Maths M.C.Q (1 Marks)

2. Many-one and into.

Solution:

f : R → R

= [x²] + [x + 1] - 3

It is many one function because in this case for two different values of x we would get the same value of f(x).

For $\text{x}=1.1,\ 1.2\in\text{R}$

f(1.1) = (1.1)² + [1.1 + 1] - 3

= [1.21] + [2.1] - 3

= 1 + 2 + 3 = 0

f(1.1) = [1.2]² + [1.2 + 1] - 3

= [1.44] + [2.2] - 3

= 1 + 2 - 3

= 0

It is into function because for the given domain we would only get the integral values of f(x).

But R is the co-domain of the given function.

That means, $\text{Co-domain}\neq\text{Range}$

Hence, the given function is into function.

Therefore, f(x) is many one and into.

4. Neither one-one nor onto.

Solution:

We have,

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

Here, $-2,2\in\text{R}$

Now, $2\neq-2$

But, f(2) = f(-2)

Therefore, function is not one-one.

And,

The minimum value of the function is 0 and maximum value is 1.

That is range of the function is [0, 1] but the co-domain of the function is given R.

Therefore, function is not onto.

$\therefore$ function is neither one-one nor onto.

2. f(x) = x + 2

Solution:

Consider, the second option i.e., f(x) = x + 2

Now, f(x₁) = f(x₂)

⇒ x₁ + 2 = x₂ + 2

⇒ x₁ = x₂

Hence, f(x) = x + 2 is one-one function.

Now, let us suppose, y = x + 2

$\text{x}=\text{y}-2\in\text{Z},\ \forall\ \text{y}\in\text{x}$

Hence, f(x) is one-one and onto.

4. All the three options.

Solution:

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

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

Then, 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, $\text{a, b}\in\text{R}$

⇒ a = b ⇒ b = a ⇒ b, $\text{a}\in\text{R}$ for all $\text{a}\in\text{A}$

So, R is symmetric on A.

2. $-\text{Sgn (x)}\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$

Solution:

Given function is f : R → (-1, 1) is defined by $\text{f(x)}=\frac{-\text{x}|\text{x}|}{1+\text{x}^2}$

Here, for mod function we will have to consider three cases as,

x < 0, x = 0, x > 0

x < 0 ⇒ |x| = -x

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

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

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

$\text{y}+\text{yx}^2=\text{x}^2$

$\text{y}=\text{x}^2-\text{yx}^2$

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

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

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

$\Rightarrow\ \text{x}=-\sqrt{\frac{|\text{y}|}{1-|\text{y}|}}\ \text{x} < 0$ 

Also you can check for the cases x = 0 and x > 0 that $\text{x}=-\sqrt{\frac{|\text{y}|}{1-|\text{y}|}}$

$-\text{Sgn (x)}\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$

2. n(A) = 1

Solution:

Given f is a constant functions.

⇒ range of f is {c}(say)

Since f is one-one

⇒ domain of A should also contain

one element.

$\therefore\text{n(A)}=1$

1. Symmetric and transitive only.

Solution:

Given that A = {a, b, c, d} then a relation R = {(a, b), (b, a), (a, a)} on A.

(a, b), (b, a) $\in\text{R}$

⇒ R is symmetric.

Also for (a, a) R is symmetric.

2. f(x) = x + 2

Solution:

1. f is not because for $\text{y}=3\in\text{Co-domain (Z)},$ there is no value of $\text{x}\in\text{Domain (Z)}$

$\text{x}^3=3$

$\Rightarrow\ \text{x}=\sqrt[3]{3}\notin\text{Z}$

⇒ f is not onto.

So, f is not a bijection.

2. Injectivity: Let x and y be two elements of the domain (Z), such that

x + 2 = y + 2

⇒ x = y

So, f is one-one.

