If c is a point at which Rolle's theorem holds for the function, f(x) = log_e((x² + α)/(7x)) in the

Q: Why can't I just use the quotient rule immediately on the original function?

You absolutely can, but in JEE Advanced, time is currency. Expanding the logarithm first turns a messy quotient rule problem into a simple sum of derivatives. It reduces the probability of a sign error significantly and keeps the algebra clean.

Q: Do I really need to verify that $c$ is in $(3, 4)$?

Yes, always. Rolle's Theorem is a statement about the existence of a point *within* the open interval. If your calculated $c$ fell outside, the theorem wouldn't apply, and you would know you made a calculation error. It is a vital sanity check.

Q: Why did we get $f''(c) = 1/12$ so easily? Did I miss something?

You didn't miss anything! That is the beauty of the problem design. The term involving $c^2$ in the numerator of the second derivative vanishes precisely because of the condition $f'(c) = 0$. It is a 'math magic' moment where the structure of the function does the heavy lifting for you.

Question

If $c$ is a point at which Rolle's theorem holds for the function, $f (x) = \log_e \left(\frac{x^2 + \alpha}{7x}\right)$ in the interval $[3, 4]$, where $\alpha \in \mathbb{R}$, then $f''(c)$ is equal to :

Options:

(A) $\frac{1}{12}$

(B) $\frac{1}{12}$

(C) $-\frac{1}{6}$

(D) $0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with the function $f(x) = \log_e \left( \frac{x^2 + \alpha}{7x} \right)$ defined on the interval $[3, 4]$. The problem demands that Rolle's Theorem holds, which serves as our golden key.

Rolle's Theorem dictates that $f(3)$ must equal $f(4)$. If these values were not equal, the function would not be constrained in a way that forces a horizontal tangent.

By setting $f(3) = f(4)$, we create an equation where the logarithms cancel out:

$$ \frac{9 + \alpha}{21} = \frac{16 + \alpha}{28} $$

Solving this algebraic relationship yields $\alpha = 12$. We have successfully unlocked the function.

### The Art of Simplification

Now, we have $f(x) = \log_e \left( \frac{x^2 + 12}{7x} \right)$. While many would reach for the quotient rule, we can utilize the logarithmic property $\log(a/b) = \log a - \log b$.

This transforms our function into:

$$ f(x) = \log_e(x^2 + 12) - \log_e(7x) $$

Differentiation now becomes a breeze. The derivative $f'(x)$ is given by:

$$ f'(x) = \frac{2x}{x^2 + 12} - \frac{1}{x} $$

This is the power of strategic simplification. Instead of wrestling with a complex quotient, we have two elegant terms that are easy to handle.

### The Hunt for the Critical Point

For Rolle's Theorem to hold, there must exist a point $c \in (3, 4)$ such that $f'(c) = 0$. Setting our derivative to zero, we obtain:

$$ \frac{2c}{c^2 + 12} - \frac{1}{c} = 0 $$

This simplifies to $c^2 - 12 = 0$, or $c^2 = 12$. We note that $c = \sqrt{12} \approx 3.46$, which sits comfortably inside our interval $(3, 4)$.

### The Second Derivative: The Final Flourish

Finally, we calculate $f''(c)$. We differentiate $f'(x)$ again to find the second derivative:

$$ f''(x) = \frac{d}{dx} \left( \frac{2x}{x^2 + 12} - \frac{1}{x} \right) $$

Using the quotient rule on the first term and the power rule on the second, we get:

$$ f''(x) = \frac{2(x^2 + 12) - 2x(2x)}{(x^2 + 12)^2} + \frac{1}{x^2} = \frac{24 - 2x^2}{(x^2 + 12)^2} + \frac{1}{x^2} $$

When we substitute $x^2 = 12$ into this expression, the numerator $24 - 2(12)$ becomes zero. The entire first term vanishes, leaving us with:

$$ f''(c) = 0 + \frac{1}{12} = \frac{1}{12} $$

The final result is **$\frac{1}{12}$**.

Animated Solution for Mathematics - Differentiation: If $c$ is a point at which Rolle's theorem holds for the function, $f (x) = lo g_{e} (\frac{x ^{2} + α}{7 x})$ in the interval $[3, 4]$ , where $α \in R$ , then $f^{''} (c)$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Function and Interval

Applying Rolle's Condition

Solving for $α$

Redefining the Function

Finding the First Derivative $f^{'} (x)$

Applying $f^{'} (c) = 0$

Solving for $c$

Verifying the Point $c$

Calculating the Second Derivative $f^{''} (x)$

Simplifying $f^{''} (x)$

Substituting $c$ into $f^{''} (x)$

Final Result

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Art of Simplification

The Hunt for the Critical Point

The Second Derivative

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Visualized Solution

Visualizing the Function and Interval

Applying Rolle's Condition

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Redefining the Function

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Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(c)=0

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Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′′(x)

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′′(x)