MATHS M.C.Q (1 Marks)

1. {(8, 11), (10, 13)}

Solution:

R is a relation from {11, 12, 13} to {8, 10, 12}, defined by y = x - 3

Now, we have,

11 - 3 = 8

13 - 3 = 10

So, R = {(13, 10), (11, 8)}

$\therefore$ R^-1 = {(10, 13), (8, 11)}

3. 12

Solution:

Given, function $ \text{f}(\text{x})=\sin \big(\frac{2\pi\text{x}}{3}\big)+\cos \big(\frac{\pi\text{x}}{2}\big)$

Now, period of $ \text{f}(\text{x})=\big(\frac{2\pi\text{x}\times{3}}{2\pi}\big)=3$

and period of $\cos \Big(\frac{\text{n}\pi}{2}\Big)=\frac{2\text{n}}{\frac{\text{n}}{2}} = \big(\frac{2\pi\text{}\times{2}}{ \pi}\big)= 2 × 2 = 4$ $$

Now, period of f(x) = LCM(3, 4) = 12 Hence, period of function $\text{f(x)} = \sin\frac{2\pi\text{x}}{3} + \cos \big(\frac{\pi\text{x}}{2}\big) + \cos$ $$is 12

3. $\text{f(xy)}\leq\text{f(x)}\text{f(y)}$

Solution:

Given, $\text{f(x)}=\sqrt{\text{x}^2+1}\ ...(\text{i})$

Replacing x by y in (i), we get

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

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

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

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

Also, replacing x by xy in (i), we get

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

Now,

$\text{x}^2\text{y}^2+1\leq\text{x}^2\text{y}^2+\text{x}^2+\text{y}^2+1$

$\Rightarrow\sqrt{\text{x}^2\text{y}^2+1}\leq\sqrt{\text{x}^2\text{y}^2+\text{x}^2+\text{y}^2+1}$

$\Rightarrow\text{f}(\text{xy})\leq\text{f(x)}\text{f(y)}$

4. $ \text{R}$

Solution:

Given,

function is $ \text{f}(\text{x}) = \frac{1}{(2 -\cos 3\text{x})}$

Since $ -1 \leq \cos \text{3x} \leq1$ for all$\text{ x }∈\text{R}$

So,$ -1 \leq 2 \cos \text{3x} \leq1$for all$\text{ x }∈\text{R}$

⇒ $\text{f}(\text{x})$ is defined for all$\text{ x }∈\text{R}$

So, domain of f(x) is R.

3. $(-\infty, -1]\cup\Big[\frac{1}{3},\infty\Big)$

Solution:

We know that, $-1\leq-\cos\text{x}\leq1$

$\Rightarrow-1\leq-\cos\text{x}\leq1$

$\Rightarrow-2\leq-2\cos\text{x}\leq2$

$\Rightarrow-1\leq-2\cos\text{x}\leq3$

Now $\text{f(x)}=\frac{1}{1-2\cos\text{x}}$ is defined if

$-1\leq-2\cos\text{x}\leq0$ or $0<1-2\cos\text{x}\leq3$

$\Rightarrow-1\geq\frac{1}{1-2\cos\text{x}}>-\infty$ or $\infty>\frac{1}{1-2\cos\text{x}}\geq\frac{1}{3}$

$\Rightarrow\frac{1}{1-2\cos\text{x}}\in(-\infty, -1]\cup\Big[\frac{1}{3},\infty\Big)$

3. $\big[-3,-2\big]\cup\big[2,3\big] $

Solution:

$\text{f(x)}=\sqrt{5|\text{x}|-\text{x}^2-16}$

For f(x) to be defined, $5|\text{x}|-\text{x}^2-6\geq0$

$\Rightarrow5|\text{x}|-\text{x}^2-6\geq0$

$\Rightarrow\text{x}^2-5|\text{x}|+6\leq0$

For x > 0, |x| = x

$\Rightarrow\text{x}^2-5\text{x}+6\leq0$

$\Rightarrow(\text{x}-2)(\text{x}-3)\leq0$

$\Rightarrow\text{x}\in[2,3]\ ...(\text{i})$

For x < 0, |x| = -x

$\Rightarrow\text{x}^2+5\text{x}+6\leq0$

$\Rightarrow(\text{x}+2)(\text{x}+3)\leq0$

$\Rightarrow\text{x}\in\big[-3,-2\big]\ ...(\text{ii})$

From (i) and (ii),

$\text{x}\in\big[-3,-2\big]\cup\big[2,3\big]$

Or, $\text{domain(f)}=\big[-3,-2\big]\cup\big[2,3\big]$

2. $ {(1, 4), (3, 4)}$

3. {3, 4, 5}

Solution:

The function $\text{f}(\text{x})= ^{7-\text{x}}\text{P}_\text{x-3}$ is defined only if x is an integer satisfying the following inequalities:

$ 7 - \text{x} ≥ 0.....(1)$

$ \text{x} - 3 ≥ 0......(2)$

$ 7 - \text{x} ≥ \text{x} - 3........(3)$

Now, from 1, we get $ \text{x} ≤ 7 ……… (4)$

from 2, we get $\text{x} ≥ 3 ……………. (5)$

and from 2, we get $\text{x} ≤ 5 ………. (6)$

From 4, 5 and 6, we get

$ 3 ≤ \text{x} ≤ 5$

So, the domain is {3, 4, 5}.