Surjectivity: Let y be an element in the co-domain (Z), such that

y = f(x)

⇒ y = x + 2

$\Rightarrow\ \text{x}=\text{y}-2\in\text{Z}$ (Domain)

⇒ f is onto.

So, f is a bijection.

3. f(x) = 2x + 1 is not onto because if we take $4\in\text{Z}$ (co domain), then 4 = f(x)

$\Rightarrow4 = 2\text{x} + 1$

$\Rightarrow 2\text{x} = 3$

$\Rightarrow\ \text{x}=\frac{3}{2}\notin\text{Z}$

So, f is not a bijection.

4. f(0) = 0² + 0 = 0

⇒ and f(-1) = (-1)² + (-1) = 1 - 1 = 0

⇒ 0 and -1 have the same image.

⇒ f is not one-one.

So, f is not a bijection.

2. $\text{f(x)}=\sin\frac{\pi\text{x}}{2}$

Solution:

​​​​​​​It is clear that f(x) is one-one.

Range of $\text{f}=\Big[\sin\frac{\pi(-1)}{2},\sin\frac{\pi(1)}{2}\Big]=\Big[\sin\frac{-\pi}{2},\sin\frac{\pi}{2}\Big]$

= A = Co-domain of f

⇒ f is onto.

So, f is a bijection.

2. Reflexive.

Solution:

Reflexivity: Since $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A},$ R is reflexive on A.

Symmetry: Since $(\text{a, b})\in\text{R}$ but $(\text{b, a})\notin\text{R,}$ is not symmetric on A.

⇒ R is not antisymmetric on A.

Also, R is not an identity relation on A.

2. $\sqrt{\text{x}}$

Solution:

Given that $\text{f(x)}=\sin^2\text{x}$ and the composite function $\text{g(f(x))}=|\sin\text{x}|$

We will do it using trial and error method.

If we take $\text{g(x)}=-\sqrt{\text{x}}$ and $\text{f(x)}=\sin^2\text{x}$

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

$=-\sin\text{x}$

Which contradicts to the $\text{g(f(x))}=|\sin\text{x}|$

Hence, we take $\text{g(x)}=\sqrt{\text{x}}$

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

$=\sqrt{\sin^2\text{x}}=|\sin\text{x}|$

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

1. $\frac{3\text{x}^2-5}{9\text{x}^4-30\text{x}^2+26}$

Solution:

Given that, f(x) = 3x² - 5 and $\text{g}(\text{x})=\frac{\text{x}}{\text{x}^2+1}$

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

= g(3x² - 5)

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

$=\frac{3\text{x}^2-5}{9\text{x}^4-30\text{x}^2+25+1}$

$=\frac{3\text{x}^2+5}{9\text{x}^4-30\text{x}^2+26}$

4. Onto but not one-one.

Solution:

$\text{M}=\begin{Bmatrix}\text{A}=\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}:\text{a, b, c, d}\in\text{R}\end{Bmatrix}$

f : M → R is given by f(A) = |A|

Injectivity: $\text{f}\begin{pmatrix}\begin{bmatrix}0&0\\0&0\end{bmatrix}\end{pmatrix}=\begin{vmatrix}0&0\\0&0\end{vmatrix}=0$

and $\text{f}\begin{pmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}\end{pmatrix}=\begin{vmatrix}1&0\\0&0\end{vmatrix}=0$

$\Rightarrow\ \text{f}\begin{pmatrix}\begin{bmatrix}0&0\\0&0\end{bmatrix}\end{pmatrix}=\text{f}\begin{pmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}\end{pmatrix}=0$

So, f is not one-one.

Surjectivity: Let y be an element of the co-domain, such that

$\text{f(A)}=-\text{y},\ \text{A}=\begin{bmatrix}\text{a} & \text{b} \\\text{c} & \text{d} \end{bmatrix}$

$\Rightarrow\ \begin{vmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{vmatrix}=\text{y}$

$\Rightarrow\ \text{ad}-\text{bc}=\text{y}$

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

$\Rightarrow\ \text{A}=\begin{bmatrix}\text{a} & \text{b} \\\text{c} & \text{d} \end{bmatrix}\in\text{M}$

⇒ f is onto.

