Relations and Functions

Total number of functions = $4^4=256$

One-one functions = $4!$

Many-one functions = 256-24 = 232

Many-one functions in which $1\notin \mathrm{f}( \mathrm{A})$

$= 3.3.3.3 = 81$

Required number of functions =

$232-81=151$

$$\begin{gathered} \text{Given that }f\left(x\right)\text{ is onto } \\ \text{therefore range of} f\left(x\right)=A \\ {f}^{\text{'}}\left(x\right)=6x^2-30x+36 \\ =6(x-2)(x-3) \\ f\left(x\right)=2x^3-15x^2+36x+7 \\ f\left(2\right)=16-60+72+7=35 \\ f\left(3\right)=54-135+108+7=34 \\ f\left(0\right) = 7 \\ \text{hence range} \in [7,35]=\mathrm{A} \\ \text{also for range of} g \left(x\right) \\ g\left(x\right)=1-\frac{1}{x^{2025}+1}\in [0,1)=B \\ S=\{0,7,8,\dots ..35\} \\ \text{hence} n\left(s\right)=30 \end{gathered}$$

\(A=\{-3,-2,-1,0,1,2,3\}, x R y \Longleftrightarrow 0 \leq x^2+2 y \leq 4, l=|R| \)

\(x= \pm 3, x^2=9: 0 \leq 9+2 y \leq 4 \Rightarrow-9 \leq 2 y \leq-5\)

\(\Rightarrow y=-3 \Rightarrow 2 \) pairs

\(x= \pm 2, x^2=4: 0 \leq 4+2 y \leq 4 \Rightarrow-4 \leq 2 y \leq 0\)

\(\Rightarrow y \in\{-2,-1,0\} \Rightarrow 6 \) pairs

\(x= \pm 1, x^2=1: 0 \leq 1+2 y \leq 4 \Rightarrow-1 \leq 2 y \leq 3\)

\(\Rightarrow y \in\{0,1\} \Rightarrow 4\) pairs

\(x=0, x^2=0: 0 \leq 2 y \leq 4\)

\(\Rightarrow y \in\{0,1,2\} \Rightarrow 3\) pairs

\( l=2+6+4+3=15\)

Reflexive on \(A\) means \((a, a) \in R \ \forall a \in A\), i.e. \(0 \leq a^2+2 a \leq 4\)

\(a=-3,-2,0,1\) satisfy it; \(a=-1,2,3\) do not satisfy it \(\Rightarrow m=3\)

\(l+m=15+3=18\)

\(
\begin{aligned}
& n(A)=4 \\
& n(B)=4
\end{aligned}
\)

Number of many one functions =
Total functions - Number of one-one function

\(
=4^4-4!=232
\)

\(
\begin{aligned}
&f(x)=\frac{x^2+2x-15}{x^2-4x+9}\\
&f'(x)=\frac{\left(x^2-4x+9\right)(2x+2)-\left(x^2+2x-15\right)(2x-4)}
{\left(x^2-4x+9\right)^2}\\
&f'(x)=\frac{-6x^2+48x-42}{\left(x^2-4x+9\right)^2}
\end{aligned}
\)

So \(f'(x)\) is \(+\)ve and \(-\)ve, hence \(f(x)\) is many-one and into.

Given \(x R y \Leftrightarrow \sec ^2 x-\tan ^2 y=1\)

Using \(\sec ^2 x-1=\tan ^2 x\), we get \(\tan ^2 x=\tan ^2 y\)

Since \(x, y \in\left[0, \frac{\pi}{2}\right)\),

we have \(\tan x \geq 0, \tan y \geq 0\), so \(\tan x=\tan y\)

Now \(\tan x\) is one-one on \(\left[0, \frac{\pi}{2}\right)\), hence \(x=y\)

So the relation becomes \(x R y \Longleftrightarrow x=y\)

Therefore \(R\) is reflexive, symmetric and transitive

$\begin{array}{l}A=\left\{\right(x,y)\in R\times R:x^2+y^2=25\},\\ \\ B=\left\{\right(x,y)\in \mathrm{ℝ}\times \mathrm{ℝ}:x^2+9y^2=144\}\end{array}$

$$x ^ {2} + 9 y ^ {2} - \left(x ^ {2} + y ^ {2}\right) = 1 4 4 - 2 5$$

Plug in $y^{2} = \frac{119}{8}$ into either equation to find x.

$$\begin{array}{l} x ^ {2} = 2 5 - \frac {1 1 9}{8} \\ x ^ {2} = \frac {2 0 0 - 1 1 9}{8} \\ x ^ {2} = \frac {8 1}{8} \\ x = \pm \sqrt {\frac {8 1}{8}}, y = \pm \sqrt {\frac {1 1 9}{8}} \\ \end{array}$$