1. $\big[-1,2\big)\cap\big[3,\infty\big)$

Solution:

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

For f(x) to be defined,

$(\text{x}-2)\neq0$

$\Rightarrow\text{x}\neq2\ ...(\text{i})$

Also,

$\frac{(\text{x}+1)(\text{x}-3)}{\text{x}-2}\geq0$

$\Rightarrow\frac{(\text{x}+1)(\text{x}-3)(\text{x}-2)}{(\text{x}-2)^2}\geq0$

$\Rightarrow(\text{x}+1)(\text{x}-3)(\text{x}-2)\geq0$

$\Rightarrow\text{x}\in\big[-1,2\big)\cup\big[3,\infty\big)\ ...(\text{ii})$

From (i) and (ii),

$\text{x}\in\big[-1,2\big)\cap\big[3,\infty\big)$

2. $ -1- \sqrt{3 ≤ \text{x} ≤ -1}$

2. $\{\cos1,\cos2,1\}$

Solution:

Since, $\text{f(x)}=\cos[\text{x}],$ where $\frac{-\pi}{2}<\text{x}<\frac{\pi}{2}$

$-\frac{\pi}{2}<\text{x}<\frac{\pi}{2}$

$\Rightarrow-1.57<\text{x}<1.57$

$\Rightarrow[\text{x}]\ \in\ \{-1,0,1,2\}$

Thus, $\cos[\text{x}]=\{\cos(-1),\cos0,\cos1,\cos2\}$

Range of $\text{f(x)}=\{\cos1,1,\cos2\}$

1. $\frac{\text{ (x}^{2} + 2)}{(\text{x}^{2} + 1)}$

Solution:

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

Given f(x) = x² + 2 and $ \text{g}(\text{x})=\frac{\text{x}}{(\text{x}\ - \ 1)}$

Now, $\text{gof(x)} = \text{g}(\text{x}^2 + 2) = \text{got}(\text{x})=\text{g}(\text{x}^2 +2)=\frac{\text{ (x}^{2} +\ 2)}{(\text{x}^{2} \ +\ 2\ -\ 1)}=\frac{(\text{x}^2\ +\ 2)}{(\text{x}^2\ +\ 1)}$

4. None of these.

Solution:

$\text{f(x)}=27\text{x}^3+\frac{1}{\text{x}^3}$

$\Rightarrow\text{f(x)}=\Big(3\text{x}+\frac{1}{\text{x}}\Big)\Big(9\text{x}^2+\frac{1}{\text{x}^2}-3\Big)$

$\Rightarrow\text{f(x)}=\Big(3\text{x}+\frac{1}{\text{x}}\Big)\Big(\Big(3\text{x}+\frac{1}{\text{x}}\Big)^2-9\Big)$

$\Rightarrow\text{f}(\alpha)=\Big(3\alpha+\frac{1}{\alpha}\Big)\Big(\Big(3\alpha+\frac{1}{\alpha}\Big)^2-9\Big)$

Since $\alpha$ and $\beta$ are the roots of $3\text{x}+\frac{1}{\text{x}}=12,$

$3\alpha+\frac{1}{\alpha}=12$ and $3\beta+\frac{1}{\beta}=12$

$\Rightarrow\text{f}(\alpha)=12\big((12)^2-9\big)$ and $\text{f}(\beta)=12\big((12)^2-9\big)$

$\Rightarrow\text{f}(\alpha)=\text{f}(\beta)=\big((12)^2-9\big)$

2. $ -1– \sqrt{3} ≤ \times ≤ -1+\sqrt{3}$

3. Domain = R - {4}, Range = {-1}

Solution:

 Given that: $\text{f(x)}=\frac{4-\text{x}}{\text{x}-4}$

We know that f(x) is defined if $\text{x}-4\neq0 \Rightarrow \text{x}\neq4$

So, the domain of f(x) is = R - {4}

Let $\text{f(x)}=\text{y}=\frac{4-\text{x}}{\text{x}-4}$

$\Rightarrow\text{yx}-4\text{y}=4-\text{x}\Rightarrow\text{yx}+\text{x}=4\text{y}+4$

$\Rightarrow\text{x}(\text{y}+1)=4\text{y}+4\Rightarrow\text{x}=\frac{4(1+\text{y})}{1+\text{y}}$

If x is real number, then $1+\text{y}\neq0\Rightarrow\text{x}\neq1$

$\therefore$ Range of f(x) = R - {-1)

3. $\Big[\frac{3}{4},1\Big]$

Solution:

Given,

$\text{f(x)}=\cos^2\text{x}+\sin^4\text{x}$

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

$\Rightarrow\text{f(x)}=\Big(\sin^2\text{x}-\frac{1}{2}\Big)^2+\frac{3}{4}$

The minimum value of $\text{f(x)}$ is $\frac{3}{4}$

Also,

$\sin^2\text{x}\leq1$

$\Rightarrow\ \sin^2\text{x}-\frac{1}{2}\leq\frac{1}{2}$

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

$\Rightarrow\ \Big(\sin^2\text{x}-\frac{1}{2}\Big)^2+\frac{3}{4}\leq\frac{1}{4}+\frac{3}{4}$

$\Rightarrow\ \text{f(x)}\leq1$

The maximum value of f(x) is 1

$\therefore\ \text{f(R)}=\Big(\frac{3}{4},1\Big)$

1. $\frac{\text{ (x}^{2} + 2)}{(\text{x}^{2} + 1)}$

Solution:

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

Given f(x) = x² + 2 and $ \text{g}(\text{x})=\frac{\text{x}}{(\text{x}\ - \ 1)}$

Now, gof(x) = g(x² + 2) = $\text{got}(\text{x})=\text{g}(\text{x}^2 +2)=\frac{\text{ (x}^{2} +\ 2)}{(\text{x}^{2} \ +\ 2\ -\ 1)}=\frac{(\text{x}^2\ +\ 2)}{(\text{x}^2\ +\ 1)}$

1. $\text{f}\Big(\frac{\pi}{2}\Big)=1$

Solution:

$\text{f(x)}=\sin[\pi^2]\text{x}+\sin[-\pi^2]\text{x}$

$\Rightarrow\text{f(x)}=\sin\big[9.8\big]\text{x}+\sin\big[-9.8\big]\text{x}$

$\Rightarrow\text{f(x)}=\sin9\text{x}-\sin10\text{x}$

$\Rightarrow\text{f}\Big(\frac{\pi}{2}\Big)=\sin9\times\frac{\pi}{2}-\sin10\times\frac{\pi}{2}$

$\Rightarrow\text{f}\Big(\frac{\pi}{2}\Big)=1-0=1$

4. None of these.

Solution:

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

Since, $|\text{x}^2-1|\neq|\text{x}-1|^2$

$\text{f(x)}^2\neq(\text{f(x)})^2$

Thus, (i) is wrong.

Since, $|\text{x}+\text{y}-1|\neq|\text{x}-1||\text{y}-1|$

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

Thus, (ii) is wrong.

