=\text{X}=\text{B}\cap\text{A}$
This case is possible only when A = X = B.
Thus, X is the only invertible element in P(X) with respect to the given operation *.
Hence, the given result is proved.
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Question 773 Marks
Let A = {1, 2, 3}, and let R2 = {(2, 2), (3, 1), (1, 3)}. Find whether or not the relations R2 on A is:
  1. Reflexive.
  2. Symmetric.
  3. Transitive.
Answer
R2 = {(2, 2), (3, 1), (1, 3)}
$\therefore\ (1,1)\notin\text{R}_2$
⇒ R1 is not reflexive.
Now, $(1,3)\in\text{R}_2$
$\Rightarrow\ (3,1)\in\text{R}_2$
$\therefore$ R2 is symmetric.
Again,
$(3,1)\in\text{R}_2$ and $(1,3)\in\text{R}_2$ but $(3,3)\notin\text{R}_2$
$\therefore$ R2 is not transitive.
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Question 783 Marks
Give examples of two one-one functions f1 and f2 from R to R, such that f1 + f2 : R → R. defined by (f1 + f2)(x) = f1(x) + f2(x) is not one-one.
Answer
We know that f1: R → R, given by f1(x) = x, and f2(x) = -x are one-one.
Proving f1 is one-one: Let f1(x) = f1(y)
⇒ x = y
So, f1 is one-one.
Proving f2 is one-one: Let f2(x) = f2(y)
⇒ -x = -y
⇒ x = y
So, f2 is one-one.
Proving (f1 + f2) is not one-one:
Given: (f1 + f2)(x) = f1(x) + f2(x)= x + (-x) =0
So, for every real number x, (f1 + f2)(x)=0
So, the image of ever number in the domain is same as 0.
Thus, (f1 + f2) is not one-one.
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Question 793 Marks
Let A = [-1, 1]. Then, discuss whether the following functions defined on A are one-one, onto or bijective:
k(x) = x2
Answer
Let k(x1) = k(x2)

$\Rightarrow\ \text{x}^2_1=\text{x}^2_2\Rightarrow\ \text{x}_1=\pm\text{x}_2$

Thus, k(x) is not one-one.

Now, let y = x2

$\Rightarrow\ \text{x}\sqrt{\text{y}}\notin\text{A},\ \forall\ \text{y}\in\text{A}$

As for $\text{y}=-1,\ \text{x}=\sqrt{-1}\notin\text{A}$

Hence, k(x) is neither one-one nor onto.

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Question 803 Marks
Let * be a binary operation on Z defined by a * b = a + b - 4 for all a, b ∈ Z.
Show that '*' is both commutative and associative.
Answer
Commutativity: Let $\text{a, b}\in\text{Z}.$ Then,
a * b = a + b - 4
= b + a - 4
= b * a
Therefore,
a * b = b * a, $\forall\ \text{a, b}\in\text{Z}$
Thus, * is commutative on Z.
Associativity: Let $\text{a, b, c}\in\text{Z}.$ Then,
a * (b * c) = a * (b + c - 4)
= a + b + c - 4 - 4
= a + b + c - 8
(a * b) * c = (a + b - 4) * c
= a + b - 4 + c - 4
= a + b + c - 8
Therefore, a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{Z}.$
Thus, * is associative on Z.
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Question 813 Marks
If f(x) = sin x and g(x) = 2x be two real functions, then describe gof and fog. Are these equal functions?
Answer
We know that

f : R → -1, 1 and g : R → R

gof clearly, the range of f is a subset of the domain of g.

gof : R → R, gof(x) = g(f(x)) = g(sinx) = 2sinx

fog clearly, the range of g is a subset of the domain of f.

fog : R → R

So,

f : R → R, fog(x) = f(g(x)) = f(2x) = sin2x

Clearly, $\text{fog}\neq\text{gof}$

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Question 823 Marks
Find gof and fog when f : R → R and g : R → R are defined by:
f(x) = x and g(x) = |x|
Answer
Given, f : R → R and g : R → R

Therefore, gof : R → R and fog : R → R

f(x) = x and g(x) = |x|

(gof)(x) = g(f(x))

= g(x)

(fog)(x) = f(g(x))

= f|x|

= |x|

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Question 833 Marks
Check whether the relation R defined on the set A = {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Answer
Reflexivity: Let a be an arbitrary element of R. Then, a = a + 1 cannot be true for all $\text{a}\in\text{A.}$

$\Rightarrow\ (\text{a, a})\notin\text{R}$

So, R is not reflexive on A.

