Problem 35 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=x+\frac{x}{1+x}+\frac{x}{(1+x)^{2}}+\ldots\) to \(\infty\), then at \(x=0\), \(f(x)\) (A) has no limit (B) is discontinuous (C) is continuous but not differentiable (D) is differentiable

Step-by-Step Solution

Answer

(D) is differentiable

Step 1: Identify the Infinite Series

The given function is expressed as an infinite series: \[ f(x) = x + \frac{x}{1+x} + \frac{x}{(1+x)^{2}} + \frac{x}{(1+x)^{3}} + \ldots \] This series continues indefinitely.

Step 2: Recognize the Geometric Series

Notice that the series terms involve powers of \( \frac{x}{1+x} \). We can write:\[ f(x) = x \left( 1 + \frac{1}{1+x} + \frac{1}{(1+x)^{2}} + \ldots \right) \] The terms in parentheses form a geometric series with the first term \(a = 1\) and common ratio \(r = \frac{1}{1+x}\).

Step 3: Apply Sum Formula for Infinite Geometric Series

The sum of an infinite geometric series is given by:\[ S = \frac{a}{1-r} \]where \(|r| < 1\). Applying this formula, the series becomes:\[ S = \frac{1}{1 - \frac{1}{1+x}} = \frac{1+x}{x} \] Thus, \( f(x) = x \times \frac{1+x}{x} = 1+x \).

Step 4: Evaluate \(f(x)\) at \(x=0\)

Substitute \(x = 0\) in \(f(x)\):\[ f(0) = 1 + 0 = 1 \] Since \(f(x)\) is defined as \(1+x\), it is both continuous and differentiable everywhere, including at \(x=0\).

Step 5: Discuss Continuity and Differentiability

The function \(f(x) = 1 + x\) is a linear function, which is continuous and differentiable at all points on the real line. This means that \(f(x)\) is continuous and differentiable at \(x = 0\).

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a constant, called the common ratio. In the infinite series presented, \[ f(x) = x + \frac{x}{1+x} + \frac{x}{(1+x)^{2}} + \frac{x}{(1+x)^{3}} + \ldots \] it has been identified that each term \( \frac{x}{(1+x)^n} \) is derived by multiplying the preceding fraction by \( \frac{1}{1+x} \). This shows that we're dealing with a geometric series where:

• The first term, \( a = 1 \)
• The common ratio, \( r = \frac{1}{1+x} \)

Recognizing this series as geometric is crucial because it allows the use of specific formulas to calculate its sum. This recognition is foundational in solving problems involving patterns of repeated multiplication, particularly in infinite contexts like the given example.

Sum Formula for Infinite Geometric Series

The sum formula for an infinite geometric series is a powerful tool that simplifies our calculations when \(|r| < 1\). The formula for the series sum is:\[ S = \frac{a}{1-r} \]where \(a\) is the first term and \(r\) the common ratio. In the context of our problem, applying this formula to the series \( 1 + \frac{1}{1+x} + \frac{1}{(1+x)^{2}} + \ldots \) allows us to find its sum by substituting \( a = 1 \) and \( r = \frac{1}{1+x} \).This leads to:\[ S = \frac{1}{1 - \frac{1}{1+x}} = \frac{1+x}{x} \]Once you obtain this, the function \( f(x) \) simplifies to:\( f(x) = x \times \frac{1+x}{x} = 1 + x \)This reveals that the seemingly complex infinite series actually defines a linear function, which is incredibly straightforward to analyze further.

Continuity and Differentiability

A function is continuous at a point if there are no "jumps" or breaks at that point — you can draw it without lifting your pencil. Differentiability adds a layer, requiring that the function also has a derivative at that point, meaning it has a well-defined tangent line and thus, a slope. For the function derived, \( f(x) = 1 + x \), it is both continuous and differentiable at any point on the real line, including \( x = 0 \).Here’s why:

• Continuity: The expression \( 1 + x \) is a linear polynomial, inherently smooth and unbroken, defining it as continuous.
• Differentiability: The derivative of \( f(x) \), which is \( f'(x) = 1 \), is constant. This shows the slope is uniform, reinforcing that the curve has no sharp corners or cusps where differentiability could fail.

Conclusively, the process demonstrates that \( f(x) \) fulfills both these properties exactly as often demanded by linear functions, confirming its well-behaved nature at all points, particularly \( x = 0 \).