Maths M.C.Q (1 Marks)

2. Reflexive and symmetric.

Solution:

Reflexivity: Let $\text{x}\in\text{R.}$ Then,

$\text{x}-\text{x}=0<1$

$\Rightarrow\ |\text{x}-\text{x}|\leq1$

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

So, R is reflexive on Z.

Symmetry: Let $\text{x, y}\in\text{R.}$ Then,

$|\text{x}-\text{y}|\leq0$

$\Rightarrow\ |-(\text{y}-\text{x})|\leq1$

$\Rightarrow\ |(\text{y}-\text{x})|\leq1$ [Since |x - y| = |y - x|]

$\Rightarrow\ (\text{y, x})\in\text{R}$ for all $\text{x, y}\in\text{Z}$

So, R is symmetric on Z.

Transitivity: Let $(\text{x, y})\in\text{R}$ and $(\text{y, z})\in\text{R}.$ Then,

$|\text{x}-\text{y}|\leq1$ and $|\text{y}-\text{z}|\leq1$

⇒ It is not always true that $|\text{x}-\text{y}|\leq1.$

$\Rightarrow\ (\text{x, z})\notin\text{R}$

So, R is not transitive on Z.

3. 16

Solution:

Total number of binary operations on a set containing n elements is

$\text{(n)}^{\text{n}^2}$ so for n = 2 we have $(2)^{2^2}=2^4=16$

2. R is reflexive and transitive but not symmetric.

3. $(6,8)\in\text{R}.$

4. f is neither one-one nor onto.

2. $\text{g}(\text{y})=\frac{4\text{y}}{4-3\text{y}}.$

Also, for any real number (y) in co-domain R, there exists $\frac{\text{y}}{3}$ in R such that $f\Big(\frac{\text{y}}{3}\Big)=3\Big(\frac{\text{y}}{3}\Big)=\text{y}$

1. f is one-one onto.

2. Is * commutative but not associative?

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

Solution:

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

$\Rightarrow\ \text{f(y)}=\text{x}$

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

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

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

$\Rightarrow\ \text{e}^{2\text{y}}-1=\text{xe}^{2\text{y}}+\text{x}$

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

$\Rightarrow\ 2\text{y}=\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$

$\Rightarrow\ \text{y}=\frac{1}{2}\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$

$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{1}{2}\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$ [From (1)]

1. Bijection.

Solution:

$\text{f(x)}=\sin^{-1}(3\text{x}-4\text{x}^3)$

$\Rightarrow\ \text{f(x)}=3\sin^{-1}\text{x}$

Injectivity: Let x and y be two elements in the domain $\Big[\frac{-1}{2},\frac{1}{2}\Big],$ such that

f(x) = f(y)

$\Rightarrow\ 3\sin^{-1}\text{x}=3\sin^{-1}\text{y}$

$\Rightarrow\ \sin^{-1}\text{x}=\sin^{-1}\text{y}$

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

So, f is one-one.

Surjectivity: Let y be any element in the co-domain, such that

f(x) = y

$\Rightarrow\ 3\sin^{-1}\text{x}=\text{y}$

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

$\Rightarrow\ \text{x}=\sin\frac{\text{y}}{3}\in\Big[\frac{-1}{2},\frac{1}{2}\Big]$

⇒ f is onto.

⇒ f is a bijection.

1. 9

Solution:

We are given that, $\text{f}(\text{x})=\begin{cases}2\text{x}:\text{x}>3\\\text{x}^2:1<\text{x}\leq3\\3\text{x}:\text{x}\leq1\end{cases}$

Now, f(-1) + f(2) + f(4) = 3(-1) + (2)² + 2 × 4

= -3 + 4 + 8

= 9

4. One-one and onto both.

Solution:

Injectivity: Let x and y be any two elements in the domain (N).

Case-1: Both x and y are even.

Let f(x) = f(y)

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

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

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

Case-2: Both x and y are odd.

Let f(x) = f(y)

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

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

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

Case-3: Let x be even and y be odd.

Then, $\text{f(x)}=\frac{-\text{x}}{2}$ and $\text{f(y)}=\frac{\text{y}-1}{2}$

Then, clearly

$\text{x}\neq\text{y}$

$\Rightarrow\ \text{f(x)}\neq\text{f(y)}$

From all the cases, f is one-one.