Symmetry: Let $(\text{a, b})\in\text{R}$

⇒ b = a + 1

⇒ -a = -b + 1

⇒ a = b - 1

Thus, $(\text{b, a})\notin\text{R}$

So, R is not symmetric on A.

Transitivity: Let (1, 2) and (2, 3) $\in\text{R}$

⇒ 2 = 1 + 1 and 3 = 2 + 1 is true.

But $3\neq1+1$

$\Rightarrow\ (1,3)\notin\text{R}$

So, R is not transitive on A.

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Question 843 Marks
The following relation are defined on the set of real numbers.
aRb if 1 + ab > 0
Find whether these relation are reflexive, symmetric or transitive.
Answer
We have aRb if 1 + ab > 0
Let R be the set of real numbers
Reflexive: Let $\text{a}\in\text{R}$
⇒ 1 + a2 > 0
⇒ aRa
⇒ R is reflexive.
Symmetric: Let aRb
⇒ 1 + ab > 0
⇒ 1 + ba > 0
⇒ bRa
⇒ R is symmetric
Transitive: Let aRb and bRc
⇒ 1 + ab > 0 and 1 + bc > 0
⇒ 1 + ac > 0
⇒ R is not transitive.
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Question 853 Marks
On the set Z of all integers a binary operation * is defined by a * b = a + b + 2 for all a, b ∈ Z. Write the inverse of 4.
Answer
To find the identity element, 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, $\forall\text{ a}\in\text{Z}$
Then,
a + e + 2 = a and e + a + 2 = a, $\forall\text{ a}\in\text{Z}$
$\text{e}=-2\in\text{Z},\ \forall\text{ a}\in\text{Z}$
Thus, -2 is the identity element in Z with respect to *.
Now,
Let b ∈ Z be the inverse of 4.
Here,
4 * b = e = b * 4
4 * b = e and b * 4 = e
Then,
4 + b + 2 = -2 and b + 4 + 2 = -2
$\text{b}=-8\in\text{Z}$
Thus, -8 is the inverse of 4.
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Question 863 Marks
If f : A → B and g : B → C are onto functions, show that gof is a onto function.
Answer
Given, f : A → B and g : B → C are onto.
Then, gof : A → C
Let us take an element z in the co-domain (C).
Now, z is in C and g : B → C is onto.
So, there exists some element y in B, such that g(y) = z .... (1)
Now, y is in B and f : A → B is onto.
So, there exists some x in A, such that f(x) = y .... (2)
From (1) and (2),
z = g(y) = g(f(x)) = (gof)(x)
So, z = (gof)(x), where x is in A.
Hence, gof is onto.
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Question 873 Marks
Let f : R → R be the function defined by f(x) = 4x - 3 for all x ∈ R. Then write f-1.
Answer
We have,
f : R → R is the function defined by f(x) = 4x - 3 for all $\text{x}\in\text{R}$
Let f(x) = y. Then,
y = 4x - 3
4x = y + 3
$\text{x}=\frac{\text{y}+3}{4}$
Therefore, $\text{f}^{-1}(\text{y})=\frac{\text{y}+3}{4}$
or, $\text{f}^{-1}(\text{x})=\frac{\text{x}+3}{4}$
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Question 883 Marks
Classify the following functions as injection, surjection or bijection:
f : Z → Z, defined by f(x) = x - 5
Answer
f : Z → Z, defined by f(x) = x - 5

Injection test: Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

x - 5 = y - 5

x = y

Therefore, f is an injection.

Surjection test: Let y be any element in the co domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x - 5 = y

x = y + 5, which is in Z.

Therefore, f is a surjection and f is a bijection.

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Question 893 Marks
Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.
Answer
Let A be a set.

