Maths M.C.Q (1 Marks)

4. Associative and commutative with an identity.

Solution:

Commutativity:

$\text{X}\triangle\text{Y}=(\overline{\text{X}}\cap\text{Y})\cup(\text{X}\cap\overline{\text{Y}})$

$=(\overline{\text{Y}}\cap\text{X})\cup(\text{Y}\cap\overline{\text{X}})$

$=\text{Y}\triangle\text{X}$

Thus,

$\text{X}\triangle\text{Y}=\text{Y}\triangle\text{X}$

Hence, $\triangle$ is commutative on A.

Let $\phi$ be the identity element for $\triangle$ on P.

$\text{A}\triangle\phi=\big(\overline{\text{A}}\cap\phi\big)\cup\big(\text{A}\cap\overline{\phi}\big)$

$=\phi\cup\text{A}$

$=\text{A}$

and,

$\phi\triangle\text{A}=\big(\overline{\phi}\cap\text{A}\big)\cup\big(\phi\cap\overline{\text{A}}\big)$

$=\text{A}\cup\phi$

$=\text{A}$

2. Reflexive, transitive but not symmetric.

Solution:

The relation S is reflexive, since for any $(\text{a, a})\in\text{S}$ the condition a2b holds,

The relation S is not symmetric since, for any ​​​​​​​$(\text{a, b}]\in\text{S}$ but $(\text{b, a})\notin\text{S}$

The relation S is transitive since, for any $(\text{a, b}]\in\text{S}$ and $(\text{b, c})\in\text{S}$

Therefore, $(\text{a, c})\notin\text{S}$

3. Bijective.

Solution:

Given function is $\text{A}=\{\text{x}:-1\leq\text{x}\leq1\}$ and f : A → A such that f(x) = x|x|

For the mod function we have to check three cases as x < 0, x = 0, x > 0.

For example, x < 0

f(x) = x|x| < 0

|x| = -x

y = -x²

$\text{x}=-\sqrt{-\text{y}}$ which is not possible for x > 0

Hence, f is onto.

⇒ f is bijection.

2. Symmetric.

​​​​​​​Solution:

Given that L denote the set of all straight lines in a plane.

A relation R be defined by lRm if is perpendicular to m for all l, m ∈ L.

R is not reflexive. R is symmetric as we can say $\text{l}\bot\text{m}$ or $\text{m}\bot\text{l}.$

1. xRy : if $\text{x}\leq\text{y}$

​​​​​​​Solution:

In the set of Z of all integers xRy : if $\text{x}\leq\text{y}$ is not an equivalence relation.

For the relation $\text{x}\leq\text{y}(\text{x, y})\in\text{R}$ but (y, x) not belongs to y as $\text{y}\geq\text{x}$ given.

Hence, it is not an equivalence relation.

3. $(\text{x}-3)^\frac{1}{3}$

Solution:

Let f^-1(x) = y

f(y) = x

⇒ y³ + 3 = x

⇒ y³ = x - 3

⇒ y = (x - 3)³

$\Rightarrow\ \text{y}=(\text{x}-3)^\frac{1}{3}$

3. x for all $\text{x}\in\text{R}-\{0,1\}$

Solution:

Domain of f: $1-\text{x}\neq0$

$\Rightarrow\ \text{x}\neq1$

Domain of f = R - {1}

Range of f: $\text{y}=\frac{1}{1-\text{x}}$

$\Rightarrow\ 1-\text{x}=\frac{1}{\text{y}}$

$\Rightarrow\ \text{x}=1-\frac{1}{\text{y}}$

$\Rightarrow\ \text{y}\neq0$

Range of f = R - {0}

So, f : R - {1} → R - {0} and f : R - {1} → R - {0}

Range of f is not a subset of the domain of f.