Now, $C = \{(x,y)\in \mathbb{Z}\times \mathbb{Z}:x^2 +y^2\leq 4\}$

Valid points are

$(-2,0),(-1,-1),(-1,0),(-1,1),(0,-2),$

$(0,-1),(0,0),(0,1),(0,2),(1,-1),(1,0),(1,1)$

$\therefore$ Total valid points in $C = 13$

$\Rightarrow$ There are 4 distinct real points in set D

$\therefore$ The number of one-one functions from D to C

$$\begin{array}{l}\Rightarrow 13P_4\Rightarrow \frac{13!}{(13-4)!}=\frac{13!}{9!}=17160\end{array}$$

$$\begin{gathered} f\left(x\right)-6f\left(\frac{1}{x}\right)=\frac{35}{3x}-\frac{5}{2}......\left(1\right) \\ f\left(\frac{1}{x}\right)-6f\left(x\right)=\left(\frac{35x}{3}-\frac{5}{2}\right) \\ 6f\left(\frac{1}{x}\right)-36f\left(x\right)=\left(\frac{35x}{3}-\frac{5}{2}\right)\times 6...\left(2\right) \\ \text{Add (1) and (2),we get} \\ -35f\left(x\right)=\frac{35}{3x}-\frac{5}{2}+70x-15 \\ -35f\left(x\right)=70x+\frac{35}{3x}-\frac{35}{2} \\ f\left(x\right)=\frac{1}{2}-2x-\frac{1}{3x} \\ \underset{x\to 0}{\lim }(\frac{1}{\alpha x}+\frac{1}{2}-2x-\frac{1}{3x})=\beta \\ =\underset{x\to 0}{\lim (}\left(\frac{1}{\alpha }-\frac{1}{3}\right)\frac{1}{x}+\frac{1}{2}-2x)=\beta \\ \alpha =3 \beta =\frac{1}{2} \\ now, \alpha +2\beta =3+1=4 \end{gathered}$$

$$\begin{gathered} \text{Given}, \mathrm{R}=\left\{\right(1,2),(2,3),(3,3\left)\right\} \\ \text{and set A} =\{1,2,3,4\} \\ \text{Now, for equivalence, minimum number} \\ \text{ of ordered pairs need to add:} \\ (1,1),(2,2),(4,4)(2,1)(3,2)(3,1)(1,3) \\ \text{i.e} 7 \text{ordered pairs.} \end{gathered}$$

\(f(x) f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)\)

\(\Rightarrow(f(x)-1)\left(f\left(\frac{1}{x}\right)-1\right)=1\)

Let \(g(x)=1-f(x)\)

Then \(g(x)\) is a quadratic polynomial and

\(g(x) g\left(\frac{1}{x}\right)=1 (x \neq 0)\)

Let \(g(x)=a x^2+b x+c\)

Then, \(\left(a x^2+b x+c\right)\left(\frac{a}{x^2}+\frac{b}{x}+c\right)=1\)

\(\left(a x^2+b x+c\right)\left(a+b x+c x^2\right)=x^2\)

\(a c x^4+b(a+c) x^3+\left(a^2+b^2+c^2\right) x^2+b(a+c) x+a c=x^2\)

Comparing coefficients:

\(a c=0, b(a+c)=0, a^2+b^2+c^2=1\)

Since \(g(x)\) is quadratic, \(a \neq 0\), so \(c=0\)

Then \(a b=0 \Rightarrow b=0\), and \(a^2=1 \Rightarrow a= \pm 1\)

Also \(f(x)<1 \Rightarrow g(x)=1-f(x)>0\) for all \(x \neq 0\), so \(a=1\)

Hence, \(g(x)=x^2\)

\(1-f(x)=x^2 \Rightarrow f(x)=1-x^2\)

Now, \(f(K)=-2 K \Rightarrow 1-K^2=-2 K\)

\(\Rightarrow K^2-2 K-1=0\)

\(K=1 \pm \sqrt{2}\)

Required sum of squares:

\((1+\sqrt{2})^2+(1-\sqrt{2})^2=6\)

Consider \(\frac{1}{\sqrt{10+3 x-x^2}}\)

For this term to be defined,

\(10+3 x-x^2>0\)

\((x-5)(x+2)<0\)

\(-2<x<5\)

Now, consider \(\frac{1}{\sqrt{x+|x|}}\)

we need \(x+|x|>0\)

If \(x \geq 0\), then \(|x|=x\), so

\(x+|x|=2 x>0 \Rightarrow x>0\)

If \(x<0\), then \(|x|=-x\), so

\(x+|x|=0\)

which is not allowed in denominator.