1. $\frac{4}3$

Solution:

Let us first find the identity element.

We know that if e is the identity element then,

$\text{a}\odot\text{e}=\text{e}$

Given $\text{a}\odot\text{e}=\frac{\text{ae}}2$

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

$\Rightarrow\text{e}=2$

Let b be the inverse of 3, then

$3\odot\text{b}=\text{e}$ 

$\Rightarrow\frac{3\text{b}}2=2$

$\Rightarrow\text{b}=\frac{4}3$

4. None of these.

Solution:

The relation R is defined as,

$\text{R}=\{(\text{x, y}):\ \text{x, y}\in\text{A}:\text{y}=3\text{x}\}$

⇒ R = {(1, 3), (2, 6), (3, 9)}

1. $\frac{20}{3}$

4. $\text{i}\phi1$

Solution:

$\because\ |2+3\text{i}|=\sqrt{13}\neq13$

$|3|\neq-3$

$|1+\text{i}|=\sqrt{2}\neq2$

and $|\text{i}|=1$

So, $(\text{i, }1)\in\phi$

4. $\text{None of these}$

Solution:

Given function is f : R → R be given by f(x) = x² - 3.

$\text{y} = \text{x}^2 - 3$

$\text{y} + 3 = \text{x}^2$

$\text{x}=\pm\sqrt{\text{y}+3}$

$\Rightarrow\ \text{y}=\pm\sqrt{\text{x}+3}$

1. Reflexive but not symmetric.

Solution:

We have,

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

As, $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$

So, R is reflexive relation.

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

So, R is not symmetric relation.

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

So, R is transitive relation.

4. Reflexive, transitive but not symmetric.

Solution:

We have,

R = {(m, n): n divides m; m, n ∈ N}

As, m divides m

$\Rightarrow\ (\text{m, m})\in\text{R}\ \forall\ \text{m}\in\text{N}$

So, R is reflexive.

Since, $(2,1)\in\text{R}$ i.e. 1 divides 2

but 2 cannot divide 1 i.e. $(2,1)\notin\text{R}$

So, R is not symmetric.

Let $(\text{m, n})\in\text{R}$ and $(\text{n, p})\in\text{R.}$ Then,

n divides m and p divides n

⇒ p divides m

$\Rightarrow\ (\text{m, p})\in\text{R}$

So, R is transitive.​​​​​​​

1. f^-1(x) = f(x)

Solution:

Given function is $\text{f}:\text{R}-\Big\{\frac{3}{5}\Big\}\rightarrow\ \text{R}$ be defined by $\text{f(x)}=\frac{3\text{x}+2}{5\text{x}-3}$

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

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

$=\frac{3\big(\frac{3\text{x}+2}{5\text{x}-3}\big)+2}{5\big(\frac{3\text{x}+2}{5\text{x}-3}\big)-3}$

After solving you will get

f(f(x)) = x

Also, f^-1(x) = f(x) you can check.

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

4. None of these.

Solution:

$\text{f}:\text{A}\rightarrow\text{B}$

$3^\text{f(x)}+2^{-\text{x}}=4$

$\Rightarrow\ 3^{\text{f(x)}}=4-2^{-\text{x}}$

Taking log on both the sides,

$\text{f(x)}\log3=\log(4-2^{-\text{x}})$

$\Rightarrow\ \text{f(x)}=\frac{\log(4-2^{-\text{x}})}{\log3}$

Logaritmic function will only be defined if $4-2^{-\text{x}}>0$

$\Rightarrow\ 4>2^{-\text{x}}$

$\Rightarrow\ 2^2>2^{-\text{x}}$

$\Rightarrow\ 2>-\text{x}$

$\Rightarrow-2<\text{x}$

$\Rightarrow\ \text{x}\in(-2,\infty)$

That means $\text{A}=\{\text{x}\in\text{R}:-2<\text{x}<\infty\}$

As we know that, $\text{f(x)}=\frac{\log(4-2^{-\text{x}})}{\log3}$

We take $\text{x}=0\in(-2,\infty)$

⇒ f(x) = 1 which does not belong to any of the options.

