(4)

Graph of the function f(x)

The sum of local minimum values at A and B

$=\frac{7}{3}–\frac{11}{72}=\frac{157}{72}$.

(1)

We have, $f\left(x\right)=\frac{2^x}{2^x+\sqrt{2}}, x\in R$

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

        $=\frac{2^x}{2^x+\sqrt{2}}+\frac{2}{2+\sqrt{2}·2^x}=1$

Now, $S=\sum _{k=1}^{81}f\left(\frac{k}{82}\right)=f\left(\frac{1}{82}\right)+f\left(\frac{2}{82}\right)+...+f\left(\frac{81}{82}\right)$

$\Rightarrow S=f\left(\frac{1}{82}\right)+...+f(1–\frac{2}{82})+f(1–\frac{1}{82})$

$=f\left(\frac{1}{82}\right)+f(1–\frac{1}{82})+f\left(\frac{2}{82}\right)+f(1–\frac{2}{82})+...+ \mathrm{upto} 40 +f\left(\frac{41}{82}\right)$

$=\underset{40 \mathrm{times}}{\underbrace{(1+1+1+...+1)}}+\frac{2^{1/2}}{2^{1/2}+\sqrt{2}}=40+\frac{1}{2}=\frac{81}{2}$.

(3)

As f(x) is onto, hence A is range of f(x).

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

for extremum, f(2) = 16 – 60 + 72 + 7 = 35; f(3) = 54 – 135 + 108 + 7 = 34; f(0) = 7; F(1) = 30.

Hence, range $\in$ [7, 35] = A

Also, for range of g(x), g(x) = $1-\frac{1}{x^{2025}+1}\in [0,1)=B$

 $\therefore$  S = {0, 7, 8, ..., 35}

Hence, n(S) = 30

(3)

Let $f_1\left(x\right)={\log }_5(18x–x^2–77)$

$\therefore 18x–x^2–77>0$

$\Rightarrow x^2–18x+77<0$

$\Rightarrow (x–11)(x–7)<0$

$\Rightarrow x\in (7,11) \therefore \alpha =7, \beta =11$

Also, let $f_2\left(x\right)={\log }_{(x–1)}\left(\frac{2x^2+3x–2}{x^2–3x–4}\right)$

$\therefore x–1>0 \Rightarrow x>1, x–1\ne 1 \Rightarrow x\ne 2$,

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

$\Rightarrow x\in (4,\infty ) \therefore \gamma =4$

Now, ${\alpha }^2+{\beta }^2+{\gamma }^2$ = 49 + 121 + 16 = 186.