Problem 110 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=x\), then the value of \(f(1)-\frac{f^{\prime}(1)}{1 !}+\frac{f^{\prime \prime}(1)}{2 !}-\frac{f^{\prime \prime \prime}(1)}{3 !}+\ldots+\frac{(-1)^{n} f^{n}(1)}{n !}\) is (A) \(2^{n}\) (B) \(2^{n-1}\) (C) 0 (D) 1

Step-by-Step Solution

Answer

The value is (C) 0.

Step 1: Understand f(x) and its Derivatives

Let's analyze the function provided: \( f(x) = x \). First, we find the derivatives of \( f(x) \): - The first derivative \( f'(x) = 1 \).- The second derivative \( f''(x) = 0 \) because the derivative of a constant is zero.- Any higher-order derivative, \( f^{(n)}(x) \) for \( n \geq 2 \), is also 0 because \( f''(x) = 0 \).

Step 2: Evaluate the Expression

Let's evaluate the expression given:\[ f(1) - \frac{f^{\prime}(1)}{1!} + \frac{f^{\prime\prime}(1)}{2!} - \frac{f^{\prime \prime \prime}(1)}{3!} + \ldots + \frac{(-1)^{n}f^{n}(1)}{n!} \]First, \( f(1) = 1 \).For \( n \geq 1 \), all derivatives \( f^{(n)}(1) \) where \( n \geq 2 \) are zero. Therefore, the expression reduces primarily to:\[ 1 - \frac{1}{1!} \]which simplifies to \( 1 - 1 = 0 \).

Step 3: Simplify Using Known Derivatives

Since higher order derivatives (\( f^{\prime\prime}, f^{\prime\prime\prime}, \ldots \)) all equal zero, the entire tail end of the expression (beyond \(\frac{1}{1!}\)) cancels to zero.This means the expression simplifies entirely as:\[ 0 \]

Step 4: Determine the Answer

The result of this entire expression is 0. After calculating, the function's series components confirm that all terms cancel out, summing to 0, given that beyond the first two terms, all derivatives are zero.

Key Concepts

Taylor Series

The Taylor Series is a powerful tool in calculus that allows us to approximate functions by a sum of terms. These terms are derived from the function's value and its derivatives at a single point, typically around zero or another point of interest. When we expand a function using its Taylor series, we are essentially predicting its behavior based on the information we have at that specific point.
The general formula for a Taylor series of a function \( f(x) \) around a point \( a \) is given by:

• \( f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \ldots \)

This series can be truncated at various levels, depending on the desired accuracy. Higher-order terms provide a more precise representation of the function. However, in many cases, enough approximation can be achieved with just a few terms.
In the original problem, the Taylor series is centered around 1, and only two terms matter since all higher-order derivatives are zero. This highlights how some functions can be fully represented with very few terms, simplifying the calculation significantly.

Derivatives

Derivatives are fundamental in calculus. They represent the rate at which a function changes at a point and are defined as the limit of the difference quotient. In other words, the derivative of a function at a point gives us the slope of the tangent line to the function at that point.
A function's first derivative, denoted as \( f'(x) \), indicates the function's rate of change. The second derivative, \( f''(x) \), shows the rate of change of the rate of change, or the function's concavity. This continues with higher-order derivatives representing further rates of change.

• First Derivative: \( f'(x) = 1 \)
• Second Derivative: \( f''(x) = 0 \)
• For \( n \geq 2 \), \( f^{(n)}(x) = 0 \)

In the exercise, knowing that higher derivatives are zero simplifies the calculation greatly. The realization that only the first two derivatives are relevant when evaluating the function expression emphasized the function's linear nature, contributing to the understanding that more complex polynomial terms would not appear in its derivative sequence.

Higher-order Derivatives

Higher-order derivatives extend the concept of a derivative beyond the first and second derivatives. They measure how a rate changes by taking the derivative of a derivative. In our particular exercise, higher-order derivatives (i.e., third derivative and above) are zero. This is a significant property of linear functions like \( f(x) = x \).
When a second derivative view remains constant or zero, it indicates a lack of curvature—suggesting the function maintains a constant slope. When we consider further derivatives:

• If a derivative sequence quickly zeros out, the function is linear or of low degree.
• For a polynomial like \( f(x) = x \), every derivative beyond the first is zero.

Understanding higher-order derivatives helps us anticipate the behavior of functions and series expansions such as the Taylor Series. In practical scenarios, higher-order derivatives solidify the series' accuracy or confirm that further terms would not add value—crucial for simplifying complex problems like the one in the exercise.