Problem 51 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(x, y \in[0,10]\), then the number of solutions \((x, y)\) of the inequation \(3^{\sec ^{2} x-1} \sqrt{9 y^{2}-6 y+2} \leq 1\) is (A) 2 (B) 4 (C) 6 (D) infinite

Step-by-Step Solution

Answer

The number of solutions is 1.

Step 1: Understanding the Inequality

We are given the inequality \(3^{\sec^2 x - 1} \sqrt{9y^2 - 6y + 2} \leq 1\). For this inequality to hold, both conditions \(3^{\sec^2 x - 1} \leq 1\) and \(\sqrt{9y^2 - 6y + 2} \leq 1\) must be satisfied.

Step 2: Solving for \(x\)

Consider the first condition: \(3^{\sec^2 x - 1} \leq 1\). This implies \(\sec^2 x - 1 \leq 0\) since any positive base like 3 raised to a negative exponent yields a value less than or equal to 1. Thus, we have \(\sec^2 x \leq 1\), or the equivalent trigonometric identity, \(\sec^2 x - 1 = \tan^2 x \leq 0\), implies \(\tan x = 0\). The solutions in \([0, 10]\) are \(x = 0\), with no additional values due to the limited range and the periodicity of \(\tan x\).

Step 3: Solving for \(y\)

Now, consider the second condition: \(\sqrt{9y^2 - 6y + 2} \leq 1\). Taking the square of both sides, we obtain \(9y^2 - 6y + 2 \leq 1\). Simplifying, we get \(9y^2 - 6y + 1 \leq 0\). This is a quadratic inequality, which factors as \((3y - 1)^2 \leq 0\). Since this is a perfect square, it equals zero at \(y = \frac{1}{3}\). Thus, \(y = \frac{1}{3}\) is the only solution for \(y\) in the range \([0, 10]\).

Step 4: Determine the Number of Solutions

Combining the results from Step 2 and Step 3, the solution pair that satisfies both conditions is \((x, y) = (0, \frac{1}{3})\). This results in exactly one solution within the given constraints.

Key Concepts

Quadratic Inequalities

Quadratic inequalities can be challenging at first glance, but understanding them can lead to mastering many math problems. A quadratic inequality involves finding the range of values that satisfy an inequality condition, typically in the form of \(ax^2 + bx + c \leq 0\) or \(ax^2 + bx + c > 0\).
In our example involving \(y\), we look at the inequality \((3y - 1)^2 \leq 0\). This inequality is a perfect square, meaning the expression inside the parentheses equals zero only when \(y = \frac{1}{3}\).
This makes quadratic inequalities straightforward when dealing with perfect squares, because they have a definitive solution rather than a range.
To solve these, follow these steps:

• Factor the quadratic expression, if possible.
• Analyze the factors to find when the product is zero.
• Determine the intervals of the inequality based on this analysis.

In some cases, like our problem, the only solution falls on a single point. Knowledge of quadratic inequalities helps understand the limitations and potential solutions available within the given domain.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. They help simplify equations and solve problems involving trigonometric functions.
In the given exercise, we use the identity \(\sec^2 x - 1 = \tan^2 x\), which simplifies the inequality \(\sec^2 x \leq 1\). This translates into \(\tan x = 0\).
The solution of \(\tan x = 0\) is straightforward within a given interval because it determines specific values of \(x\). Known solutions occur at multiples of \(\pi\), specifically:

• \(x = n\pi\), where \(n\) is an integer.

Within the range \([0, 10]\), and accounting for periodicity, we find \(x = 0\) as the solution.
These identities not only simplify the calculation but also emphasize the relationships between different trigonometric functions. Understanding these identities allows interpreting and solving trigonometric inequalities effectively.

Polar Coordinates

Polar coordinates offer a different yet powerful way to describe a point's location compared to the Cartesian coordinate system.
In polar coordinates, a point is described by the distance from the origin, \(r\), and the angle \(\theta\) from the positive x-axis. This system is particularly useful in contexts involving circular or rotational symmetry.
While polar coordinates weren't directly used to solve the textbook problem, understanding this concept can translate skills learned here to similar problems utilizing different coordinate systems. Points on a plane can be transformed between Cartesian and polar coordinates effectively, offering versatility in geometry:

• From Cartesian to polar: - \(r = \sqrt{x^2 + y^2}\)
• \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\)
• From polar to Cartesian: - \(x = r\cos\theta\)
• \(y = r\sin\theta\)

Mastering both systems gives a comprehensive toolset for addressing various types of problems. Recognizing how trigonometric functions relate to polar coordinates enhances both problem-solving skills and spatial understanding.