(1)

For f(x) to be defined we have,

$1–{\log }_4(x^2–9x+18)>0 i.e., x^2–9x+18<4$

Also, $x^2–9x+18>0$

$\Rightarrow (x–3)(x–6)>0$

$\Rightarrow x\in (–\infty ,3)\cup (6,\infty )$          ... (i)

Now, $x^2–9x+18<4$

$\Rightarrow x^2–9x+14<0$

$\Rightarrow (x–2)(x–7)<0$

$\Rightarrow x\in (2,7)$          ... (ii)

From equation (i) & (ii), we get

$x\in (2,3)\cup (6,7)=(\alpha ,\beta )\cup (\gamma ,\delta )$          [Given]

Hence, $\alpha +\beta +\gamma +\delta =2+3+6+7=18$.

(4)

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

Put x = 3, f(3) + 3f(8) = 12          ... (i)

Put x = 8, f(8) + 3f(3) = 32          ... (ii)

Adding (i) and (ii), we get f(3) + f(8) = 11.

(1)

Given $f,g : (1,\infty )\to R$, $f\left(x\right)=\frac{2x+3}{5x+2}$, $g\left(x\right)=\frac{2–3x}{1–x}$

also, we have $fog : [2,4]\to R$

Now, $g\left(2\right)=\frac{2–6}{1–2}=4, g\left(4\right)=\frac{2–12}{1–4}=\frac{10}{3}$

$\therefore f\left(g\right(2\left)\right)=\frac{8+3}{20+2}=\frac{1}{2}, f\left(g\right(4\left)\right)=\frac{20+9}{50+6}=\frac{29}{56}$

i.e., $\alpha =\frac{1}{2} \mathrm{and} \beta =\frac{29}{56}$

Then, $\frac{1}{\beta –\alpha }=\frac{1}{{\displaystyle \frac{29}{56}}–{\displaystyle \frac{1}{2}}}=\frac{1}{{\displaystyle \frac{1}{56}}}=56$.

(3)

We Have, $A : x^2+y^2=25$           ... (i)

$B : \frac{x^2}{144}+\frac{y^2}{16}=1$          ... (ii)

$C : x^2+y^2\le 4$          ... (iii)

Solving (i) and (ii), we get

$x^2+9(25–x^2)=144 \Rightarrow –8x^2=–81 \Rightarrow x=\pm \frac{9}{2\sqrt{2}}$

From (i), $\frac{81}{8}+y^2=25 \Rightarrow y^2=25 –\frac{81}{8} \Rightarrow y=\pm \frac{\sqrt{119}}{2\sqrt{2}}$

As, $D=A\cap B$

$=\left\{\right(\frac{9}{2\sqrt{2}},\frac{\sqrt{119}}{2\sqrt{2}}),(\frac{9}{2\sqrt{2}},–\frac{\sqrt{119}}{2\sqrt{2}}),(\frac{–9}{2\sqrt{2}},\frac{\sqrt{119}}{2\sqrt{2}}),(–\frac{9}{2\sqrt{2}},–\frac{\sqrt{119}}{2\sqrt{2}}\left)\right\}$

$\Rightarrow n\left(D\right)=4$

Also, $C=\left\{\right(x,y)\in Z\times Z : x^2+y^2\le 4\}$

={(0, 2), (0, –2), (2, 0), (–2, 0), (1, 1), (–1, –1), (–1, 1), (1, –1), (0, 1), (0, –1), (1, 0), (–1, 0), (0, 0)}.

$\Rightarrow n\left(C\right)=13$

$\therefore$   Total number of one-one function from D to C = ${}^{13}{P}_4$ = 17160.

(2)

$y=\frac{5–x}{x^2–3x+2}, x\ne 1,2$

$\Rightarrow y{x}^2–3xy+2y+x–5=0$

$\Rightarrow y{x}^2+(–3y+1)x+(2y–5)=0$

Case I : If y = 0

$\Rightarrow$ x = 5

Case II : if y $\ne$ 0

For real solutions, $D\ge 0$

$\Rightarrow {(–3y+1)}^2–4\left(y\right)(2y–5)\ge 0$

$\Rightarrow 9y^2+1–6y–8y^2+20y\ge 0$

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

$\Rightarrow {(y+7)}^2–48\ge 0$

$\Rightarrow |y+7|\ge 4\sqrt{3} \Rightarrow y+7\ge 4\sqrt{3} \mathrm{or} y+7\le –4\sqrt{3}$

$\Rightarrow y\ge 4\sqrt{3}–7 \mathrm{or} y\le –4\sqrt{3}–7$

From Case I and Case II, we have

    $y\in (–\infty ,–4\sqrt{3}–7]\cup [4\sqrt{3}–7,\infty )$

$\therefore \alpha =–4\sqrt{3}–7 \mathrm{and} \beta =4\sqrt{3}–7$

$\Rightarrow {\alpha }^2+{\beta }^2={(–4\sqrt{3}–7)}^2+{(4\sqrt{3}–7)}^2=2(48+49)=194$.

