(A)
In a triangle $X Y Z$ let a,b,and c be the lengths
of the sides opposite to the angles X,Y and Z
respectively. If $2 (a^{2} - b^{2}) = c^{2} a n d λ = \frac{\sin (X - Y)}{\sin Z}$
,then possible value(s) of n for which $\cos (n π λ) = 0$
is (are)
(P) 1
(B)
In a triangle $X Y Z$ let a,b,and c be the lengths
of the sides opposite to the angles X,Y and Z
respectively. If $1 + \cos 2 X - 2 \cos 2 Y = 2 \sin X \sin Y,$
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 $β \hat{i} + (1 - β) \hat{j}$ be
the position vectors of X,Yand Z with respect
to the origin O, respectively.If the distance of
Z from the bisector of acute angle of $\vec{O X}$ with
$\vec{O Y}$ is $\frac{3}{\sqrt{2}}$ then possible value(s) of $|β|$ is (are)
(R) 3
(D)
Suppose that $F (α)$ denotes the area of the
region bounded by $x = 0, x = 2, y^{2} = 4 x a n d y = |α x - 1| + |α x - 2| + α x$
,where $α \in \{0, 1\}$ ,Then the value(s) of $F (α) + \frac{8}{3} \sqrt{2}$
,when $α = 0$ and $α = 1$ is (are)

(S) 5
(T) 6

Difficult

Solution

$\begin{array}{l} (A) 2 (a^{2} - b^{2}) = c^{2} \\ \Rightarrow 2 (\sin^{2} X - \sin^{2} Y) = \sin^{2} Z \\ \Rightarrow 2 \sin (X + Y) \sin (X - Y) = \sin^{2} Z \\ \Rightarrow \frac{\sin (X - Y)}{\sin Z} = \frac{1}{2} \\ \Rightarrow λ = \frac{1}{2} \\ \cos (n π λ) = 0 \Rightarrow \cos \frac{n π}{2} = 0 \\ n = 1, 3, 5 \end{array}$
$\begin{array}{l} (B) 1 + \cos 2 X - 2 \cos 2 Y = 2 \sin X \sin Y \\ \Rightarrow 1 + 1 - 2 \sin^{2} X - 2 (1 - 2 \sin^{2} Y) = 2 \sin X \sin Y \\ \Rightarrow \sin^{2} X + \sin X \sin Y - 2 \sin^{2} Y = 0 \\ \frac{\sin X}{\sin Y} = - 2 o r 1 \\ but - 2 is not possible \\ \Rightarrow \frac{\sin X}{\sin Y} = 1 \\ \Rightarrow \frac{a}{b} = 1 \end{array}$
$\begin{array}{l} (C) Bisectors of \vec{O X} & \vec{O Y} are x = y or x + y = 0 \\ ∴ |\frac{β - (1 - β)}{\sqrt{2}}| = \frac{3}{\sqrt{2}} \\ \Rightarrow |2 β - 1| = 3 \\ \Rightarrow |β| = 1, 2 \end{array}$
(D) For $α = 1$
$y = |x - 1| + |x - 2| + x$ $= \{\begin{cases} 3 - x; x < 1 \\ 1 + x; 1 \leq x < 2 \\ 3 x - 3; x \geq 2 \end{cases}$
$\begin{array}{l} ∴ A r e a = F (1) = \frac{1}{2} (2 + 3) \times 1 + \frac{1}{2} (2 + 3) \times 1 -_{0}^{2} 2 \sqrt{x} d x = 5 - \frac{8 \sqrt{2}}{3} \\ \Rightarrow F (1) + \frac{8 \sqrt{2}}{3} = 5 \end{array}$
$F o r α = 0, y = 3$ .
$\begin{array}{l} ∴ A r e a = F (0) = 2 \times 3 - \int_{0}^{2} 2 \sqrt{2} d x = 6 - \frac{8 \sqrt{2}}{3} \\ \Rightarrow F (0) + \frac{8 \sqrt{2}}{3} = 6 \end{array}$

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