Match the following:
Column I
(A) In R², if the magnitude of the projection vector of the vector αi

Q: Wait, why do we reject the negative root in the vector part?

We don't! The absolute value $|4\alpha - 2| = 6$ gives us two cases: $4\alpha - 2 = 6$ and $4\alpha - 2 = -6$. Both are valid, leading to $\alpha = 2$ and $\alpha = -1$. Always solve both branches of the modulus!

Q: Why can't we just use energy conservation?

This isn't physics! But in math, we use 'Conservation of Logic.' For the piecewise function, the 'energy' (value) must be conserved at the junction $x=1$ (continuity), and the 'momentum' (slope) must be conserved (differentiability).

Q: How do I handle the complex power $4n+3$?

Think of it as $4n+3 \equiv n \pmod{3}$. Since $\omega^3 = 1$, any power of $\omega$ is just $\omega^0, \omega^1,$ or $\omega^2$. It's all about the remainder!

Question

Match the following:

List-I:

P. In $R^2$, if the magnitude of the projection vector of the vector $\alpha \hat{i} + \beta \hat{j}$ on $\sqrt{3}\hat{i} + \hat{j}$ is $\sqrt{3}$ and if $\alpha = 2 + \sqrt{3}\beta$, then possible value of $|\alpha|$ is/are

Q. Let $a$ and $b$ be real numbers such that the function $f(x) = \begin{cases} -3ax^2 - 2, & x < 1 \ bx + a^2, & x \ge 1 \end{cases}$ if differentiable for all $x \in R$. Then possible value of $a$ is (are)

R. Let $\omega 
eq 1$ be a complex cube root of unity. If $(3 - 3\omega + 2\omega^2)^{4n+3} + (2 + 3\omega - 3\omega^2)^{4n+3} + (-3 + 2\omega + 3\omega^2)^{4n+3} = 0$, then possible value (s) of $n$ is (are)

S. Let the harmonic mean of two positive real numbers $a$ and $b$ be 4. If $q$ is a positive real number such that $a, 5, q, b$ is an arithmetic progression, then the value(s) of $|q - a|$ is (are)

List-II:

1. 1

2. 2

3. 3

4. 4

5. 5

Sigma Percentile · Accepted Answer

### The JEE Marathon: A Journey Through Four Mathematical Landscapes

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a four-part expedition. The JEE Advanced is not merely about speed; it is about the elegance of your approach.

We have four distinct terrains to cross: Vectors, Calculus, Complex Numbers, and Progressions. Let us take a breath and conquer them one by one.

### Part 1: The Geometry of Shadows (Vectors)

Imagine you are standing in a 2D plane. You have a vector $\vec{v} = \alpha \hat{i} + \beta \hat{j}$ and you are projecting it onto $\vec{u} = \sqrt{3}\hat{i} + \hat{j}$.

The projection is essentially the 'shadow' of $\vec{v}$ on $\vec{u}$. The formula is our compass:

$$	ext{Projection} = \frac{|\vec{v} \cdot \vec{u}|}{|\vec{u}|}$$

Given the magnitude is $\sqrt{3}$, we set up our equation:

$$\frac{|\sqrt{3}\alpha + \beta|}{2} = \sqrt{3}$$

This simplifies to $|\sqrt{3}\alpha + \beta| = 2\sqrt{3}$. With the constraint $\alpha = 2 + \sqrt{3}\beta$, we substitute $\beta = \frac{\alpha - 2}{\sqrt{3}}$.

Suddenly, the complexity collapses into a simple modulus equation:

$$|4\alpha - 2| = 6$$

Solving this gives us $\alpha = 2$ or $\alpha = -1$. Thus, $|\alpha|$ is either $1$ or $2$. We have successfully navigated the first terrain.

### Part 2: The Calculus Bridge (Continuity and Differentiability)

Now, we step into the realm of functions. We have a piecewise function $f(x)$ that must be differentiable everywhere.

The trap here is thinking differentiability is enough. Differentiability is a luxury that requires the foundation of continuity. At the junction $x=1$, the function must be 'glued' together.

Equating the Left Hand Limit and Right Hand Limit, we get:

$$-3a - 2 = b + a^2$$

Then, we ensure the 'smoothness' by equating the derivatives:

$$-6a = b$$

Substituting the second into the first, we arrive at the quadratic $a^2 - 3a + 2 = 0$. The roots **$a=1$** and **$a=2$** are our keys to the gate. We have bridged the gap.

