Class 11th

$P\left(\frac{15}{2}{t}^2,15t\right)$

C (-15,0)

Q (-30,0)

tanθ = $\frac{PN}{QN}$

=> $\frac{1}{t}=\frac{15t}{30+\frac{15}{2}{t}^2}$

=> t = 2

$Tangent=y=\frac{1}{2}\left(x+30\right)$

$T_{\left(P\right)}:\frac{x{x}_1}{8}+\frac{y{y}_1}{4}=1,$

where $\frac{x_{1}2}{8}+\frac{y_{1}2}{4}=1...........\left(i\right)$

given : slope = 2

=> x₁ = -4y         . (ii)

(i) & (ii) => P $\left(-\frac{8}{3},\frac{2}{3}\right)$

$e=\frac{1}{\sqrt[]{2}}$         

A = ar (PSS') = $\frac{4}{3}$

$\left(5-e^2\right)A=6$

Kindly consider the following figure

$S-\frac{1}{x-1}=\frac{-2^{101}}{{\left(x^{2^{100}}\right)}^2-1}$

For x = 2, S = 1 $\frac{2^{101}}{4^{101}-1}$

Centre (0, 1)

Radius = $\sqrt[]{2}$

$f\left(x\right)=\frac{1-x}{1+x},0<x<1$ (on simplification)

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

${\left(1-x\right)}^{2}f\text{'}\left(x\right)=-2{\left(f\left(x\right)\right)}^2$

Given equation

$\Rightarrow \frac{cosx}{1+sinx}=\frac{2\left|sinx\right|cosx}{\left|co{s}^{2}x-si{n}^{2}x\right|}$

$\Rightarrow \left|1-2si{n}^{2}x\right|=2\left|sinx\right|\left(1+sinx\right)$

Put sin x = t

then equation,

$\left|1-2t^2\right|=2\left|t\right|\left(1+t\right),t\in \left(-1,1\right)-\left\{\pm \frac{1}{\sqrt[]{2}}\right\}$

Case I : For $\ge \frac{1}{\sqrt[]{2}}$ $\frac{}{\text{exception 25:}}$

the equation has no solution

Case II : For $0\le t<\frac{1}{\sqrt[]{2}}$ $\frac{}{\text{exception 25:}}$

Equation t = $\frac{\sqrt[]{5}-1}{4}\Rightarrow x=18°$ $\frac{\text{exception 25:}}{}$

Case III : $For\frac{-1}{\sqrt[]{2}}<t<0$ $\frac{}{\text{exception 25:}}$

Equation t = $\frac{-1}{2}\Rightarrow x=-30°$ $\frac{}{}$

Case IV : For $T<-\frac{1}{\sqrt[]{2}}$ $\frac{}{\text{exception 25:}}$

Equation t = $\frac{\sqrt[]{5}+1}{4}\Rightarrow$ $\frac{\text{exception 25:}}{}$ x = 54°

Sum of solutions = 18° - 30° - 54° = 66°

Total number = 20 + 8 + 24 = 52

$x^2+y^2+{\left(x-1\right)}^2+y^2+x^2+{\left(y-1\right)}^2+{\left(x-1\right)}^2+{\left(y-1\right)}^2=18$

$\Rightarrow x^2+y^2-x-y-\frac{7}{2}=0$

$r^2=\frac{1}{4}+\frac{1}{4}+\frac{7}{2}=4$

$LHS={\displaystyle \sum _{r=1}^{15}r\left(r!\right)}$

$=\sum \left(r+1-1\right)r!={\displaystyle \sum _{r=1}^{15}\left(r+1\right)!-r!}$

$=\left(2!-1!\right)+\left(3!-2!\right)+....+\left(16!-15!\right)$

$\begin{array}{l}=16!-1\\ {=}^{16}{P}_{16}-1\end{array}$