Maths M.C.Q (1 Marks)

4. Transitive but not symmetric.

Solution:

We have,

R = {(a, b): a is brother of b}

Let $(\text{a, b})\in\text{R}.$ Then,

a is brother of b.

but b is not necessary brother of a (As, b can be sister of a)

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

So, R is not symmetric.

Also,

Let $(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$

⇒ a is brother of b and b is brother of c

⇒ a is brother of c

$\Rightarrow\ (\text{a, c})\in\text{R}$

So, R is transitive.

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

3. {2, 4, 6}

Solution:

The relation R is defined as R = x, y: $\text{x, y}\in\text{N}$ and x + 2y = 8

⇒ R = x, y: $\text{x, y}\in\text{N}$ and $\text{y}=\frac{8-\text{x}}{2}$

Domain of R is all values of $\text{x}\in\text{N}$ satisfying the relation R.

Also, there are only three values of x that result in y, which is a natural number.

These are {2, 6, 4}.

4. $\frac{4}{\text{a}}$

Solution:

Let e be the identity element in Q⁺ with respect to * such that

a * e = a = e * a, $\forall\text{ a}\in\text{Q}^+$

a * e = a and e * a = a, $\forall\text{ a}\in\text{Q}^+$

$\frac{\text{ae}}2=\text{a}$ and $\frac{\text{ea}}2=\text{a}$, $\forall\text{ a}\in\text{Q}^+$

$\text{e}=2\in\text{Q}^+, \forall\text{ a}\in\text{Q}^+$

Thus, 2 is the identity element in Q⁺ with respect to *.

Let $\text{ a}\in\text{Q}^+$ and $\text{ b}\in\text{Q}^+$ be the inverse of a.

Then,

a * e = a = e * a

a * b = e and b * a = e

$\frac{\text{ab}}2=2$ and $\frac{\text{ba}}2=2$

$\text{b}=\frac{4}{\text{a}}\in\text{Q}^+$

Thus, $\frac{4}{\text{a}}$ is the inverse of $\text{ a}\in\text{Q}^+$.

4. $\Big(\frac{\text{x}-7}{2}\Big)^\frac{1}{3}$

3. An equivalence relation.

Solution:

We observe the following properties of relation R.

Reflexivity: Let $(\text{a, b})\in\text{N}\times\text{N}$

$\Rightarrow\ \text{a, b}\in\text{N}$

$\Rightarrow\ \text{a}+\text{b}=\text{b}+\text{a}$

$\Rightarrow\ (\text{a, b})\in\text{R}$

So, R is reflexive on N × N.

Symmetry: Let $(\text{a, b}),\ (\text{c, d})\in\text{N}\times\text{N}$ such that (a, b)R(c, d)

$\Rightarrow\ \text{a}+\text{d}=\text{b}+\text{c}$

$\Rightarrow\ \text{d}+\text{a}=\text{c}+\text{b}$

$\Rightarrow\ (\text{d, c}),\ (\text{b, a})\in\text{R}$

So, R is symmetric on N × N.

Transitivity: Let $(\text{a, b}),\ (\text{c, d}),\ (\text{e, f})\in\text{N}\times\text{N}$ such that (a, b)R(c, d) and (c, d)R(e, f)

⇒ a + d = b + c and c + f = d + e

⇒ a + d + c + f = b + c + d + e

⇒ a + f = b + e

⇒ (a, b)R(e, f)

So, R is transitive on N × N.

Hence, R is an equivalence relation on N.​​​​​​​

4. f is neither an injection nor a surjection.

Solution:

f : R → R

$\text{f(x)}=\frac{\text{e}^{|\text{x}|}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$

For x = -2 and -3 $\in\text{R}$

$\text{f(-2)}=\frac{\text{e}^{|-2|}-\text{e}^2}{\text{e}^{-2}+\text{e}^2}$

$=\frac{\text{e}^2-\text{e}^2}{\text{e}^{-2}+\text{e}^2}$

$=0$

Hence, for different values of x we are getting same values of f(x)

That means, the given function is many one.

Therefore, this function is not injective.

For x < 0

f(x) = 0

For x > 0

$\text{f(x)}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$

$=\frac{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{}e^{-\text{x}}}-\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$

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

The value of $\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$ is always positive.

Therefore, the value of f(x) is always less than 1.

Numbers more than 1 are not included in the range but they are included in co-domain.

As the codomain is R.

$\therefore\ \text{Co-domain}\neq\text{Range}$

Hence, the given function is not onto.

Therefore, this function is not surjective.

1. $\frac{3}{160}$

1. Reflexive but not symmetric.

Solution:

Given that, A = {1, 2, 3}

and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

$\because\ (1,1), (2,2),(3,3)\in\text{R}$

Hence, R is reflexive.

