Problem 8 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(e^{x 2}<\theta<\pi / 2\), then (A) \(\cos \log \theta<\log \cos \theta\) (B) \(\cos \log \theta>\log \cos \theta\) (C) \(\cos \log \theta \leq \log \cos \theta\) (D) none of these

Step-by-Step Solution

Answer

Option B: \(\cos \log \theta > \log \cos \theta\)

Step 1: Understand the Given Condition

The condition given in the question is \(e^{x^2}<\theta<\frac{\pi}{2}\). This implies that \(\theta\) is between \(e^{x^2}\) and \(\frac{\pi}{2}\). We need to deduce the behavior of the expressions \(\cos \log \theta\) and \(\log \cos \theta\) under this condition.

Step 2: Analyze \(\log \theta\)

Since \(\theta > e^{x^2}\), it follows that \(\log \theta > \log e^{x^2} = x^2\). Given that \(\theta < \frac{\pi}{2}\), \(\log \theta\) is between \(x^2\) and \(\log\frac{\pi}{2}\). Thus, \(\log \theta\) is positive.

Step 3: Evaluate \(\cos \log \theta\)

The function \(\cos \) of a strictly positive number is evaluated. Since \(\log \theta\) is less than \(\frac{\pi}{2}\) but positive, it implies \(\cos \log \theta\) will also be positive and less than 1.

Step 4: Analyze \(\log \cos \theta\)

For \(\theta < \frac{\pi}{2}\) and since \(\theta\) is between 0 and \(\pi/2\), \(\cos \theta > 0\). Therefore, \(\log \cos \theta\) will be negative because \(\cos \theta < 1\) and thus its logarithm is negative.

Step 5: Compare \(\cos \log \theta\) and \(\log \cos \theta\)

From Steps 3 and 4, we deduce that \(\cos \log \theta\) is positive while \(\log \cos \theta\) is negative. Hence, \(\cos \log \theta > \log \cos \theta\).

Step 6: Choose the Correct Answer

Based on the comparison made in the previous step, the inequality \(\cos \log \theta > \log \cos \theta\) holds true. Thus, option (B) is the correct answer.

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where variables occur as exponents. In the given exercise, we encounter the expression \( e^{x^2} \), where the exponent is the square of \( x \). Exponential functions have unique properties: they are always positive and grow rapidly as the exponent increases.
Key points to remember about exponential functions:

• The base (e.g., \( e \)— the natural base, approximately equal to 2.718) is raised to a variable power.
• Exponential growth means that small changes in the variable result in large changes in the function value.

These functions are critical in scientific models due to their large-scale growth patterns. Understanding where exponential functions stand in inequalities involves recognizing their always-positive nature. For instance, \( e^{x^2} \) sets the lower boundary \( \theta > e^{x^2} \) in the initial condition.

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions. These functions help us find the power to which a base number must be raised to obtain a certain value. In our exercise, we encounter expressions like \( \log \theta \) and \( \log \cos \theta \).
Key features of logarithmic functions include:

• Logarithms transform multiplicative relationships into additive ones.
• They grow slower than exponential functions and can handle a wide range of values with ease.

In our solution, it is crucial to understand that \( \log \theta \) must be positive as long as \( \theta > 1 \). However, \( \log \cos \theta \) is negative because \( \cos \theta \) is less than 1 (since \( \theta < \frac{\pi}{2} \)). Recognizing these attributes helps us contrast the inequalities given in the problem.

Trigonometric Functions

Trigonometric functions involve relationships between the angles and sides of triangles. In this example, we focus on the cosine function. The cosine function, \( \cos(\theta) \), is particularly important because it oscillates between -1 and 1.
When examining \( \cos \log \theta \), understanding the nature of \( \log \theta \) is key. Since \( \log \theta \) defines the angle and is positive yet less than \( \frac{\pi}{2} \), \( \cos \log \theta \) remains positive, between 0 and 1.Important aspects of the cosine function:

• It is periodic with period \( 2\pi \).
• The cosine of an angle in the first quadrant \( (0, \frac{\pi}{2}) \) is always positive.

In the given inequality, \( \cos \log \theta > \log \cos \theta \), we see that a positive cosine value is contrasted with a negative log value, explaining why this particular inequality holds true.