Let f : S → S where S = (0, ∞) be a twice differentiable function such that f(x + 1) = xf(x). If g :

Q: Why can't we just differentiate $f(x + 1) = xf(x)$ directly? Why the log?

Excellent question. If you differentiate $f(x + 1) = xf(x)$ directly using the product rule, you get $f'(x + 1) = f(x) + xf'(x)$. This introduces $f(x)$ and $f'(x)$ in a way that doesn't easily isolate the second derivative. By taking the log, we convert the product $xf(x)$ into a sum $\ln x + \ln f(x)$, which makes the differentiation process linear and elegant.

Q: How do I know when to stop the telescoping sum?

Look at your target: $|g''(5) - g''(1)|$. This tells you the range of $x$ you need to consider. Since you have $g''(x + 1) - g''(x)$, you need to bridge the gap from $1$ to $5$. This requires $x = 1, 2, 3, 4$. The indices of the telescoping sum are dictated by the boundaries of the expression you are trying to evaluate.

Q: Is there a deeper meaning to this function $f(x)$?

Yes! The functional equation $f(x + 1) = xf(x)$ is the defining property of the Gamma function, $\Gamma(x)$. The function $g(x) = \ln \Gamma(x)$ is known as the log-gamma function. Its second derivative is related to the Trigamma function. You are essentially exploring the properties of one of the most important functions in higher mathematics!

Question

Let $f : S 	o S$ where $S = (0, \infty)$ be a twice differentiable function such that $f(x + 1) = xf(x)$. If $g : S 	o R$ be defined as $g(x) = \log_e f(x)$, then the value of $|g''(5) - g''(1)|$ is equal to :

Options:

(A) $\frac{205}{144}$

(B) $\frac{197}{144}$

(C) $\frac{187}{144}$

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex functional equation: $f(x + 1) = xf(x)$. At first glance, it looks like a simple recursive definition, but it hides a deep, structural beauty.

In JEE Advanced, we often encounter these multiplicative relationships. The secret to breaking them down is to transform them into something linear using the natural logarithm.

By defining $g(x) = \ln f(x)$, we can rewrite the equation as $\ln f(x + 1) = \ln(x f(x))$. Using the property $\ln(ab) = \ln a + \ln b$, this becomes $\ln f(x + 1) = \ln x + \ln f(x)$.

Substituting $g(x)$ back in, we get the elegant difference equation:
$$g(x + 1) - g(x) = \ln x$$

This is our bridge. We have moved from a multiplicative world to an additive one.

### The Derivative Dance: From Difference to Differential

Now that we have $g(x + 1) - g(x) = \ln x$, the path to the second derivative becomes clear. We need to differentiate this equation twice.

First, differentiating with respect to $x$, we get:
$$g'(x + 1) - g'(x) = \frac{d}{dx}(\ln x) = \frac{1}{x}$$

This is our first derivative relation. But the problem asks for the second derivative, so we must differentiate once more.

Differentiating again, we arrive at:
$$g''(x + 1) - g''(x) = \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}$$

This is the core relation that will allow us to solve the problem. It tells us exactly how the second derivative changes as we increment $x$.

### The Telescoping Symphony: The Final Cancellation

We are tasked with finding $|g''(5) - g''(1)|$. This is where the **telescoping sum** technique shines.

Let's write out the relation $g''(x + 1) - g''(x) = -\frac{1}{x^2}$ for $x = 1, 2, 3, 4$:

For $x = 1$: $g''(2) - g''(1) = -\frac{1}{1^2}$

For $x = 2$: $g''(3) - g''(2) = -\frac{1}{2^2}$

For $x = 3$: $g''(4) - g''(3) = -\frac{1}{3^2}$

For $x = 4$: $g''(5) - g''(4) = -\frac{1}{4^2}$

When we add these four equations, notice the magic: $g''(2)$ cancels with $-g''(2)$, $g''(3)$ cancels with $-g''(3)$, and $g''(4)$ cancels with $-g''(4)$.

We are left with:
$$g''(5) - g''(1) = -\left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2}\right)$$

Calculating the sum:
$$- \left(1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16}\right) = - \left(\frac{144 + 36 + 16 + 9}{144}\right) = -\frac{205}{144}$$

The absolute value is **$\frac{205}{144}$**. You have just navigated a classic JEE Advanced problem by transforming a functional equation into a beautiful, telescoping sum.

Animated Solution for Mathematics - Differentiation: Let $f : S \to S$ where $S = (0, \infty)$ be a twice differentiable function such that $f (x + 1) = x f (x)$ . If $g : S \to R$ be defined as $g (x) = lo g_{e} f (x)$ , then the value of $∣ g^{''} (5) - g^{''} (1) ∣$ is equal to :

Select Answer:

Visualized Solution

Analyze the Functional Equation

Apply Logarithms to the Equation

Substitute $g (x)$ into the Relation

First Differentiation

Second Differentiation

Evaluate at $x = 1$

Evaluate at $x = 2$

Evaluate at $x = 3$

Evaluate at $x = 4$

Summing the Equations (Telescoping)

Final Calculation

Final Answer

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

The Derivative Dance

The Telescoping Symphony

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Select Answer:

Visualized Solution

Analyze the Functional Equation

Apply Logarithms to the Equation

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First Differentiation

Second Differentiation

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