Surjectivity: Co-domain of f = Z = {......, -3, -2, -1, 0, 1, 2, 3, ......}

Range of $\text{f}=\Big\{....,\ \frac{-3-1}{2},\ \frac{-(-2)}{2},\ \frac{-1-1}{2},\ \frac{0}{2},\ \frac{1-1}{2},\ \frac{-2}{2},\ \frac{3-1}{2},\ ....\Big\}$

⇒ Range of f = {....., -2, 1, -1, 0, 0, -1, 1, .....}

⇒ Range of f = {....., -2, -1, 0, 1, 2, ......}

⇒ Co-domain of f = Range of f

⇒ f is onto.

2. * defined by $\text{a}*\text{b}=\frac{\text{a + b}}2$ is a binary operation on Q.

Solution:

For option a, if we take 3 and 2 then

$3*2=\frac{5}2\in\text{Z}$. So, option a is not true.

For option b, if we take any two numbers a and b

then $\frac{\text{a + b}}2$ belongs to Q for $\text{a, b}\in\text{Q}$.

So, option b is correct.

For option d, if we take 2, 3 then $2-3=-1\in\text{N}$.

So, option d is not true.

Option c is not true.

2. is given by $\frac{\text{x}+5}{3}$

Solution:

Given function is f : R → R is given by f(x) = 3x - 5

To find f^-1(x)

y = f(x)

⇒ y = 3x - 5

⇒ y + 5 = 3x

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

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

3. x

Solution:

We are given that, f : [0, 1] → [0, 1] be defined by $\text{f}(\text{x})=\begin{cases}\text{x, if x is rational}\\1-\text{x, if x is irrational}\end{cases}$

Now, $(\text{fof})\text{x} = \text{f}(\text{f(x)})$

$=\text{x}$

4. f is not defined.

Solution:

Given that, $\text{f}(\text{x})=\frac{1}{\text{x}}\ \forall\ \text{x}\in\text{R}.$

For x = 0, f(x) is not defined.

Hence, f(x) is a not defined function.

1. $\frac{9}8$

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,

$\frac{\text{ae}}{3}=\text{a}\text{ and }\frac{\text{ea}}{3}=\text{a},\forall\text{ a}\in\text{Q}^+$

e = 3, $\forall\text{ a}\in\text{Q}^+$

Thus, 3 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 * b = e = b * a

a * b = e and b * a = e

$\therefore\ \frac{\text{ab}}3=3\text{ and }\frac{\text{ba}}3=3$

$\text{b}=\frac{9}{\text{a}}\in\text{Q}^+$

Thus, $\frac{9}{\text{a}}$ is the inverse of $\text{a}\in\text{Q}^+$.

Given: $\text{a}*\text{b}=\frac{\text{ab}}3$

$4*6=\frac{4\times6}3=8$

Now,

$\text{a}^{-1}=\frac{9}{\text{a}}$

$(4*6)^{-1}=8^{-1}$

$=\frac{9}8$

1. Commutative but not associative.

Solution:

We are given that, a * b = 1 + ab ∀ a, b ∈ R

Consider, a * b = ab + 1

= ba + 1

= b * a

Hence, * is a communicative binary operation.

Also, a * (b * c) = a * (bc + 1) $[\because$ b * c = bc + 1$]$

= a(bc + 1) + 1

= a + abc + 1

Now, (a * b) * c = (ab + 1) * c $[\because$ a * b = ab + 1$]$

= (1 + ab)c + 1

= c + abc +1

Now, $\text{a}+\text{abc}+1\neq\text{c}+\text{abc}+1$

$\Rightarrow\ \text{a}\ ^*\ (\text{b}\ ^* \ \text{c})\neq(\text{a}\ ^* \ \text{b})\ ^* \ \text{c}$

Therefore, * is not associative.

Hence, * is communicative but not associative.

