Matching Column I Expressions with Corresponding Values in Column II

To solve the matching problem, we analyze each item in Column I step by step:

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(A)

Let $\vec{u} = \alpha\hat{i} + \beta\hat{j}$ and $\vec{v} = \sqrt{3}\hat{i} + \hat{j}$.

The magnitude of the projection of $\vec{u}$ onto $\vec{v}$ is given by: $\left| \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|} \right| = \frac{|\sqrt{3}\alpha + \beta|}{\sqrt{(\sqrt{3})^2 + 1^2}} = \frac{|\sqrt{3}\alpha + \beta|}{2}$

Given that this magnitude equals $\sqrt{3}$: $\frac{|\sqrt{3}\alpha + \beta|}{2} = \sqrt{3} \implies |\sqrt{3}\alpha + \beta| = 2\sqrt{3}$

We are also given the relation $\alpha = 2 + \sqrt{3}\beta \implies \beta = \frac{\alpha - 2}{\sqrt{3}}$. Substituting $\beta$ into the equation: $\left| \sqrt{3}\alpha + \frac{\alpha - 2}{\sqrt{3}} \right| = 2\sqrt{3} \implies \left| \frac{3\alpha + \alpha - 2}{\sqrt{3}} \right| = 2\sqrt{3}$ $|4\alpha - 2| = 6$

This gives two cases:

1. $4\alpha - 2 = 6 \implies 4\alpha = 8 \implies \alpha = 2 \implies |\alpha| = 2$
2. $4\alpha - 2 = -6 \implies 4\alpha = -4 \implies \alpha = -1 \implies |\alpha| = 1$

Thus, the admissible values of $|\alpha|$ are $1$ and $2$. $\mathbf{(A) \to (P), (Q)}$

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(B)

The function $f(x)$ is given as:

$$f(x) = \begin{cases} -3ax^2 - 2, & x < 1 \\ bx + a^2, & x \ge 1 \end{cases}$$

For $f(x)$ to be differentiable for all $x \in \mathbb{R}$, it must be continuous and differentiable at $x = 1$.

1. Continuity at $x = 1$: $\lim_{x \to 1^-} f(x) = f(1) \implies -3a(1)^2 - 2 = b(1) + a^2$ $-3a - 2 = b + a^2 \implies b = -a^2 - 3a - 2 \quad \text{--- (1)}$

2. Differentiability at $x = 1$: $f'(1^-) = f'(1^+) \implies \left. \frac{d}{dx}(-3ax^2 - 2)\right|_{x=1} = \left. \frac{d}{dx}(bx + a^2)\right|_{x=1}$ $-6a(1) = b \implies b = -6a \quad \text{--- (2)}$

Equating $(1)$ and $(2)$: $-6a = -a^2 - 3a - 2 \implies a^2 - 3a + 2 = 0$ $(a - 1)(a - 2) = 0 \implies a = 1 \text{ or } a = 2$

Thus, the admissible values of $a$ are $1$ and $2$. $\mathbf{(B) \to (P), (Q)}$

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(C)

Since $\omega \ne 1$ is a non-real cube root of unity, we have $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.

Let us simplify the terms inside the parentheses:

• $3 - 3\omega + 2\omega^2 = 2(1 + \omega + \omega^2) + 1 - 5\omega = 1 - 5\omega$
• $2 + 3\omega - 3\omega^2 = 2(1 + \omega + \omega^2) + \omega - 5\omega^2 = \omega - 5\omega^2 = \omega(1 - 5\omega)$
• $-3 + 2\omega + 3\omega^2 = 2(1 + \omega + \omega^2) - 5 + \omega^2 = \omega^2 - 5 = \omega^2(1 - 5\omega) \quad (\text{since } -5\omega^3 = -5)$

Let $A = 1 - 5\omega \neq 0$. The given equation becomes: $A^{4n+3} + (\omega A)^{4n+3} + (\omega^2 A)^{4n+3} = 0$ $A^{4n+3} \left( 1 + \omega^{4n+3} + \omega^{2(4n+3)} \right) = 0$

Since $A \neq 0$: $1 + \omega^{4n+3} + \omega^{2(4n+3)} = 0$

The sum $1 + \omega^k + \omega^{2k} = 0$ holds if and only if $k$ is not a multiple of $3$. Here, $k = 4n + 3 \equiv n \pmod 3$. Hence, the equation holds if and only if: $n \not\equiv 0 \pmod 3$

Testing the values of $n$ from Column II:

• For $n = 1$: $1 \not\equiv 0 \pmod 3$ (Valid)
• For $n = 2$: $2 \not\equiv 0 \pmod 3$ (Valid)
• For $n = 3$: $3 \equiv 0 \pmod 3$ (Invalid)
• For $n = 4$: $4 \not\equiv 0 \pmod 3$ (Valid)
• For $n = 5$: $5 \not\equiv 0 \pmod 3$ (Valid)

$\mathbf{(C) \to (P), (Q), (S), (T)}$

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(D)

The harmonic mean of two positive real numbers $a$ and $b$ is $4$: $\frac{2ab}{a+b} = 4 \implies ab = 2(a+b) \quad \text{--- (3)}$

The sequence $a, 5, q, b$ forms an arithmetic progression with common difference $d$: $d = 5 - a$ $q = 5 + d = 5 + (5 - a) = 10 - a$ $b = q + d = (10 - a) + (5 - a) = 15 - 2a$

Substituting $b = 15 - 2a$ into $(3)$: $a(15 - 2a) = 2(a + 15 - 2a)$ $15a - 2a^2 = 30 - 2a$ $2a^2 - 17a + 30 = 0$ $(2a - 5)(a - 6) = 0 \implies a = \frac{5}{2} \text{ or } a = 6$

Now, calculate $|q - a| = |(10 - a) - a| = |10 - 2a|$:

• For $a = 6$: $|q - a| = |10 - 2(6)| = |-2| = 2$ (with $q = 4 > 0, b = 3 > 0$)
• For $a = \frac{5}{2}$: $|q - a| = \left|10 - 2\left(\frac{5}{2}\right)\right| = |5| = 5$ (with $q = \frac{15}{2} > 0, b = 10 > 0$)

Thus, the possible values of $|q - a|$ are $2$ and $5$. $\mathbf{(D) \to (Q), (T)}$

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Final Correct Matching:

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