Problem 115 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\mathrm{f} 0\frac{\sin x}{x}\) (B) \(\frac{2}{\pi}<\frac{\sin x}{x}\) (C) \(\frac{\sin x}{x}<1\) (D) \(\frac{\sin x}{x}>1\)

Step-by-Step Solution

Answer

(C) \( \frac{\sin x}{x} < 1 \) is true for \( 0 < x < \frac{\pi}{2} \).

Step 1: Understanding the Problem

We need to determine which inequality involving \( \frac{\sin x}{x} \) is valid for \( 0 < x < \frac{\pi}{2} \). The potential inequalities compare \( \frac{\sin x}{x} \) to fractions like \( \frac{2}{\pi} \) and 1.

Step 2: Analyzing \( \frac{\sin x}{x} \) Behavior

For \( x \) close to 0, the function \( \sin x \approx x \), so \( \frac{\sin x}{x} \approx 1 \). Also, \( \sin x \leq x \) in general, meaning \( \frac{\sin x}{x} \leq 1 \) for \( 0 < x < \frac{\pi}{2} \). This suggests option (C) might be true.

Step 3: Checking Each Option

Since \( \frac{\sin x}{x} \leq 1 \), option (D) is incorrect. To check (A) and (B), recall that \( \frac{\sin x}{x} \) approaches 1 as \( x \) approaches 0. Therefore, \( \frac{2}{\pi} \approx 0.6366 \) is less than 1, making (A) unlikely and (B) possible only between \( 0 < x < \frac{\pi}{2} \), outside of very small \( x \) values.

Step 4: Conclusion

From the above analysis, the inequality \( \frac{\sin x}{x} < 1 \) holds true consistently for \( 0 < x < \frac{\pi}{2} \), confirming that option (C) is correct.

Key Concepts

Sine Function

When we talk about the sine function, it is crucial to understand what it means. The sine function, often denoted as \( \sin x \), is a fundamental trigonometric function.

• It relates an angle \( x \) within a right triangle to the ratio of the side opposite the angle to the hypotenuse.
• The sine function shows periodic behavior, meaning it repeats values in a regular cycle.
• Its graph, the sine wave, oscillates above and below the horizontal axis, ranging from -1 to 1.

For the purposes of understanding trigonometric inequalities, we note that for small values of \( x \) close to zero, \( \sin x \) is approximately equal to \( x \) (expressed as \( \sin x \approx x \)). This is because near zero, the arc length and the linear distance along the x-axis are nearly the same. Understanding this property is key to tackling problems involving inequalities like \( \frac{\sin x}{x} \).

Limits and Continuity

Limits and continuity are fundamental concepts in calculus, underpinning much of the work with functions including trigonometric ones. In this context:

• A limit examines what value a function approaches as the input (in our case \( x \)) approaches a specified point.
• Continuity means that small changes in \( x \) lead to small changes in \( \sin x \). There's no abrupt jumps or gaps.

For \( \lim_{{x \to 0}} \frac{\sin x}{x} \), we find that it tends towards 1 from both the left and right, indicating continuity and consistency in behavior. This knowledge assists in validating inequalities involving \( \frac{\sin x}{x} \).Using limits, we can confirm that for \( 0 < x < \frac{\pi}{2} \), \( \frac{\sin x}{x} \) is indeed less than 1, which is representative of the behavior as \( x \) ascends to \( \frac{\pi}{2} \) while \( \sin x \) reaches its maximum value of 1.

Inequality Proofs

Inequalities are powerful tools in mathematics used to demonstrate relative sizes without providing precise quantities. Inequality proofs allow us to ascertain which values a function may assume:

• We may use direct substitution or known identities to simplify our terms.
• Visualizing the problem through a graph is often helpful, showing \( \sin x \)'s path compared to \( x \).
• The squeeze theorem, which can tighten a certain function between two routes approaching the same value, is useful here.

In the problem, the inequality \( \frac{\sin x}{x} < 1 \) was justified by leveraging the fact that \( \sin x \leq x \) for \( 0 < x < \frac{\pi}{2} \).By proving these inequalities, mathematicians can confirm assumptions about behavior and characteristics, essential in fields as varied as physics and engineering.