Then $\text{I}_\text{A}=\{(\text{a, a});\text{ a}\in\text{A}\}$ is the identity relation on A.

Hence, every identity relation on a set is reflexive by defination.

Converse:

Let A = {(a, b, c)} be a set.

Let R = {(a, a), (b, b), (c, c), (a, b)} be a relation defined on A.

Clearly R is reflexive on set A, but it is not identity relation on set A as $(\text{a, b})\in\text{R}$

Hence, a reflexive relation need not be identity relation.

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Question 903 Marks
Find gof and fog when f : R → R and g : R → R are defined by:
f(x) = x2 + 2x - 3 and g(x) = 3x - 4
Answer
Given, f : R → R and g : R → R
Therefore, gof : R → R and fog : R → R
f(x) = x2 + 2x - 3 and g(x) = 3x - 4
Now, gof(x) = g(f(x)) = g(x2 + 2x - 3)
$\therefore$ gof(x) = 3(x2 + 2x - 3) - 4
⇒ gof(x) = 3x2 + 6x - 13
and, fog(x) = f(g(x)) = f(3x - 4)
$\therefore$ fog(x) = (3x - 4)2 + 2(3x - 4) - 3
= 9x2 + 16 - 24x + 6x - 8 - 3
$\therefore$ fog(x) = 9x2 - 18x + 5
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Question 913 Marks
Classify the following functions as injection, surjection or bijection:
f : R → R, defined by f(x) = sin2x + cos2x
Answer
f : R → R, defined by f(x) = sin2x + cos2x
f(x) = sin2x + cos2x = 1
So, f(x) = 1 for every x in R.
So, for all elements in the domain, the image is 1.
So, f is not an injection.
Range of f = {1}
Co-domain of f = R
Both are not same.
So, f is not a surjection and f is not a bijection.
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Question 923 Marks
If f(x) = |x|, prove that fof = f.
Answer
We have, f(x) = |x|
We assume the domain of f = R
Range of $\text{f}=(0,\infty)$
$\therefore$ Range of f $\subset$ domain of f
$\therefore$ fof exists.
Now,
fof(x) = f(f(x)) = f(|x|) = ||x|| = f(x)
$\therefore$ fof = f
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Question 933 Marks
Let * be a binary operation on Q - {-1} defined by a * b = a + b + ab for all a, b ∈ Q - {-1}. Then,
Show that '*' is both commutative and associative on Q - {-1}.
Answer
We have,
a * b = a + b + ab for all a, b ∈ Q - {-1}
Commutativity: Let a, b ∈ Q - {-1}
⇒ a * b = a + b + ab = b + a + ba = b * a
⇒ a * b = b * a
⇒ '*' is commutative on Q - {-1}
Associativity: Let a, b, c ∈ Q - {-1}, then
⇒ (a * b) * c = (a + b + ab) * c
= a + b + ab + c + ac + bc + abc .......(i)
and, a * (b * c) = a * (b + c + bc)
= a + b + c + bc + ab + ac + abc .....(ii)
From (i) and (ii)
(a * b) * c = a * (b * c)
⇒ * is associative on Q - {-1}
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Question 943 Marks
State with reasons whether the following functions have inverse:
h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Answer
h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

Here, different elements of the domain have different images in the co-domain.

⇒ h is one-one.

Also,

each element in the co-domain has a pre-image in the domain.

⇒ h is onto.

⇒ h is bijection.