Since, $|\text{|x|}-1\neq||\text{x}-1||=|\text{x}-1|$

$\text{f(|x|)}\neq|\text{f(x)}|$

Thus, (iii) is wrong.

Hence, none of the given options is the answer.

3. $ ±5$

Solution:

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

⇒ $\text{y}=\text{x}^{2}+1$

⇒ $ \text{y}-1=\text{x}^{2}$

⇒ $ \text{x} = ±\sqrt{(\text{y} – 1)}$

⇒ $\text{f}^{-1} \text{x} = ±\sqrt{(\text{x} – 1)}$

Now, $\text{f}^{-1} (26) = ±\sqrt{(\text{26} – 1)}$

⇒ $\text{f}^{-1} (26) = ±\sqrt{(\text{25} )}$

⇒ $\text{f}^{-1} (26) = ± {5 }$

3. {1}

Solution:

A = {1, 2, 3} and B = {1, 4, 6, 9}

R is a relation from A to B defined by: x is greater than y.

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

$\therefore$ Range (R) = {1}

2. $\{(\text{x, y}):\text{y}=\text{|x|},\text{x, y}\in\text{R}\}$

Solution:

For every value of $\text{x}\in\text{R},$ there is a unique value $\text{y} \in\text{R}$

i.e., there is a unique image for all values of $\text{x}\in\text{R},$

Also, values of x occur only once in the ordered pairs.

Thus, it is a function.

3. $\text{A}= \text{ B}$

1. $ \text{R}$

Solution:

Since $ \tan^1$ x exists if $\text{x}\in(-\infty,\infty)$

So, (2x + 1) is defined if

$ -\infty < 2\text{x} + 1 <\infty$

$\Rightarrow-\infty < \times <\infty$

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

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

So, domain of $ \tan^-1(2\text{x}+1)$ is R.

4. [0, 3]

Solution:

We know, square root is always non-negative.$\sqrt{9-\text{x}^2}>0$.

So, the range of the function is set of positive real numbers from 0 to 3.

2.  $\big[-1,-\sqrt{3},-1+\sqrt{3}\big]$

Solution:

$\text{f(x)}=\sqrt{2-2\text{x}-\text{x}^2}$

Since, $2-2\text{x}-\text{x}^2\geq0$

$\text{x}^2+2\text{x}-2\leq0$

$\Rightarrow\text{x}^2-2\text{x}-2+1-1\leq0$

$\Rightarrow(\text{x}-1)^2-\big(\sqrt{3}\big)^2\leq0$

$\Rightarrow\big[\text{x}-\big(1-\sqrt{3}\big)\big]\big[\text{x}-\big(1+\sqrt{3}\big)\big]\leq0$

$\Rightarrow\big(-1-\sqrt{3}\big)\leq\text{x}\leq(-1+\sqrt{3})$

Thus, domain $(\text{f})=\big[-1-\sqrt{3},-1+\sqrt{3}\big]$ 

1. $(\text{R}\times\text{P}) \cap(\text{R}\times\text{Q})$

2. Domain = R, Range $= ( –\infty, 2]$

Solution:

We have, f(x) = 2 - |x - 5|

Clearly, f(x) is defined for all $\text{x}\in\text{R}.$

$\therefore$ Domain of f = R

Now, $|\text{x}-5|\geq0,\forall\text{x}\in\text{R}$

$\Rightarrow-|\text{x}-5|\leq0$

$\Rightarrow2-|\text{x}-5|\leq2$

$\therefore\text{f(x)}\leq2$

$\therefore$ Range of $\text{f}=(-\infty, 2]$

2. $\Big[-1,\frac{1}{3}\Big]$

Solution:

We know that $-1\leq\cos\text{x}\leq1$ for all $\text{x}\in\text{R}$

Now,

$-1\leq\cos\text{x}\leq1$

$\Rightarrow-1\leq\cos\text{x}\leq1$

$\Rightarrow-2\leq-2\cos\text{x}\leq2$

$\Rightarrow-1\leq1-2\cos\text{x}\leq3$ (Adding 1 ro each term)

But,

$\cos\text{x}\neq\frac{1}{2}$

$\Rightarrow1-2\cos\text{x}\in\big[-1,3\big]-\{0\}$

$\Rightarrow\frac{1}{1-2\cos\text{x}}\in\big(-\infty,-1\big]\cap\Big[\frac{1}{3},\infty\Big)$

$\therefore\ \text{Range of }\text{f(x)}=\big(-\infty,-1\big]\cap\Big[\frac{1}{3},\infty\Big)$

Disclaimer: The range of the function does not matches with either of the given options. The range matches with option (c) if it is given as $\big(-\infty,-1\big]\cap\Big[\frac{1}{3},\infty\Big)$

3. $0$

3. f(x) + f(1 - x) = 1

Solution:

$\text{f(x)}=\frac{4^{\text{x}}}{4^{\text{x}}+2}\text{ x }\in\text{R},$

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

$=\frac{4}{2\times4^{\text{x}}+4}$

$=\frac{2}{4^{\text{x}}+2}$

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

$=\frac{4^{\text{x}}+2}{4^{\text{x}}+2}=1$

1. 1

Solution:

Given,

$\text{f(x)}=\frac{\sin^{4}\text{x}+\cos^2\text{x}}{\sin^2\text{x}+\cos^4\text{x}}$

On dividing the numerator and denominator by $\cos^4\text{x},$ we get

$\text{f(x)}=\frac{\tan^4\text{x}+\sec^2\text{x}}{1+\tan^2\text{x}\sec^2\text{x}}$

$=\frac{1+\tan^4\text{x}+\tan^2\text{x}}{1+\tan^2\text{x}(1+\tan^2\text{x})}$

$=\frac{1+\tan^{4}\text{x}+\tan^{2}\text{x}}{1+\tan^{4}\text{x}+\tan^{2}\text{x}}=1$ $(\text{For every x}\in\text{R})$

$\text{For x}=2002,$

We have,

$\text{f}(2002)=1$

4. None of these

Solution:

Given, $ \text{f}(\text{x}) = \cos(2\text{x} + 5)$

Period of $\text{f}(\text{x})=\frac{2\pi}{5}$

Since f(x) is a periodic function with period $\text{f}(\text{x})=\frac{2\pi}{5}$ so it is not injective.