1. $\frac{3+4\text{x}}{6\text{x}-4}$

3. $(-\infty,4]$

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

$\Rightarrow\ \text{x}-2=\pm\sqrt{4-\text{y}}$

$\Rightarrow\ \text{x}=2\pm\sqrt{4-\text{y}}$

This is defined only when

$4-\text{y}\geq0$

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

X = Range of $\text{f}=(-\infty,4]$

1. 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, $\forall\text{ a}\in\text{Q}-\{-1\}$

Then,

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

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

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

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

$\Rightarrow\ \ f[f(\text{x})]=\Big[3-[f(\text{x})^3\Big]^{\frac{1}{3}}=\left[3-\Big\{(3-\text{x}^3)^{\frac{1}{3}}\Big\}^3\right]^{\frac{1}{3}}$

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

Therefore, option (C) is correct.

3. x.

2. $\text{S}\subset\text{R}$

​​​​​​​Solution:

Given that R is the largest relation on A and S is any relation on A.

We know that R is always subset of A × A.

Hence, $\text{S}\subset\text{R}.$

4. $\text{none of these}.$

Solution:

We have $\text{f}(\text{x})=\frac{2\text{x}-1}{2}$ and g(x) = x + 2

$\text{gof}\Big(\frac{3}{2}\Big)=\text{g}\Big(\text{f}\Big(\frac{3}{2}\Big)\Big)$

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

$=\text{g}(1)=1+2=3$

4. -1

Solution: Given function is $\text{f(x)}=\frac{\alpha\text{x}}{\text{x}+1},\ \text{x}\neq-1$

Also f(f(x)) = x

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

$\frac{\alpha\big(\frac{\alpha\text{x}}{\text{x}+1}\big)}{\frac{\alpha\text{x}}{\text{x}+1}+1}=\text{x}$

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

$\alpha^2=\alpha\text{x}+\text{x}+1$

$\alpha^2=(\alpha+1)\text{x}+1$

Comparing on both sides,

$\alpha+1=0\Rightarrow\ \alpha=-1$​​​​​​​

1. Commutative but not associative.

Solution:

Commutativity:

Let $\text{a, b}\in\text{R}$

a * b = ab + 1

= ba + 1

= b * a

Therefore,

a * b = b * a, $\forall\text{ a, b}\in\text{R}$

Therefore, * is commutative on R.

Associativity:

Let $\text{ a, b, c}\in\text{R}$

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

$\therefore$ a * (b * c) $\neq$ (a * b) * c

For example: a = 1, b = 2 and c = 3 [which belong to R]

Now,

1 * (2 * 3) = 1 * (6 + 1)

= 1 * 7

= 7 + 1

= 8

(1 * 2) * 3 = (2 + 1) * 3

= 3 * 3

= 9 + 1

= 10

⇒ 1 * (2 * 3) $\neq$ (1 * 2) * 3

Therefore, $\exists$ a = 1, b = 2 and c = 3 which belong to R such that

a * (b * c) $\neq$ (a * b) * c

Hence, * is not associative on R.

4. One-one and onto.

Solution:

Given function is,

$\text{f(n)}=\frac{\text{n}-1}{2}$ for n is odd

$=-\frac{\text{n}}{2}$ for n is even

For n is odd,

If f(n) = f(m) then

$\frac{\text{n}-1}{2}=\frac{\text{m}-1}{2}$

⇒ n = m

Also, for n is even if f(n) = f(m) then n = m

Hence, f is one-one.

Also, each element of y is associated with at least one element of x,

f is onto.