Problem 88 - A Complete Resource Book in Mathematics for JEE Main 2017

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

The range of values of \( k \) is (A) \( \left(1-\frac{\pi}{2}, \frac{\pi}{2}-1\right) \).

Step 1: Understand the Function

Consider the function \( f(x) = x - \sin x \). We need to determine the range of this function over the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).

Step 2: Examine Function Behavior

The function \( f(x) = x - \sin x \) is continuous and differentiable over the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). At the endpoints, \( f\left( -\frac{\pi}{2} \right) = -\frac{\pi}{2} + 1 \) and \( f\left( \frac{\pi}{2} \right) = \frac{\pi}{2} - 1 \).

Step 3: Calculate the Derivative

The derivative of \( f(x) \) is \( f'(x) = 1 - \cos x \). This derivative tells us the rate at which \( f(x) \) changes. \( \cos x \) ranges between \(-1\) and \(1\), so \( f'(x) \) ranges between \(0\) and \(2\).

Step 4: Analyze Monotonicity

Since \( f'(x) \geq 0 \) over \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), \( f(x) \) is non-decreasing across the interval. Therefore, the minimum value is at \( x = -\frac{\pi}{2} \) and the maximum value is at \( x = \frac{\pi}{2} \).

Step 5: Determine Function Range

From Step 2, we calculated \( f\left( -\frac{\pi}{2} \right) = 1 - \frac{\pi}{2} \) and \( f\left( \frac{\pi}{2} \right) = \frac{\pi}{2} - 1 \). These are the endpoint values of \( f(x) \).

Step 6: Match Range with Options

The range of values \( k \) for which the function \( f(x) = k \) has a unique root will be in the open interval \( \left(1-\frac{\pi}{2}, \frac{\pi}{2}-1\right) \). Thus, A \( \left(1-\frac{\pi}{2}, \frac{\pi}{2}-1\right) \) is the correct answer.

Key Concepts

Continuity and Differentiability

In mathematics, a function is considered continuous when there are no jumps or breaks in its graph over given intervals. Differentiability means the function has a well-defined derivative at every point in its domain. These concepts are important when analyzing trigonometric equations to determine their behavior. The function in our exercise, \( f(x) = x - \sin x \), is both continuous and differentiable over the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
Understanding continuity helps us ensure that the function doesn't have any sudden jumps. This means we can achieve smoothness in drawing the graph of \( f(x) \). Differentiability furthers our understanding by allowing us to analyze how \( f(x) \) changes at any given point.
The derivative of the function, \( f'(x) = 1 - \cos x \), allows us to explore how the small changes in \( x \) affect the function's output. This smooth change is vital when looking for unique roots of the equation as it ensures a predictable behavior of \( f(x) \) over an interval without abrupt changes.

Monotonic Functions

Monotonic functions are functions that are either entirely non-increasing or non-decreasing over a given interval. In our exercise, we found the derivative of \( f(x) = x - \sin x \) to be \( f'(x) = 1 - \cos x \). Since the cosine function, \( \cos x \), can vary from \(-1\) to \(1\), the derivative \( f'(x) \) ranges from \(0\) to \(2\).
The fact that \( f'(x) \geq 0 \) on the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\) indicates that the function \( f(x) \) is non-decreasing. This monotonic behavior is critical because it guarantees that within this interval, the function will not decrease. It allows us to confidently determine that roots, if they exist, are unique within this range.
Such information is crucial when analyzing and solving trigonometric equations because it simplifies the problem of finding a unique value that satisfies the equation. Since the significant observation that the function is always increasing or stable across the interval helps in predicting its crossing behavior at particular values.

Range of Trigonometric Functions

The range of a function is the set of possible output values it can produce. When dealing with trigonometric functions, the range can severely affect how equations behave and where roots are located. In the given exercise, we have been tasked with finding the range of \( f(x) = x - \sin x \) over \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
The sine function, \( \sin x \), oscillates between \(-1\) and \(1\). This oscillation impacts any equations involving sine, including \( x - \sin x \). At the endpoints, \( f(-\frac{\pi}{2}) = 1 - \frac{\pi}{2} \) and \( f(\frac{\pi}{2}) = \frac{\pi}{2} - 1 \), these values define the boundaries of the function's range over this interval.
Thus, to find values \( k \) for which \( x - \sin x = k \) has a unique solution, we find that \( k \) must lie within \( (1-\frac{\pi}{2}, \frac{\pi}{2}-1) \). With this conclusion, understanding the range helps clarify which values yield unique roots, making problem-solving much more straightforward when dealing with trigonometric equations.