(A) CS₁ and CS₂ divide the triangle into three of equal area so that S₁ and S₂ are points of trisection of AB. Thus S₁ = (0, 0), S₂ = (2, – 3). Equation of CS₁ is y – x = 0. Slope of CS₂ = – 4. The line through (0, 0) drawn parallel to CS₂ is y + 4x = 0

Required equation is  y² + 3xy – 4x² = 0 ⇒ λ + μ = 3 + 4 = 7

(B) The required area is bounded by the lines y = ± x  and the parabola x² = – (y – 2) having the vertex at (0, 2) and passing through (±1, 1). By symmetry about the y-axis,

$\mathrm{Area}=2[\frac{1}{2}+\underset{1}{\overset{2}{\int }}\sqrt{2-\mathrm{y}}\mathrm{dy}]=\frac{7}{3}\equiv \frac{\mathrm{p}}{\mathrm{q}}$

⇒ p – q = 7 – 3 = 4

(C) With y = xv, xv' = $\frac{1+{\mathrm{v}}^2}{2\mathrm{v}}-\mathrm{v}=\frac{1-{\mathrm{v}}^2}{2\mathrm{v}}\Rightarrow \frac{2\mathrm{v}\mathrm{dv}}{1-{\mathrm{v}}^2}=\frac{\mathrm{dx}}{\mathrm{x}}$

whose solution is x(1 – v²) = c. When x = 2, v = $\mathrm{v}=\frac{1}{2}\mathrm{so}\mathrm{that}\mathrm{c}=\frac{3}{2}.$

The equation of the curve is x² – y² = $\frac{3\mathrm{x}}{2}\Rightarrow {(\mathrm{x}-\frac{3}{2})}^2-{\mathrm{y}}^2=\frac{9}{16}.$

This is a rectangular hyperbola with eccentricity $\sqrt{2}$.

$\therefore$ e⁴ = 4 

(D) $\mathrm{P}=\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{Lim}}\frac{{\left[\right({\mathrm{n}}^2+1^2\left)\right({\mathrm{n}}^2+2^2).......({\mathrm{n}}^2+{\mathrm{n}}^2\left)\right]}^{\mathrm{m}/\mathrm{n}}}{{\mathrm{n}}^{2\mathrm{m}}}$

In $\mathrm{P}=\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{Lim}}\frac{\mathrm{m}}{\mathrm{n}}\left[\sum _{\mathrm{r}=1}^{\mathrm{n}}\mathrm{\ell n}({\mathrm{n}}^2+{\mathrm{r}}^2)-2\mathrm{m}\mathrm{\ell n}\mathrm{n}\right]$

In $\mathrm{P}=\mathrm{m}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{Lim}}\left[\sum _{\mathrm{r}=1}^{\mathrm{n}}\frac{\mathrm{\ell n}({\mathrm{n}}^2+{\mathrm{r}}^2)-2\mathrm{n}\mathrm{\ell n}\mathrm{n}}{\mathrm{n}}\right]$

In $\mathrm{P}=\mathrm{m}[\mathrm{\ell n}2+\frac{\mathrm{\pi }}{2}-2]$

$\mathrm{P}={\mathrm{e}}^{\mathrm{m}[\ln 2+\frac{\mathrm{\pi }}{2}-2]}={\mathrm{e}}^{\mathrm{m}\ln 2}·{\mathrm{e}}^{-2\mathrm{m}}·{\mathrm{e}}^{\frac{\mathrm{m\pi }}{2}}$

$=\frac{2^{\mathrm{m}}{\mathrm{e}}^{\frac{\mathrm{m\pi }}{2}}}{{\mathrm{e}}^{2\mathrm{m}}}=\frac{{\left(2{\mathrm{e}}^{\mathrm{\pi }/2}\right)}^{\mathrm{m}}}{{\mathrm{e}}^{2\mathrm{m}}}={\left(\frac{2\sqrt{{\mathrm{e}}^{\mathrm{\pi }}}}{{\mathrm{e}}^2}\right)}^{\mathrm{m}}$

⇒ a = 2