Problem 49 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)=\sqrt{x-1}+\sqrt{x+24-10 \sqrt{x-1}} ; 1

Step-by-Step Solution

Answer

The derivative \( f'(x) \) is 0. (Option A)

Step 1: Simplify Expression

The function given is \( f(x) = \sqrt{x-1} + \sqrt{x + 24 - 10 \sqrt{x-1}} \). Notice that the second term \( \sqrt{x + 24 - 10 \sqrt{x-1}} \) can be rewritten. Let \( u = \sqrt{x-1} \), so \( u^2 = x-1 \) or \( x = u^2 + 1 \). Substituting \( x = u^2 + 1 \) into the expression gives \( \sqrt{u^2 + 1 + 24 - 10u} = \sqrt{u^2 - 10u + 25} \). This simplifies further because \( u^2 - 10u + 25 = (u - 5)^2 \). Thus, it becomes \( \sqrt{(u-5)^2} = |u-5| \). For \( u = \sqrt{x-1} \), since \( u = \sqrt{x-1} > 0 \), it simplifies to \( 5 - u \). The expression for \( f(x) \) thus becomes \( f(x) = u + (5 - u) = 5 \).

Step 2: Differentiate the Simplified Function

Since \( f(x) = 5 \), a constant function, the derivative of any constant is 0. Thus, the derivative \( f'(x) \) is 0. Differentiating \( f(x) = 5 \) with respect to \( x \) directly gives \( f'(x) = 0 \).

Key Concepts

Function Differentiation

Function differentiation is a cornerstone of calculus that helps us to understand the rate at which one quantity changes with respect to another. When dealing with any function, the differentiation process involves finding its derivative. This derivative represents a function that shows the rate of change.
For the function, it's like finding the slope of the tangent to the function's curve at any point. To differentiate a function, we apply rules such as the power rule, the product rule, or the chain rule, depending on the form of the function.

• The power rule is used for polynomials.
• The product rule applies to the product of two functions.
• The chain rule is essential when a function is composed of another function.

Tackling complex functions often requires simplifying the expression first to make the differentiation process smoother. This was the case in our problem, where simplifying the function helped us to confirm that it ultimately reduces down to just a constant value.

Derivative of a Constant

Understanding the derivative of a constant is an essential aspect of calculus. A constant function is where the output value does not change, regardless of the input value. Mathematically, if a function is defined as a constant, say 5, its derivative with respect to any variable is zero.
This is because the rate of change of a constant value is zero. There is no variation or slope; it remains flat across any interval. In our specific problem, after simplifying, we found that the function reduced to a constant, namely 5. Hence, its derivative was 0.

So why is this important? Recognizing this rule can save you a lot of time. Whenever you are tasked with differentiating, ask yourself if the function is constant. If it is, you have the derivative right away! In our original task, this realization simplified the problem significantly.

Simplifying Expressions

Simplifying expressions is a powerful mathematical technique that can make complex calculations much more manageable. This involves rewriting the expression in a simpler or more efficient form without changing its values.
In calculus, simplifying before differentiating can reveal hidden details or make processes more straightforward. Look at our challenge: the initial complex expression seemed formidable, but by substituting variables and re-evaluating terms, we simplified the function down to a constant.

• First, isolate and rearrange parts of the expression that can be simplified.
• Second, look for substitutions. This can involve setting variables for common terms.
• Finally, calculate or simplify using algebraic identities or approximations.

This approach breaks down complicated problems into more palatable parts. So, whenever you're faced with a complex expression, simplifying it can be your best first step.