Problem 6 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

The function \(f(x)=\log (1+x)-\frac{2 x}{2+x}\) is increasing on \(\quad\) [EAMCET-2002] (a) \((-1, \infty)\) (b) \((-\infty, 0)\) (c) \((-\infty, \infty)\) (d) None of these

Step-by-Step Solution

Answer

Function is increasing on \((-\infty, 0) \cup (0, \infty)\). The answer is (d) None of these.

Step 1: Find the derivative of the function

To determine where the function is increasing, we need to find the first derivative of the function and analyze its sign. The function given is:\[ f(x) = \log(1+x) - \frac{2x}{2+x} \]First, find the derivative of each part separately:- Derivative of \( \log(1+x) \) with respect to \( x \) is \( \frac{1}{1+x} \).- For \( \frac{2x}{2+x} \), use the quotient rule: \((\frac{d}{dx}[\frac{u}{v}] = \frac{vu' - uv'}{v^2})\) where \( u = 2x \) and \( v = 2+x \). Variables' derivatives:- \( u' = 2 \), \( v' = 1 \)Substitute values into the quotient rule:\[ \frac{d}{dx}[\frac{2x}{2+x}] = \frac{(2+x)(2) - (2x)(1)}{(2+x)^2} = \frac{4 + 2x - 2x}{(2+x)^2} = \frac{4}{(2+x)^2} \]Combining both parts, we have:\[ f'(x) = \frac{1}{1+x} - \frac{4}{(2+x)^2} \]

Step 2: Determine where the derivative is positive

The function \( f(x) \) is increasing where \( f'(x) > 0 \). Solve:\[ \frac{1}{1+x} - \frac{4}{(2+x)^2} > 0 \]Multiply through by \((1+x)(2+x)^2\) to eliminate fractions:\[ (2+x)^2 > 4(1+x) \]Expand and simplify:\[ (4 + 4x + x^2) > (4 + 4x) \]Cancel terms on both sides:\[ x^2 > 0 \]The inequality \(x^2 > 0\) holds for all \(x eq 0\). Thus \(f'(x) > 0\) for all \(x eq 0\).

Step 3: Identify the intervals of increase

Since \( f'(x) > 0 \) for all \( x eq 0 \), the function is increasing everywhere except at \( x = 0 \). Therefore, the intervals on which the function is increasing are \((-\infty, 0)\) and \((0, \infty)\). There is no single interval that covers both, except for the disjoint union excluding zero.

Key Concepts

Derivative

The **derivative** of a function measures how the function's value changes as the input changes. It's a fundamental concept in calculus. To find the derivative of a function, we use differentiation. For the function given, we break it down into two parts:

• The derivative of \(\log(1+x)\) with respect to \(x\), which is \(\frac{1}{1+x}\).
• The derivative of \(\frac{2x}{2+x}\) using the quotient rule.

Understanding derivatives helps us analyze a function's behavior, like where it's increasing or decreasing.

Increasing Function

An **increasing function** is one where, as the input value rises, the output value also rises. We determine whether a function is increasing by checking its first derivative. If the derivative, \(f'(x)\), is positive over an interval, then the function is increasing on that interval.
For the exercise function, we found the derivative and analyzed its sign. \(f'(x) > 0\) on the intervals \((-\infty, 0)\)and \((0, \infty)\).
This tells us the function is increasing everywhere except at \(x = 0\).

Quotient Rule

The **quotient rule** is a technique used in calculus to find the derivative of a quotient of two functions. If you have a function \(\frac{u}{v}\), its derivative is \(\frac{vu' - uv'}{v^2}\).
For our problem, we applied the quotient rule to \(\frac{2x}{2+x}\).

• First, we found the derivatives \(u' = 2\)and \(v' = 1\).
• Then, substituted into the formula, which gave us \(\frac{4}{(2+x)^2}\).

Using the quotient rule helps accurately calculate derivatives when dealing with fractions of functions.

Inequality

In mathematics, an **inequality** compares two values, showing if one is less than, greater than, or not equal to the other. To find where the function is increasing, we needed the inequality \(f'(x) > 0\) to hold true.
This led us to solve the equation \((2+x)^2 > 4(1+x)\).
Once expanded and simplified, it became \(x^2 > 0\), indicating that \(f'(x)\) is greater than zero everywhere except at \(x = 0\).
Understanding and solving inequalities is crucial in analyzing the behavior of functions.