(3)

Here, n(A) = 4, n(B) = 4

Total number of functions from A to B = $4^4$ = 256

Number of one-one functions from A to B = ${}^{4}P_4$ = 4! = 24

Number of many-one functions from A to B = 256 – 24 = 232

Number of many-one function for which $1\notin f\left(A\right)=3^4$ = 81

$\therefore$   Required number of many-one functions = 232 – 81 = 151.

(4)

Given, $g\left(x\right)=\frac{x^4–2x^3+3x^2–2x+2}{2x^2–2x+1}$

Here, $Dg\in R$          [$\because 2x^2–2x+1>0$]

Also, $f\left(x\right)={\log }_{e}x \Rightarrow D_f\in (0,\infty )$           $\therefore D_{fog}\Rightarrow g\left(x\right)>0$

$\Rightarrow \frac{x^4–2x^3+3x^2–2x+2}{2x^2–2x+1}>0 \Rightarrow x^4–2x^3+3x^2–2x+2>0$

The above expression is always positive for any value of x.

$\therefore \mathrm{Domain} \mathrm{of} fog \in R$.

(1)

$f\left(x\right)=\frac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32}$

       $=\frac{2^2(2^x+4)}{2[{\left(2^x\right)}^2+8\times 2^x+16]}=\frac{2(2^x+4)}{{(2^x+4)}^2}=\frac{2}{2^x+4}$

$\Rightarrow f(4–x)=\frac{2}{2^{4–x}+4}=\frac{2^x}{2(2^x+4)}$

     $f\left(x\right)+f(4–x)=\frac{2}{2^x+4}+\frac{2^x}{2(2^x+4)}=\frac{1}{2}$

$\therefore f\left(\frac{1}{15}\right)+f\left(\frac{59}{15}\right)=f\left(\frac{2}{15}\right)+f\left(\frac{58}{15}\right)$

     $=f\left(\frac{3}{15}\right)+f\left(\frac{57}{15}\right)=...=f\left(\frac{29}{15}\right)+f\left(\frac{31}{15}\right)=\frac{1}{2}$

        $f\left(\frac{30}{15}\right)=f\left(2\right)=\frac{2}{4+4}=\frac{1}{4}$

       $8\left[f\right(\frac{1}{15})+f(\frac{2}{15})+...+f(\frac{59}{15}\left)\right]$

      $=8[\frac{1}{2}+\frac{1}{2}+... \mathrm{upto} (29 \mathrm{terms})+\frac{1}{4}]=8(\frac{29}{2}+\frac{1}{4})=118$

(3)

Given function is $f\left(x\right)=\frac{2^x–2^{–x}}{2^x+2^{–x}}$

$f\left(x\right)=\frac{2^x–{\displaystyle \frac{1}{2^x}}}{2^x+\frac{1}{2^x}}=\frac{2^{2x}–1}{2^{2x}+1}=1–\frac{2}{2^{2x}+1}$

$\Rightarrow f\text{'}\left(x\right)=\frac{2}{{(2^{2x}+1)}^2}\times 2\times 2^{2x}\times {\log }_e\left(2\right)>0$

$\Rightarrow$ f(x) is always increasing.

Hence it is one-one.

Since $f(–\infty )=–1$ and $f(\infty )=1 \Rightarrow f\left(x\right)\in (–1,1)\ne (–\infty ,1)$

Thus, the function f(x) is one-one but not onto.

(1)

Given, $f\left(x\right)=(2+3a)x^2+\left(\frac{a+2}{a–1}\right)x+b, a\ne 1$          ... (i)

Also, $f(x+y)=f\left(x\right)+f\left(y\right)+1–\frac{2}{7}xy$          ... (ii)

Put x = y = 0 in (ii), we get

f(0) = f(0) + f(0) + 1 – 0 $\Rightarrow$ f(0) = –1          ... (iii)

Using (i), f(0) = b = –1

Put y = – x in (ii), we get

$f(x–x)=f\left(x\right)+f(–x)+1+\frac{2}{7}{x}^2$

$\Rightarrow –1=(2+3a)x^2+\left(\frac{a+2}{a–1}\right)x–1+(2+3a)x^2–\left(\frac{a+2}{a–1}\right)x–1+1+\frac{2}{7}{x}^2$

$\Rightarrow \left(2\right(2+3a)+\frac{2}{7})x^2=0 \Rightarrow a=\frac{–5}{7}$

$\therefore f\left(x\right)=\frac{–1}{7}{x}^2–\frac{3}{4}x–1=\frac{–1}{28}(4x^2+21x+28)$

Now, $28\sum _{i=1}^5\left|f\right(i\left)\right|=28\left[\right|f\left(1\right)|+|f\left(2\right)|+...+|f\left(5\right)\left|\right]$

        $=28\times \frac{1}{28}\left[4\right(1^2+2^2+...+5^2)+21(1+2+...+5)+28(5\left)\right]$

         = 675