Problem 2 - Algebra for JEE ADVANCED

Question

A rod of fixed length \(k\) slides along the coordinate axes. If it meets the axes at \(A(a, 0)\) and \(B(0, b)\), then the minimum value of \(\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}\) is a. 0 b. 8 c. \(k^{2}-4+\frac{4}{k^{2}}\) d. \(k^{2}+4+\frac{4}{k^{2}}\)

Step-by-Step Solution

Answer

The minimum value of the expression is 8, so the answer is b.

Step 1: Understand the Problem

We are given a rod of length \(k\) that meets the coordinate axes at points \(A(a, 0)\) and \(B(0, b)\). We need to find the minimum value of the expression \(\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}\) given the constraint \(a^2 + b^2 = k^2\).

Step 2: Use Constraint to Relate Variables

Given that the rod length is constant, we have \(a^2 + b^2 = k^2\). We can use this relationship to express one variable in terms of the other, for example, \(b^2 = k^2 - a^2\).

Step 3: Substitute and Simplify

Substitute \(b\) in the expression \(\left(b+\frac{1}{b}\right)^{2}\): We have \(b = \sqrt{k^2 - a^2}\) and \(b + \frac{1}{b}\), where \(b eq 0\). Substitute in the expression to get \(\left(\sqrt{k^2-a^2} + \frac{1}{\sqrt{k^2-a^2}}\right)^2\).

Step 4: Use AM-GM Inequality

By the AM-GM inequality, we know \((a + \frac{1}{a}) \geq 2\) and \((b + \frac{1}{b}) \geq 2\). This means \(\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq 2^2 + 2^2 = 8\).

Step 5: Determine if Minimum is Achievable

The minimum value of 8 is achievable when both \((a + \frac{1}{a}) = 2\) and \((b + \frac{1}{b}) = 2\). This implies \(a = 1\) and \(b = 1\) since \(a + \frac{1}{a} = 2 \implies a = 1\). Therefore, the rod's length condition \(a^2 + b^2 = k^2\) becomes \(1^2 + 1^2 = 2 = k^2\). Hence, the minimum value is realized when \(k = \sqrt{2}\). Finally, the minimum value of the expression is 8.

Step 6: Choose the Correct Answer

Comparing the solution to the provided options, 8 matches option b. Thus, the minimum value of the given expression is 8.

Key Concepts

Rod Length Constraint

In the given problem, we have a rod with a fixed length, denoted by the constant variable \(k\). This rod slides along the x and y axes, forming endpoints on the axes, specifically at point \(A(a, 0)\) on the x-axis, and point \(B(0, b)\) on the y-axis. The length of the rod introduces a crucial constraint due to the Pythagorean theorem from geometry:

• The coordinates \((a, 0)\) and \((0, b)\) are connected by the rod, resulting in the equation \(a^2 + b^2 = k^2\).

This constraint allows us to express one variable in terms of the other, making it possible to manipulate and minimize expressions linked to these coordinates. Understanding this relationship is key in addressing optimization problems like the one in this exercise.

AM-GM Inequality

The Arithmetic Mean-Geometric Mean (AM-GM) Inequality is a pivotal inequality in mathematics, and it's highly useful for optimization and minimization tasks. It states that for any non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. For any non-zero real number \(x\), this means:

• \((x + \frac{1}{x}) \geq 2\), with equality when \(x = 1\).

In our problem, we apply this inequality twice, for both expressions \((a + \frac{1}{a})\) and \((b + \frac{1}{b})\) since we want to minimize the sum of their squares. By ensuring that each term is at its minimum, achieved when both \(a = 1\) and \(b = 1\), we can achieve the lowest possible value of the original expression. The AM-GM Inequality thus simplifies and guides us towards finding the minimum value.

Minimization Problem

In mathematics, solving a minimization problem involves finding a value that makes a given function reach its lowest point. Here, our focus is on determining the minimum value of the expression:

• \(\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}\).

Given the constraint \(a^2 + b^2 = k^2\), and applying the AM-GM inequality, we deduced that the combined squares cannot go below 8. This is found by setting both \(a\) and \(b\) to 1. Thus, the minimization problem elegantly converges to establishing that the least value is 8. Thorough understanding of both the constraint and properties of the expressions helped in reaching this conclusion efficiently.

Coordinate Axes Intersection

The role of coordinate axes intersection in this problem gives a geometric visualization to the given equation. The rod, intersecting the x-axis and y-axis at points \((a, 0)\) and \((0, b)\) respectively, serves as a visual cue for the application of algebraic principles. These intersections essentially translate a theoretical algebraic concept into a tangible spatial scenario:

• The points \((a,0)\) and \((0,b)\) highlight the maximum extent on x and y, respectively, within which the rod can remain under a given constraint.
• From understanding this intersection, it’s easier to align geometric concepts (like the Pythagorean theorem) with algebraic constraints.

This intersection not only serves as a concrete representation but also aids in the comprehensive understanding of constraints and optimizations involved in coordinate geometry.