Hence, \(x>0\)

Therefore, the domain is:

\((-2,5) \cap(0, \infty)=(0,5)\)

So, \(a=0, b=5\)

\((1+a)^2+b^2=(1+0)^2+5^2=1+25=26\)

\(y=\frac{5-x}{x^2-3 x+2}\)

\(y x^2+(1-3 y) x+(2 y-5)=0\)

For some real \(x\), this quadratic in \(x\) must have real roots, so

\(D \geq 0\)

\((1-3 y)^2-4 y(2 y-5) \geq 0\)

\(y^2+14 y+1 \geq 0\)

\(y^2+14 y+1=(y+7)^2-48 \geq 0\)

\(y \leq-7-4 \sqrt{3} \) or \(y \geq-7+4 \sqrt{3}\)

Hence, \(\alpha=-7-4 \sqrt{3}, \beta=-7+4 \sqrt{3}\)

Now, \(\alpha+\beta=-14, \alpha \beta=1\)

\(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\)

\(\alpha^2+\beta^2=196-2=194\)

$f^{\text{'}}\left(x\right)=6x^2-30x+36=6\left(x^2-5x+6\right)$
$=6(x-2)(x-3)$

$$\begin{gathered} f\left(0\right)=7 \\ f\left(2\right)=16-60+72+7=35 \\ f\left(3\right)=54-135+108+7=34 \\ \therefore A=\left[7,35\right] \\ \mathrm{g}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^{2025}}{1+{\mathrm{x}}^{2025}}=1-\frac{1}{1+{\mathrm{x}}^{2025}} \\ {\mathrm{x}}^{2025}\ge 0\Rightarrow \therefore 1+{\mathrm{x}}^{2025}\ge 1 \\ \Rightarrow 0<\frac{1}{1+{\mathrm{x}}^{2025}}\le 1 \\ \Rightarrow 1>1-\frac{1}{1+{\mathrm{x}}^{2025}}\ge 0 \therefore \mathrm{B}=[0,1) \\ \therefore \mathrm{S}=\{0,7,8,\dots \dots ,35\}, n\left(S\right)=30 \end{gathered}$$

Set $y = f(x) = \dfrac{5-x}{x^2-3x+2}$.

Rearranging gives a quadratic in $x$:

$yx^2 + (1-3y)x + (2y-5) = 0$

For real $x$ to exist (note $x=1,2$ are excluded as they don't satisfy this equation), the discriminant must be non-negative: $D = (1-3y)^2 - 4y(2y-5) \ge 0$

$1 - 6y + 9y^2 - 8y^2 + 20y \ge 0$

$y^2 + 14y + 1 \ge 0$

The roots of $y^2 + 14y + 1 = 0$ are the boundary values $\alpha$ and $\beta$.

By Vieta's formulas: $\alpha + \beta = -14, \quad \alpha\beta = 1$

Therefore: ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$$=(-14{)}^2-2(1)=196-2=\boxed{194}$

$\begin{array}{l}A:\left|\right|x-3|-3|\le 1\\ \Rightarrow -1\le |x-3|-3\le 1\\ \Rightarrow 2\le |x-3|\le 4\\ \Rightarrow 2\le (x-3)\le 4\text{ or }-4\le (x-3)\le -2\end{array}$

$\begin{array}{l}\Rightarrow 5\le x\le 7\text{ or}-1\le x\le 1\\ A=\{-1,0,1,5,6,7\}\end{array}$

For B:

\(\frac{(x-2)(x-4)}{x-1} \ln |x-2|=0\).

Domain \(x \neq 1,2\)
Roots: \(x-4=0 \Rightarrow x=4 \)

\( \ln |x-2|=0 \Rightarrow |x-2|=1 \Rightarrow x=3 \text{ or } x=1 (rejected)\)
Set \(B\) has \(2\) elements.

\( B=\{3,4\}\)

Number of onto functions from $A \text{to} B=2^6–2=62$

$$\begin{gathered} xRy\Rightarrow {\sec }^{2}x-{\tan }^{2}y=1 \\ xRx\Rightarrow {\sec }^{2}x-{\tan }^{2}x=1 \\ \Rightarrow R\text{ is reflexive } \\ xRy\Rightarrow yRx \\ \Rightarrow {\sec }^{2}x-{\tan }^{2}y=1 \\ {\sec }^{2}y-{\tan }^{2}x=\left(1+{\tan }^{2}y\right)-\left({\sec }^{2}x-1\right) \\ =2-{\sec }^{2}x+{\tan }^{2}y \\ =2-\left({\sec }^{2}x-{\tan }^{2}y\right)=2-1=1 \\ \Rightarrow R\text{ is symmetric } \\ xRy\Rightarrow yRz \\ \Rightarrow {\sec }^{2}x-{\tan }^{2}y=1 \\ {\sec }^{2}y-{\tan }^{2}z=1 \\ \text{ Add }\Rightarrow {\sec }^{2}x+{\sec }^{2}y-{\tan }^{2}y-{\tan }^{2}z=2 \\ \Rightarrow {\sec }^{2}x+\left(1\right)-{\tan }^{2}z=2 \\ \Rightarrow xRz \\ \Rightarrow R\text{ is transitive. } \end{gathered}$$

Given, \( f(x)=\frac{2^{2 x}-1}{2^{2 x}+1} \)
\( f(x)=1-\frac{2}{2^{2 x}+1}\)

On differentiating, we get

\(\mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{\left(2^{2 \mathrm{x}}+1\right)^2} \cdot 2 \cdot 2^{2 \mathrm{x}} \cdot \ln 2 \)

Since \(\mathrm{f}^{\prime}(\mathrm{x})>0\)

so \(\mathrm{f}(\mathrm{x})\) is increasing function

\( \therefore \mathrm{f}(-\infty)=-1 \)
\( f(\infty)=1 \)
\( \therefore \mathrm{f}(\mathrm{x}) \in(-1,1) \neq \) co-domain

so function is one-one but not onto.

