(96)

We have, $f\left(\theta \right)=\frac{{\sin }^4\theta +3{\cos }^2\theta }{{\sin }^4\theta +{\cos }^2\theta }$

$=1+\frac{2{\cos }^2\theta }{{\sin }^4\theta +{\cos }^2\theta }=1+\frac{2{\cos }^2\theta }{1+{\cos }^4\theta -{\cos }^2\theta }$

$=1+\frac{2}{{\displaystyle \frac{{\displaystyle 1}}{{\displaystyle {\cos }^2\theta }}}{\displaystyle +}{\cos }^2\theta -1}$

Now, ${\cos }^2\theta +\frac{1}{{\cos }^2\theta }\ge 2$                    $[\because \mathrm{A}.\mathrm{M}.\ge \mathrm{G}.\mathrm{M}.]$

$\Rightarrow \frac{1}{{\cos }^2\theta }+{\cos }^2\theta -1\ge 1\Rightarrow {\cos }^2\theta +\frac{1}{{\cos }^2\theta }-1\in [1,\infty )$

$\Rightarrow \frac{1}{{\cos }^2\theta +\frac{1}{{\cos }^2\theta }-1}\in (0,1]$

When $\cos \theta =0,f\left(\theta \right)=1$

$\therefore$    $f\left(\theta \right)\in [1,3]\Rightarrow \alpha =1,\beta =3$

Sum of infinite G.P. with first term 64 and common ratio

$\frac{1}{3}=\frac{64}{1-\frac{1}{3}}=32\times 3=96$

(1)

Let $a,ar$ and $a{r}^2$ be the three sides of the triangle. 

Now, $a+ar>a{r}^2$

$\Rightarrow r^2-r-1<0\Rightarrow r\in (\frac{{\displaystyle 1-\sqrt{5}}}{{\displaystyle 2}},\frac{{\displaystyle 1+\sqrt{5}}}{{\displaystyle 2}})$

So, $3\left[r\right]+[-r]=3+(-2)=1$

(9)

$8=3+\frac{1}{\left(4\right)}(3+p)+\frac{1}{{\left(4\right)}^2}(3+2p)+\frac{1}{{\left(4\right)}^3}(3+3p)+\dots \infty$          ...(i)

Multiplying both sides by $\frac{1}{4},$ we get  $2=\frac{3}{4}+\frac{3+p}{{{\displaystyle (}{\displaystyle 4}{\displaystyle )}}^2}+\frac{3+2p}{{{\displaystyle (}{\displaystyle 4}{\displaystyle )}}^3}+\dots +\infty$                                      ...(ii)

Subtracting (ii) from (i), we get $6=3+\frac{p}{4}+\frac{p}{4^2}+\dots$

$\Rightarrow 3=p[\frac{{\displaystyle 1}}{{\displaystyle 4}}+\frac{{\displaystyle 1}}{{\displaystyle {\left(4\right)}^2}}+\frac{{\displaystyle 1}}{{\displaystyle {\left(4\right)}^3}}+\dots +\infty ]$

$\Rightarrow 3=p\left[\frac{{\displaystyle \frac{{\displaystyle 1}}{{\displaystyle 4}}}}{{\displaystyle 1-\frac{{\displaystyle 1}}{{\displaystyle 4}}}}\right]$                    $[\because S_{\infty }=\frac{a}{1-r}$ for infinite geometric series $]$

$\Rightarrow 3=p[\frac{{\displaystyle 1}}{{\displaystyle 4}}\times \frac{{\displaystyle 4}}{{\displaystyle 3}}]\Rightarrow p=9$

(25)

We have, 

${(x+3)}^{n-1}+{(x+3)}^{n-2}(x+2)+{(x+3)}^{n-3}{(x+2)}^2+\dots +{(x+2)}^{n-1}$

$\therefore$   $\sum _{r=0}^n{\alpha }_r=4^{n-1}+4^{n-2}\times 3+4^{n-3}\times 3^2+\dots +3^{n-1}$

$=4^{n-1}[1+\frac{{\displaystyle 3}}{{\displaystyle 4}}+{\left(\frac{{\displaystyle 3}}{{\displaystyle 4}}\right)}^2+\dots +{\left(\frac{{\displaystyle 3}}{{\displaystyle 4}}\right)}^{n-1}]$

$=4^{n-1}\times \frac{{\displaystyle 1-{\left(\frac{{\displaystyle 3}}{{\displaystyle 4}}\right)}^n}}{{\displaystyle 1-\frac{{\displaystyle 3}}{{\displaystyle 4}}}}=4^{n-1}(1-{\left(\frac{{\displaystyle 3}}{{\displaystyle 4}}\right)}^n)\left(4\right)$

$=4^n-3^n=\beta -\gamma \Rightarrow b=4,\gamma =3$

$\therefore$    ${\beta }^2+{\gamma }^2=16+9=25$