$(1,2)\in\text{R}$ but $(2,1)\notin\text{R}$

Hence, R is not symmetric.

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

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

Hence, R is transitive.

4. Neither one-one nor onto.

Solution:

Injectivity: Let x and y be two elements in the domain (R), such that

f(x) = f(y)

$\frac{\text{x}^2-8}{\text{x}^2+2}=\frac{\text{y}^2-8}{\text{y}^2+2}$

⇒ (x² - 8)(y² + 2) = (y² - 8)(x² + 2)

⇒ x²y² + 2x² - 8y² - 16 = x²y² + 2y² - 8x² - 16

⇒ 10x² = 10y²

⇒ x² = y²

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

So, f is not one-one.

Surjectivity: $\text{f}(-1)=\frac{(-1)^2-8}{(-1)^2+2}=\frac{1-8}{1+2}=\frac{-7}{3}$

and $\text{f(1)}=\frac{(1)^2-8}{(1)^2+2}=\frac{1-8}{1+2}=\frac{-7}{3}$

$\Rightarrow\ \text{f}(-1)=\text{f}(1)=\frac{-7}{3}$

⇒ f is not onto.

1. Reflexive.

Solution:

We have,

$\text{R} = \big\{(\text{x, y}):\text{x}-\text{y}+\sqrt{2}$ $$ is an irrational number, $\text{x, y}\in\text{R}\big\}$

As, $\text{x}-\text{x}+\sqrt{2}=\sqrt{2},$ which is an irrational number

$\Rightarrow\ (\text{x, x})\in\text{R}$

So, R is reflexive relation.

Since, $\Big(\sqrt{2},2\Big)\in\text{R}$

i.e. $\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-2,$ which is an irrational number

but $2-\sqrt{2}+\sqrt{2}=2,$ which is a rational number

$\Rightarrow\ \Big(2,\sqrt{2}\Big)\notin\text{R}$

So, R is not symmetric relation.

Also, $\Big(\sqrt{2},2\Big)\in\text{R}$ and $\Big(2,2\sqrt{2}\Big)\in\text{R}$

$\Rightarrow\ \Big(\sqrt{2},2\sqrt{2}\Big)\notin\text{R}$

So, R is not transitive relation.

3. $\frac{5\text{x}+4}{3}$

3. $\text{hofog}=\text{hogof}$

Solution:

hogof(x) = h(f(g(x)))

= h(f([x]))

= h(sin^-1[x])

= 2sin^-1[x]

= 2 × 0 = 0

f(x) = sin^-1x

hogof(x) = hogo(x) = 0

4. None of these.

Solution:

$\text{f(x)}=\text{x}+\sqrt{\text{x}^2}=\text{x}\pm\text{x}=0\text{ or }2\text{x}$

⇒ Each element of the domain has 2 images.

f is not a function.

3. Equivalence.

Solution:

Consider that aRb, if a is congruent to b, $\forall\ \text{a, b}\in\text{T}$

Then, $\text{aRa}\Rightarrow\ \text{a}\cong\text{a},$

Which is true for all $\text{a}\in\text{T}$

So, R is reflexive, ....(i)

Let $\text{aRb}\Rightarrow\ \text{a}\cong\text{b}$

$\Rightarrow\ \text{b}\cong\text{a}\Rightarrow\ \text{b}\cong\text{a}$

$\Rightarrow\ \text{bRa}$

So, R is symmetric. ...(ii)

Let aRb and bRc

$\Rightarrow\ \text{a}\cong\text{b}\text{ and }\text{b}\cong\text{c}$

$\Rightarrow\ \text{a}\cong\text{c}\Rightarrow\ \text{aRc}$

So, R is transitive. .....(iii)

Hence, R is equivalence relation.

1. 10⁵

Solution:

Let e be the identity element in Q⁺ with respect to * such that

a * e = a = e * a, $\forall\text{ a}\in\text{Q}^+$

a * e = a and e * a = a, $\forall\text{ a}\in\text{Q}^+$

$\frac{\text{ae}}{100}=\text{a}\text{ and }\frac{\text{ea}}{100}=\text{a},\forall\text{ a}\in\text{Q}^+$

$\text{e}=100,\forall\text{ a}\in\text{Q}^+$

Thus, 100 is the identity element in Q⁺ with repect to *.

0.1 * b = e = b * 0.1

0.1 * b = e and b * 0.1 = e

$\frac{(0.1)\text{b}}{100}=100\text{ and }\frac{\text{b}(0.1)}{100}=100$

$\text{b}=\frac{100\times100}{0.1}$

$=10^5\in\text{Q}^+$

Thus, 10⁵ is the inverse of 0.1.

1. A bijection.

Solution:

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

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.

