Mathematics - Properties of Triangles Question with Solution | TestHub

$\left(\mathrm{A}\right)$ $\mathrm{\lambda }=\frac{\sin \left(\mathrm{X}-\mathrm{Y}\right)}{\sin \mathrm{Z}}$
$\frac{\mathrm{sinX}\mathrm{cosY}-\mathrm{cosX}\sin Y}{\mathrm{sinZ}}$
$\frac{\mathrm{a}\mathrm{cosY}-\mathrm{b}\mathrm{cosX}}{c}$
$\frac{\mathrm{a}\left(\frac{{\mathrm{a}}^2+{\mathrm{c}}^2-{\mathrm{b}}^2}{2\mathrm{a}\mathrm{c}}\right)-\mathrm{b}\left(\frac{{\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2}{2\mathrm{b}\mathrm{c}}\right)}{\mathrm{c}}=\mathrm{\lambda }$
$\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{c}}^2}=\mathrm{\lambda }$
As $2\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)={\mathrm{c}}^2$
$\therefore \mathrm{\lambda }=\frac{1}{2}$
$\therefore \cos \left(\frac{\mathrm{n}\mathrm{\pi }}{2}\right)=0$
$\therefore \mathrm{n}=1,\mathrm{ }3,\mathrm{ }5$
$\left(\mathrm{B}\right)$ $1+\cos 2\mathrm{X}-2\cos 2\mathrm{Y}=2\sin \mathrm{X}\sin \mathrm{Y}$
Solving
$4{\sin }^2\mathrm{Y}-2{\sin }^2\mathrm{X}=2\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{X}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{Y}$
$4{\mathrm{b}}^2-2{\mathrm{a}}^2=2\mathrm{a}\mathrm{b}$
$4-2{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^2=\frac{2\mathrm{a}}{\mathrm{b}}$

$\frac{\mathrm{a}}{\mathrm{b}}=\mathrm{X}$

${\mathrm{X}}^2+\mathrm{X}-2=0$

$\mathrm{X}=1$

$\frac{\mathrm{a}}{\mathrm{b}}=1$

$\mathrm{x}=-2$

$\frac{\mathrm{a}}{\mathrm{b}}=-2$ (rejected)

$\left(\mathrm{C}\right)$ $\mathrm{O}\mathrm{Y}\equiv \mathrm{y}=\sqrt{3}\mathrm{x}$

$\mathrm{O}\mathrm{X}\equiv \mathrm{y}=\frac{1}{\sqrt{3}}\mathrm{x}$

$\therefore$ Equation of bisector of $\mathrm{O}\mathrm{X},\mathrm{ }\mathrm{O}\mathrm{Y}$

$\mathrm{y}=\mathrm{x}$

$\therefore \left|\frac{\mathrm{x}-\mathrm{y}}{\sqrt{2}}\right|=\frac{3}{\sqrt{2}}$

$\frac{\mathrm{\beta }-\left(1-\mathrm{\beta }\right)}{\sqrt{2}}=\pm \frac{3}{\sqrt{2}}$

$2\mathrm{\beta }-1=\pm 3$

$\mathrm{\beta }=2$

$\mathrm{\beta }=-1$

$\left|\mathrm{\beta }\right|=1,\mathrm{ }2$

$\left(\mathrm{D}\right)$

$\therefore$ Area $=6-{\int }_0^{2}2\sqrt{x}\mathrm{ }dx$

$=6-\frac{8\sqrt{2}}{3}$

$\therefore$ Required value $=6-\frac{8\sqrt{2}}{3}+\frac{8\sqrt{2}}{3}=6$

Area$=$ Shaded region $-$ area under parabola.

$=5-\frac{8\sqrt{2}}{3}$

$\therefore$ Required value $=5$