2. $\frac{1}{2}\log_{10}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$

2. $[1,\infty)$

Solution:

Given that, $\text{f}(\text{x})=\text{x}^2-4\text{x}+5,$

Let $\text{y}=\text{x}^2-4\text{x}+5$

$\Rightarrow\ \text{y}=\text{x}^2-4\text{x}+4+1$

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

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

$\Rightarrow\ \text{x}-2=\sqrt{\text{y}-1}$

$\Rightarrow\ \text{x}=2+\sqrt{\text{y}-1}$

$\therefore\ \text{y}-1\geq0,\ \text{y}\geq1$

Range $=[1,\infty)$

3. $\{0,\pm3,\pm4,\pm5\}$

Solution:

As aRb ⇔ a < b

does not satisfy reflexive and symmetric relation.

1. $\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$

Solution:

f(x) = -x² + 6x - 8, is a polynomial function and the domain of polynomial function is real number.

$\therefore\ \text{x}\in\text{R}$

f(x) = -x² + 6x - 8

= -(x² - 6x + 8)

= -(x² - 6x + 9 - 1)

= -(x - 3)² + 1

Maximum value of -(x - 3)² woud be 0

$\therefore$ Maximum value of -(x - 3)² + 1 woud be 1

$\therefore\ \text{f(x)}\in(-\infty,1]$

We can see from the given graph that function is symmetrical about x = 3 and the given function is bijective.

So, x would be either $(-\infty,3]\text{ or }[3,\infty)$

The correct option which satisfy A and B both is:

$\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$

3. {1}

Solution:

Here, $\text{R}=\text{x, y}:\text{x}\in\text{A}$ and $\text{y}\in\text{B}:\text{x}>\text{y}$ ⇒ R = 2, 1, 3, 1

Thus, Range of R = {1}

1. Onto but not one-one.

Solution:

Given function is

$\text{f(x)}=\frac{\text{x}}{2}$ if x is even

= 0 if x is odd

For f(3) = 0 and f(4) = 0

⇒ f(3) = f(4)

But, $3\neq4$

Hence, it is not one-one.

$\text{x}\in\text{R}\Rightarrow\ \text{y}\in\text{R}$

Here, Domain = range of f

Hence, it is onto.

4. Neither injective nor surjective.

Solution:

Given function is f : R → R, f(x) = x²

If f(x) = f(y) then

x² = y²

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

Hence, it is not one-one or injective.

f(x) = y

y = x²

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

But co-domain is R.

Hence, it is not onto or surjective.

3. Not commutative.

Solution:

Let $\text{a, b}\in\text{Z}$

a * b = 3a + b

b * a = 3b + a

Thus, a * b $\neq$ b * a

If a = 1 and b = 2,

1 * 2 = 3(1) + 2

= 5

2 * 1 = 3(2) + 1

= 7

1 * 2 $\neq$ 2 * 1

Thus, * is not commutative on Z.

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

Solution:

1. Range of $\text{f}=\Big[\frac{-1}{2},\frac{1}{2}\Big]\neq\text{A}$

So, f is not a bijection.

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

So, g is a bijection.

3. h(-1) = |-1| = 1

And h(1) = |1| = 1

⇒ -1 and 1 have the same images.

So, h is not a bijection.

4. k(-1) = (-1)² = 1

And k(1) = (1)² = 1

⇒ -1 and 1 have the same images.

So, k is not a bijection.

3. Transitive.

Solution:

Reflexivity: Since $(1, 1)\notin\text{B,}$ B is not reflexive on A.

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

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

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

​​​​​​​Solution:

Let f^-1(x) = y

​​​​​​​$\Rightarrow\ \text{f(y)} = \text{x}$

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

$\Rightarrow\ \text{y}^2 + 1 = \text{xy}$

$\Rightarrow\ \text{y}^2 - \text{xy} + 1 = 0$

$\Rightarrow\ \text{y}^2-2\times\text{y}\times\frac{\text{x}}{2}+\big(\frac{\text{x}}{2}\big)^2-\big(\frac{\text{x}}{2}\big)^2+1=0$

$\Rightarrow\ \text{y}^2-2\times\text{y}\times\frac{\text{x}}{2}+\big(\frac{\text{x}}{2}\big)^2=\frac{\text{x}^2-1}{4}$

$\Rightarrow\ \Big(\text{y}-\frac{\text{x}}{2}\Big)^2=\frac{\text{x}^2-1}{4}$

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

$\Rightarrow\ \text{y}=\frac{\text{x}}{2}+\frac{\sqrt{\text{x}^2-4}}{2}$

$\Rightarrow\ \text{y}=\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$

$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$