1. f^-1og^-1

Solution:

Given that, f : A → B and g : B → C be the bijective functions.

$(\text{f}^{-1}\text{o}\text{g}^{-1})\text{o}(\text{gof})=\text{f}^{-1}\text{o}(\text{g}^{-1}\text{ogof})$

$=\text{f}^{-1}\text{o}(\text{g}^{-1}\text{og})\text{of}$ (As composition of functions is associative)

$=(\text{f}^{-1}\text{o}\text{I}_{\text{B}}\text{of})$ (where I_B is identity function on B)

$=(\text{f}^{-1}\text{o}\text{I}_{\text{B}})\text{of}$

$=\text{f}^{-1}\text{of}$

$=\text{I}_{\text{A}}$

Thus $(\text{gof})^{-1}=\text{f}^{-1}\text{og}^{-1}$

2. $\pm2$

2. $\frac{4\text{x}+7}{2}$

2. $[1,\infty)$

​​​​​​​Solution:

Since f is a bijection, co-domain of f = range of f

⇒ B = range of f

Given: f(x) = x² - 4x + 5

Let f(x) = y

⇒ y = x² - 4x + 5

⇒ x² - 4x + (5 - y) = 0

$\because$ Discrimant, $\text{D}=\text{b}^2-4\text{ac}\geq0,$

$(-4)^2-4\times1\times(5-\text{y})\geq0$

$\Rightarrow\ 16-20+4\text{y}\geq0$

$\Rightarrow\ 4\text{y}\geq4$

$\Rightarrow\ \text{y}\geq1$

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

⇒ Range of $\text{f}=[1,\infty)$

$\Rightarrow\ \text{B}=[1,\infty)$

4. $\text{g}(\text{x})=\Big(\frac{\text{x}^\frac{1}{3}-\text{b}}{\text{a}}\Big)^\frac{1}{2}$

4. {1, 2, 3}

Solution:

We know that

$7-\text{x}>0;\ \text{x}-3\geq0$ and $7-\text{x}\geq\text{x}-3$

$\Rightarrow\ \text{x}<7;\ \text{x}\geq3$ and $2\text{x}\leq10$

$\Rightarrow\ \text{x}<7;\ \text{x}\geq3$ and $\text{x}\leq5$

Therefore, x = 3, 4, 5

Range of $\text{f}=\Big\{^{(7-3)}\text{P}_{(3-3)},\ ^{(7-4)}\text{P}_{(4-3)},\ ^{(5-3)}\text{P}_{(7-5)}\Big\}$

= {4P₀, 3P₁, 2P₂}

= {1, 3, 2}

= {1, 2, 3}

4. An equivalence relation.

Solution:

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

Then,

aa = a² > 0 $\Rightarrow\ \text{a, }\forall$

So, S is reflexive on R.

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

Then,

$\text{a, b}\in\text{S}\Rightarrow\ \text{ab}\geq0$ $\Rightarrow\ \text{ba}\geq0\Rightarrow\ \text{ba}\geq0\Rightarrow\ \text{b, a}\in\text{S}\ \forall\ \text{a, b}\in\text{R}$

So, S is symmetric on R.

Transitivity: If $\text{a, b, b, c}\in\text{S}\Rightarrow\ \text{ab}\geq0$ and $\text{bc}\geq0\Rightarrow\ \text{ab}\times\text{bc}\geq0$ $\Rightarrow\ \text{ac}\geq0$

$\text{b}^2\geq0\Rightarrow\ \text{a, c}\in\text{S}$ for all $\text{a, b, c}\in\text{set R}$

Hence, S is an equivalence relation on R.

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

Solution:

We have, $\text{f}(\text{x})=\frac{3\text{x}+2}{5\text{x}-3}=\text{y}\ (\text{let})$

$\Rightarrow\ 3\text{x}+2=5\text{xy}-3\text{y}$

$\Rightarrow\ \text{x}(3-5\text{y})=-3\text{y}-2$

$\Rightarrow\ \text{x}=\frac{3\text{y}+2}{5\text{y}-3}$

$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{3\text{x}+2}{5\text{x}-3}$

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

3. An equivalence relation.

Solution:

R = {(a, b): a = b and a, b $\in\text{A}$}

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

Here,

a = a

$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{A}$

So, R is reflexive on A.

Symmetry: Let $\text{a, b}\in\text{A}$ such that $ (\text{a, b})\in\text{R}.$ Then,

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

$\Rightarrow\ \text{a}=\text{b}$

$\Rightarrow\ \text{b}=\text{a}$

$\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a}\in\text{A}$

So, R is symmetric on A.

