Question

Column Matching:

$$\begin{array}{cc}\text{Column (I)}& \text{Column (II)}\\ \begin{array}{cc}\text{(A)}& \text{In a triangle}△XYZ,\text{let}a,b\text{and}c\text{be the}\\ & \text{lengths of the sides opposite to the angles}\\ & X,Y\text{and}Z,\text{respectively. If}2(a^2-b^2)=c^2\\ & \text{and}\lambda =\frac{\sin (X-Y)}{\sin Z},\text{then possible values}\\ & \text{of}n\text{for which}\cos \left(n\pi \lambda \right)=0\text{is (are)}\end{array}& \text{(P) 1}\\ & \\ \begin{array}{cc}\text{(B)}& \text{In a triangle}△XYZ,\text{let}a,b\text{and}c\text{be}\\ & \text{the lengths of the sides opposite to the}\\ & \text{angles}X,Y\text{and}Z,\text{respectively. If}\\ & 1+\cos 2X-2\cos 2Y=2\sin X\sin Y,\text{then}\\ & \text{possible value(s) of}\frac{a}{b}\text{is (are)}\end{array}& \text{(Q) 2}\\ & \\ \begin{array}{cc}\text{(C)}& \text{In}{R}^2,\text{let}\sqrt{3}\hat{i}+\hat{j},\hat{i}+\sqrt{3}\hat{j}\text{and}\beta \hat{i}+(1-\beta )\hat{j}\\ & \text{be the position vectors of}X,Y\text{and}Z\text{with}\\ & \text{respect to the origin}O,\text{respectively. If the}\\ & \text{distance of}Z\text{from the bisector of the acute}\\ & \text{angle of}\overrightarrow{OX}\text{with}\overrightarrow{OY}\text{is}\frac{3}{\sqrt{2}},\text{then possible}\\ & \text{value(s) of}|\beta |\text{is (are)}\end{array}& \text{(R) 3}\\ & \\ \begin{array}{cc}\text{(D)}& \text{Suppose that}F\left(\alpha \right)\text{denotes the area of the}\\ & \text{region bounded by}x=0,x=2,y^2=4x\\ & \text{and}y=|\alpha x-1|+|\alpha x-2|+\alpha x,\text{where}\\ & \alpha \in \{0,1\}.\text{Then the value(s) of}F\left(\alpha \right)+\frac{8}{3}\sqrt{2}\text{,}\\ & \text{when}\alpha =0\text{and}\alpha =1,\text{is (are)}\end{array}& \text{(S) 5}\\ & \\ & \text{(T) 6}\\ & \end{array}$$
Option $\text{(D)}$ matches with which of the elements of right hand column?

• A. $P$
• B. $Q$
• C. $R$
• D. $S$
• E. $T$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $P$
C $R$
D $S$


$\text{Given,}2(a^2–b^2)=c^2\text{From sine rule,}(ak{)}^2={\sin }^{2}X\Rightarrow 2({\sin }^{2}X-{\sin }^{2}Y)={\sin }^2(Z)\Rightarrow 2\sin (X-Y\left)\sin \right(X+Y)={\sin }^2(Z)$

$\text{Also,}\sin (X+Y)=\sin (\pi -Z)=\sin \left(Z\right)$
$\therefore \frac{\sin (X-Y)}{\sin \left(Z\right)}=\frac{1}{2}\Rightarrow \lambda =\frac{1}{2}\cos \left(n\pi \lambda \right)=0\Rightarrow \cos \frac{n\pi }{2}=0\therefore n=1,3,5$