⇒ h has an inverse and it is given by,

h - 1 = {(7, 2), (9, 3), (11, 4), (13, 5)}

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Question 953 Marks
If the mapping f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3}, given by f = {(1, 2), (3, 5), (4, 1)} and g= {(2, 3), (5, 1), (1, 3)}, then write fog.
Answer
We have,
f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3}, are given by f = {(1, 2), (3, 5), (4, 1)} and g= {(2, 3), (5, 1), (1, 3)} respectively.
As,
fog(2) = f(g(2)) = f(3) = 5,
fog(5) = f(g(5)) = f(1) = 2,
fog(1) = f(g(1)) = f(3) = 5,
Therefore, fog : {1, 2, 5} → {1, 2, 5} is given by fog = {(2, 5), (5, 2), (1, 5)}
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Question 963 Marks
Find gof and fog when f : R → R and g : R → R are defined by:
f(x) = x2 + 8 and g(x) = 3x3 + 1
Answer
Given, f : R → R and g : R → R
Therefore, gof : R → R and fog : R → R
f(x) = x2 + 8 and g(x) = 3x3 + 1
(gof)(x) = g(f(x))
= g(x2 + 8)
= 3(x2 + 8)3 + 1
(fog)(x) = f(g(x))
= f(3x3 + 1)
= (3x3 + 1)2 + 8
= 9x6 + 6x3 + 1 + 8
= 9x6 + 6x3 + 9
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Question 973 Marks
Write the composition table for the binary operation multiplication modulo 10 (×10) on the set S = {2, 4, 6, 8}.
Answer
2 ×104 = Remainder obtained by dividing 2 × 4 by 10 = 8
4 ×106 = Remainder obtained by dividing 4 × 6 by 10 = 4
2 ×108 = Remainder obtained by dividing 2 × 8 by 10 = 6
3 ×104 = Remainder obtained by dividing 3 × 4 by 10 = 2
Therefore, the composition table is as follows:
×10 2 4 6 8
2 4 8 2 6
4 8 6 4 2
6 2 4 6 8
8 6 2 8 4
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Question 983 Marks
Let * be a binary operation on Q - {-1} defined by a * b = a + b + ab for all a, b ∈ Q - {-1}. Then,
Show that every element of Q − {−1} is invertible. Also, find the inverse of an arbitrary element.
Answer
We have,
a * b = a + b + ab for all a, b ∈ Q - {-1}
Let b be the inverse of a ∈ Q - {-1}
Then, a * b = b * a = e [e is the identity element]
⇒ a + b + ab = e
⇒ a + b + ab = 0
⇒ b(1 + a) = -a
$\Rightarrow\text{b}=\frac{-\text{a}}{1+\text{a}}$ $\begin{bmatrix}\because\ \frac{-\text{a}}{1+\text{a}}\neq-1\text{ because if }\frac{-\text{a}}{1+\text{a}}=-1\\\Rightarrow\text{a}=1+\text{a}\Rightarrow1=0\text{ Not possible}\end{bmatrix}$
$\text{b}=\frac{-\text{a}}{1+\text{a}}$ is the inversre of a with respect to *.
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Question 993 Marks
Three relation R4 is defined in set A = {a, b, c} as follows:
R4 = {(a, b), (b, c), (c, a)}
Find whether or not the relation R4 on A is:
  1. Reflexive.
  2. Symmetric.
  3. Transitive.
Answer
R4 is reflexive: Here, $\text{a}\notin\text{R}_4,\ \text{b},\ \text{b}\notin\text{R}_4,\ \text{c},\ \text{c}\notin\text{R}_4$
Therefore, R4 is not reflexive.
Symmetric: Here, $\text{a, b}\notin\text{R}_4,$ but $\text{b, b}\notin\text{R}_4$
Therefore, R4 is not symmetric.
Transitive: Here, $\text{a, b}\notin\text{R}_4,\ \text{b, c}\in\text{R}_4$
But $\text{a, c}\notin\text{R}_4$
Therefore, R4 is not transitive.
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Question 1003 Marks
The following relation are defined on the set of real numbers.
aRb if a - b > 0
Find whether these relations are reflexive, symmetric or transitive.
Answer
aRb if a - b > 0

Let R be the set of real numbers.

Reflexivity: Let $\text{a}\in\text{R}$

$\Rightarrow\ \text{a}-\text{a}=0$

$\Rightarrow\ (\text{a, a})\notin\text{R}$

$\therefore$ R is not reflexive.

Symmetric: Let aRb

⇒ a - a > 0

⇒ b - a < 0

$\therefore$ R is not symmetric.

Transitive: Let aRb and bRc

⇒ a - a > and b - c > 0

⇒ a - c > 0

⇒ aRc

$\therefore$ R is transitive.

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3 Marks - Page 2 - Maths STD 12 Science Questions - Vidyadip