Problem 1 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)=\sqrt{x^{2}-10 x+25}\), then the derivative of \(f(x)\) on the interval \([0,7]\) is (A) 1 (B) \(-1\) (C) 0 (D) Does not exist

Step-by-Step Solution

Answer

The derivative does not exist over \([0,7]\).

Step 1: Simplify the Expression

First, let's simplify the given function. The function is \( f(x) = \sqrt{x^2 - 10x + 25} \). Notice that the expression under the square root is a perfect square trinomial. It can be rewritten as \((x-5)^2\). Thus, \( f(x) = \sqrt{(x-5)^2} \).

Step 2: Analyze the Absolute Value

Since \( \sqrt{(x-5)^2} = |x-5| \), we need to consider the absolute value function. The absolute value function \(|x-5|\) is piecewise defined as:- \( x-5 \) if \( x \geq 5 \)- \( -(x-5) \) if \( x < 5 \).

Step 3: Determine the Derivative for Each Case

Let's find the derivative for both cases of the piecewise function:- For \( x \geq 5 \), \( f(x) = x-5 \). The derivative is \( f'(x) = 1 \).- For \( x < 5 \), \( f(x) = -(x-5) = -x+5 \). The derivative is \( f'(x) = -1 \).

Step 4: Evaluate on the Interval \([0,7]\)

On the interval \([0,7]\), we have two subintervals to consider: 1. \([0,5)\): Here, \(f'(x) = -1\).2. \([5,7]\): Here, \(f'(x) = 1\).Since the derivative changes from \(-1\) to \(1\) at \(x = 5\), the derivative is not continuous at \(x = 5\), and hence does not exist over the interval \([0,7]\).

Key Concepts

Understanding the Absolute Value Function

The absolute value function is a fundamental concept in mathematics that measures the distance of a number from zero on the number line. It is denoted by two vertical bars, like this: \(|x|\). In simple terms, the absolute value of a number is always non-negative, meaning it can't be less than zero. This is because it represents a distance, and distances cannot be negative. When applied to expressions, the absolute value \(|x-5|\) translates to two potential output values based on a condition:

• If \(x \geq 5\), \(|x-5| = x-5\).
• If \(x < 5\), \(|x-5| = -(x-5) = 5-x\).

By using these definitions, we can easily handle expressions involving absolute values by considering different scenarios or branches. This way, every x-value has its respective condition to check. Keep this in mind, as it shows that this function changes its behavior at a key point—right here at \(x = 5\).

The Role of Piecewise Functions

Piecewise functions allow us to describe a function in segments or pieces. Each segment has its own rule or expression that applies to certain intervals of the domain. For example, our function can be split into segments based on its behavior around a critical point, like the \(x = 5\) in the absolute value function \(|x-5|\).To better understand, think of a piecewise function as several "small" functions combined to form a single "big" function. With respect to the piecewise section of this exercise, the function \(|x-5|\) becomes:

• The line \(x-5\) when \(x \geq 5\).
• The line \(-x+5\) when \(x < 5\).

This feature helps pinpoint exactly where the function changes its expression. Such functions are especially handy when analyzing real-world situations involving different scenarios or inputs, as they can vary in response to the conditions they operate under. This flexibility can, however, lead to situations where the derivative of such functions isn’t smooth or continuous across the entire domain. That's precisely what happens here at \(x = 5\), making the overall derivative not exist over the interval.

Exploring the Perfect Square Trinomial

A perfect square trinomial is a special type of quadratic expression that can be factored into an identical binomial multiplied by itself. It takes the form \(a^2 + 2ab + b^2 = (a+b)^2\). Recognizing a perfect square trinomial is crucial in simplifying square roots, which is exactly what we did in our function \(f(x) = \sqrt{x^2 - 10x + 25}\).Observing that the expression under the square root is \(x^2 - 10x + 25\), we can identify it as \(x^2 - 2(5)x + 5^2\), which perfectly matches the \(a^2 - 2ab + b^2\) form. Thus, the trinomial simplifies to \((x-5)^2\). This simplification step is of utmost importance because:

• It aids in rewriting the original equation into a form that deals directly with absolute value: \(|x-5|\).
• It streamlines the process of finding derivatives by reducing the number of terms directly involved.

Whenever you identify a perfect square trinomial, use this opportunity to simplify further complex expressions to their most basic form. This aids significantly in problem-solving, particularly when you delve into derivatives and other calculus operations.