Total number of functions = $4^4=256$

One-one functions = $4!$

Many-one functions $= 256-24 = 232$

Many-one functions in which $1\notin \mathrm{f}( \mathrm{A})$

$= 3.3.3.3 = 81$

Required number of functions =

$232-81=151$

$$\begin{gathered} f\left(x\right)=\frac{2x}{2^x+\sqrt{2}} \\ f\left(x\right)+f(1-x)=1 \& f(1/2)=\frac{2^{1/2}}{2^{1/2}+\sqrt{2}}=1/2 \\ \sum _{k=1}^{81}f\left(\frac{k}{82}\right)=f\left(\frac{1}{82}\right)+f\left(\frac{2}{82}\right)+f\left(\frac{3}{32}\right)+\cdots f\left(\frac{81}{82}\right) \\ =40+f\left(\frac{1}{2}\right)=40+\frac{1}{2}=\frac{81}{2} \end{gathered}$$

For equivalence it must be transitive, symmetric and reflexive all.
For reflexive \(\rightarrow \quad(1,1),(2,2),(4,4)\)
For symmetric \(\rightarrow \quad(2,1),(3,2)\)
For transitive \(\rightarrow \quad(1,3),(3,1)\)
Total 7 pairs has to be added to make it's an equivalence relation.

Let \(t=e^{\sin x}>0\)

\(t-\dfrac{2}{t}=2\)

\(t^2-2t-2=0\)

Since \(t>0\)

\(t=1+\sqrt{3}\)

\(\sin x=\ln(1+\sqrt{3})\)

\(\ln(1+\sqrt{3})>1\)

\(-1\le \sin x\le 1\)

the equation has no real solution

We are given:

$f\left(x\right)=\frac{\sqrt{x^2-25}}{4-x^2}+{\log }_{10}\left(x^2+2x-15\right)$

We are told the domain is:

$(-\infty ,\alpha )\cup [\beta ,\infty )$

We need to find:

${\alpha }^2+{\beta }^3$

Step-by-step (short):
1. Square root condition:

$\sqrt{x^2-25}\text{ is defined when }{x}^2-25\ge 0\Rightarrow x\le -5\text{ or }x\ge 5$

2. Denominator condition:

$4-x^2\ne 0\Rightarrow x\ne \pm 2$

3. Log condition:

${\log }_{10}\left(x^2+2x-15\right)\text{ defined when }{x}^2+2x-15>0$

Factor:

$(x+5)(x-3)>0\Rightarrow x<-5\text{ or }x>3$

Now combine all conditions:
- Square root: \(x \leq-5\) or \(x \geq 5\)
- Log: \(x<-5\) or \(x>3\)
- Denominator: \(x \neq \pm 2\)

So:
- For left side: \(x<-5\)

- For right side: Must satisfy \(x \geq 5\) and \(x>3 \Rightarrow x \geq 5\)

Hence, domain is:

$(-\infty ,-5)\cup [5,\infty )\Rightarrow \alpha =-5,\beta =5$

Now calculate:

\[
\alpha^2+\beta^3=(-5)^2+(5)^3=25+125=150
\]

Final Answer:
(c) 150

$$\begin{gathered} \text{ Let }S\text{ denote the set of all triangles in a plane.} \\ \text{Let }R\text{ be the relation on }S\text{ defined by }\left({\Delta }_1,{\Delta }_2\right)\in R \\ \Rightarrow \text{ triangle }{\Delta }_1\cong {\Delta }_2. \\ \text{ ( }i\text{ ) Let any triangle }\Delta \in S\text{, we have } \\ \Delta \cong \Delta \\ \Rightarrow (\Delta ,\Delta )\in R\forall \Delta \in S \\ \Rightarrow R\text{ is reflexive on }S\text{. } \end{gathered}$$

$$\begin{gathered} \text{ (ii) Let }{\Delta }_1,{\Delta }_2\in S\text{, such that }\left({\Delta }_1,{\Delta }_2\right)\in R\text{, then } \\ {\Delta }_1\cong {\Delta }_2 \\ \Rightarrow {\Delta }_2\cong {\Delta }_1 \\ \Rightarrow \left({\Delta }_2,{\Delta }_1\right)\in R \\ \Rightarrow R\text{ is symmetric } \end{gathered}$$

$$\begin{gathered} \text{ (iii) Again, let }{\Delta }_1,{\Delta }_2,{\Delta }_3\in S\text{ such that }\left({\Delta }_1,{\Delta }_2\right)\in R\text{ and } \\ \left({\Delta }_2,{\Delta }_3\right)\in R \\ \therefore {\Delta }_1\cong {\Delta }_2\cong {\Delta }_3 \\ \therefore \left({\Delta }_1,{\Delta }_3\right)\in R \\ \Rightarrow R\text{ is transitive. } \end{gathered}$$

