Problem 6 - A Complete Resource Book in Mathematics for JEE Main

Question

The left-hand derivative of \(f(x)=[x] \sin (\pi x)\) at \(x=k\), \(k\) an integer and \([x]=\) greatest integer \(\leq x\), is (A) \((-1)^{k}(k-1) \pi\) (B) \((-1)^{k-1} \cdot(k-1) \pi\) (C) \((-1)^{k} \cdot k \pi\) (D) \((-1)^{k-1} \cdot k \pi\).

Step-by-Step Solution

Answer

The left-hand derivative is option (A): \((-1)^{k}(k-1) \pi\).

Step 1: Understand the Problem

We're asked to find the left-hand derivative of the function \( f(x) = [x] \sin(\pi x) \) at \( x = k \) where \( k \) is an integer. The function involves the greatest integer function \( [x] \).

Step 2: Definition of Left-Hand Derivative

The left-hand derivative at \( x = k \) is defined as: \[ f'_-(k) = \lim_{h \to 0^-} \frac{f(k+h) - f(k)}{h}. \] We'll calculate this expression for \( f(x) = [x] \sin(\pi x) \).

Step 3: Evaluate \( f(k) \)

At \( x = k \), where \( k \) is an integer, the greatest integer function gives \([k] = k\). Therefore, \( f(k) = k \sin(\pi k) = k \cdot 0 = 0 \).

Step 4: Evaluate \( f(k+h) \) for Negative \( h \)

For \( x = k + h \) where \( h < 0 \), we have \([k+h] = k-1\) because \( k+h \) is slightly less than \( k \). Thus, \( f(k+h) = (k-1) \sin(\pi(k+h)). \)

Step 5: Calculate the Derivative Expression

Substitute \( f(k) = 0 \) and \( f(k+h) = (k-1) \sin(\pi(k+h)) \) into the derivative expression: \[ f'_-(k) = \lim_{h \to 0^-} \frac{(k-1) \sin(\pi(k+h)) - 0}{h}. \]

Step 6: Simplify Using Sine Identity

Using the identity \( \sin(\pi(k+h)) = \sin(\pi k + \pi h) = \sin(\pi k)\cos(\pi h) + \cos(\pi k)\sin(\pi h) \) and \( \sin(\pi k) = 0 \), it simplifies to \( \cos(\pi k)\sin(\pi h) \).

Step 7: Evaluation of the Limit

The expression simplifies to: \[ f'_-(k) = \lim_{h \to 0^-} \frac{(k-1)\cos(\pi k)\sin(\pi h)}{h}. \] As \( h \to 0^- \), \( \sin(\pi h) \sim \pi h \), so the limit becomes: \[ f'_-(k) = (k-1)\cos(\pi k)\pi = (k-1)(-1)^k\pi. \]

Step 8: Choose the Correct Option

The answer \( f'_-(k) = (k-1)(-1)^k\pi \) corresponds to the option (A): \((-1)^{k}(k-1) \pi\).

Key Concepts

Greatest Integer Function

The greatest integer function, denoted as \([x]\), gives the largest integer that is less than or equal to a number \(x\). This function is also known as the floor function. It's important in various mathematical fields due to its ability to 'remove' the fractional part of a number.

When analyzing a function like \(f(x) = [x] \sin(\pi x)\), the greatest integer function affects how the function behaves at different points. For any real number \(x\), \([x]\) will output an integer value.

Here's how it works:

• For any integer \(a\), \([a] = a\).
• For a non-integer like 3.7, \([3.7] = 3\).
• If \(h\) is a small negative number, \([k + h] = k - 1\) when \(k\) is an integer.

These properties are crucial when evaluating piecewise expressions or dealing with limits, especially in calculus.

Trigonometric Identity

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the occurring variables. They help simplify complex trigonometric expressions.

In this exercise, the identity used is:

• \( \sin(\pi(k+h)) = \sin(\pi k + \pi h) \)

This can be expanded using the angle sum identity:

• \( \sin(A+B) = \sin A \cos B + \cos A \sin B \)

So in our case, it becomes:

• \( \sin(\pi(k+h)) = \sin(\pi k)\cos(\pi h) + \cos(\pi k)\sin(\pi h) \)

Because \( \sin(\pi k) = 0 \) (since \( k\) is an integer), this identity simplifies to \( \cos(\pi k)\sin(\pi h) \).

Identities like this one are instrumental when simplifying terms within limits and derivatives, making calculations more straightforward.

Limit Evaluation

Limit evaluation is an essential concept in calculus, as it helps us understand the behavior of functions as they approach certain points. When evaluating a limit, especially in derivatives, we often look at how a function behaves as it approaches a specific point from the left (or right).

The left-hand derivative is evaluated using the formula:

• \( f'_-(k) = \lim_{h \to 0^-} \frac{f(k+h) - f(k)}{h} \)

In the context of our exercise, the limit helps compute the left-hand derivative at an integer \(k\). By looking at \(f(k+h)\), where \(h < 0\), and \(f(k)\), we can use the greatest integer function to simplify the expression as \(h\) approaches zero.

With the substitution of \(f(k) = 0\) and the simplified expression \( (k-1)\cos(\pi k)\sin(\pi h) \), it's crucial to recognize that as \( h \) becomes infinitesimal, \( \sin(\pi h) \sim \pi h \).

This lets us resolve the limit to attain the left-hand derivative, showing how precise limit handling is vital in such calculations.

Function Analysis

Function analysis involves breaking down the components and behavior of a mathematical function. In this context, it is crucial to understand how each part of the function \( f(x) = [x] \sin(\pi x) \) contributes to the overall behavior of the function.

Here's a breakdown of its analysis:

• **Greatest Integer Function**: It causes \( [x] \) to be constant between integers, causing jumps at integer points.
• **Sine Component**: The \( \sin(\pi x) \) oscillates between -1 and 1, repeating every interval of 2.

The combination of these two components means that the function's value is entirely determined by both the integer part of \(x\) and the sine function, whose value depends on \(x\).

Analyzing the function around integer points \(k\), we observe that at \(x = k\), \(f(x)\) drops to zero since \( \sin(\pi k) = 0\). As \(x\) approaches \(k\) from the left, \([x] = k-1\), leading to \( f(x) = (k-1) \sin(\pi x) \), which results in the calculation of derivatives at these points being affected.

Overall, understanding each component helps in evaluating specific behaviors and properties such as derivatives.