Transitive: Let $\text{a, b, c}\in\text{A}$ such that $ (\text{a, b})\in\text{R}$ and $ (\text{b, c})\in\text{R}.$ Then,

$ (\text{a, b})\in\text{R}\Rightarrow\ \text{a}=\text{b}$

and $ (\text{b, c})\in\text{R}\Rightarrow\ \text{b}=\text{c}$

$\Rightarrow\ \text{a}=\text{c}$

$\Rightarrow\ (\text{a, c})\in\text{R}$ for all $\text{a}\in\text{A}$

So, R is transitive on A.

Hence, R is an equivalence relation on A.

4. $0$

Solution:

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

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

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

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

e(1 - a) = 0, $\forall\text{ a}\in\text{Q}-\{-1\}$ $[\because \text{a}\neq1]$

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

4. 2

Solution:

The number of commutative binary operations on a set of n elements is $\text{n}\frac{\text{n}(\text{n}-1)}{2}$.

Therefore,

Number of commutative binary operations an a set of 2 elements $=2\frac{2(2-1)}{2}=2^1$

$=2$

4. 5

Solution:

Given that, A = {1, 2, 3}

Now, number of equivalence relations as follows

R₁ = {(1, 1), (2, 2), (3, 3)}

R₂ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}

R₃ = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}

R₄ = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}

R₅ = {(1, 2, 3) ⇔ A × A = A²}

$\therefore$ Maximum number of equivalence relation is '5'.

1. $\frac{\pi}{4}$

Solution:

Given that, f(x) = tanx

Let y = tanx ⇒ x = tan^-1y

⇒ f^-1(x) = tan^-1x ⇒ f^-1(1) = tan^-11

$\Rightarrow\ \tan^{-1}\tan\frac{\pi}{4}=\frac{\pi}{4}\ \Big[\because\ \tan\frac{\pi}{4}=1\Big]$

4. $\pi$

3. 2

Solution:

Given that, $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{2}\ \forall\ \text{a, b}\in\text{Q}\sim\{0\}$

Let e be the identity element for *

$\therefore\ \text{a}\ ^*\ \text{e}=\frac{\text{ae}}{2}(\text{a}\ *\ \text{e}=\text{e}\ *\ \text{a}=\text{a})$

$\Rightarrow\ \text{a}=\frac{\text{ae}}{2}$

$\Rightarrow\ \text{e}=2$

1. 2x - 3

Solution:

We will solve this problem by the trial-and-error method.

Let us check option (a) first.

If f(x) = 2x - 3

$\frac{1}{2}(\text{gof})(x)=\text{g(f(x))}$

$=\frac{1}{2}\text{g}(2\text{x}-3)$

$=\frac{1}{2}\big[(2\text{x}-3)^2+(2\text{x}-3)-2\big]$

$=\frac{1}{2}[4\text{x}^2+9-12\text{x}+2\text{x}-3-2]$

$=\frac{1}{2}[4\text{x}^2-10\text{x}+4]$

$=2\text{x}^2-5\text{x}+2$

The given condition is satisfied by (a).

2. $\frac{1}2$

Solution:

Let e be the identity element in Q⁺ with respect to * such that

a * e = a = e * a, $\forall\text{ a}\in\text{Q}^+$

a * e = a and e * a = a, $\forall\text{ a}\in\text{Q}^+$

Then,

$\frac{\text{ae}}{2}=\text{a}\text{ and }\frac{\text{ea}}{2}=\text{a},\forall\text{ a}\in\text{Q}^+$

e = 2, $\forall\text{ a}\in\text{Q}^+$

Thus, 2 is the identity element in Q⁺ with respect to *.

Let $\text{b}\in\text{Q}^+$ be the inverse of 8. Then,

8 * b = e = b * 8

8 * b = e and b * 8 = e

$\frac{(8)\text{b}}2=2\text{ and }\frac{\text{b}(8)}2=2$ $[\because\ \text{e}=2]$

b = 12

Thus, $\frac{1}2$ is the inverse of 8.

2. One-one but not onto.

Solution:

Given function is $\text{f(x)}=\frac{\text{x}}{\text{x}+1}$ on $\text{f}:[0,\infty)\rightarrow\ \text{R}$

If f(x) = f(y)

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

⇒ xy + x = xy + y

⇒ x = y

Hence, f is one-one.

If y = f(x)

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

⇒ xy + y = x

⇒ xy - x = -y

x(y - 1) = -y

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

It is not onto.

1. $\frac{2\text{x}-5}{3}$

1. 9

Solution:

We have,

$\text{f(x)}=\begin{cases}2\text{x},&\text{if x}>3\\\text{x}^2,&\text{if }1<\text{x}\leq3\\3\text{x},&\text{if x}\leq1\end{cases}$

Now,

f(-1) + f(2) + f(4)

= 3(-1) + 2² + 2(4)

= -3 + 4 + 8

= 9