$$\begin{gathered} \text{Given, } \\ f\left(x\right)={\log }_{e}x \\ \text{then} D_f\in \left(0,\infty \right) \\ \text{and} g\left(x\right)=\frac{x^4-2x^3+3x^2-2x+2}{2x^2-2x+1} \\ \text{then} D_g\in R \\ \text{Now, for} fog\left(x\right)= f\left(g\left(x\right)\right) \\ ={\log }_e\left(\frac{x^4-2x^3+3x^2-2x+2}{2x^2-2x+1}\right) \\ \text{Now, Domain for} fog\left(x\right) \\ \frac{x^4-2x^3+3x^2-2x+2}{2x^2-2x+1}>0 \forall x\in R \\ \text{Therefore }{\log }_e\left(\frac{x^4-2x^3+3x^2-2x+2}{2x^2-2x+1}\right)\text{ is defined ∀}x\in R \end{gathered}$$

Number of functions which are not onto
\(=(\)Total number of functions possible\()-(\)Number of onto functions \() \)
\( =3^4-\left(3^4-{ }^3 C_1(3-1)^4+{ }^3 C_2(3-2)^4\right) \)
\(= 3(2)^4-3(1) \)
\(=48-3=45\)

\((f\circ g^{-1})'(2)=f'(g^{-1}(2))(g^{-1})'(2).\)

Since

\(g(0)=2\)

\(g^{-1}(2)=0\)

\(f'(0)=\frac{1}{2}\) and \(g'(0)=2\)

\((g^{-1})'(2)=\frac{1}{g'(0)}=\frac{1}{2}\)

\((f\circ g^{-1})'(2)=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}.\)

$$\begin{gathered} f\left(x\right)=5 x-3 \\ {\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{\mathrm{x}+3}{5} \\ \text{Now}, \\ g\left(x\right)=x^2+3 \\ {gof}^{-1}\left(3\right)=g\left(f^{-1}\left(3\right)\right) \\ =g\left(\frac{6}{5}\right)=\frac{111}{25} \end{gathered}$$

$$\begin{gathered} \text{Given, } \\ f\left(x\right)=(3a+2)x^2+\left(\frac{a+2}{a-1}\right)x+b \\ f(x+y)=f\left(x\right)+f\left(y\right)+1-\frac{2}{7}xy ...\left(i\right) \\ \text{put y} = \frac{1}{2}, \text{we get} \\ f(x+\frac{1}{2})=f\left(x\right)+f\left(\frac{1}{2}\right)+1-\frac{2}{7}x\frac{1}{2} \\ f(x+\frac{1}{2})=f\left(x\right)+f\left(\frac{1}{2}\right)+1-\frac{1}{7}x ...\left(ii\right) \\ \text{Now, put x=y=0} \\ \text{then f}\left(0\right)=2f\left(0\right)+1 \\ f\left(0\right)=-1 \\ \text{ So, }f\left(0\right)=0+0+b=-1 \\ \Rightarrow b=-1 \\ \text{ In (1) Put }y=-x \\ \Rightarrow f\left(0\right)=f\left(x\right)+f(-x)+1+\frac{2}{7}{x}^2 \\ -1=2(3a+2)x^2+2b+1+\frac{2}{7}{x}^2 \\ -1=\left(2(3a+2)+\frac{2}{7}\right)x^2+1-2 \\ \Rightarrow 6a+4+\frac{2}{7}=0 \\ a=-\frac{5}{7} \\ \text{put the value of a and b, we get} \\ \text{ So }f\left(x\right)=-\frac{1}{7}{x}^2-\frac{3}{4}x-1 \\ \Rightarrow \left|\mathrm{f}\right(\mathrm{x}\left)\right|=\frac{1}{28}\left|4{\mathrm{x}}^2+21\mathrm{x}+28\right| \\ \text{ Now, } \\ 28\sum _{i=1}^5\left|f\right(i\left)\right|=28\left(\left|f\left(1\right)+f\left(2\right)+\dots +f\left(5\right)\right|\right) \\ 28\sum _{i=1}^5\left|f\right(i\left)\right|=28\times \frac{1}{28}\times 675=675 \end{gathered}$$

\(f(x)=\frac{2^x}{2^x+\sqrt{2}}\)

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

Let \(2^x=t\). Then \(2^{1-x}=\frac{2}{t}\), so

\(f(1-x)=\frac{2 / t}{2 / t+\sqrt{2}}=\frac{2}{2+\sqrt{2} t}=\frac{\sqrt{2}}{\sqrt{2}+t}\)

Hence, \(f(x)+f(1-x)=\frac{t}{t+\sqrt{2}}+\frac{\sqrt{2}}{\sqrt{2}+t}=1\)

