Problem 19 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A particle starts at the origin and moves along the \(x\)-axis in such a way that its velocity at the point \((x, 0)\) is given by the formula \(\frac{d x}{d t}=\cos ^{2} \pi x\). Then the particle never reaches the point on (A) \(x=\frac{1}{4}\) (B) \(x=\frac{3}{4}\) (C) \(x=\frac{1}{2}\) (D) \(x=1\)

Step-by-Step Solution

Answer

The particle never reaches the point \(x=\frac{1}{2}\).

Step 1: Understand the Problem

We need to determine at which point the particle never reaches, given its velocity function along the x-axis. The velocity function is \(\frac{dx}{dt} = \cos^2(\pi x)\). We are asked whether the particle never reaches one of the given points: \(x=\frac{1}{4}\), \(x=\frac{3}{4}\), \(x=\frac{1}{2}\), \(x=1\).

Step 2: Analyze the Velocity Function

The velocity function \(\cos^2(\pi x)\) determines how the position \(x\) changes over time. For the particle to reach a point, the integral of the velocity function from the starting point (the origin) to that point must converge. If \(\cos^2(\pi x)\) ever equals zero at any of these points, it implies infinite time to reach that point.

Step 3: Evaluate \(\cos^2(\pi x)\) at Given Points

Evaluate the velocity function at each point:- For \(x=\frac{1}{4}\), \(\cos^2(\pi \cdot \frac{1}{4}) = \cos^2(\frac{\pi}{4}) = \frac{1}{2}\).- For \(x=\frac{3}{4}\), \(\cos^2(\pi \cdot \frac{3}{4}) = \cos^2(\frac{3\pi}{4}) = \frac{1}{2}\).- For \(x=\frac{1}{2}\), \(\cos^2(\pi \cdot \frac{1}{2}) = \cos^2(\frac{\pi}{2}) = 0\).- For \(x=1\), \(\cos^2(\pi \cdot 1) = \cos^2(\pi) = 1\).

Step 4: Determine the Point the Particle Never Reaches

The function \(\cos^2(\pi x)\) evaluates to zero at \(x=\frac{1}{2}\). If the velocity is zero, the particle requires infinite time to reach it because it momentarily stops moving at \(x=\frac{1}{2}\). Hence, the particle never reaches \(x = \frac{1}{2}\).

Key Concepts

Velocity Function

The velocity function is a fundamental concept when considering particle motion along a path. In this context, the velocity function describes how fast the particle is moving and in which direction, specifically along the x-axis. Here, the function is given by \( \frac{dx}{dt} = \cos^2(\pi x) \). This formula illustrates how the velocity changes as the particle moves along its trajectory.

When dealing with velocity functions, it is crucial to understand that velocity is the rate of change of the position. Therefore, this particular function tells us how the position \( x \) changes over time \( t \). Key points to remember include:

• Velocity can be positive or negative, indicating direction along the x-axis.
• When the velocity is zero, the particle comes to a stop.
• The value of the velocity function at a specific point directly affects whether or not the particle will reach certain points on its path.

Understanding how \( \cos^2(\pi x) \) behaves at different values of \( x \) is vital for predicting the motion of the particle.

Integral Convergence

Integral convergence is a key concept when determining whether a particle will reach a certain point along its path. This involves considering the integral of the velocity function over a particular interval.

For a particle to reach a point \( x \), the integral of the velocity function from the starting position to \( x \) must converge. Think of the integral as the total 'time spent moving', so if the integral goes to infinity, it indicates infinite time to reach that point.

• If \( \int_0^a \cos^2(\pi x) \, dx \) converges for \( x = a \), the particle can reach \( a \).
• If \( \cos^2(\pi x) = 0 \) at some \( x \), the particle stops moving, and reaching the point requires infinite time (integral divergence).

In our problem, when evaluating \( \cos^2(\pi x) \), we find it is zero at \( x = \frac{1}{2} \), meaning the integral diverges, indicating the particle cannot reach this point. This is why integral convergence plays a crucial role in understanding particle motion.

Trigonometric Functions

Trigonometric functions, like \( \cos(\theta) \), are essential in calculus problems involving periodic behavior. In this exercise, we deal with \( \cos^2(\pi x) \), which is a transformation of the basic cosine function.

Key properties of trigonometric functions essential for solving problems include:

• They are periodic, with cosine having a period of \( 2\pi \).
• The square of a cosine function, \( \cos^2(\theta) \), ranges from 0 to 1.
• Knowing specific values: for instance, \( \cos(\pi/2) = 0 \) and \( \cos(0) = 1 \).

In our exercise, \( \cos^2(\pi x) \) is particularly insightful. For \( x = \frac{1}{2} \), \( \cos(\pi/2) = 0 \), explaining why the particle never reaches that point. Understanding these behaviors aids in predicting how functions impact physics-based scenarios, like particle movements on a path.