Problem 98 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Given \(f(x)=x^{2}, \quad x \geq 0\) \(=x, \quad x<0\) and \(\begin{aligned} g(x) &=\frac{1}{x}, & & x \geq 1 \\ &=1, & & x<1 . \end{aligned}\) Determine the following functions:- i. \(h(x)=x \times g(x)\); ii. \(\phi(x)=f(x)+g(x)\) iii. \(\psi(x)=f(x) \times g(x)\) \\{Ans. i. \(h(x)=1, \quad x \geq 1\) ii. \(\quad \phi(x)=x^{2}+\frac{1}{x}, \quad x \geq 1\) \(=x^{2}+1, \quad 0 \leq x<1\) \(=x+1, \quad x<0 ;\) iii. \(\quad \psi(x)=x, \quad x \geq 1\) \(=x^{2}, \quad 0 \leq x<1\) \(=x, \quad x<0\\}\)

Step-by-Step Solution

Answer

Ans. i. \(h(x)=1, \quad x \geq 1\) \(h(x)=x, \quad x<1\) ii. \(\phi(x)=x^{2}+\frac{1}{x}, \quad x \geq 1\) \(\phi(x)=x^{2}+1, \quad 0 \leq x<1\) \(\phi(x)=x+1, \quad x<0\) iii. \(\psi(x)=x, \quad x \geq 1\) \(\psi(x)=x^{2}, \quad 0 \leq x<1\) \(\psi(x)=x, \quad x<0\)

Step 1: For \(x \geq 1\)

In this case, \(g(x) = \dfrac{1}{x}\). So, \(h(x) = x \times \dfrac{1}{x} = 1\).

Step 2: For \(x < 1\)

In this case, \(g(x) = 1\). So, \(h(x) = x \times 1 = x\). Putting it all together, \(h(x) = 1\) if \(x \geq 1\) and \(h(x) = x\) if \(x < 1\). ii. Determine \(\phi(x) = f(x) + g(x)\): We will analyze the domains for f(x) and g(x), and calculate 𝜑(x) accordingly.

Step 3: For \(x \geq 1\)

In this case, \(f(x) = x^{2}\) and \(g(x) = \dfrac{1}{x}\). So, \(\phi(x) = x^{2} + \dfrac{1}{x}\).

Step 4: For \(0 \leq x < 1\)

In this case, \(f(x) = x^{2}\) and \(g(x) = 1\). So, \(\phi(x) = x^{2} + 1\).

Step 5: For \(x < 0\)

In this case, \(f(x) = x\) and \(g(x) = 1\). So, \(\phi(x) = x + 1\). Putting it all together, \(\phi(x) = x^{2} + \dfrac{1}{x}\) if \(x \geq 1\), \(\phi(x) = x^{2} + 1\) if \(0 \leq x < 1\), and \(\phi(x) = x + 1\) if \(x < 0\). iii. Determine \(\psi(x) = f(x) \times g(x)\): We will analyze the domains for f(x) and g(x), and calculate 𝜓(x) accordingly.

Step 6: For \(x \geq 1\)

In this case, \(f(x) = x^{2}\) and \(g(x) = \dfrac{1}{x}\). So, \(\psi(x) = x^{2} \times \dfrac{1}{x} = x\).

Step 7: For \(0 \leq x < 1\)

In this case, \(f(x) = x^{2}\) and \(g(x) = 1\). So, \(\psi(x) = x^{2} \times 1 = x^{2}\).

Step 8: For \(x < 0\)

In this case, \(f(x) = x\) and \(g(x) = 1\). So, \(\psi(x) = x \times 1 = x\). Putting it all together, \(\psi(x) = x\) if \(x \geq 1\), \(\psi(x) = x^{2}\) if \(0 \leq x < 1\), and \(\psi(x) = x\) if \(x < 0\).

Key Concepts

Function Operations

When working with piecewise functions, an essential skill is performing operations on these functions. Operations such as addition, multiplication, and composition can change the function's expression based on the defined intervals. For the given piecewise functions, let's break down how these operations are conducted:

• Addition: When you add two piecewise functions, ensure you are adding the correct expressions according to their respective intervals. For example, for the function \( \phi(x) = f(x) + g(x) \), you add \( f(x) \) and \( g(x) \) for each interval they are defined on. This requires careful identification of the correct expressions for each domain segment.
• Multiplication: Similar to addition, multiplying functions requires you to consider the individual expressions defined over specific intervals. For example, \( \psi(x) = f(x) \times g(x) \) involves multiplying the expressions of \( f(x) \) and \( g(x) \) within their respective intervals.

Practicing these operations helps solidify understanding of function behavior across intervals and is crucial for solving more complex problems later.

Domain Analysis

Understanding the domain of a piecewise function is crucial for correctly evaluating each part of the function. A piecewise function can switch its expression based on the value of \( x \), meaning evaluating its domain requires attention to detail.

• Identify intervals: Begin by examining each piece of the function to see over what range of \( x \) it is active. For instance, in the function \( g(x) \), two conditions are given: \( g(x) = \frac{1}{x} \) for \( x \geq 1 \) and \( g(x) = 1 \) for \( x < 1 \).
• Match intervals with operations: Once intervals are determined, apply them correctly while performing function operations. This ensures that the operations you perform correctly correspond to \( x \)'s values. For example, adding or multiplying values within incorrect intervals results in wrong evaluations of the function.

Thus, domain analysis enables you to understand where each part of the piecewise function is defined and how to correctly apply operations to these parts.

Calculus

Calculus relies heavily on understanding how functions behave, specifically in terms of limits, continuity, and differentiability—which can be complex for piecewise functions.

• Continuity: A function is continuous if there are no breaks, jumps, or holes in its graph within its domain. Evaluating a piecewise function's continuity requires checking the limits on either side of the intervals. For instance, with \( f(x) \) and \( g(x) \), you would derive similar continuous expressions at the transition points, like \( x = 1 \).
• Differentiability: A function is differentiable if it has a defined derivative at all points within its domain, excluding potential corner cases or cusps in the graph. For piecewise functions, examine each piece individually and check for smooth transitions at their endpoints.
• Limits: When dealing with expressions like \( \phi(x) \) or \( \psi(x) \), consider evaluating the limits of these functions as \( x \) approaches boundary points to better understand their behavior at transitions.

Grasping these calculus concepts enables you to appreciate how piecewise functions behave under transformation and assists in evaluating their properties effectively.