The function f is not surjective also as its range [-1, 1] is a proper subset of its co-domain R.

1. either P or Q

Solution:

If either set P or set Q is a null set then $\text{P}\times\text{Q}$ is an empty set.

i.e. if P is $ \phi$ or Q is $ \phi$ then $ \text{P} ×\text{Q}=\phi $.

3. $\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$

Solution:

Given that: $\text{f(x)}=\sqrt{1+\text{x}^2}$

$\Rightarrow\text{f(xy)}=\sqrt{1+\text{x}^2\text{y}^2}$

and $\text{f(x)}.\text{f(y)}=\sqrt{1+\text{x}^2}.\sqrt{1+\text{x}^2}=\sqrt{1+\text{x}^2 +\text{y}^2+\text{x}^2\text{y}^2}$

$\therefore\sqrt{1+\text{x}^2\text{y}^2}\leq\sqrt{1+\text{x}^2 +\text{y}^2+\text{x}^2\text{y}^2}$

$\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$

4. $\cos(\text{x}+\text{y})$

Solution:

$\text{f(x)}=\cos(\log_\text{e}\text{x})$

$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)=\cos\Big(\log_\text{e}\Big(\frac{1}{\text{x}}\Big)\Big)$

$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)=\cos\Big(-\log_\text{e}(\text{x})\Big)$

$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)=\cos\Big(\log_\text{e}(\text{x})\Big)$

Similarly,

$\text{f}\Big(\frac{1}{\text{y}}\Big)=\cos(\log_\text{e}\text{y})$

Now,

$\text{f(xy)}=\cos(\log_\text{e}\text{xy})=\cos\big(\log_\text{e}\text{x}+\log_\text{e}\text{y}\big)$

and

$\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)=\cos\Big(\log_\text{e}\frac{\text{x}}{\text{y}}\Big)=\cos\big(\log_\text{e}\text{x}-\log_\text{e}\text{y}\big)$

$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=\cos\big(\log_\text{e}\text{x}+\log_\text{e}\text{y}\big)+\cos\big(\log_\text{e}\text{x}-\log_\text{e}\text{y}\big)$

$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=2\cos\big(\log_\text{e}\text{x}\big)\cos(\log_\text{e}\text{y})$

$\Rightarrow\frac{1}{2}\Big[\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}\Big]=\cos\big(\log_\text{e}\text{x}\big)\cos\big(\log_\text{e}\text{y}\big)$

$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)\text{f}\Big(\frac{1}{\text{y}}\Big)-\frac{1}{2}\Big\{\text{f(xy)}+\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)\Big\}=\cos\big(\log_\text{e}\text{x}\big)\cos\big(\log_\text{e}\text{y}\big)\\\ \ -\cos\big(\log_\text{e}\text{x}\big)\cos\big(\log_\text{e}\text{y}\big)=0$

Disclaimer: The question in the book has some error, so none of the options are matching with the solution. The solution is created according to the question given in the book.

1. [1, -1]

Solution:

As, $\text{|x|}=\begin{cases}\text{x},\ \text{x}\geq0\\-\text{x}<0\end{cases}$

So, $\text{f(x)}=\frac{\text{x}}{|\text{x}|}$

When, $\text{x}<0\text{ i.e.,}\text{ x}\in\big[-4,0\big)$

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

and when, $\text{x}>0\text{ i.e., x}\in\big(0,4\big]$

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

So, range $\text{f}=\{-1,1\}$

3. (-3, 3)

Solutuion:

We know radical cannot be negative.

So, $9-\text{x}^2,\geq 0$

$ (3 - \text{x}) (3 + \text{x}) ≥$

$\Rightarrow (\text{x} - 3) (\text{x} + 3) ≤ 0$

$ \Rightarrow \text{x} ∈ [-3,3].$

4. $\text{R}-\{0\}$

Solution:

$\text{f(x)}=\log|\text{x}|$

For f(x) to be defined,

$|\text{x}|>0,$ which is always true.

But $|\text{x}|\neq0$

$\Rightarrow\text{x}\neq0$

Thus, $\text{domain(f)}=\text{R}-\{0\}$

3. {-3, 3}

Solution:

If f : A → B, such that $\text{y}\in\text{B},$ then $\text{f}^{-1}(\text{y})=\{\text{x}\in\text{A}:\text{f(x)}=\text{y}\}$

In other words, f^-1{y} is the set of pre-images of y.

Let $\text{f}^{-1}\{9\}=\text{x}$

Then, $\text{f(x)}=9$

$\Rightarrow\text{x}^2=9$

$\Rightarrow\text{x}=\pm3$

$\therefore\ \text{f}^{-1}\{9\}=\{-3,3\}$

1. $\Big\{-1, \frac{4}{3}\Big\}$

Solution:

We have, f(x) = 3x² – 1 and g(x) = 3 + x

f(x) = g(x)

⇒ 3x² – 1 = 3 + x

⇒ 3x² – x - 4 = 0

⇒ (3x - 4)(x + 1) = 0

$\therefore\text{x}=-1, \frac{4}{3}$

4. None of these.

Solution:

Given,

$\text{f(x)}=\cos(\log\text{x})$

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

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

Similarly,

$\text{f}(\text{y}^2)=\cos(2\log(\text{y}))$

Now,

$\text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)=\cos\Big(\log\Big(\frac{\text{x}^2}{\text{y}^2}\Big)\Big)=\cos\big(\log\text{x}^2-\log\text{y}^2\big)$

and $\text{f}(\text{x}^2\text{y}^2)=\cos(\log\text{x}^2\text{y}^2)=\cos\big(\log\text{x}^2+\log\text{y}^2\big)$

$\Rightarrow\ \text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\text{f}(\text{x}^2\text{y}^2)=\cos\big((2\log\text{x}-2\log\text{y})\big)+\cos\big((2\log\text{x}-2\log\text{y})\big)$

$\Rightarrow\ \text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\text{f}(\text{x}^2\text{y}^2)=2\cos(2\log\text{x})\cos(2\log\text{y})$