Now, \(f\left(\frac{k}{82}\right)+f\left(1-\frac{k}{82}\right)=1\)

\(f\left(\frac{1}{82}\right)+f\left(\frac{81}{82}\right)=1, \ldots, f\left(\frac{40}{82}\right)+f\left(\frac{42}{82}\right)=1\)

These give 40 pairs, so sum \(=40\)

Middle term:

\(f\left(\frac{41}{82}\right)=f\left(\frac{1}{2}\right)=\frac{2^{1 / 2}}{2^{1 / 2}+\sqrt{2}}=\frac{\sqrt{2}}{2 \sqrt{2}}=\frac{1}{2}\)

Therefore, \(\sum_{k=1}^{81} f\left(\frac{k}{82}\right)=40+\frac{1}{2}=\frac{81}{2}\)

$f(x) + 3f\left(\frac{24}{x}\right) = 4x$

Substitute $x = 3$ into the equation: $f(3) + 3f(8) = 12$ --- (1)

Substitute $x = 8$ into the original equation: $f(8) + 3f(3) = 32$ --- (2)

Add equations (1) and (2) together: $(f(3) + 3f(8)) + (f(8) + 3f(3)) = 12 + 32$ $4f(3) + 4f(8) = 44$

Divide by 4: $f(3)+f(8)=11$

For the square root in the denominator to be defined, the expression inside must be positive:

$$10 + 3x - x^2 > 0$$

Rearrange and factor the quadratic:

$$-x^2 + 3x + 10 > 0 \implies x^2 - 3x - 10 < 0 \implies (x-5)(x+2) < 0$$

The quadratic is negative between its roots $x=-2$ and $x=5$. Thus, the domain for this term is $(-2, 5)$.

Again, the expression inside the square root must be positive:

$$|x| + x > 0$$

• If $x \geq 0$: $|x| + x = x + x = 2x$. For this to be positive, $x > 0$.
• If $x < 0$: $|x| + x = -x + x = 0$, which is not allowed.

Thus, the domain for this term is $(0, \infty)$.

The domain of $f(x)$ is the overlap of the two domains:

$$(-2,5) \cap (0, \infty) = (0,5)$$

Given the domain is $(a,b)$, we have $a=0$ and $b=5$.

$$(1+0{)}^2+5^2=1+25=26$$

Given: \(f(x)=\log _e\left(\frac{2 x-3}{5+4 x}\right)+\sin ^{-1}\left(\frac{4+3 x}{2-x}\right)\)

For the logarithmic term:

\(\frac{2 x-3}{4 x+5}>0\)

\(\Rightarrow x \in\left(-\infty,-\frac{5}{4}\right) \cup\left(\frac{3}{2}, \infty\right)\)

For the inverse sine term:

\(\left|\frac{4+3 x}{2-x}\right| \leq 1 \Rightarrow|4+3 x| \leq|2-x|\)

\((4+3 x)^2 \leq(2-x)^2\)

\(2 x^2+7 x+3 \leq 0\)

\((2 x+1)(x+3) \leq 0\)

\(x \in\left[-3,-\frac{1}{2}\right]\)

Now intersect both conditions:

\(\left[-3,-\frac{1}{2}\right] \cap\left[\left(-\infty,-\frac{5}{4}\right) \cup\left(\frac{3}{2}, \infty\right)\right]=\left[-3,-\frac{5}{4}\right)\)

So, \(\alpha=-3, \beta=-\frac{5}{4}\)

\(\alpha^2+4 \beta=(-3)^2+4\left(-\frac{5}{4}\right)=9-5=4\)

$$\begin{gathered} \text{Given that }f\left(x\right)\text{ is onto } \\ \text{therefore range of} f\left(x\right)=A \\ {f}^{\text{'}}\left(x\right)=6x^2-30x+36 \\ =6(x-2)(x-3) \\ f\left(x\right)=2x^3-15x^2+36x+7 \\ f\left(2\right)=16-60+72+7=35 \\ f\left(3\right)=54-135+108+7=34 \\ f\left(0\right) = 7 \\ \text{hence range} \in [7,35]=\mathrm{A} \\ \text{also for range of} g \left(x\right) \\ g\left(x\right)=1-\frac{1}{x^{2025}+1}\in [0,1)=B \\ S=\{0,7,8,\dots ..35\} \\ \text{hence} n\left(s\right)=30 \end{gathered}$$

$$\begin{gathered} f\left(x\right)=5 x-3 \\ {\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{\mathrm{x}+3}{5} \\ \text{Now}, \\ g\left(x\right)=x^2+3 \\ {gof}^{-1}\left(3\right)=g\left(f^{-1}\left(3\right)\right) \\ =g\left(\frac{6}{5}\right)=\frac{111}{25} \end{gathered}$$

