The maximum value of logx/x is

D. 1/e. Step 1: Let f(x)=logx/x for x>0. Then f′(x)=1-logxx^2. Step 2: f′(x)=0 gives logx=1, so x=e. Also f′>0 for x e, so x=e is a maximum. Step 3: f(e)=loge/e=1/e. Check: f(1)=0 and f(e^2)=2e^2, which is smaller than 1/e (see figure). Trap: Option (B) is 2/e, twice the true maximum; the graph never rises that high.

Question 5 of 34 · Applications of Derivatives

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The maximum value of $\frac{l o g x}{x}$ is

Answer (Detailed Solution Below)View answer ▾

Answer: Option D — $\frac{1}{e}$

Detailed Solution

Let $f (x) = \frac{l o g x}{x}$ for $x > 0$ . Then $f' (x) = \frac{1 - l o g x}{x ^{2}}$ .

$f' (x) = 0$ gives $l o g x = 1$ , so $x = e$ . Also $f' > 0$ for $x < e$ and $f' < 0$ for $x > e$ , so $x = e$ is a maximum.

$f (e) = \frac{l o g e}{e} = \frac{1}{e}$ .

✓ Check

f (1) = 0

and

f (e^{2}) = \frac{2}{e ^{2}}

, which is smaller than

\frac{1}{e}

(see figure).

⚠ Common trap

Option (B) is

\frac{2}{e}

, twice the true maximum; the graph never rises that high.

Quick tip

$\frac{d}{d x} (\frac{l o g x}{x}) = 0 \Rightarrow x = e, maximum value = \frac{1}{e}$

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Applications of Derivatives — Preparation

This question tests your understanding of Applications of Derivatives, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

This topic uses derivatives to solve practical and geometric problems. It covers the equations of tangents and normals to a curve, the rate of change of quantities (such as how fast the area of a circle grows as its radius increases), approximations using differentials, and increasing and decreasing functions. It also covers maxima and minima, using the first derivative test and the second derivative test, and word problems on optimisation such as finding the largest area for a fixed perimeter. Rolle's theorem and the mean value theorem are included as well. For maxima and minima, set f'(x) = 0 to find the critical points, then check f''(x): a negative value means a maximum and a positive value means a minimum. Remember that the slope of the tangent at a point is f'(x) and the slope of the normal is -1/f'(x). Practise the applications of derivatives questions below, each with a step-by-step solution.

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