$\Rightarrow\frac{1}{2}\bigg[\Big(\text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\text{f}(\text{x}^2\text{y}^2)\bigg]=\cos(2\log)\cos(2\log\text{y})$

$\Rightarrow\ \text{f(x}^2)\text{f}(\text{y}^2)-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\text{f}\big(\text{x}^2\text{y}^2\big)\Big\}\\\ =\cos(2\log)\cos(2\log\text{y})-\cos(2\log)\cos(2\log\text{y})=0$

1.  $ (-2, \infty )$

1. $\text{f}(\alpha)=\text{f}(\beta)=-9$

Solution:

$\text{f(x)}=64\text{x}^3+\frac{1}{\text{x}^3}$

$\Rightarrow\text{f(x)}=\Big(4\text{x}+\frac{1}{\text{x}}\Big)\Big(16\text{x}^2+\frac{1}{\text{x}^2}-4\Big)$

$\Rightarrow\text{f(x)}=\Big(4\text{x}+\frac{1}{\text{x}}\Big)\Big(\Big(4\text{x}+\frac{1}{\text{x}}\Big)^2-12\Big)$

$\Rightarrow\text{f}(\text{a})=\Big(4\alpha+\frac{1}{\alpha}\Big)\Big(\Big(4\alpha+\frac{1}{\alpha}\Big)^2-12\Big)$ and $\text{f}(\beta)=\Big(4\beta+\frac{1}{\beta}\Big)\Big(\Big(4\beta+\frac{1}{\beta}\Big)^2-12\Big)$

Since $\alpha$ and $\beta$ are the roots of $4\text{x}+\frac{1}{\text{x}}=3,$

$4\alpha+\frac{1}{\alpha}=3$ and $4\beta+\frac{1}{\beta}=3$

$\Rightarrow\text{f}(\alpha)=3\big((3)^2-12\big)=-9$ and $\text{f}(\beta)=3\big((3)^2-12\big)=-9$

$\Rightarrow\text{f}(\alpha)=\text{f}(\beta)=-9$

4. $[0,4]$

Solution:

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

Clearly, f(x) assumes real values if

$4\text{x}-\text{x}^2\geq0$

$\Rightarrow\text{x}(4-\text{x})\geq0$

$\Rightarrow-\text{x}(\text{x}-4)\geq0$

$\Rightarrow\text{x}(\text{x}-4)\leq0$

$\Rightarrow\text{x}\in[0,4]$

Hence, domain $(\text{f})=[0,4]$

3. $ (-1, 2) \ \cup\ (3, \infty)$

4. None os these.

Solution:

$3\text{f(x)}+5\text{f}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{x}}-3\ ...(\text{i})$

Multiplying (1) by 3,

$15\text{f}\Big(\frac{1}{\text{x}}\Big)+9\text{f(x)}=\frac{3}{\text{x}}-9\ ...(\text{ii})$

Replacing x by $\frac{1}{\text{x}}$ in (i)

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

Multiplying by 5

$15\text{f}\Big(\frac{1}{\text{x}}\Big)+25\text{f(x)}=5\text{x}-15\ ...(\text{iii})$

Solving (ii) and (iii),

$-16\text{f(x)}=\frac{3}{\text{x}}-5\text{x}+6$

$\Rightarrow\text{f(x)}=\frac{1}{16}\Big(-\frac{3}{\text{x}}+5\text{x}-6\Big)$

Disclaimer: The question in the book has some error, so, none of the options are matching with the solution. The solution is created according to the question given in the book.

4. $\text{i}\ \phi\ 1$

Solution:

We have,

$|\text{i}|=\sqrt{1^2+0^2}=1$

Thus, $\text{i }\phi\ 1$ satisfies $\text{x}\ \phi\text{ y}\Leftrightarrow|\text{x}|=\text{y}$

3. R

Solution:

Given, function $\text{f}(\text{x})=\frac{\text{x}}{(1+\text{x}^2)}$

Since f(x) is defined for all real values of x

So, domain(f) = R.

1. $-\frac{7}{4}$

Solution:

$2\text{f(x)}-3\text{f}\Big(\frac{1}{\text{x}}\Big)=\text{x}^2\ ...(\text{i})$ $(\text{x}\neq0)$

Replacing x by $\frac{1}{\text{x}}$

$2\text{f}\Big(\frac{1}{\text{x}}\Big)-3\text{f(x)}=\frac{1}{\text{x}^2}\ ...(\text{ii})$

Solving equations (i) & (ii)

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

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

Thus, $\text{f(2)}=\frac{-1}{5}\Big(\frac{3}{4}+2\times4\Big)$

$=\frac{-1}{5}\Big(\frac{3+32}{4}\Big)$

$=-\frac{7}{4}$

3. {-1, 1}

Solution:

$\text{f(x)}=\frac{\text{x}}{|\text{x}|}$

Let $\text{y}=​​\frac{\text{x}}{|\text{x}|}$

For x > 0, |x| = x

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

For x < 0, = -x

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

Thus, range of f(x) is {-1, 1}

1. R – {3, –2}

Solution:

Given that: $\text{f(x)}=\frac{\text{x}^2+2\text{x}+1}{\text{x}^2-\text{x}-6}$

f(x) is defined if $\text{x}^2-\text{x}-6\neq0$

$\Rightarrow\text{x}^2-3\text{x}+2\text{x}-6\neq0$

$\Rightarrow(\text{x}-3)(\text{x}+2)\neq0$

$\Rightarrow\text{x}\neq-2,\text{x}\neq3$

So, the domain of f(x) = R - {-2, 3}

4. None of these.

Solution:

Given,

$\text{f(x)}=\cos(\log\text{x})$

$\therefore\ \text{f(y)}=\cos(\log\text{y})$

Now,

$\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)=\cos\Big(\cos\Big(\frac{\text{x}}{\text{y}}\Big)\Big)=\cos(\log\text{x}-\log\text{y})$

and

$\text{f(xy)}=\cos(\log\text{xy})=\cos(\log\text{x}+\log\text{y})$

$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=\cos(\log\text{x}-\log\text{y})+\cos(\log\text{x}+\log\text{y})$

$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=2\cos(\log\text{x})\cos(\log\text{y})$