$$\begin{gathered} xRy\Rightarrow {\sec }^{2}x-{\tan }^{2}y=1 \\ xRx\Rightarrow {\sec }^{2}x-{\tan }^{2}x=1 \\ \Rightarrow R\text{ is reflexive } \\ xRy\Rightarrow yRx \\ \Rightarrow {\sec }^{2}x-{\tan }^{2}y=1 \\ {\sec }^{2}y-{\tan }^{2}x=\left(1+{\tan }^{2}y\right)-\left({\sec }^{2}x-1\right) \\ =2-{\sec }^{2}x+{\tan }^{2}y \\ =2-\left({\sec }^{2}x-{\tan }^{2}y\right)=2-1=1 \\ \Rightarrow R\text{ is symmetric } \\ xRy\Rightarrow yRz \\ \Rightarrow {\sec }^{2}x-{\tan }^{2}y=1 \\ {\sec }^{2}y-{\tan }^{2}z=1 \\ \text{ Add }\Rightarrow {\sec }^{2}x+{\sec }^{2}y-{\tan }^{2}y-{\tan }^{2}z=2 \\ \Rightarrow {\sec }^{2}x+\left(1\right)-{\tan }^{2}z=2 \\ \Rightarrow xRz \\ \Rightarrow R\text{ is transitive. } \end{gathered}$$

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

\(f^{\prime}(x)=\frac{(4 x-3)\left(3 x^2+x+3\right)-(6 x+1)\left(2 x^2-3 x+2\right)}{\left(3 x^2+x+3\right)^2}\)

\(=\frac{11 x^2-11}{\left(3 x^2+x+3\right)^2}\)

\(f^{\prime}(x)=0\) gives \(11\left(x^2-1\right)=0\)

\(x=1\) and \(x=-1\)

Since \(f^{\prime}(x)\) changes sign at these points

function is not one-to-one

Let \(\frac{2 x^2-3 x+2}{3 x^2+x+3}=y\)

\((3 y-2) x^2+(y+3) x+(3 y-2)=0\)

\(\mathrm{D} \geq 0\)

\((7 y-1)(5 y-7) \leq 0\)

\(y \in\left[\frac{1}{7}, \frac{7}{5}\right]\)

Since range of \(f\) is not equal to co-domain

Hence \(f(x)\) is not onto

Let $A$ be the set of all functions $f:\mathbb{Z}\to\mathbb{Z}$ and define

$$R = \{(f,g): f(0)=g(1)\ \text{and}\ f(1)=g(0)\}.$$

Check each property:

Reflexive
For reflexivity we need $(f,f)\in R$ for every $f$. That means

$$f(0)=f(1) \quad\text{and}\quad f(1)=f(0),$$

so $f(0)=f(1)$ must hold for all $f$.
But there are functions with $f(0)\neq f(1)$ (e.g. $f(n)=n$).
So $R$ is not reflexive.

Symmetric
Suppose $(f,g)\in R$. Then

$$f(0)=g(1),\quad f(1)=g(0).$$

This is equivalent to

$$g(0)=f(1),\quad g(1)=f(0),$$

which are exactly the conditions for $(g,f)\in R$.
So $R$ is symmetric.

Transitive
Need: if $(f,g)\in R$ and $(g,h)\in R$, then $(f,h)\in R$.

Write

$$f(0)=a,\ f(1)=b.$$

Then $(f,g)\in R$ implies

$$g(0)=b,\ g(1)=a.$$

Next, $(g,h)\in R$ implies

$$h(0)=a,\ h(1)=b.$$

So $f(0)=a,\ f(1)=b,\ h(0)=a,\ h(1)=b$.
For $(f,h)\in R$, we need

$$f(0)=h(1)\ (a=b),\quad f(1)=h(0)\ (b=a),$$

which is only true if $a=b$. Pick $a\neq b$ (e.g. $a=0,b=1$) to get

$$(f,g)\in R,\ (g,h)\in R,\ \text{but}\ (f,h)\notin R.$$

Thus $R$ is not transitive.

Hence $R$ is symmetric but neither reflexive nor transitive.

\(\begin{aligned}
& \mathrm{f}(\mathrm{~g}(\mathrm{x}))=\operatorname{lng}(\mathrm{x}) \\
& =\ln \left(\frac{4}{2 x^2-2 x+1}\right) \\
& 2 \mathrm{x}^2-2 \mathrm{x}+1>0 \\
& \forall \mathrm{x} \in \mathrm{R} \\
& \operatorname{Dom}=\mathrm{R}
\end{aligned}\)

Reflexive

for $(a, b) R (a, b)$
$\Rightarrow \mathrm{ab}-\mathrm{ab}=0$ is divisible by \(5.\)
So $(\mathrm{a},\mathrm{b})\mathrm{R}(\mathrm{a},\mathrm{b})\forall \mathrm{a},\mathrm{b}\in \mathrm{Z}$
∴ R is reflexive
Symmetric

For \((\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})\)
If \(\mathrm{ad}-\mathrm{bc}\) is divisible by \(5.\)
Then \(\mathrm{bc}-\mathrm{ad}\) is also divisible by \(5.\)
$\Rightarrow (\mathrm{c},\mathrm{d})\mathrm{R}(\mathrm{a},\mathrm{b})\forall \mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\in \mathrm{Z}$
\(\therefore R\) is symmetric.

