Question

Match Column 1 with suitable options from Column 2

Answer 1

Element A:
$\lambda =\frac{sin(X-Y)}{sinZ}=\frac{sinXcosY-sinYcosX}{sinZ}$
Using sine and cosine law,
$\lambda =\frac{a\cdot \frac{a^2+c^2-b^2}{2ac}-b\cdot \frac{b^2+c^2-a^2}{2bc}}{c}=\frac{a^2-b^2}{c^2}=\frac{1}{2}$
$\therefore cos\left(n\pi \lambda \right)=cos\left(\frac{n\pi }{2}\right)=0$ only when $n$ is an odd integer.
So, $\text{1,3,5}$
Element B:
$1+cos2X-2cos2Y=2sinXsinY$
Using $cos2a=1-2si{n}^{2}a$ and sine law
$1+1-2si{n}^{2}X-2(1-2si{n}^{2}Y)=2sinXsinY$
$\Rightarrow 2si{n}^{2}Y-si{n}^{2}X=sinXsinY$
$\Rightarrow 2\frac{sinY}{sinX}-\frac{sinX}{sinY}=1$
At solving, we get $\frac{sinX}{sinY}=1$
$\therefore \frac{sinX}{sinY}=\frac{a}{b}=1$
Element C:
$\overrightarrow{OX}:\sqrt{3}\hat{i}+\hat{j}$
$\overrightarrow{OY}:\hat{i}+\sqrt{3}\hat{j}$
$\overrightarrow{OZ}:\beta \hat{i}+(1-\beta )\hat{j}$
Bisector of acute angle between $\overrightarrow{OX}$ and $\overrightarrow{OY}$ is $\overrightarrow{OP}:\hat{i}+\hat{j}$
So, distance of point $(\beta ,1-\beta )$ from line $x=y$ is
$\left|\frac{2\beta -1}{\sqrt{2}}\right|=\frac{3}{\sqrt{3}}$
$2\beta =1\pm 3\Rightarrow |\beta |=4$ or $2$.
Element D:
Case 1: $\alpha =0$
Area bounded by $x=0,x=2,y^2=4x$ and $y=3$ is
$F\left(0\right)=3\times 2-{\int }_0^{2}2\sqrt{x}=6-\frac{8\sqrt{2}}{3}$
Case 2: $\alpha =1$
Area bounded by $x=0,x=2,y^2=4x$
$y=\{\begin{array}{c}3(x-1);x\ge 2\\ x+1;1\le x<2\\ 3-x;x<1\end{array}$ is
$F\left(1\right)=4+2\times \frac{1}{2}-{\int }_0^{2}2\sqrt{x}=5-\frac{8\sqrt{2}}{3}$
$\therefore F\left(0\right)+\frac{8\sqrt{2}}{3}=6$ and $F\left(1\right)+\frac{8\sqrt{2}}{3}=5$