Let f be a real valued function, defined on R - {-1, 1} and given by f(x) = 3 loge |(x-1)/(x+1)| - 2

Q: Wait, why do we ignore the $(x-1)^2$ term when using the Wavy Curve Method?

That is a brilliant observation! We don't 'ignore' it; we recognize that for all $x$ in the domain, $(x-1)^2$ is strictly positive. Since multiplying or dividing an inequality by a positive number does not change the direction of the inequality sign, we can safely remove it from our sign analysis without losing any information. Just remember to keep the restriction $x \neq 1$ in your final answer!

Q: Why can't I just differentiate the whole thing using the chain rule immediately?

You absolutely can! However, the chain rule on $\ln|u|$ is $u'/u$. If you don't simplify the log first, you will end up with a very messy fraction that is difficult to simplify later. By using $\ln(a/b) = \ln a - \ln b$, you turn a complex quotient into two simple terms, making the derivative much cleaner and more elegant.

Q: Why is the interval for $x=1$ excluded even though it's not a critical point?

Great catch! The domain of the function is defined as $\mathbb{R} - \{-1, 1\}$. Even if the derivative suggests the function is 'increasing' at a certain point, if the function itself doesn't exist at that point, it cannot be part of the interval of increase. Always check your domain constraints at the very end of every problem.

Q: How do I know if I should use a closed bracket or an open bracket for the critical point $x = 1/2$?

Since the question asks where the function is increasing, we look for $f'(x) \ge 0$. At $x = 1/2$, the derivative is exactly zero. A function is considered increasing on an interval if its derivative is non-negative. Therefore, we include the point $x = 1/2$ with a closed bracket.

Q: Is there a trick to the Wavy Curve Method when there is a squared term in the denominator?

Yes! When you have a factor raised to an even power, like $(x-1)^2$, the sign of the expression does not change as you cross that root. However, since we are excluding $x=1$ from the domain entirely, we simply treat it as a 'hole' in our number line. The sign change only depends on the factors with odd powers, like $(2x-1)$ and $(x+1)$.

Question

Let $f$ be a real valued function, defined on $\mathbb{R} - \{-1, 1\}$ and given by $f(x) = 3 \log_e \left| \frac{x-1}{x+1} \right| - \frac{2}{x-1}$. Then in which of the following intervals, function $f(x)$ is increasing?

Options:

(A) (-\infty, -1) \cup ([\frac{1}{2}, \infty) - \{1\})

(B) (-\infty, \infty) - \{-1, 1\}

(C) (-1, \frac{1}{2}]

(D) (-\infty, \frac{1}{2}] - \{-1\}

Sigma Percentile · Accepted Answer

### The Architecture of Change: Unlocking the Function

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dissect a function that might look intimidating at first glance: $f(x) = 3 \ln \left| \frac{x-1}{x+1} ight| - \frac{2}{x-1}$.

When you see a function like this, it is easy to feel overwhelmed by the combination of logarithmic and rational terms. But remember, in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

### Phase 1: The Power of Simplification

Before we even think about calculus, let us simplify our life. We have the term $3 \ln \left| \frac{x-1}{x+1} ight|$.

Using the fundamental property of logarithms, $\ln \left( \frac{a}{b} ight) = \ln a - \ln b$, we can rewrite our function as:

$$f(x) = 3 \ln |x - 1| - 3 \ln |x + 1| - \frac{2}{x - 1}$$

Suddenly, the mountain of a problem has become a series of small, manageable hills. We are no longer dealing with a complex quotient inside a logarithm; we are dealing with simple, differentiable components.

This is the first secret of a master problem solver: **Simplify before you differentiate.**

### Phase 2: The Dance of Derivatives

Now, we enter the realm of calculus. We want to know where the function is increasing, which means we need to find where the slope, $f'(x)$, is non-negative ($f'(x) \ge 0$).

Let us differentiate term by term:

1. The derivative of $3 \ln |x-1|$ is $\frac{3}{x-1}$.
2. The derivative of $-3 \ln |x+1|$ is $-\frac{3}{x+1}$.
3. The derivative of $-\frac{2}{x-1}$ is $-2 \cdot \left( -\frac{1}{(x-1)^2} ight) = \frac{2}{(x-1)^2}$.

Combining these, we get:

$$f'(x) = \left( \frac{3}{x-1} - \frac{3}{x+1} ight) + \frac{2}{(x-1)^2}$$

### Phase 3: The Algebraic Convergence

This is where many students stumble, but you are not going to. Let us find a common denominator for the first two terms. The common denominator is $(x-1)(x+1) = x^2 - 1$.

Thus:

$$f'(x) = \frac{3(x+1) - 3(x-1)}{x^2-1} + \frac{2}{(x-1)^2} = \frac{6}{x^2-1} + \frac{2}{(x-1)^2}$$

To combine these, we need a common denominator for the whole expression, which is $(x-1)^2(x+1)$. Multiplying the terms appropriately, we arrive at the beautiful, simplified derivative:

$$f'(x) = \frac{6(x-1) + 2(x+1)}{(x-1)^2(x+1)} = \frac{8x - 4}{(x-1)^2(x+1)} = \frac{4(2x - 1)}{(x-1)^2(x+1)}$$

### Phase 4: The Final Verdict

We are almost there. We need $f'(x) \ge 0$. Since $(x-1)^2$ is always positive (for $x 
eq 1$), it does not affect the sign of the expression.

We are left with the inequality:

$$\frac{2x - 1}{x + 1} \ge 0$$

Using the Wavy Curve Method, we identify the critical points at $x = -1$ and $x = 1/2$. Testing the intervals, we find that the expression is positive in $(-\infty, -1)$ and $[1/2, \infty)$.

However, we must respect the domain of the original function, which forbids $x = 1$. Therefore, we must exclude $1$ from our interval.

Our final solution is **$(-\infty, -1) \cup

Animated Solution for Mathematics - Differentiation: Let $f$ be a real valued function, defined on $R - {- 1, 1}$ and given by $f (x) = 3 lo g_{e} \frac{x - 1}{x + 1} - \frac{2}{x - 1}$ . Then in which of the following intervals, function $f (x)$ is increasing?

Select Answer:

Visualized Solution

Analyze the Function $f (x)$

Condition for Increasing Function

Simplify using Log Properties

Differentiate Logarithmic Terms

Differentiate Rational Term

Combine First Two Terms

Simplify the Partial Derivative

Find Common Denominator

Final Expression for $f^{'} (x)$

Analyze the Sign of $f^{'} (x)$

Identify Critical Points

Apply Wavy Curve Method

Final Conclusion

The Sigma Insight: Monotonicity

The Architecture of Change

Phase 1

Phase 2

Phase 3

Phase 4

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