Match the Statements/Expressions in Column I with the Statements / Expressions in Column II and indi

Q: Wait, why can't we just use differentiation for Part A? Is it wrong?

Not at all! Differentiation is perfectly valid and will lead you to the correct answer. However, in competitive exams like JEE, time is your most valuable resource. Using the AM-GM inequality allows you to find the minimum value in seconds by recognizing the reciprocal structure of the terms, which is much faster than finding the derivative and setting it to zero.

Q: Why does $(AB)^T = -AB$ imply that $k$ must be an odd integer?

Great question! We found that $(AB)^T = -AB$. The problem states $(AB)^T = (-1)^k AB$. By comparing these two expressions, we get $-AB = (-1)^k AB$. This implies that $(-1)^k = -1$. Since $(-1)^k$ is $-1$ only when $k$ is an odd integer, $k$ must be odd.

Q: How do we handle the log inequality $1 < 2^{-k + \log_2 3} < 2$?

The key is to take the logarithm base 2 of all parts of the inequality. Since the base 2 is greater than 1, the inequality sign remains the same. So, $\log_2(1) < \log_2(2^{-k + \log_2 3}) < \log_2(2)$, which simplifies to $0 < -k + \log_2 3 < 1$. From here, you can easily isolate $k$.

Q: Why is the expression in Part D always an even integer?

After converting the equation to $\cos(\frac{\pi}{2} - \theta) = \cos \phi$, we get the general solution $\frac{\pi}{2} - \theta = 2n\pi \pm \phi$. Rearranging this gives $\theta \pm \phi - \frac{\pi}{2} = -2n\pi$. When you divide this by $\pi$, you get $-2n$, which is always an even integer for any integer $n$.

Question

Match the Statements/Expressions in Column I with the Statements / Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the $4 	imes 4$ matrix given in the ORS.

List-I:

P. The minimum value of $\frac{x^2 + 2x + 4}{x + 2}$ is

Q. Let A and B be $3 	imes 3$ matrices of real numbers, where A is symmetric, B is skew-symmetric, and $(A+B)(A-B) = (A-B)(A+B)$. If $(AB)^t = (-1)^k AB$, where $(AB)^t$ is the transpose of the matrix AB, then the possible values of $k$ are

R. Let $a = \log_3 \log_3 2$. An integer $k$ satisfying $1 < 2^{-k + 3^{-a}} < 2$, must be less than

S. If $\sin 	heta = \cos \phi$, then the possible values of $\frac{1}{\pi} \left( 	heta \pm \phi - \frac{\pi}{2} ight)$ are

List-II:

1. 0

2. 1

3. 2

4. 3

Sigma Percentile · Accepted Answer

### The Grand Symphony of Concepts

Imagine standing before a mountain of problems, where each peak represents a different domain of mathematics. Today, we are not just climbing one; we are traversing the entire range.

This matching problem is a masterclass in synthesis, pulling together calculus, matrix algebra, logarithms, and trigonometry. Let us embark on this journey, one step at a time.

### Part A: The Elegance of Optimization

We begin with a rational function: 
$$f(x) = \frac{x^2 + 2x + 4}{x + 2}$$

The instinct might be to reach for the quotient rule of differentiation. While that path is correct, it is a long and winding road. Instead, let us look for the hidden symmetry.

By rewriting the numerator as $(x+2)^2 - 2(x+2) + 4$, we can split the fraction into:
$$f(x) = (x+2) + \frac{4}{x+2} - 2$$

Now, the problem transforms. We are looking for the minimum of a sum of two terms whose product is constant. This is the perfect stage for the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

The sum $(x+2) + \frac{4}{x+2}$ is always greater than or equal to:
$$2\sqrt{(x+2) \cdot \frac{4}{x+2}} = 4$$

Subtracting the constant 2, we find the minimum value is $4 - 2 = 2$. It is a beautiful example of how algebraic insight can bypass brute-force calculus.

### Part B: The Dance of Matrices

Next, we step into the world of linear algebra. We are given two matrices, $A$ and $B$, where $A$ is symmetric ($A^T = A$) and $B$ is skew-symmetric ($B^T = -B$).

The condition $(A+B)(A-B) = (A-B)(A+B)$ is our compass. Expanding both sides, we get:
$$A^2 - AB + BA - B^2 = A^2 + AB - BA - B^2$$

The quadratic terms $A^2$ and $B^2$ vanish, leaving us with $2BA = 2AB$, or simply $AB = BA$. This tells us the matrices commute.

Now, consider $(AB)^T$. By the reversal law of transposes, this is $B^T A^T$. Substituting our known properties, we get $(-B)A = -BA$.

Since $AB = BA$, this is $-AB$. The problem states $(AB)^T = (-1)^k AB$. Comparing these, $(-1)^k = -1$, which forces $k$ to be an odd integer. Thus, $k$ can be 1 or 3.

### Part C: The Logarithmic Labyrinth

Now, we face the logarithmic challenge. Given $a = \log_3 \log_3 2$, we find $3^a = \log_3 2$. We need $3^{-a}$, which is the reciprocal:
$$3^{-a} = \frac{1}{\log_3 2} = \log_2 3$$

Our inequality is $1 < 2^{-k + \log_2 3} < 2$. Taking the logarithm base 2 across the inequality, we get:
$$\log_2 1 < -k + \log_2 3 < \log_2 2$$

This simplifies to $0 < -k + \log_2 3 < 1$. Rearranging for $k$, we find:
$$\log_2 3 - 1 < k < \log_2 3$$

Since $\log_2 3 \approx 1.58$, we have $0.58 < k < 1.58$. The only integer in this range is $k = 1$. Since $1 < 2$ and $1 < 3$, $k$ matches with 2 and 3.

### Part D: The Harmony of Trigonometry

Finally, we arrive at the trigonometric equation $\sin 	heta = \cos \phi$. To solve this, we must speak the same language.

Using the complementary angle identity, we write $\cos(\frac{\pi}{2} - 	heta) = \cos \phi$. The general solution for $\cos x = \cos y$ is $x = 2n\pi \pm y$.

Applying this, we get:
$$\frac{\pi}{2} - 	heta = 2n\pi \pm \phi$$

Rearranging to match the expression in the question, we have $	heta \pm \phi - \frac{\pi}{2} = -2n\pi$. Dividing by $\pi$, we get $-2n$, which is always an even integer. Thus, the expression matches with 0 and 2.

We have navigated the peaks and valleys of this problem, and the view from the top is clear. **Mathematics is not just about solving; it is about seeing the connections.**

Animated Solution for Mathematics - Matrices and Determinants: Match the Statements/Expressions in Column I with the Statements / Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Matching Problem Overview

Part A: Function Setup

Part A: Algebraic Manipulation

Part A: AM-GM Inequality

Part B: Matrix Properties

Part B: Matrix Commutativity

Part B: Finding $k$

Part C: Logarithmic Simplification

Part C: Solving the Inequality

Part C: Finding Integer $k$

Part D: Trigonometric Equation

Part D: General Solution

Part D: Final Match

The Sigma Insight: Algebraic Operations on Matrices

The Grand Symphony of Concepts

Part A

Part B

Part C

Part D

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