$\Rightarrow\frac{1}{2}\Big[\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}\Big]=\cos(\log\text{x})\cos(\log\text{y})$

$\Rightarrow\text{f(x})\text{f}(\text{y})-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f}\big(\text{x}\text{y}\big)\Big\}\\=\cos(\log\text{x})\cos(\log\text{y})-\cos(\log\text{x})\cos(\log\text{y})=0$

3. 3f(x)

Solution:

$\text{f(x)}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$ and $\text{g(x)}=\frac{3\text{x}+\text{x}^3}{1+3\text{x}^2}$

Now,

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

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

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

Then, $\text{f}(\text{g(x)})=\log=\log\Big(\frac{1+\text{g(x)}}{1-\text{g(x)}}\Big)$

$=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^3$

$=3\text{f}(\text{x})$

3. {2, 4, 6}

Solution:

x + 2y = 8

⇒ x = 8 - 2y

For y = 1, x = 6

y = 2, x = 4

y = 3, x = 2

Then R = {(2, 3), (4, 2), (6, 1)}

$\therefore$ Domain of R = {2, 4, 6}

1. {-1, 1}

Solution:

$\text{f(x)}=\frac{\text{x}+2}{|\text{x}+2|},\text{ x}\neq-2$

Let $\text{y}=\frac{\text{x}+2}{|\text{x}+2|}$

For |x + 2| > 0

Or x > -2

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

For |x + 2| < 0

Or x < -2

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

Thus, y = {-1, 1}

Or range f(x) = {-1, 1}

1. $ \frac{1}{2} \big[\text{}f(\text{2x}) +\text{ f} \text{(2y)}\big]$

4. $\text{x}\in\big[2,4\big)$

Solution:

The given equation is $[\text{x}^2]-5[\text{x}]+6=0$

$[\text{x}^2]-5[\text{x}]+6=0$

$\Rightarrow[\text{x}^2\big]-3\big[\text{x}\big]-2\big[\text{x}\big]+6=0$

$\Rightarrow\big[\text{x}\big]\big([\text{x}]-3\big)-2\big([\text{x]}-3\big)=0$

$\Rightarrow\big([\text{x}]-2)\big([\text{x}]-3)=0$

$\Rightarrow\big[\text{x}\big]-2=0$ or $[\text{x}]-3=0$

$\Rightarrow[\text{x}]=2$ or $[\text{x}]=3$

$\Rightarrow\text{x}\in\big[2,3\big)$ or $\big[3,4\big)$

$\Rightarrow\text{x}\in\big[2,4\big)$

1. $\frac{1}{2}\big[\text{f(2}\text{x})+\text{f}(2\text{y})\big]$

Solution:

Given,

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

Now,

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

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

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

$\Rightarrow\text{f}(\text{x}+\text{y})\text{f}(\text{x}-\text{y})=\frac{1}{2}\big[\text{f}(2\text{x})+\text{f}(2\text{y})\big]$

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

4. Signum function

Solution:

$\text{f}(\text{x})={\frac{\text{|x|}}{\text{x}}}$ {for $ \text{x} ≠ 0$ and 0 for $ \text{x} = 0$}.

Function is {(-3, -1), (-2, -1), (-1, 1), (0, 0), (1, 1), (2, 1), (3, 1), …….}

This is signum function.

3. 2f(x)

Solution:

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

Then, $\text{f}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)=\log\Bigg(\frac{1+\frac{2\text{x}}{1+\text{x}^2}}{1-\frac{2\text{x}}{1+\text{x}^2}}\Bigg)$

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

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

$=2\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$

$=2(\text{f(x)})$

2. $\{(\text{x, y}):\text{x},\text{ y}\in\text{R},\text{y}^2=\text{x}\}$

Solution:

y² = x gives two values of y for a value of x

i.e. there are two images for a value of x.

For example: (2)² = 4 and (-2)² = 4

Thus, it is not a function.

2. $(2, \infty)$

1. $\big[4,\infty\big)$

Solution:

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

For f(x) to be defined, $\text{x}-4\geq0$ 

$\Rightarrow\text{x}-4\geq0$

$\Rightarrow\text{x}\geq4\ ...(\text{i})$

Also, $\text{x}-3-2\sqrt{\text{x}-4}\geq0$

$\Rightarrow\text{x}-3-2\sqrt{\text{x}-4}\geq0$

$\Rightarrow\text{x}-3\geq2\sqrt{\text{x}-4}$

$\Rightarrow(\text{x}-3)^2\geq\big(2\sqrt{\text{x}-4}\big)^2$

$\Rightarrow\text{x}^2+9-6\text{x}\geq4(\text{x}-4)$

$\Rightarrow\text{x}^2-10\text{x}+25\geq0$

$\Rightarrow\big(\text{x}-5\big)^2\geq0,$ which is always true.

Similarly, $\text{x}-3+2\sqrt{\text{x}-4}\geq0$ is always true.

Thus, domain $(\text{f(x)})=\big[4,\infty)$

3. 0

Solution:

Given, $\text{f(x)}=\cos(\log\text{x})$

Then, $\text{f(x)}\text{f(4)}-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{4}\Big)+\text{f}(4\text{x})\Big\}$

$=\cos(\log\text{x})\cos(\log4)+\frac{1}{2}\Big\{\cos\Big(\log\frac{\text{x}}{4}\Big)+\cos(\log4\text{x})\Big\}$

$=\frac{1}{2}\big[\cos(\log\text{x}+\log4\big)+\cos(\log\text{x}-\log4)\big]\\-\frac{1}{2}\Big\{\cos\Big(\log\frac{\text{x}}{4}\Big)+\cos(\log4\text{x})\Big\}$

$=\frac{1}{2}\Big\{\cos(\log4\text{x})+\cos\Big(\log\frac{\text{x}}{4}\Big)-\cos\Big(\log\frac{\text{x}}{4}\Big)-\cos(\log4\text{x})\Big\}$

$=\frac{1}{2}\times0=0$

3. $\phi$

Solution:

