Part (A)

1. Understand the problem: We are given two vectors, $\vec{v} = \alpha \widehat i + \beta \widehat j$ and $\vec{u} = \sqrt 3 \widehat i + \widehat j$. The magnitude of the projection of $\vec{v}$ on $\vec{u}$ is $\sqrt{3}$. We are also given a relation between $\alpha$ and $\beta$, which is $\alpha = 2 + \sqrt 3 \beta$. We need to find the possible values of $|\alpha|$.

2. Use the formula for the magnitude of projection: The magnitude of the projection of vector $\vec{v}$ on vector $\vec{u}$ is given by $\left| \frac{\vec{v} \cdot \vec{u}}{|\vec{u}|} \right|$.

3. Calculate the dot product and magnitude:

   • $\vec{v} \cdot \vec{u} = (\alpha)(\sqrt{3}) + (\beta)(1) = \sqrt{3}\alpha + \beta$
   • $|\vec{u}| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2$

4. Set up the equation: We are given that the magnitude of the projection is $\sqrt{3}$. $\left| \frac{\sqrt{3}\alpha + \beta}{2} \right| = \sqrt{3}$ $|\sqrt{3}\alpha + \beta| = 2\sqrt{3}$ This leads to two cases: Case 1: $\sqrt{3}\alpha + \beta = 2\sqrt{3}$ Case 2: $\sqrt{3}\alpha + \beta = -2\sqrt{3}$

5. Solve the system of equations: We have the second equation: $\alpha = 2 + \sqrt{3}\beta$, which can be rewritten as $\beta = \frac{\alpha - 2}{\sqrt{3}}$.

   • Case 1: Substitute $\beta$ into $\sqrt{3}\alpha + \beta = 2\sqrt{3}$. $\sqrt{3}\alpha + \frac{\alpha - 2}{\sqrt{3}} = 2\sqrt{3}$ Multiplying by $\sqrt{3}$, we get: $3\alpha + (\alpha - 2) = 6 \implies 4\alpha = 8 \implies \alpha = 2$ So, $|\alpha| = 2$.
   • Case 2: Substitute $\beta$ into $\sqrt{3}\alpha + \beta = -2\sqrt{3}$. $\sqrt{3}\alpha + \frac{\alpha - 2}{\sqrt{3}} = -2\sqrt{3}$ Multiplying by $\sqrt{3}$, we get: $3\alpha + (\alpha - 2) = -6 \implies 4\alpha = -4 \implies \alpha = -1$ So, $|\alpha| = |-1| = 1$.

6. Conclusion for (A): The possible values of $|\alpha|$ are 1 and 2. This corresponds to (p) and (q) in Column II. So, (A) $\to$ p, q.

Part (B)

1. Understand the problem: We have a piecewise function $f(x)$ which is differentiable for all $x \in R$. We need to find the possible values of $a$. $f(x) = \begin{cases} -3ax^2 - 2, & x < 1 \\ bx + a^2, & x \ge 1 \end{cases}$

2. Apply the condition of continuity: For $f(x)$ to be differentiable at $x=1$, it must be continuous at $x=1$. This means the left-hand limit (LHL) must equal the right-hand limit (RHL).

   • LHL: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (-3ax^2 - 2) = -3a - 2$
   • RHL: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (bx + a^2) = b + a^2$
   • Equating LHL and RHL: $-3a - 2 = b + a^2$ (Equation 1)

3. Apply the condition of differentiability: The left-hand derivative (LHD) must equal the right-hand derivative (RHD) at $x=1$.

   • The derivative is $f'(x) = \begin{cases} -6ax, & x < 1 \\ b, & x > 1 \end{cases}$
   • LHD: $f'_-(1) = -6a(1) = -6a$
   • RHD: $f'_+(1) = b$
   • Equating LHD and RHD: $b = -6a$ (Equation 2)

4. Solve for a: Substitute Equation 2 into Equation 1. $-3a - 2 = (-6a) + a^2$ $a^2 - 3a + 2 = 0$ Factoring the quadratic equation: $(a-1)(a-2) = 0$ The possible values for $a$ are $a=1$ and $a=2$.

5. Conclusion for (B): The possible values of $a$ are 1 and 2. This corresponds to (p) and (q) in Column II. So, (B) $\to$ p, q.

