Match Column List Questions on Triangle Geometry Vectors and Area Bounded by Curves

To match the expressions in Column I with their correct corresponding values in Column II, we evaluate each part step-by-step:

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Part (A)

In $\Delta XYZ$, using the Sine Rule, the side lengths $a, b, c$ opposite to angles $X, Y, Z$ are given by: $a = 2R \sin X, \quad b = 2R \sin Y, \quad c = 2R \sin Z$

Given the relation: $2(a^2 - b^2) = c^2$

Substituting the sine expressions: $2\left(\sin^2 X - \sin^2 Y\right) = \sin^2 Z$

Using the trigonometric identity $\sin^2 X - \sin^2 Y = \sin(X - Y)\sin(X + Y)$ and the property $X + Y + Z = \pi \implies \sin(X + Y) = \sin Z$: $2 \sin(X - Y) \sin Z = \sin^2 Z$

Since $\sin Z \neq 0$ for a non-degenerate triangle, we divide both sides by $\sin Z$: $2 \sin(X - Y) = \sin Z \implies \lambda = \frac{\sin(X - Y)}{\sin Z} = \frac{1}{2}$

Now, we are given: $\cos(n \pi \lambda) = 0 \implies \cos\left(\frac{n\pi}{2}\right) = 0$

For $\cos\left(\frac{n\pi}{2}\right) = 0$, $\frac{n\pi}{2}$ must be an odd multiple of $\frac{\pi}{2}$, which implies $n$ must be an odd integer ($n = 1, 3, 5, \dots$).

From Column II, the odd integers are $1, 3,$ and $5$.

Thus, (A) matches with (P), (R), and (S).

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Part (B)

Given the trigonometric identity for $\Delta XYZ$: $1 + \cos 2X - 2\cos 2Y = 2\sin X \sin Y$

Rewriting using double-angle identities $1 + \cos 2X = 2\cos^2 X$ and $\cos 2Y = 1 - 2\sin^2 Y$: $2\cos^2 X - 2(1 - 2\sin^2 Y) = 2\sin X \sin Y$

Dividing by $2$ and using $\cos^2 X = 1 - \sin^2 X$: $(1 - \sin^2 X) - 1 + 2\sin^2 Y = \sin X \sin Y$ $2\sin^2 Y - \sin X \sin Y - \sin^2 X = 0$

Dividing the entire equation by $\sin^2 Y$ (since $\sin Y \neq 0$) and letting $t = \frac{\sin X}{\sin Y} = \frac{a}{b} > 0$: $2 - t - t^2 = 0 \implies t^2 + t - 2 = 0$ $(t + 2)(t - 1) = 0$

Since $t = \frac{a}{b} > 0$, we get: $\frac{a}{b} = 1$

Thus, (B) matches with (P).

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Part (C)

The position vectors are: $\overrightarrow{OX} = \sqrt{3}\hat{i} + \hat{j}, \quad \overrightarrow{OY} = \hat{i} + \sqrt{3}\hat{j}, \quad \overrightarrow{OZ} = \beta\hat{i} + (1 - \beta)\hat{j}$

The magnitudes of $\overrightarrow{OX}$ and $\overrightarrow{OY}$ are both equal to $2$. The unit vectors along $\overrightarrow{OX}$ and $\overrightarrow{OY}$ are: $\hat{u}_X = \frac{\sqrt{3}\hat{i} + \hat{j}}{2}, \quad \hat{u}_Y = \frac{\hat{i} + \sqrt{3}\hat{j}}{2}$

The vector along the acute angle bisector is: $\hat{u}_X + \hat{u}_Y = \frac{\sqrt{3}+1}{2}(\hat{i} + \hat{j})$

Since this direction is along $\hat{i} + \hat{j}$ and passes through the origin $O(0,0)$, the line representing the acute angle bisector is: $x - y = 0$

The distance $d$ of point $Z(\beta, 1-\beta)$ from the line $x - y = 0$ is given as $\frac{3}{\sqrt{2}}$: $d = \frac{|\beta - (1 - \beta)|}{\sqrt{1^2 + (-1)^2}} = \frac{|2\beta - 1|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$

This simplifies to: $|2\beta - 1| = 3$

Solving for $\beta$:

1. $2\beta - 1 = 3 \implies \beta = 2 \implies |\beta| = 2$
2. $2\beta - 1 = -3 \implies \beta = -1 \implies |\beta| = 1$

Thus, the possible values of $|\beta|$ are $1$ and $2$.

Thus, (C) matches with (P) and (Q).

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Part (D)

Let $g(x) = |\alpha x - 1| + |\alpha x - 2| + \alpha x$. We need to compute the area $F(\alpha)$ bounded by $x = 0$, $x = 2$, $y^2 = 4x \implies y = 2\sqrt{x}$ (for $y \ge 0$), and $y = g(x)$.

Case 1: When $\alpha = 0$

$g(x) = |-1| + |-2| + 0 = 3$

Since $3 \ge 2\sqrt{x}$ for $x \in [0, 2]$, the area $F(0)$ is: $F(0) = \int_{0}^{2} (3 - 2\sqrt{x}) \, dx = \left[ 3x - \frac{4}{3}x^{3/2} \right]_{0}^{2} = 6 - \frac{8\sqrt{2}}{3}$

Therefore: $F(0) + \frac{8}{3}\sqrt{2} = 6$

Case 2: When $\alpha = 1$

$g(x) = |x - 1| + |x - 2| + x$

Breaking the domain $x \in [0, 2]$:

• For $x \in [0, 1]$: $g(x) = (1 - x) + (2 - x) + x = 3 - x$
• For $x \in [1, 2]$: $g(x) = (x - 1) + (2 - x) + x = x + 1$

For all $x \in [0, 2]$, $g(x) \ge 2\sqrt{x}$. The area $F(1)$ is: $F(1) = \int_{0}^{2} g(x) \, dx - \int_{0}^{2} 2\sqrt{x} \, dx$

Calculating $\int_{0}^{2} g(x) \, dx$: $\int_{0}^{1} (3 - x) \, dx + \int_{1}^{2} (x + 1) \, dx = \left[ 3x - \frac{x^2}{2} \right]_{0}^{1} + \left[ \frac{x^2}{2} + x \right]_{1}^{2} = \frac{5}{2} + \frac{5}{2} = 5$

Thus: $F(1) = 5 - \frac{8\sqrt{2}}{3} \implies F(1) + \frac{8}{3}\sqrt{2} = 5$

Hence, the possible values of $F(\alpha) + \frac{8}{3}\sqrt{2}$ are $5$ and $6$.

Thus, (D) matches with (S) and (T).

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Final Match:

• (A) $\rightarrow$ (P), (R), (S)
• (B) $\rightarrow$ (P)
• (C) $\rightarrow$ (P), (Q)
• (D) $\rightarrow$ (S), (T)