Question

If \(f(x)\) is differentiable for all \(x\in\mathbb{R}\) and satisfies the relation \[ x=\lim_{n\rightarrow\infty}\frac{[1^{2}(f(x))^{x}]+[2^{2}(f(x))^{x}]+\dots+[n^{2}(f(x))^{x}]}{n^{3}}, \] where \([\cdot]\) denotes the greatest integer function, then \(f^{\prime}(x)\) is equal to:

• A. $\frac{1}{3x^{2}}\ln x$
• B. $3x^{1/x}(1-\ln 3x)$
• C. $(3x)^{\frac{1}{x}}\left[\frac{1-\ln 3x}{x^{2}}\right]$
• D. $(3x)^{\frac{1}{x}}\frac{(\ln 3x+1)}{x^{2}}$

Hint:

Whenever an integration or limit problem features a summation over $n^3$ containing $[r^2 \cdot K]$, you can conceptually treat the floor brackets as transparent. The sum $\sum r^2/n^3$ converges cleanly to $\int_0^1 x^2 dx = 1/3$, which immediately reduces the entire limit to $K/3$!

Answer 1

The Correct Option is C

Concept: We resolve the floor brackets inside the infinite limit using the Sandwich Theorem. For any real number $y$, the greatest integer function satisfies the bounding tracking property $y - 1 < [y] \le y$. Once the functional form of $f(x)$ is isolated from the limit, we compute its derivative using logarithmic differentiation. Step 1: Apply the Sandwich Theorem to eliminate the floor brackets.
Let us use the standard boundary inequality $y - 1 < [y] \le y$ for each individual term in the numerator sum: \[ \sum_{r=1}^{n} \left( r^2 (f(x))^x - 1 \right) < \sum_{r=1}^{n} [r^2 (f(x))^x] \le \sum_{r=1}^{n} r^2 (f(x))^x \] Factor out the constant variable block $(f(x))^x$ from the discrete summation loops: \[ (f(x))^x \left( \sum_{r=1}^{n} r^2 \right) - n < \sum_{r=1}^{n} [r^2 (f(x))^x] \le (f(x))^x \left( \sum_{r=1}^{n} r^2 \right) \]

Step 2: Evaluate the infinite limit of the bounding expressions.
Recall the standard polynomial identity for the sum of squares: $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6} \approx \frac{2n^3}{6} = \frac{n^3}{3}$. Divide across by the denominator $n^3$ and take the limit as $n \to \infty$: \[ \lim_{n \to \infty} \frac{(f(x))^x \cdot \frac{n^3}{3}}{n^3} = \frac{(f(x))^x}{3} \] By the Sandwich Theorem, the interior expression converges exactly to this same limit profile, yielding the functional equation: \[ x = \frac{(f(x))^x}{3} \quad \Rightarrow \quad (f(x))^x = 3x \quad \cdots (3) \]

Step 3: Isolate function $f(x)$ using logarithms.
Take the natural logarithm ($\ln$) on both sides of equation (3) to bring down the exponent: \[ \ln\left((f(x))^x\right) = \ln(3x) \quad \Rightarrow \quad x \cdot \ln(f(x)) = \ln(3x) \] Divide both sides by $x$ to completely isolate the logarithmic function layer: \[ \ln(f(x)) = \frac{\ln(3x)}{x} \quad \Rightarrow \quad f(x) = e^{\frac{\ln(3x)}{x}} = (3x)^{\frac{1}{x}} \quad \cdots (4) \]

Step 4: Differentiate expression using the quotient rule.
Differentiate $\ln(f(x)) = \frac{\ln(3x)}{x}$ implicitly with respect to $x$ using the standard calculus quotient rule: \[ \frac{1}{f(x)} \cdot f'(x) = \frac{\frac{1}{3x} \cdot 3 \cdot x - \ln(3x) \cdot 1}{x^2} = \frac{1 - \ln(3x)}{x^2} \] Multiply both sides by $f(x)$ to solve for the explicit derivative function $f'(x)$: \[ f'(x) = f(x) \cdot \left[ \frac{1 - \ln(3x)}{x^2} \right] \] Substitute $f(x) = (3x)^{\frac{1}{x}}$ from equation (4) into this expression: \[ f'(x) = (3x)^{\frac{1}{x}} \left[ \frac{1 - \ln(3x)}{x^2} \right] \] This matches option (C) perfectly.