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

For f(x) to be defined,

$\text{x}+2\neq0$

$\Rightarrow\text{x}\neq-2\ ...(\text{i})$

And $1+\text{x}\neq0$

$\Rightarrow\text{x}\neq-1\ ...(\text{ii})$

Also, $\frac{\text{x}-2}{\text{x}+2}\geq0$

$\Rightarrow\frac{(\text{x}-2)(\text{x}-2)}{(\text{x}-2)^2}\ge0$

$\Rightarrow(\text{x}-2)(\text{x}+2)\geq0$

$\Rightarrow\text{x}\in(\infty,-2)\cup\big[2,\infty\big)\ ...(\text{iii})$

And $\frac{1-\text{x}}{1+\text{x}}\geq0$

$\Rightarrow\frac{(1-\text{x})(1+\text{x})}{(1+\text{x})^2}\geq0$

$\Rightarrow(1-\text{x})(1+\text{x})\geq0$

$\Rightarrow\text{ x}\in\big(-\infty,-1\big)\cup\big[1,\infty\big)\ ...(\text{iv})$

From (i), (ii), (iii) and (iv) we get

$\text{x }\in\phi$

Thus, domain $(\text{f(x)})=\phi$

1. $(–\infty, –1) \cup (1, 4]$

Solution:

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

f(x) is defined if $4 - \text{x}\geq 0$and $\text{x}^2-1>0$

$\Rightarrow\text{x}\leq4$ and $(\text{x}+1)(\text{x}-1)>0$

$\Rightarrow\text{x}\leq4$ and $(\text{x}<-1 \ \text{or} \ \text{x}>1)$

$\therefore$ Domain of $\text{f}=(-\infty, -1)\cup(1, 4]$

4. 12

Solution:

Period of sin$ \sin\Big(\frac{\pi\text{x}}{2}\Big)=\frac{2\pi}{\big(\frac{\pi\text{x}}{2}\big) }=4=\frac{2\pi}{\big(\frac{\pi\text{x}}{2}\big)}=4$ 

Period of cos$ \cos\Big(\frac{\pi\text{x}}{3}\Big)=\frac{2\pi}{\big(\frac{\pi\text{x}}{3}\big) }=6 $ 

Period of tan$ \sin\Big(\frac{\pi\text{x}}{2}\Big)=\frac{2\pi}{\big(\frac{\pi\text{x}}{2}\big) }=4$

So, period of f(x) = LCM (4, 6, 4) = 12

3. $\Big\{-\frac{1}{2}\Big\}$

Solution:

Given,

$\text{f(x)}=\begin{cases}-1,&\text{for}-2\leq\text{x}\leq0\\\text{x}-1,&\text{for }0\leq\text{x}\leq2\end{cases}$

We know,

$|\text{x}|\geq0$

$\Rightarrow\text{f(|x|)}=|\text{x}|-1\ ...(\text{i})$

Also,

If $\text{x}\leq0,$ then $|\text{x}|=-\text{x}\ ...(\text{ii})$

$\{\text{x}\in[-2,2]:\text{x}\leq0\text{ and }\text{f}(\text{|x|})=\text{x}\}$​​​​​​​

$=\{\text{x}:|\text{x}|-1=\text{x}\}$ [Using(i)]

$=\{\text{x}:-\text{x}-1=\text{x}\}$ [Using (ii)]

$=\Big\{\text{x}:2\text{x}=\frac{-1}{2}\Big\}$

$=\Big\{\text{x}:\text{x}=\frac{-1}{2}\Big\}$

$=\Big\{-\frac{1}{2}\Big\}$

1. $ \frac{(\text{x}^{2} + 2)}{(\text{x}^{2} + 1)}$

Solution:

Given $ \text{f}(\text{x}) = \text{x}^2 + 2$ and $\text{gof}(\text{x}) =\text{g}(\text{x}^{2} - 1)$

Now, $\text{gof}(\text{x}) =\text{g}(\text{x}^{2} + 2)$

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

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

3. {-2, -1, 0, 1, 2}

Solution:

$\text{R}=\{(\text{x, y}):\text{x, y}\in\text{Z},\text{ x}^2+\text{y}^2\leq4\}$

We know that,

$(-2)^2+0^2\leq4$

$\Rightarrow(2)^2+0^2\leq4$

$\Rightarrow(-1)^2+0^2\leq4$

$\Rightarrow(1)^2+0^2\leq4$

$\Rightarrow(-1)^2+(1)^2\leq4$

$\Rightarrow0^2+0^2\leq4$

$\Rightarrow(1)^2+(1)^2\leq4$

$\Rightarrow(-1)^2+(-1)^2\leq4$

Hence, domain(R) = {-2, -1, 0, 1, 2}

2. [-a, a]

Solution:

We have $\text{f(x)}\sqrt{\text{a}^2-\text{x}^2}$

Clearly f(x) is defined, if ${\text{a}^2-\text{x}^2}\geq0$

$\Rightarrow\text{x}^2\leq\text{a}^2$

$\Rightarrow-\text{a}\leq\text{x}\leq\text{a} \ [\therefore\text{a}>0]$

$\therefore$ Domain of f is [-a, a]

4. $\text{x}\in\text{R},\text{ x}\neq0$

Solution:

Given,

$\text{f(x)}=\text{x},\text{ g(x)}=\frac{1}{\text{x}}$ and $\text{h(x)}=\text{f(x)}\text{g(x)}$

Now,

$\text{h(x)}=\text{x}\times\frac{1}{\text{x}}=1$

We observe that the domain of f is R and the domain of g is R - {0}

$\therefore\ \text{Domain of h}=\text{Domain of f }\cap\text{ Domain of g}\\\ \ \ =\text{R }\cap\big[\text{R}-\{0\}\big]=\text{R}-\{0\}$

$\Rightarrow\text{x}\in\text{R},\text{ x}\neq0$ 

1. 0.5

Solution:

$\text{e}^{\text{f(x)}}=\frac{10+\text{x}}{10-\text{x}}$

$\Rightarrow\text{ f(x)}=\log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)\ ...(\text{i})$

$\Rightarrow\ \text{f(x)}=\text{kf}\Big(\frac{200\text{x}}{100+\text{x}^2}\Big)$

$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\Bigg(\frac{10+\frac{200\text{x}}{100+\text{x}^2}}{10-\frac{200\text{x}}{100+\text{x}^2}}\Bigg)$ {from (1)}

