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Solve the following.
Let , where . Show that . The graph of has exactly one maximum point at P. Find the -coordinate of…
Let $f(x) = \frac{\log 5x}{kx}$, where $x > 0, k \in \mathbb{R}^+$.
(a)[1.0]
Show that $f'(x) = \frac{1-\log 5x}{kx^2}$.
(b)[1.0]
The graph of $f$ has exactly one maximum point at P. Find the $x$-coordinate of P.
(c)[1.0]
Find $f''(x)$.
(d)[1.0]
Find the value of $x$ for which $f''(x)$ vanishes.
Answer
Answer (a)
Official answer key$f(x) = \frac{\log 5x}{kx}$, $x > 0$, $k \in R^+$.
$= \frac{1}{k}\cdot\frac{\log 5x}{x}$
Therefore, $f'(x) = \frac{1}{k}\cdot\frac{x\cdot\frac{d}{dx}(\log 5x) - \log 5x\cdot\frac{d}{dx}(x)}{x^2}$
$= \frac{1}{k}\cdot\frac{x\cdot\frac{1}{5x}\cdot 5 - \log 5x}{x^2}$
$f'(x) = \frac{1 - \log 5x}{kx^2}$ Hence, proved.
- $f(x) = \frac{1}{k}\cdot\frac{\log 5x}{x}$
- $f'(x) = \frac{1}{k}\cdot\frac{x\cdot\frac{d}{dx}(\log 5x) - \log 5x\cdot\frac{d}{dx}(x)}{x^2}$ (quotient rule)
- $= \frac{1}{k}\cdot\frac{x\cdot\frac{1}{5x}\cdot 5 - \log 5x}{x^2}$
- $f'(x) = \frac{1 - \log 5x}{kx^2}$ Hence, proved.
Answer (b)
Official answer key$x$-coordinate of P is $\frac{e}{5}$
Final answer: $\frac{e}{5}$
Answer (c)
Official answer key$f''(x) = \frac{2\log 5x - 3}{kx^3}$
Final answer: $\frac{2\log 5x - 3}{kx^3}$
Answer (d)
Official answer key$\frac{1}{5}e^{3/2}$
Final answer: $\frac{1}{5}e^{3/2}$
From ISC 2025 Practice Mathematics, question 93.