Problem 228 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \begin{aligned} &\text { Given }\\\ &\begin{aligned} f(x) &=x^{2} \sin \frac{1}{x}, & & x \neq 0 \\ &=0, & & x=0 \end{aligned} \end{aligned} $$

Step-by-Step Solution

Answer

The domain of the given function \(f(x)\) is all real numbers since the function is defined for \(x \neq 0\) as \(f(x) = x^2 \sin(\frac{1}{x})\) and for \(x = 0\) as \(f(0) = 0\). The complete piecewise function is: \[ f(x) = \begin{cases} x^2 \sin(\frac{1}{x}), & x \neq 0 \\ 0, & x = 0 \end{cases} \]

Step 1: Determine the function's domain

First, we need to determine the function's domain. Since \(f(x) = x^2 \sin(\frac{1}{x})\) for non-zero values of x and f(0) is defined, the domain of the function is all real numbers.

Step 2: Case 1: x ≠ 0

For values of x that are not equal to 0, the function is defined as \(f(x) = x^2 \sin(\frac{1}{x})\). In this part, the exercise is already showing the proper expression for f(x), so we don't need to make any changes.

Step 3: Case 2: x = 0

For the value of x = 0, the function is defined as \(f(0) = 0\). This tells us that the function has a specific value for this singular point in its domain.

Step 4: Final Piecewise Function

Now that we have analyzed both cases, we can combine them back into a single piecewise function: \[ f(x) = \begin{cases} x^2 \sin(\frac{1}{x}), & x \neq 0 \\ 0, & x = 0 \end{cases} \] This piecewise function represents the given function \(f(x)\) for the entire domain of real numbers.

Key Concepts

Function Domain

The domain of a function is simply the set of all possible input values (usually denoted as \(x\)) for which the function is defined. In simpler terms, it's the collection of all numbers that you can plug into a function without encountering any undefined behavior, such as division by zero or taking the square root of a negative number. In the case of the function \(f(x) = x^2 \sin(\frac{1}{x})\), for \(x eq 0\), and \(f(0) = 0\), the function is defined for all real numbers. This is because:

• When \(x eq 0\), the expression \(x^2 \sin(\frac{1}{x})\) is perfectly valid since we are not dividing by zero.
• When \(x = 0\), the function simply takes on the value zero, which is defined, complying with the piecewise structure of the function.

So, the domain of this piecewise function is all real numbers, from negative infinity to positive infinity, \((-\infty, \infty)\). It's always important to check points where the function definition changes, like at \(x = 0\) in this case.

Trigonometric Functions

Trigonometric functions, such as \(\sin\), \(\cos\), and \(\tan\), play a significant role in this function. Specifically, \(\sin(\frac{1}{x})\) is used when \(x eq 0\). Let's delve into some key properties of these functions:

• The sine function, \(\sin(\theta)\), oscillates between -1 and 1 for any angle \(\theta\). This periodic behavior is crucial to understand how the function behaves when \(x eq 0\).
• As \(x\) gets very small (but not zero), \(\frac{1}{x}\) becomes very large, leading the angle \(\theta\) to shoot up to infinity, causing the sine function to oscillate wildly.

Given that \(\sin(\frac{1}{x})\) is multiplied by \(x^2\), a very interesting behavior arises around \(x=0\). Despite the massive changes in \(\sin(\frac{1}{x})\), the function \(f(x) = x^2 \sin(\frac{1}{x})\) is kept in check since \(x^2\) approaches zero faster, smoothing out the oscillation.

Limit of a Function

A limit helps us understand the behavior of a function as the input approaches a certain value. For this piecewise function, we need to investigate the limit of \(f(x) = x^2 \sin(\frac{1}{x})\) as \(x\) approaches 0 from both positive and negative directions.

• The expression \(\sin(\frac{1}{x})\) oscillates between -1 and 1 regardless of how large \(\frac{1}{x}\) becomes.
• As \(x \to 0\), \(x^2\) gets closer and closer to zero, dominating the behavior of the product \(x^2 \sin(\frac{1}{x})\).
• Thus, the limits \(\lim_{x \to 0^+} x^2 \sin(\frac{1}{x})\) and \(\lim_{x \to 0^-} x^2 \sin(\frac{1}{x})\) both equate to zero.

The function \(f(x)\) is therefore continuous at \(x = 0\), confirming that \(f(0) = 0\) is correctly defined. The limit gives us valuable insight, showing how the function behaves around a point that might otherwise seem problematic.