$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\Big(\frac{1000+10\text{x}^2+200\text{x}}{1000+10\text{x}^2-200\text{x}}\Big)$

$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\bigg(\frac{(\text{x}+10)^2}{(\text{x}-10)^2}\bigg)$

$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=2\text{k}\log_\text{e}\frac{(\text{x}+10)}{(\text{x}+10)}$

$\Rightarrow\ 1=2\text{k}$

$\Rightarrow\ \text{k}=\frac{1}{2}=0.5$

1. {(3, 3), (3, 1), (5, 2)}

Solution:

A = {1, 2, 3}, B = {1, 3, 5}

R = {(1, 3), (2, 5), (3, 3)}

$\therefore$ R^-1 = {(3, 3), (3, 1), (5, 2)}

1.  $(-\infty,-3]\cup(2,5)$

Solution:

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

For f(x) to be defined,

$(2-\text{x})(\text{x}-5)\neq0$

$\Rightarrow\text{x}\neq2,5\ ...(\text{i})$

Also, $\frac{(\text{x}+3)}{(2-\text{x})(\text{x}-5)}\geq0$

$\Rightarrow\frac{(\text{x}+3)(2-\text{x})(\text{x}-5)}{(2-\text{x})^2(\text{x}-5)^2}\geq0$

$\Rightarrow(\text{x}+3)(\text{x}-2)(\text{x}-5)\leq0$

$\Rightarrow\text{x}\in\big(-\infty,-3\big]\cap(2,5)\ ...(\text{ii})$

From (i) and (ii)

$\text{x}\in\big(-\infty,-3\big]\cup(2,5)$ 

2. {1, 2}

Solution:

In each ordered pair of A × B, first element belongs to set A and second element belongs to set B.

1, $ 2 ∈\text{A} $ so, $ \text{A} = {1, 2}.$

2. 1

Solution:

Let T is a positive real number.

Let f(x) is periodic with period T.

Now, f(x + T) = f(x), for all $\text{x} \in \text{R}$

⇒ x + T - [x + T] = x - [x], for all $ \text{x} \in \text{R}$

⇒ [x + T] - [x] = T, for all $ \text{x} \in \text{R}$

Thus, there exist T > 0 such that f(x + T) = f(x) for all $ \text{x} \in \text{R}$

Now, the smallest value of T satisfying f(x + T) = f(x) for all $ \text{x} \in \text{R}$ is 1

So, f(x) = x - [x] has period 1

3. $\text{R}\subseteq\text{A}\times\text{B}$

Solution:

If R is a relation from set A to set B, then R is always a subset of A × B.

2. $\big[0,\infty\big)$

Solution:

$\text{f(x)}=|\text{x}-1|\geq0\ \forall\text{ x}\in\text{R}$

Thus, range $=\big[0,\infty\big)$

1. Domain $= [1, \infty),$ Range $= [0, \infty)$

 Solution:

We have, $\text{f(x)}=\sqrt{\text{x}-1}$

Clearly, f(x) is defined if $\text{x}-1\geq0$

$\Rightarrow\text{x}\geq1$

$\therefore$ Domain of $\text{f}=[1, \infty)$

Now for $\text{x}\geq1,\text{x}-1\geq0$

$\Rightarrow\sqrt{\text{x}-1}\geq1$

⇒ Range of $= [0, \infty)$

3. $(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{A})=(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{C})$

3. $\text{R}-\Big\{\frac{1}{2},1\Big\}$

Solution:

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

Let, $\text{y}=\frac{\text{x}^2-\text{x}}{\text{x}^2+2\text{x}}$ $\big[\text{Also},\text{ x}\neq0\big]$

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

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

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

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

Here, $1-\text{y}\neq0$

Or, $\text{y}\neq1$

Also, $\text{x}\neq0$

$\Rightarrow\frac{2\text{y}+1}{1-\text{y}}\neq0$

$\Rightarrow\text{y}\neq-\frac{1}{2}$

Thus, range $\text{(f)}=\text{R}-\Big\{-\frac{1}{2},1\Big\}$

1. $ 2\text{f}(\text{x} ) \text{ f}(\text{y})$

1. 2^mn

Solution:

Given, n(A) = m

n(B) = n

$\therefore$ n(A × B) = mn

Then, the number of relations from A to is 2^mn

3. 3

Solution:

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

$\text{f}(\text{f}(\text{f(2)}))$

$=\text{f}\Big(\text{f}\Big(\frac{2+1}{2-1}\Big)\Big)$

$=\text{f}(\text{f}(3))$

$=\text{f}\Big(\frac{3+1}{3-1}\Big)$

$=\text{f}(2)=3$

3. $ ​​\text{R} \subseteq \text{A}\times{\text{B}}$

4. $\big[1,3\big]$

Solution:

$\text{f(x)}=\sqrt{\text{x}-1}+\sqrt{3-\text{x}}$

For f(x) to be defined,

$(\text{x}-1)\geq0$

$\Rightarrow\text{x}\geq1\ ...(\text{i})$

and $(3-\text{x})\geq0$

From (i) and (ii),

$\text{x}\in[1,3]$

3. 3

Solution:

Given thatn + 2n + 3n + …. + 99n

$ =\text{ n} × (1 + 2 + 3 + …….. + 99)$

$=\frac{\text{n} \times99 \times100}{2}$

$=\text{n} × 99 × 50$

 $= \text{n} × 9 × 11 × 2 × 25$

To make it perfect square we need 2 × 11

So n = 2 × 11 = 22.

Now n² = 22 × 22 = 484

So, the number of digit in n² = 3.

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

Solution:

Given, A finite set with n elements

Its Cartesian product with itself will have n² elements.

$\therefore$ Number of relations on $\text{A}=2^{\text{n}^2}$

1. 4

Solution:

Period of $\sin \Big(\frac{\pi‎\text{x}}{2}\Big) = 2\pi\Big(\frac{\pi}{2}\Big) = 4$

Period of $\cos \Big(\frac{\pi‎\text{x}}{2}\Big) = 2\pi\Big(\frac{\pi}{2}\Big) = 4$

So, period of f(x) = LCM (4, 4) = 4