Problem 35 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \begin{aligned} &\text { Given }\\\ &\begin{aligned} f(x) &=x^{p} \cos \frac{1}{x}, \quad x \neq 0 \\ &=0, \quad x=0 \end{aligned} \end{aligned} $$ $$ \text { What conditions should be imposed on } p \text { so that } $$ $$ \text { (ii) } f(x) \text { may be differentiable at } x=0 $$

Step-by-Step Solution

Answer

The function is differentiable at \(x = 0\) for \(p > 1\).

Step 1: Define the Function

Let's consider the function \(f(x) = x^{p} \cos \frac{1}{x}\) for \(x \neq 0\) and \(f(x) = 0\) for \(x = 0\).

Step 2: Compute Right-hand Derivative

We first compute the derivative from the right as \(x\) approaches 0. By definition, \[f^{\'}_{+}(0) = \lim_{{h \to 0+}} \frac{{f(h)-f(0)}}{h} = \lim_{{h \to 0+}} h^{p-1}\cos\frac{1}{h}\]

Step 3: Examine the Limit of Step 2

Now we have the limit of \(h^{p-1}\) times oscillating term \(\cos(1/h)\) value as \(h\) approaches 0. Here, \(\cos\frac{1}{h}\) fluctuates between -1 and 1. Therefore, the entire limit will exist and be finite if we have \(h^{p-1}\) going to 0 as \(h\) approaches 0, or in other words, if \(p-1 > 0\). Therefore \(p>1\) is the condition for right-hand derivative to exist.

Step 4: Compute Left-hand Derivative

We now compute the derivative from the left as \(x\) approaches 0. By definition, \[f^{\'}_{-}(0) = \lim_{{h \to 0-}} \frac{{f(h)-f(0)}}{h} = \lim_{{h \to 0-}} (-h)^{p-1}\cos\frac{1}{-h}\]

Step 5: Examine the Limit of Step 4

Similar to Step 3, here \((-h)^{p-1}\) needs to go to 0 as \(h\) approaches 0, in order for the left-hand derivative to exist. As \(h \to 0-\), \(-h = |h|\) remains positive and behaves exactly like Step 3. Thus the condition remains the same: \(p>1\).

Step 6: Comparing Left and Right Hand Derivatives

We have seen in Step 3 and Step 5, that the condition on \(p\) to have both right-hand and left-hand derivatives exist (and be finite) at \(x=0\), is equivalent to \(p>1\).

Key Concepts

Limit of a Function

Understanding the limit of a function is vital in calculus, especially when dealing with continuity and differentiability. Conceptually, the limit describes the behavior of a function as the input, often denoted by x, approaches a particular value. It doesn't matter what the function’s value is at that point; what's important is where the values are heading towards as x gets close to that point.

For instance, in the exercise, the limit comes into play when determining the right-hand and left-hand derivatives at x = 0. The expressions involving h to the power of p-1, multiplied by the oscillating term cos(1/h), only make sense as h approaches 0, not at h = 0. If the limits of these expressions as h approaches 0 from the right or left exist and are equal, the function may be differentiable at that point. The condition that p > 1 ensures that the function's growth rate towards zero is slow enough that the limits exist and are finite, satisfying an essential criterion for differentiability.

Moreover, if the limit does not exist, or if the function behaves differently as x approaches from the right as opposed to the left (right-hand limit vs. left-hand limit), it hints at a discontinuity or non-differentiability at that point.

Right-hand Derivative

The right-hand derivative plays a critical role in determining a function's differentiability at a point. It measures the instantaneous rate of change of the function as it approaches a point from the right. In simple terms, it's like taking a magnifying glass and focusing on how the function behaves just to the right of our point of interest.

Applying Right-hand Derivative

In the given exercise, we investigate the right-hand derivative of the function when x is near zero. We do this by taking the limit of the difference quotient as h, which represents a small increment to the right of 0, approaches zero from the positive side. The reason we examine this limit is to guarantee that the slope of the function (its derivative) isn't heading towards infinity or does not oscillate unpredictably as we zoom into x = 0 from the right.

The condition found, p > 1, is critical because it dictates that the function's increase or decrease rate as we approach x = 0 from the right is moderate, allowing for a meaningful derivative to be calculated. This ensures the existence of the right-hand derivative and an assurance that right at x = 0, the function is smoothly transitioning with respect to its immediate right side.

Left-hand Derivative

Similarly, the left-hand derivative of a function is defined as the rate of change of the function as the input approaches the point of interest from the left. It's a mirror concept to the right-hand derivative but focused on the behavior of the function just to the left of our point.

Examining Left-hand Derivative

For our function in the exercise, the left-hand derivative at x = 0 is investigated similarly to the right-hand derivative but as h approaches zero from the negative side. This close examination is crucial to ensure the slope of the function just to the left of x = 0 is not erratic or growing without bounds.

What's interesting in this exercise is that the same condition that secures the existence of the right-hand derivative, p > 1, also secures the existence of the left-hand derivative. This means that as h approaches zero from either side, the moderated growth rate ensures that the slope — the derivative at x = 0 — is defined and stable. It's essential to note that for a function to be differentiable at a point, both the left-hand and right-hand derivatives must exist and be equivalent. The exercise highlights that by satisfying the condition p > 1, the function maintains consistent behavior from both directions, solidifying its differentiability at the origin.