Problem 88 - A Complete Resource Book in Mathematics for JEE Main

Question

If the equation \(x-\sin x=k\) has a unique root in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), then the range of values of \(k\) are (A) \(\left(1-\frac{\pi}{2}, \frac{\pi}{2}-1\right)\) (B) \(\left[1-\frac{\pi}{2}, \frac{\pi}{2}-1\right]\) (C) \(\left[0, \frac{\pi}{2}+1\right]\) (D) None of these

Step-by-Step Solution

Answer

Option (A) \((1-\frac{\pi}{2}, \frac{\pi}{2}-1)\) is correct.

Step 1: Define the function

First, we define the function \( f(x) = x - \sin x \). Now, we need to analyze \( f(x) \) in the interval \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).

Step 2: Determine the function endpoints

Calculate the values of the function \( f(x) \) at the endpoints of the interval: - For \( x = -\frac{\pi}{2} \), \( f\left(-\frac{\pi}{2}\right) = -\frac{\pi}{2} - (-1) = 1 - \frac{\pi}{2} \). - For \( x = \frac{\pi}{2} \), \( f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} - 1 \).

Step 3: Analyze the behavior of the function

The function \( \sin x \) is continuous and differentiable over the interval, so \( f(x) = x - \sin x \) is continuous as well. Since it is a combination of continuous functions, \( f(x) \) is continuous over \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).

Step 4: Evaluate the derivative

Calculate the derivative \( f'(x) = 1 - \cos x \). Since \( \cos x \) has a range from 0 to 1 within \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \), \( f'(x) \geq 0 \). This derivative suggests that the function is non-decreasing over the interval.

Step 5: Determine the range of \( k \)

For the equation \( x - \sin x = k \) to have a unique root, \( k \) must satisfy \( 1 - \frac{\pi}{2} < k < \frac{\pi}{2} - 1 \) because these are the values of \( f(x) \) at \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \), respectively.

Step 6: Conclusion

The range of values for \( k \) ensuring a unique root of \( x - \sin x = k \) in the interval is \( \left( 1 - \frac{\pi}{2}, \frac{\pi}{2} - 1 \right) \). Therefore, the correct option is (A).

Key Concepts

Unique Root

When dealing with functions, one concept that comes up often is the idea of a "unique root." A unique root is essentially a solution to an equation that is the only one within a given interval. For example, if we have the equation \(x - \sin x = k\) and wish to find its unique root, it means we want to identify the value of \(x\) which makes the equation hold true, and there's no other \(x\) in that interval providing the same satisfaction.

In the exercise, this unique root is found within the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\). Since the function \(f(x)\) is continuous and differentiable, and the derivative \(f'(x) = 1 - \cos x \geq 0\) ensures algebraically that the slope of \(f(x)\) does not decrease, it suggests the function could either rise or stay constant but never drop. Thus, any horizontal line \(k\) will intersect the graph of \(f(x)\) just once within this range, confirming a unique solution.

Trigonometric Functions

Trigonometric functions play a central role in our problem. The function given is a combination of basic trigonometric function \(\sin x\) and linear function \(x\). Understanding how these trigonometric components work helps in analyzing that function.

The function \(\sin x\) oscillates between -1 and 1, so within \([-\frac{\pi}{2}, \frac{\pi}{2}]\), it smoothly transitions from -1 at \(-\frac{\pi}{2}\) to 1 at \(\frac{\pi}{2}\). This key property assists, as it shows that over the given interval, \(\sin x\) is continuous and differentiable. Knowing this feature allows us to properly understand and compute the derivative of \(f(x) = x - \sin x\), indicating the continuation and predictable nature of the interval.

Interval Analysis

Interval analysis is about inspecting how a function behaves within a certain range, and it was crucial for this problem. By evaluating \(f(x) = x - \sin x\) at both the endpoints \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), we discover the extremal values that can be achieved by \(f(x)\), which are \(1 - \frac{\pi}{2}\) and \(\frac{\pi}{2} - 1\) respectively.

On examining \(f'(x)\), the non-decreasing nature is concluded based on the derivative being non-negative overall. This information allows for the decision that every value \(k\) within this output interval \((1 - \frac{\pi}{2}, \frac{\pi}{2} - 1)\) could be possible for the definition of any unique root. Ultimately, by considering both endpoint behavior and derivative signs, we determine \(k\)'s correct range, ensuring only one intersection point for each \(k\) and proving a unique solution exists.