Problem 14 - A Complete Resource Book in Mathematics for JEE Main 2017

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

The inequality given is \( \log _{1 / 2} \sin \theta > \log _{1 / 2} \cos \theta \). Note the base of the logarithm here is \( \frac{1}{2} \), which is less than 1. This means the logarithmic function is decreasing, so the inequality will reverse when removing the logarithms.

Step 2: Remove the Logarithms

Since the base of the logarithm is less than 1, the inequality reverses when removing the logs. So, the inequality becomes \( \sin \theta < \cos \theta \). We aim to solve this for \( \theta \) in the interval \([0, 2\pi]\).

Step 3: Transform the Inequality

Rewrite the inequality \( \sin \theta < \cos \theta \) as \( \frac{\sin \theta}{\cos \theta} < 1 \) or \( \tan \theta < 1 \). This form might be easier to manage, especially when considering known values of \( \tan \theta \).

Step 4: Solve \( \tan \theta < 1 \)

Since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), analyze when \( \tan \theta < 1 \). Knowing that \( \tan \theta = 1 \) occurs at \( \theta = \frac{\pi}{4} + n\pi \), investigate intervals between these points within our domain of \([0, 2\pi]\).

Step 5: Determine the Interval Where \( \tan \theta < 1 \)

\( \tan \theta < 1 \) implies angles where \( \frac{\pi}{4} < \theta < \frac{5\pi}{4} \), because between \( \theta = 0 \), \( \tan \theta \) increases to 1 at \( \frac{\pi}{4} \), becomes greater than 1, and then decreases back to 1 at \( \frac{5\pi}{4} \). Thus, in the first period, \( \theta < \frac{\pi}{4} \).

Step 6: Conclude the Interval

In the interval \([0, 2\pi]\), the portion that satisfies \( \tan \theta < 1 \) before it reaches \( \tan \theta = 1 \) again at \( \frac{5\pi}{4} \) is \( [0, \frac{\pi}{4}) \). Therefore, we exclude \( \frac{\pi}{4} \) where \( \tan \theta = 1 \).

Step 7: Compare with Given Options

The interval \( \theta \in [0, \frac{\pi}{4}) \) is equivalent to option (C) \((0, \frac{\pi}{4})\). This matches our derived solution interval.

Key Concepts

trigonometric functions

Understanding trigonometric functions is crucial for tackling problems involving angles, especially when dealing with expressions like \( \sin \theta \) and \( \cos \theta \). These functions are fundamental in describing the relationships between different sides of a right-angled triangle with respect to an angle \( \theta \). Here are a few key points to remember:

• \( \sin \theta \) represents the ratio of the opposite side to the hypotenuse.
• \( \cos \theta \) represents the ratio of the adjacent side to the hypotenuse.
• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), expressing the relationship between sine and cosine.

Knowing these basics helps you manipulate trigonometric identities effectively. In this exercise, we leveraged the property \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) to express the inequality \( \sin \theta < \cos \theta \) as \( \tan \theta < 1 \). This transformation can simplify the problem, helping us identify when one trigonometric function becomes greater than another within a specific interval.

logarithmic properties

Logarithmic properties play a significant role when solving inequalities involving trigonometric functions. In this particular exercise, the logs involved have a base of \( \frac{1}{2} \), which is less than 1. Understanding the behavior of logarithms with such bases is vital:

• For bases less than 1, the logarithmic function is decreasing. This means as the input increases, the output decreases.
• When removing logs from both sides of an inequality, remember to reverse the inequality sign because of the decreasing nature of the function.

By applying these properties, we were able to transform the original logarithmic inequality \( \log _{1 / 2} \sin \theta > \log _{1 / 2} \cos \theta \) into \( \sin \theta < \cos \theta \). This highlights the critical step of reversing the inequality when working with bases less than 1, an essential concept that simplifies the solving process in this context.

inequality solving techniques

Inequality solving techniques often involve transforming and simplifying expressions to make them more manageable. Working with trigonometric inequalities is no exception. The step-by-step approach taken in the original solution exemplifies how to handle these cases:

• Transform the inequality into a form that is easier to analyze, like \( \tan \theta < 1 \).
• Identify key points where the inequality switches direction, such as when \( \tan \theta = 1 \).
• Analyze intervals between these key points to determine where the inequality holds true.

In this example, we began by rewriting \( \sin \theta < \cos \theta \) as \( \frac{\sin \theta}{\cos \theta} < 1 \), leading us to work with \( \tan \theta \). Recognizing that \( \tan \theta = 1 \) at \( \theta = \frac{\pi}{4} \) helped us explore the intervals where this inequality remains valid. By focusing on these specific phases of change, we accurately determined the solution interval \( (0, \frac{\pi}{4}) \), falling in line with choice (C), enhancing our problem-solving toolkit for similar challenges.