Question

	Column I		Column II
(A)	In a triangle $\Delta XYZ$, let a, b and c be the lengths of the sides opposite to the angles $X,Y$ and $Z$, respectively. If $2(a^2–b^2)=c^2$ and $\lambda =\frac{\sin (X-Y)}{\sin Z}$, then possible values of n for which $\cos \left(n\pi \lambda \right)=0$ is (are)	(p)	1
(B)	In a triangle $\Delta XYZ$, let a, b and c be the lengths of the sides opposite to the angles X, Y and Z respectively. If $1+\cos 2X–2\cos 2Y=2\sin X\sin Y$, then possible value(s) of $\frac{a}{b}$ is (are)	(q)	2
(C)	In $R^2$, let $\sqrt{3}\hat{i}+\hat{j},\hat{i}+\sqrt{3}\hat{j}$ and be the position vectors of X, Y and Z with respect to the origin O, respectively. If the distance of Z from the bisector of the cute angle of $\overrightarrow{OX}$ with $\overrightarrow{OY}$ is $\frac{3}{\sqrt{2}}$, then possible value(s) of $|\beta |$ is (are)	(r)	3
(D)	Suppose that $F\left(\alpha \right)$ denotes the area of the region bounded by $x=0,x=2,y2=4x$ and $y=|\alpha x–1|+|\alpha x–2|+\alpha x,$ where $\alpha \in 0,1.$ Then the value(s) of $F\left(\alpha \right)+\frac{8}{3}\sqrt{2}$, when $\alpha =0$ and $\alpha =1,$ is (are)	(s)	5
		(t)	6

Answer 1

A(P, R, S), B(P), C(P, Q), D(S, T)


(A)
$\lambda =\frac{\sin \left(XY\right)}{\sin Z}$
$\frac{\sin X\cos Y\cos X\sin Z}{\sin Z}$
$\frac{a\cos Yb\cos X}{C}$
$\frac{a\left(\frac{a^2+c^2{b}^2}{2ac}\right)b\left(\frac{b^2+c^2{a}^2}{2bc}\right)}{c}=\lambda$
$\frac{a^2{b}^2}{c^2}=\lambda$
as $2(a^2–b^2)=c^2$
$\therefore$ $\lambda =\frac{1}{2}$
$\therefore$ $\cos \left(\frac{n\pi }{2}\right)=0$
$\therefore$ $n=1,3,5$

(B)

$1+\cos 2X–2\cos 2Y=2\sin X\sin Y$
Solving
$4{\sin }^{2}Y–2{\sin }^{2}X=2\sin X\sin Y$
$4b^2–2a^2=2ab$
$42{\left(\frac{a}{b}\right)}^2=\frac{2a}{b}$

$X^2+X–2=0$
$X=1$ $\frac{a}{b}=1$
$X=–2$ $\frac{a}{b}=2$ (rejected)

(C) $OY\equiv y=\sqrt{3}x$
$OX\equiv y=\frac{1}{\sqrt{3}}x$
$\therefore$ Equation of bisector of OX, OY
$y=x$
$\therefore$ $\left|\frac{xy}{\sqrt{2}}\right|=\frac{3}{\sqrt{2}}$
$\frac{\beta \left(1\beta \right)}{\sqrt{2}}=\pm \frac{3}{\sqrt{2}}$
$2\beta –1=\pm 3$
$\beta =2\beta =–1$
$|\beta |=1,2$


$\therefore$ $\text{Area}=6{\int }_0^{2}2\sqrt{x}dx$
$=6\frac{8\sqrt{2}}{3}$
$\therefore$ 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