Domain (fof) = {x : $\text{x}\in$ domain of f and $\text{f(x)}\in$ domain of f}

Domain (fof) $=\big\{\text{x}:\text{x}\in\text{R}-\{1\}\text{ and }\frac{1}{1-\text{x}}\in\text{R}-\{1\}\big\}$

Domain (fof) $=\big\{\text{x}:\text{x}\neq1\text{ and }\frac{1}{1-\text{x}}\neq1\big\}$

Domain (fof) $=\{\text{x}:\text{x}\neq1\text{ and }1-\text{x}\neq1\}$

Domain (fof) $=\{\text{x}:\text{x}\neq1\text{ and }\text{x}\neq0\}$

Domain (fof) = R - {0, 1}

(fof)(x) = f(f(x))

$=\text{f}\Big(\frac{1}{1-\text{x}}\Big)=\frac{1}{1-\frac{1}{1-\text{x}}}=\frac{1-\text{x}}{1-\text{x}-1}$

$=\frac{1-\text{x}}{-\text{x}}=\frac{\text{x}-1}{\text{x}}$

For range of fof, $\text{x}\neq0$

Now, fof : R → {0, 1} → R - {0} and f : R - {1} → R - {0}

Range of fof is not a subset of domain of f.

Domain (fo(fof)) $=\{\text{x}:\text{x}\in$ domain of fof and (fof)(x) $\in$ domain of f$\}$

Domain (fo(fof)) $=\Big\{\text{x}:\text{x}\in\text{R}-\{0,1\}\text{ and }\frac{\text{x}-1}{\text{x}}\in\text{R}-\{1\}\Big\}$

Domain (fo(fof)) $=\Big\{\text{x}:\text{x}\neq0,1\text{ and }\frac{\text{x}-1}{\text{x}}\neq1\Big\}$

Domain (fo(fof)) $=\{\text{x}:\text{x}\neq0,1\text{ and }\text{x}-1\neq\text{x}\}$

Domain (fo(fof)) $=\{\text{x}:\text{x}\neq0,1\text{ and }\text{x}\in\text{R}\}$

Domain (fo(fof)) = R - {0, 1}

​​​​​​​Domain (fo(fof)) = f((fof)(x))

$=\text{f}\Big(\frac{\text{x}-1}{\text{x}}\Big)$

$=\frac{1}{1-\frac{\text{x}-1}{\text{x}}}$

$=\frac{\text{x}}{\text{x}-\text{x}+1}$

$=\text{x}$

So, (fo(fof))(x) = x, where $\text{x}\neq0,1$

3. f(x) = -x - 1, g(x) = x - 1

Solution:

​​​​​​​Since f is invertible, range of f = co-domain of f = x

So, we need to find the range of f to find X.

For finding the range, let f(x) = y

⇒ 4x - x² = y

⇒ x² - 4x = -y

⇒ x² - 4x + 4 = 4 - y

⇒ (x - 2)² = 4 - y

$\Rightarrow\ \text{x}-2=\pm4-\text{y}$

$\Rightarrow\ \text{x}=2\pm4-\text{y}$

This is defined only when $4-\text{y}\geq0$

$\Rightarrow\ \text{y}\leq4,$

X = Range of f $=(-\infty,4]$

3. $\begin{bmatrix}\frac{2}{13}&\frac{-3}{13}\\\frac{3}{13}&\frac{2}{13}\end{bmatrix}$

Solution:

Let the identity of $\begin{bmatrix}2&3\\-3&2\end{bmatrix}$ be $\begin{bmatrix}\text{e}&\text{f}\\-\text{f}&\text{e}\end{bmatrix}$, then

$\begin{bmatrix}2&3\\-3&2\end{bmatrix}*\begin{bmatrix}\text{e}&\text{f}\\-\text{f}&\text{e}\end{bmatrix}=\begin{bmatrix}2&3\\-3&3\end{bmatrix}$

⇒ 2e - 3f = 2 →(1)

2f + 3e = 3 →(2)

Solving (1) and (2) we get e = 1 and f = 0

So, the identity is $\begin{bmatrix}1&0\\0&1\end{bmatrix}$.

Let the inverse be $\begin{bmatrix}\text{a}&\text{b}\\-\text{b}&\text{a}\end{bmatrix}$, then

$\begin{bmatrix}2&3\\-3&2\end{bmatrix}*\begin{bmatrix}\text{a}&\text{b}\\-\text{b}&\text{a}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

⇒ 2a - 3b = 1 →(1)

2b + 3a = 0 →(2)

Solving (1) and (2), we get $\text{a}=\frac{2}{13}$ and $\text{b}=\frac{-3}{13}$

So, the inverse of $\begin{bmatrix}2&3\\-3&2\end{bmatrix}$ is $\begin{bmatrix}\frac{2}{13}&\frac{-3}{13}\\\frac{3}{13}&\frac{2}{13}\end{bmatrix}$

2. Onto but not one-one.

Solution:

Given function is f(x) = (x - 1)(x - 2)(x - 3)

If f(x) = f(y) then

(x - 1)(x - 2)(x - 3) = (y - 1)(y - 2)(y - 3)

⇒ f(1) = f(2) = f(3) = 0

It is not one-one.

y = f(x)

$\text{x}\in\text{R}$ also $\text{y}\in\text{R}$ hence f is onto.