Part (C)

1. Understand the problem: We are given an equation involving complex cube roots of unity, $\omega$, and an integer $n$. We need to find the possible values of $n$. The equation is: ${\left( {3 - 3\omega + 2{\omega ^2}} \right)^{4n + 3}} + {\left( {2 + 3\omega - 3{\omega ^2}} \right)^{4n + 3}} + {\left( { - 3 + 2\omega + 3{\omega ^2}} \right)^{4n + 3}} = 0$

2. Simplify the terms: Let the bases be $X, Y, Z$.

   • $X = 3 - 3\omega + 2\omega^2$
   • $Y = 2 + 3\omega - 3\omega^2$
   • $Z = -3 + 2\omega + 3\omega^2$ Let's check the relationship between them. Consider $X\omega$: $X\omega = (3 - 3\omega + 2\omega^2)\omega = 3\omega - 3\omega^2 + 2\omega^3 = 3\omega - 3\omega^2 + 2(1) = Y$ Now consider $Y\omega$: $Y\omega = (2 + 3\omega - 3\omega^2)\omega = 2\omega + 3\omega^2 - 3\omega^3 = 2\omega + 3\omega^2 - 3(1) = Z$ So, we have $Y = X\omega$ and $Z = X\omega^2$. (Also, $X+Y+Z = X(1+\omega+\omega^2) = 0$.)

3. Substitute into the equation: The given equation becomes: $X^{4n+3} + (X\omega)^{4n+3} + (X\omega^2)^{4n+3} = 0$ $X^{4n+3} \left[ 1 + \omega^{4n+3} + (\omega^2)^{4n+3} \right] = 0$ Since $X = 3 - 3\omega + 2\omega^2 \neq 0$, the term in the brackets must be zero. $1 + \omega^{4n+3} + \omega^{8n+6} = 0$

4. Simplify the powers of $\omega$: Using $\omega^3=1$.

   • $\omega^{4n+3} = \omega^{4n} \cdot \omega^3 = (\omega^4)^n \cdot 1 = (\omega)^n = \omega^n$
   • $\omega^{8n+6} = \omega^{8n} \cdot \omega^6 = (\omega^8)^n \cdot (\omega^3)^2 = (\omega^2)^n \cdot 1 = \omega^{2n}$ The condition becomes: $1 + \omega^n + \omega^{2n} = 0$.

5. Find the condition on n: The equation $1+z+z^2=0$ holds if and only if $z$ is a complex cube root of unity other than 1. So, $\omega^n$ must be equal to $\omega$ or $\omega^2$. This means $\omega^n \neq 1$. This condition is satisfied if and only if $n$ is not a multiple of 3.

6. Conclusion for (C): From the values in Column II {1, 2, 3, 4, 5}, the possible values of $n$ are those not divisible by 3, which are 1, 2, 4, 5. This corresponds to (p), (q), (s), (t). So, (C) $\to$ p, q, s, t.

Part (D)

1. Understand the problem: The harmonic mean (HM) of two positive real numbers $a$ and $b$ is 4. Also, $a, 5, q, b$ form an arithmetic progression (AP). We need to find the possible values of $|q-a|$.

2. Use the HM formula: $\frac{2ab}{a+b} = 4 \implies ab = 2(a+b)$ (Equation 1)

3. Use the AP properties: Let the common difference be $d$.

   • The terms are $a, a+d, a+2d, a+3d$.
   • Second term: $a+d = 5 \implies a = 5-d$.
   • Third term: $q = a+2d = (5-d)+2d = 5+d$.
   • Fourth term: $b = a+3d = (5-d)+3d = 5+2d$.

4. Solve for d: Substitute the expressions for $a$ and $b$ in terms of $d$ into the HM equation.

   • $a+b = (5-d) + (5+2d) = 10+d$
   • $ab = (5-d)(5+2d) = 25 + 10d - 5d - 2d^2 = 25+5d-2d^2$
   • Substituting into Equation 1: $25+5d-2d^2 = 2(10+d)$ $25+5d-2d^2 = 20+2d$ $2d^2 - 3d - 5 = 0$
   • Factoring the quadratic: $(2d-5)(d+1) = 0$ This gives two possible values for the common difference: $d = \frac{5}{2}$ or $d = -1$.
   • We must check that $a,b,q$ are positive. For $d=5/2$, $a=2.5, q=7.5, b=10$. All positive. For $d=-1$, $a=6, q=4, b=3$. All positive. Both values of $d$ are valid.

5. Calculate $|q-a|$: The value we need is $|q-a|$. $|q-a| = |(5+d) - (5-d)| = |2d|$

   • If $d = \frac{5}{2}$, $|q-a| = |2(\frac{5}{2})| = 5$.
   • If $d = -1$, $|q-a| = |2(-1)| = 2$.

6. Conclusion for (D): The possible values of $|q-a|$ are 2 and 5. This corresponds to (q) and (t) in Column II. So, (D) $\to$ q, t.

Final Matching

• (A) $\to$ p, q
• (B) $\to$ p, q
• (C) $\to$ p, q, s, t
• (D) $\to$ q, t

This corresponds to Option A.