Problem 110 - A Complete Resource Book in Mathematics for JEE Main 2017

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 [2003] (A) \(2^{n}\) (B) \(2^{n-1}\) (C) 0 (D) 1

Step-by-Step Solution

Answer

The value is 0, which corresponds to option (C).

Step 1: Define the Function Derivatives

The function given is \(f(x) = x\). Calculate the derivatives: Since \(f(x) = x\), the first derivative \(f'(x) = 1\), the second derivative \(f''(x) = 0\), and all higher derivatives are also zero because the derivative of a constant (1) is zero.

Step 2: Evaluate Derivatives at 1

Evaluate these derivatives at \(x = 1\): \(f(1) = 1\), \(f'(1) = 1\), \(f''(1) = 0\), \(f'''(1) = 0\), etc.

Step 3: Understand the Series

The expression given in the problem is a series: \(f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \frac{f'''(1)}{3!} + \ldots + \frac{(-1)^{n} f^{n}(1)}{n!}\). This is a Taylor series centered around \(x=1\) but alternating in sign.

Step 4: Substitute Derivatives into the Series

Substitute the evaluated derivatives into the series: The series becomes \(1 - \frac{1}{1!} + \frac{0}{2!} - \frac{0}{3!} + \ldots + \frac{0}{n!}\) because all higher order derivatives (from second onwards) are zero.

Step 5: Simplify the Series

Simplify the series: \(1 - 1 + 0 + 0 + \ldots + 0 = 0\). Hence, the value of the series is 0.

Key Concepts

function derivatives

Understanding function derivatives is key to analyzing how a function behaves. When we consider a function like \( f(x) = x \), finding its derivative helps us understand its rate of change. The first derivative, denoted as \( f'(x) \), gives us the slope of the function at any point. In this case, since the function is a straight line, the slope is constant, and \( f'(x) = 1 \).

But what happens when we look further? Calculating second or higher derivatives means examining the rate of change of the rate of change. Here, \( f''(x) = 0 \), suggesting that the rate of change itself isn’t changing as \( f(x) = x \) is a linear function. In essence:

• The first derivative \( f'(x) = 1 \) indicates a constant slope or rate of change.
• Second and higher derivatives are zero, confirming the function's linearity.

These derivatives lay the groundwork for understanding the behavior of more complex functions.

higher order derivatives

Higher order derivatives refer to taking a derivative of a derivative. It helps to reveal the curvature and behavior of more complex functions. For \( f(x) = x \), we've seen that everything beyond the first derivative results in zero.

In mathematical terms:

• \( f''(x) = 0 \) tells us that the function has no curvature.
• The third derivative \( f'''(x) \) and so on, remain zero, further affirming the lack of change in slope.

When tackling questions that involve higher order derivatives, it’s important to note:

• These help to analyze the behavior of polynomial or more intricate functions.
• Each derivative gives an incrementally finer view of the function’s nuances.

Understanding higher order derivatives is particularly helpful in physics and engineering disciplines, where changes in acceleration or other complex curves need to be modeled.

alternating series

An alternating series involves terms that alternate in sign. This setup is quite common in mathematical series, especially when approximating functions.

In our exercise, the expression can be seen as an alternating series which follows the pattern of changing signs between terms. It's a Taylor Series expanded to approximate function values around a specific point, in this case, around \( x = 1 \).

Why is this important? Check out these highlights:

• Alternating signs can simplify series, especially for convergence purposes.
• The alternating series test can confirm if a series converges.

In practical terms, while our example led to a simple outcome due to zero higher order derivatives, often, alternating series have meaningful convergence properties which make them useful in computations and numerical methods.