3. 0

Solution:

As, the number of bijection from A into B can only be possible when provided $\frac{7}{(\text{A})}>\frac{7}{(\text{B})}$

But here n(A) < n(B)

So, the number of bijection.

i.e. one-one and onto mapping from A to B.

2. 1

Solution:

When, -1 < x < 0

Then, g(x) = 1 + x - [x]

= 1 + x - (-1) = 2 + x

$\therefore$ f(g(x)) = 1

When, x = 0

Then, g(x) = 1 + x - [x]

= 1 + x - 0 = 1 + x

$\therefore$ f(g(x)) = 1

When, x > 1

Then, g(x) = 1 + x - [x]

= 1 + x - 1 = x

$\therefore$ f(g(x)) = 1

Therefore, for each interval f(g(x)) = 1

3. $\begin{bmatrix}\frac{1}2&\frac{1}2\\\frac{1}2&\frac{1}2\end{bmatrix}$

Solution:

Let $\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}\in\text{G}$ and $\begin{bmatrix}\text{e}&\text{e}\\\text{e}&\text{e}\end{bmatrix}\in\text{G}$ such that

$\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}\begin{bmatrix}\text{e}&\text{e}\\\text{e}&\text{e}\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}$

$\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}\begin{bmatrix}\text{e}&\text{e}\\\text{e}&\text{e}\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}$

$\begin{bmatrix}2\text{ex}&2\text{ex}\\2\text{ex}&2\text{ex}\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}$

$2\text{ex}=\text{x}$

$\text{e}=\frac{1}2\in\text{R}-\{0\}$

Thus, $\begin{bmatrix}\frac{1}2&\frac{1}2\\\frac{1}2&\frac{1}2\end{bmatrix}\in\text{G}$, is the identity element in G.

3. $\frac{\text{x}}{\text{x}-1}$

2. $-\frac{\text{a}}{\text{a}-1}$

Solution:

Let e be the identity element in R - {1} with respect to * such that

a * e = a = e * a, $\forall\text{ a}\in\text{R}-\{1\}$

a * e = a and e * a = a, $\forall\text{ a}\in\text{R}-\{1\}$

Then,

a + e + ae = a and e + a + ea = a, $\forall\text{ a}\in\text{R}-\{1\}$

e(1 + a) = 0, $\forall\text{ a}\in\text{R}-\{1\}$

$\text{e}=0\in\text{R}-\{1\}$

Thus, 0 is the identity element in R - {1} with respect to *.

Let $\text{a}\in\text{R}-\{1\}$ and $\text{b}\in\text{R}-\{1\}$ be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

⇒ a + b + ab = 0 and b + a + ba = 0

$\Rightarrow\text{b}(1+\text{a})=-\text{a}\in\text{R}-\{1\}$

$\Rightarrow\text{b}=\frac{-\text{a}}{\text{a}-1}\in\text{R}-\{1\}$

Thus, $\frac{-\text{a}}{\text{a}-1}$ is the inverse of $\text{a}\in\text{R}-\{1\}$.

2. $2^\text{n}-2$

Solution:

The number of functions from a set with n number of elements into a set with 2 number of elements = 2ⁿ

But two functions can be many-one into function.

4. Many one and into.

Solution:

Graph of the given function is as follows:

A line parallel to X-axis is cutting the graph at two different values.

Therefore, for two different values of x we are getting the same value of y.

That means it is many one function.

From the given graph we can see that the range is $[2,\infty)$ and R is the co-domain of the given function.

Hence, Co-dornain = Range

Therefore, the given function is into.

1. Neither commutative nor associative.

Solution:

Let $\text{a, b}\in\text{Z}$, then

a * b = a - b

b * a = b - a

⇒ a * b $\neq$ b * a

Substraction is not commutative.

