Problem 63 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Does the equation \(\sin x-x+1=0\) has a root?

Step-by-Step Solution

Answer

By analyzing the function \(f(x) = \sin x - x + 1\), we found that the function is continuous and changes its sign from positive at \(x = 0\) to negative at \(x = \pi\). According to the Intermediate Value Theorem, there must be a root for the equation \(\sin x - x + 1 = 0\) within the interval \([0, \pi]\). Therefore, the equation has a root.

Step 1: Identify the function

First, let's rewrite the given equation as a single function: \(f(x) = \sin x - x + 1\). Now, we want to find a change of sign in the values of \(f(x)\) for different values of \(x\).

Step 2: Analyze the function at x = 0

We start by analyzing the value of \(f(x)\) at \(x = 0\). We have: \[f(0) = \sin 0 - 0 + 1 = 0 + 0 + 1 = 1\] At \(x = 0\), the function is positive (specifically, \(f(0) = 1\)).

Step 3: Find a suitable interval where the function is negative

Since we want to find an interval where the function changes its sign, let's try several values for \(x > 0\). We know that \(\sin x\) is always between -1 and 1, i.e., \(-1 \leq \sin x \leq 1\). Let's consider \(x = \pi\). We know that \(\sin \pi = 0\), so \[f(\pi) = \sin \pi - \pi + 1 = -\pi + 1\] Since \(\pi \approx 3.14\), we have that \(f(\pi) = -3.14 + 1 = -2.14\), which is negative.

Step 4: Apply the Intermediate Value Theorem

We have found that the function \(f(x)\) is positive at \(x = 0\) and negative at \(x = \pi\), and its continuous because \(\sin x\) is continuous and any linear function, like \(x-1\), is continuous as well. According to the Intermediate Value Theorem, since the function is continuous over the interval \([0, \pi]\) and changes its sign from positive to negative, there must be a root for the equation \(\sin x - x + 1 = 0\) within this interval. Therefore, the equation has a root.

Key Concepts

Understanding Continuous Functions

A continuous function is a fundamental concept in calculus and analysis. Essentially, a function is continuous if there are no breaks, holes, or jumps in its graph. This means you can draw the graph of a function without lifting your pen off the paper.
In mathematical terms, a function \(f(x)\) is continuous at a point \(x = a\) if the following conditions are satisfied:

• \(f(a)\) is defined.
• The limit of \(f(x)\) as \(x\) approaches \(a\) exists.
• The value of the function at \(a\) equals the limit as \(x\) approaches \(a\): \(\lim_{x \to a}f(x) = f(a)\).

Continuous functions are essential because they allow us to apply various important theorems, such as the Intermediate Value Theorem. In the context of the problem "\(\sin x - x + 1 = 0\)," since \(\sin x\) and \(-x + 1\) are both continuous, \(f(x) = \sin x - x + 1\) is also continuous.

The Sine Function

The sine function, \(\sin x\), is one of the primary trigonometric functions. It's known for its wavelike pattern that repeats every \(2\pi\) radians.

• It oscillates between -1 and 1.
• It starts from 0 at \(x = 0\), peaks at \(x = \pi/2\) (sin\(\pi/2=1\)), goes back to 0 at \(x = \pi\), and hits its lowest at \(x = 3\pi/2\) (sin\(3\pi/2=-1\)).
• It completes a full cycle from 0 back to 0 across \(2\pi\).

In solving our equation, \(\sin x - x + 1 = 0\), the sine function's properties are key. Knowing that \(\sin x\) fluctuates, it helps us understand why \(f(x)\) might change sign, providing a potential root.

Root Finding Techniques

Root finding involves identifying values of \(x\) for which \(f(x)\) equals zero. This is because the roots correspond to the function's intersections with the x-axis.
One intuitive way to find roots is by plotting the function and visually checking for sign changes. However, analytical methods such as using the Intermediate Value Theorem (IVT) are more precise.
The IVT states: If a function \(f\) is continuous on the interval \([a, b]\) and \(f(a)\) and \(f(b)\) have opposite signs, then there is at least one root \(c\) in \((a, b)\) such that \(f(c) = 0\).In our problem, we checked values at \(x = 0\) and \(x = \pi\). Observing that \(f(0)\) was positive and \(f(\pi)\) was negative, we confirmed the presence of a root between \(0\) and \(\pi\). This is a straightforward application of the IVT.

Exploring Trigonometric Equations

Trigonometric equations involve trigonometric functions, such as sine or cosine, and are commonly expressed like the one in our problem, \(\sin x - x + 1 = 0\). Solving these equations often requires understanding the periodic nature of trigonometric functions.

• They often have an infinite number of solutions because trigonometric functions repeat their values in cycles.
• To find specific solutions or roots, we often restrict the interval to one or two periods of the function (e.g., \([0, 2\pi]\) for sine and cosine).

The process is typically about reducing the equation using trigonometric identities or numerical methods, and then applying continuous function properties and theorems, such as the IVT, to pinpoint solutions. In our case, simplifying the equation helped us examine the function's behavior and find the interval likely to contain a root, successfully concluding the exercise.