Problem 77 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=\left\\{\begin{array}{cc}x^{p} \cos \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{array}\right.\), then at \(x=0, f(x)\) is (A) continuous if \(p>0\) (B) differentiable if \(p>1\) (C) continuous if \(p>1\) (D) differentiable if \(p>0\)

Step-by-Step Solution

Answer

Correct answers are (A) continuous if \(p>0\) and (B) differentiable if \(p>1\).

Step 1: Find the Continuity Condition

For a function to be continuous at a point, the limit as it approaches that point from both sides must equal the function's value at that point. Here, we need \(\lim_{{x \to 0}} f(x)\) to equal \(f(0) = 0\). Consider \(\lim_{{x \to 0}} x^p \cos \frac{1}{x}\). Since \(\cos\frac{1}{x}\) oscillates between \(-1\) and \(1\), the expression \(x^p \cos \frac{1}{x}\) is bounded by \([-x^p, x^p]\). Thus, \(\lim_{{x \to 0}} x^p \cos \frac{1}{x} = 0\) if \(p > 0\). Hence, \(f(x)\) is continuous at \(x = 0\) for \(p > 0\).

Step 2: Check Differentiability Condition

For differentiability, the limit that defines the derivative must exist. The derivative at \(x = 0\) is given by \(f'(0) = \lim_{{h \to 0}} \frac{f(h) - f(0)}{h} = \lim_{{h \to 0}} \frac{h^p \cos \frac{1}{h}}{h}\). Simplifying, this results in \(\lim_{{h \to 0}} h^{p-1} \cos \frac{1}{h}\). Since \(\cos\frac{1}{h}\) is bounded by \(-1\) and \(1\), \(h^{p-1} \cos \frac{1}{h}\) is bounded by \([-h^{p-1}, h^{p-1}]\). For the limit to be zero as \(h \to 0\), we require \(p-1 > 0\), hence \(p > 1\). Thus, \(f(x)\) is differentiable at \(x = 0\) for \(p > 1\).

Step 3: Identify Correct Option

Based on our analysis: - Option (A) is correct since \(f(x)\) is continuous for \(p > 0\).- Option (B) is correct as \(f(x)\) is differentiable for \(p > 1\).- Option (C) is incorrect, as continuity requires \(p > 0\), not \(p > 1\).- Option (D) is incorrect because differentiability requires \(p > 1\), not \(p > 0\).

Key Concepts

Continuity

In calculus, continuity refers to a function's behavior at a point. A function is said to be continuous at a point if its limit from both sides equals its value at that point. For the function given in the exercise, we want to check continuity at \(x = 0\).
To achieve this, we need \(\lim_{{x \to 0}} f(x)\) to be equal to \(f(0) = 0\). Given the function \(f(x) = x^{p} \cos \frac{1}{x}\) for \(x eq 0\), and \(f(x) = 0\) for \(x = 0\), continuity at \(x = 0\) depends on making \(\lim_{{x \to 0}} x^p \cos \frac{1}{x} = 0\).

• The term \(\cos \frac{1}{x}\) oscillates between \(-1\) and \(1\), meaning \(x^p \cos \frac{1}{x}\) can be bounded in the interval \([-x^p, x^p]\).
• For \(\lim_{{x \to 0}} x^p = 0\) to hold, \(p\) must be greater than 0. If \(p > 0\), then \(x^p \cos \frac{1}{x}\) tends towards zero.

This finding confirms that \(f(x)\) is continuous at \(x = 0\) when \(p > 0\). This insight directly answers option (A) from the exercise.

Differentiability

Differentiability involves determining whether a function has a derivative at a certain point. A function is differentiable at a point if its derivative exists at that point. In this exercise, we are concerned with differentiability at \(x = 0\).
The derivative at a point is derived from the limit representation: \[f'(0) = \lim_{{h \to 0}} \frac{f(h) - f(0)}{h}\,\] where in this case, translates to \[\lim_{{h \to 0}} \frac{h^p \cos \frac{1}{h}}{h} = \lim_{{h \to 0}} h^{p-1} \cos \frac{1}{h}.\]

• The expression \(\cos \frac{1}{h}\) oscillates within \(-1\) to \(1\), so \(h^{p-1} \cos \frac{1}{h}\) is bounded within \([-h^{p-1}, h^{p-1}]\).
• To ensure this limit reaches zero, \(p-1\) must be greater than \(0\), implying \(p > 1\).

Thus, the function is differentiable at \(x = 0\) only if \(p > 1\), and this directly justifies option (B) as correct in the exercise.

Limits

The concept of limits is foundational to both continuity and differentiability. It measures the behavior of a function as the input approaches a particular point. In analyzing the problem, limits play a crucial role at \(x = 0\).
For continuity, we calculated \(\lim_{{x \to 0}} x^p \cos \frac{1}{x}\). Since \(\cos\) oscillates between \(-1\) and \(1\), the bounded nature helped demonstrate continuity if \(p > 0\) as it ensured approaching zero for the limit.

• For differentiability, the focus shifted to the expression \(\lim_{{h \to 0}} h^{p-1} \cos \frac{1}{h}\). The challenge with oscillatory functions is maintaining bounded limits leading to negligible values as \(h\) nears zero.
• By requiring \(p-1 > 0\), we safeguard that the limit indeed zeroes out, confirming differentiability.

Thus, mastering limits clarifies when a function smoothly transitions through a point and when it can possess a derivative, tying back to options (A) and (B) in our analysis.