### Part 3: The Complex Dance (Roots of Unity)

This is where the beauty of symmetry shines. We are given a massive expression involving $\omega$. Let $A = 3 - 3\omega + 2\omega^2$, $B = 2 + 3\omega - 3\omega^2$, and $C = -3 + 2\omega + 3\omega^2$.

If you sum them, $A+B+C = 0$. Even more elegantly, $B = A\omega$ and $C = A\omega^2$.

The equation becomes:

$$A^k(1 + \omega^k + \omega^{2k}) = 0$$

For this to hold, $1 + \omega^k + \omega^{2k}$ must be zero, which happens only when $k$ is not a multiple of $3$. Since $k = 4n+3$, we find $n$ cannot be a multiple of $3$. The elegance of complex numbers never ceases to amaze.

### Part 4: The Progression Puzzle (Harmonic and Arithmetic)

Finally, we arrive at the harmonic mean. Given $a, 5, q, b$ in AP, we define the common difference $d = 5-a$.

This allows us to express $q = 10-a$ and $b = 15-2a$. The Harmonic Mean formula becomes our final battleground:

$$\frac{2ab}{a+b} = 4$$

Substituting our expressions, we solve the quadratic equation:

$$2a^2 - 17a + 30 = 0$$

This yields $a=2.5$ or $a=6$. Calculating $|q-a|$ for both cases gives us **$5$** and **$2$**.

We have traversed all four landscapes. Remember, the JEE is not about memorizing formulas; it is about seeing the patterns. You have done well today.

Animated Solution for Mathematics - Vector Algebra: Match the following:

List-I

List-II

Select Matching Pairs:

Visualized Solution

Multi-Concept Challenge Overview

Part A: Vector Projection Setup

Part A: Solving for $∣ α ∣$

Part B: Continuity Condition

Part B: Differentiability and Solving

Part C: Complex Roots Insight

Part C: Power Condition for Zero

Part D: Arithmetic Progression Relations

Part D: Harmonic Mean Substitution

Part D: Final Calculation

Final Matching and Conclusion

The Sigma Insight: Scalar (Dot) Product

The JEE Marathon

Part 1

Part 2

Part 3

Part 4

Similar Questions

Let the vectors $a, b, c$ be such that $∣ a ∣ = 2, ∣ b ∣ = 4$ and $∣ c ∣ = 4$ . If the projection of $b$ on $a$ is equal to the projection of $c$ on $a$ and $b$ is perpendicular to $c$ , then the value of $∣ a + b - c ∣$ is

Let $a = 2 \hat{i} + \hat{j} - \hat{k}$ and $b = \hat{i} + 2 \hat{j} + \hat{k}$ be two vectors. Consider a vector $c = α a + β b, α, β \in R$ . If the projection of $c$ on the vector $(a + b)$ is $32$ , then the minimum value of $(c - (a \times b)) \cdot c$ equals

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Let $a$ and $b$ be two unit vectors such that the angle between them is $\frac{π}{3}$ If $λ a + 2 b$ and $3 a - λ b$ are perpendicular to each other, then the number of values of $λ$ in $[- 1, 3]$ is:

Let $a = 2 \hat{i} + λ_{1} \hat{j} + 3 \hat{k}$ , $b = 4 \hat{i} + (3 - λ_{2}) \hat{j} + 6 \hat{k}$ and $c = 3 \hat{i} + 6 \hat{j} + (λ_{3} - 1) \hat{k}$ be three vectors such that $b = 2 a$ and $a$ is perpendicular to $c$ . Then a possible value of $(λ_{1}, λ_{2}, λ_{3})$ is :-

For $p > 0$ , a vector $v_{2} = 2 \hat{i} + (p + 1) \hat{j}$ is obtained by rotating the vector $v_{1} = 3 p \hat{i} + \hat{j}$ by an angle $θ$ about origin in counter clockwise direction. If $tan θ = \frac{( α 3 - 2 )}{( 4 3 + 3 )}$ , then the value of $α$ is equal to

Let $a$ and $b$ be the vectors of the same magnitude such that $\frac{∣ a + b ∣ + ∣ a - b ∣}{∣ a + b ∣ - ∣ a - b ∣} = 2 + 1$ . Then $\frac{∣ a + b ∣ ^{2}}{∣ a ∣ ^{2}}$ is :

Let $u, v, w$ be such that $∣ u ∣ = 1, ∣ v ∣ = 2, ∣ w ∣ = 3$ . If the projection $v$ along $u$ is equal to that of $w$ along $u$ and $v, w$ are perpendicular to each other then $∣ u - v + w ∣$ equals