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Question: Suppose that $f$ is a function on the interval $[1, 3]$ such that $-1 \leq f(x) \leq 1$ for all $x$ ...

Suppose that ff is a function on the interval [1,3][1, 3] such that −1≤f(x)≤1-1 \leq f(x) \leq 1 for all xx and ∫f(x)dx=0\int f(x) dx = 0. Find the maximum value of ∫13f(x)xdx\int_1^3 \frac{f(x)}{x} dx be?

A

log⁡163\log \frac{16}{3}

B

log⁡43\log \frac{4}{3}

C

1+log⁡431 + \log \frac{4}{3}

D

None of these

Answer

log⁡43\log \frac{4}{3}

Explanation

Solution

To maximize the integral ∫13f(x)xdx\int_1^3 \frac{f(x)}{x} dx, given the constraints, we should set f(x)=1f(x) = 1 where 1x\frac{1}{x} is large and f(x)=−1f(x) = -1 where 1x\frac{1}{x} is small. This leads to a step function:

f(x)={1if 1≤x≤a−1if a<x≤3f(x) = \begin{cases} 1 & \text{if } 1 \le x \le a \\ -1 & \text{if } a < x \le 3 \end{cases}

The condition ∫13f(x)dx=0\int_1^3 f(x) dx = 0 implies:

∫1a1 dx+∫a3(−1) dx=0\int_1^a 1 \, dx + \int_a^3 (-1) \, dx = 0 a−1−(3−a)=0a - 1 - (3 - a) = 0 2a−4=02a - 4 = 0 a=2a = 2

Thus, f(x)={1if 1≤x≤2−1if 2<x≤3f(x) = \begin{cases} 1 & \text{if } 1 \le x \le 2 \\ -1 & \text{if } 2 < x \le 3 \end{cases}

Now, calculate the integral:

∫13f(x)xdx=∫121xdx−∫231xdx=[ln⁡x]12−[ln⁡x]23=(ln⁡2−ln⁡1)−(ln⁡3−ln⁡2)=2ln⁡2−ln⁡3=ln⁡4−ln⁡3=ln⁡43\int_1^3 \frac{f(x)}{x} dx = \int_1^2 \frac{1}{x} dx - \int_2^3 \frac{1}{x} dx = [\ln x]_1^2 - [\ln x]_2^3 = (\ln 2 - \ln 1) - (\ln 3 - \ln 2) = 2 \ln 2 - \ln 3 = \ln 4 - \ln 3 = \ln \frac{4}{3}