Transitive:

If \((a, b) R(c, d)\)
⇒ $ad-bc$ divisible by \(5\) and \((c, d) R(e, f)\)
⇒ $cf-de$ divisible by \(5\)
\(\mathrm{ad}-\mathrm{bc}=5 \mathrm{k}_1 \mathrm{k}_1\) and \(\mathrm{k}_2\) are integers
\(c f-d e=5 k_2\)
\(\mathrm{afd}-\mathrm{bcf}=5 \mathrm{k}_1 \mathrm{f}\)
\(\mathrm{bcf}-\mathrm{bde}=5 \mathrm{k}_2 \mathrm{b}\)
\(afd - bde =5\left(k_1 f+k_2 b\right)\)
\(d(a f-b e)=5\left(k_1 f+k_2 b\right)\)
\(\Rightarrow \mathrm{af}- be\) is not divisible by \(5\) for every \(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f} \in \mathrm{Z}\).
\(\therefore \mathrm{R}\) is not transitive.

Reflexive

for $(a, b) R (a, b)$
$\Rightarrow \mathrm{ab}-\mathrm{ab}=0$ is divisible by \(5.\)
So $(\mathrm{a},\mathrm{b})\mathrm{R}(\mathrm{a},\mathrm{b})\forall \mathrm{a},\mathrm{b}\in \mathrm{Z}$
∴ R is reflexive
Symmetric

For \((\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})\)
If \(\mathrm{ad}-\mathrm{bc}\) is divisible by \(5.\)
Then \(\mathrm{bc}-\mathrm{ad}\) is also divisible by \(5.\)
$\Rightarrow (\mathrm{c},\mathrm{d})\mathrm{R}(\mathrm{a},\mathrm{b})\forall \mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\in \mathrm{Z}$
\(\therefore R\) is symmetric.

Transitive:

If \((a, b) R(c, d)\)
⇒ $ad-bc$ divisible by \(5\) and \((c, d) R(e, f)\)
⇒ $cf-de$ divisible by \(5\)
\(\mathrm{ad}-\mathrm{bc}=5 \mathrm{k}_1 \mathrm{k}_1\) and \(\mathrm{k}_2\) are integers
\(c f-d e=5 k_2\)
\(\mathrm{afd}-\mathrm{bcf}=5 \mathrm{k}_1 \mathrm{f}\)
\(\mathrm{bcf}-\mathrm{bde}=5 \mathrm{k}_2 \mathrm{b}\)
\(afd - bde =5\left(k_1 f+k_2 b\right)\)
\(d(a f-b e)=5\left(k_1 f+k_2 b\right)\)
\(\Rightarrow \mathrm{af}- be\) is not divisible by \(5\) for every \(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f} \in \mathrm{Z}\).
\(\therefore \mathrm{R}\) is not transitive.

$$\begin{gathered} f\left(x\right).f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right) \\ f\left(x\right)=1\pm x^2 \\ f\left(1\right)=2 \left(given\right) \\ f\left(x\right)=1+x^2 \\ f\left(k\right)=-2k \\ 1+k^2=-2k \\ {(k+1)}^2=0 \\ k=-1 \end{gathered}$$

\(f(x)+3 f\left(\frac{24}{x}\right)=4 x, x \neq 0\)

Now, \(f(3)+3 f(8)=12\)

and \(f(8)+3 f(3)=32\)

From \(f(3)=12-3 f(8)\)

\(\Rightarrow f(8)+3(12-3 f(8))=32\)

\(\Rightarrow f(8)=\frac{1}{2}\)

\(f(3)=12-3 \cdot \frac{1}{2}=12-\frac{3}{2}=\frac{21}{2}\)

\(f(3)+f(8)=\frac{21}{2}+\frac{1}{2}=11\)

$$\begin{gathered} For Reflexive: x\in Z \\ x+x = 2x \to Even \\ (x,x)\in R \forall x\in Z \\ So, R is reflexive \\ For Symmetric: x,y \in Z \\ (x,y)\in R \to x+y is even \\ y+x is also even \\ it means (y,x)\in R \forall x,y\in Z \\ So, R is symmetric. \\ For Transitive: x,y,w\in Z \\ (x,y)\in R \to x+y is even\Rightarrow x+y=2\lambda \\ (y+w)\in R \to (y+w) is even\Rightarrow y+w=2\mu \\ (x+y)+(y+w)\to even \\ x+w+2y =2\lambda +2\mu \\ x+w = 2(\lambda +\mu -y) \\ (x, w)\in R \forall x,y,w\in Z \\ So, R is transitive. \\ Hence, R is an equivalence relation. \end{gathered}$$

The argument of the outer log must be positive.

$1 - \log_4(x^2 - 9x + 18) > 0$

$\log_4(x^2 - 9x + 18) < 1$

$x^2 - 9x + 18 < 4^1$