Problem 14 - A Complete Resource Book in Mathematics for JEE Main

Question

The solution of the inequality \(\log _{1 / 2} \sin \theta>\log _{1 / 2} \cos \theta\) in \([0,2 p]\) is (A) \(\left(0, \frac{\pi}{2}\right)\) (B) \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\) (C) \(\left(0, \frac{\pi}{4}\right)\) (D) none of these

Step-by-Step Solution

Answer

The solution is (C) \((0, \frac{\pi}{4})\).

Step 1: Understand the Inequality Base

The given inequality is \(\log _{1 / 2} \sin \theta > \log _{1 / 2} \cos \theta\). Since the base of the logarithm \( \frac{1}{2} \) is less than 1, the logarithm is a decreasing function. This means that \( \sin \theta < \cos \theta \) will satisfy the given inequality.

Step 2: Identify the Trigonometric Interval

We need to determine the interval where \(\sin \theta < \cos \theta\). This occurs when the angle \(\theta\) is in the first quadrant, specifically when \( \theta \in \left(0, \frac{\pi}{4}\right) \) inside the interval \([0, \pi]\), since \(\sin \theta\) increases from 0 to \(\frac{1}{\sqrt{2}}\) and \(\cos \theta\) decreases from 1 to \(\frac{1}{\sqrt{2}}\).

Step 3: Verify Solutions within the Given Range

Since the task is to find the solution in the interval \([0, 2\pi]\), we need to analyze whether the condition \(\sin \theta < \cos \theta\) persists beyond \(\pi\). However, in the interval \(\left(0, \frac{\pi}{4}\right)\), it satisfies the condition as mentioned above for the principal range.

Key Concepts

Logarithmic Functions

Logarithmic functions are mathematical expressions that help us understand how quantities scale together. Imagine you have a number and you want to know the power that you need to raise a base to, in order to get that number. That's exactly what a logarithmic function does!

• The notation for a logarithm is typically \(\log_b(x)\), where \(b\) is the base and \(x\) is the number you're taking the log of.
• In this problem, we use base \(\frac{1}{2}\), which is less than 1, making the logarithm a decreasing function.

This is crucial in solving inequalities because the direction of the inequality changes when we deal with bases less than 1. In simpler terms, if \(\log_{1/2}(a) > \log_{1/2}(b)\), it actually means that \(a < b\). This behavior flips the comparison and provides a unique way to solve the inequality!

Trigonometric Functions

Trigonometric functions relate angles of triangles to the lengths of the triangle's sides. Two primary trigonometric functions we encounter are sine \(\sin\) and cosine \(\cos\).

• \(\sin \theta\): This function represents the ratio of the length of the side opposite the angle \(\theta\) to the hypotenuse in a right triangle.
• \(\cos \theta\): This represents the ratio of the adjacent side to the hypotenuse for the angle \(\theta\).

The behavior of these functions over specific intervals helps us solve trigonometric inequalities. For example:

• In the first quadrant (\(0\) to \(\frac{\pi}{2}\)), both sine and cosine start positive but behave oppositely: \(\sin\) starts small and grows, whereas \(\cos\) starts large and decreases.
• This means that within \(\left(0, \frac{\pi}{4}\right)\), \(\sin \theta < \cos \theta\).

Recognizing these relationships allows us to compare and solve inequalities involving trigonometric expressions.

Inequalities in Mathematics

Mathematical inequalities tell us the relative size or order of two values. They express the notion that one value is less than, greater than, or not equal to another.

• For example, the inequality \(\log _{1 / 2} \sin \theta > \log _{1 / 2} \cos \theta\) relates two trigonometric expressions within a given range of \(\theta\).
• When we deal with logarithms that have a base less than 1, the inequality direction flips, leading to \(\sin \theta < \cos \theta\).

Solving inequalities involves not just finding when the numerical conditions are met, but also understanding within which intervals these conditions are true. For trigonometric inequalities, this means determining specific angle ranges. In this problem, the solution is within \(\left(0, \frac{\pi}{4}\right)\). Learning to navigate these mathematical relationships is critical for understanding complex concepts in algebra and calculus.