(a * b) * c

= (a - b) * c

= a - b - c

a * (b * c)

= a * (b - c)

= a - b + c

⇒ (a * b) * c $\neq$ a * (b * c)

Substraction is not associative.

1. $\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$

3. Not associative.

Solution:

a * b = a² + b²

b * a = b² + a²

⇒ a * b = b * a

So * is commutative.

Now

(a * b) * c

= (a² + b²) * c

= (a² + b²)² + c²

a * (b * c)

= a * (b² + c²)

= a² + (b² +c²)²

⇒ (a * b) * c $\neq$ a * (b * c)

So * is not associative.

4. [0, 1)

Solution:

As f is surjective, range of f = co-domain of f

⇒ A = range of f

$=\frac{\text{x}^2}{\text{x}^2+1},$

$\text{y}=\frac{\text{x}^2}{\text{x}^2+1}$

$\Rightarrow\ \text{y}(\text{x}^2+1)$

$\Rightarrow\ \text{x}^2=\frac{-\text{y}}{(\text{y}-1)}$

$\Rightarrow\ \text{x}=\sqrt{\frac{\text{y}}{(1-\text{y})}}$

$\Rightarrow\ \frac{\text{y}}{(1-\text{y})}\geq0$

$\Rightarrow\ \text{y}\in[0,1)$

⇒ Range of f = [0, 1)

⇒ A = [0, 1)

4. {2, 3, 4, 5}

Solution:

Given that relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : xRy ⇔ x is relatively prime to y. R can be written as,

{(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}

Here we can see that domain means x element which is $2\leq\text{x}\leq5.$

Hence, {2, 3, 4, 5}

3. {0, 2}

Solution:

Since (fog)(x) = (gof)(x),

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

$\Rightarrow\ \text{f}(2^\text{x})=\text{g}(\text{x}^2)$

$\Rightarrow\ \big(2^{\text{x}}\big)^{2}=2^{\text{x}^2}$

$\Rightarrow\ 2^{2\text{x}}=2^{\text{x}^2}$

$\Rightarrow\ \text{x}^2=2\text{x}$

$\Rightarrow\ \text{x}^2-2\text{x}=0$

$\Rightarrow\ \text{x}(\text{x}-2)=0$

$\Rightarrow\ \text{x}=0, 2$

$\Rightarrow\ \text{x}\in\{0,2\}$

2. $\big\{\text{n}\pi+\frac{\pi}{4}:\text{n}\in\text{Z}\big\}$

Solution:

We have, f : R → R is given by

$\text{f(x)}=\tan\text{x}$

$\Rightarrow\ \text{f}^{-1}(\text{x})=\tan^{-1}\text{x}$

$\therefore\ \text{f}^{-1}(1)=\tan^{-1}1=\big\{\text{n}\pi+\frac{\pi}{4}:\text{n}\in\text{Z}\big\}$

1. Symmetric.

Solution:

Given R is the relation over the set of all straight lines in a plane such that $\text{l}_1\text{Rl}_2\Leftrightarrow\text{l}_1\bot\text{l}_2.$

It is symmetric relation as we can say either or $\text{l}_2\bot\text{l}_1.$

4. None of these.

Solution:

* is not clouser because when a = 1 and b = 2,

$\text{a}*\text{b}=\frac{\text{a}}{\text{b}}=\frac{1}{2}\in\text{Z}$

* is not commutative because when a = 1, b = 2 and c = 3,

$1*(2*3)=1*\Big(\frac{2}3\Big)$

$=\frac{1}{\big(\frac{2}{3}\big)}$

$=\frac{3}2$

$(1*2)*3=\frac{1}2*3$

$=\frac{\big(\frac{1}2\big)}{3}$

$=\frac{1}6$

Thus,

$1*(2*3)\neq(1*2)*3$

2. R is reflexive and transitive but not symmetric.

Solution:

Reflexivity: Clearly, ​​​​​​​$(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$

So, R is reflexive on A.

Symmetry: Since, $1,2\in\text{R},$ but $2,1\notin\text{R,}$ R is not symmetric on A.

Transitivity: Since, $1,3,3,2\in\text{R}$ and $1,2\in\text{R},$ R is transitive on A.