If f(x):[1,3]→[−1,1] satisfies ∫13f(x)dx=0, then the maximum value of ∫13f(x)xdx is

1−f(x)x,1+f(x)x≥0∀x∈[1,3] ∫121−f(x)xdx≥∫121−f(x)2dx ∫231+f(x)xdx≤∫231+f(x)2dx⇒−∫231+f(x)xdx≥−∫231+f(x)2dxAdding equation (1) and (2), we get ln 2−ln(32)≥∫23f(x)xdxEquality holds for ln 2−ln(32)≥∫23f(x)xdx

If f(x):[1,3]→[−1,1] satisfies ∫13f(x)dx=0, then the maximum value of ∫13f(x)xdx is

1−f(x)x,1+f(x)x≥0∀x∈[1,3] ∫121−f(x)xdx≥∫121−f(x)2dx ∫231+f(x)xdx≤∫231+f(x)2dx⇒−∫231+f(x)xdx≥−∫231+f(x)2dxAdding equation (1) and (2), we get ln 2−ln(32)≥∫23f(x)xdxEquality holds for ln 2−ln(32)≥∫23f(x)xdx

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If $f (x) : [1, 3] \to [- 1, 1]$ satisfies $\int_{1}^{3} f (x) d x = 0$ , then the maximum value of $\int_{1}^{3} \frac{f (x)}{x} d x$ is

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$\ln (\frac{3}{2})$

$\frac{8}{3}$

$\ln (\frac{4}{3})$

$\ln (\frac{9}{2})$

answer is B.

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Detailed Solution

$\frac{1 - f (x)}{x}, \frac{1 + f (x)}{x} \geq 0 \forall x \in [1, 3] \int_{1}^{2} \frac{1 - f (x)}{x} d x \geq \int_{1}^{2} \frac{1 - f (x)}{2} d x \int_{2}^{3} \frac{1 + f (x)}{x} d x \leq \int_{2}^{3} \frac{1 + f (x)}{2} d x \Rightarrow - \int_{2}^{3} \frac{1 + f (x)}{x} d x \geq - \int_{2}^{3} \frac{1 + f (x)}{2} d x$
Adding equation (1) and (2), we get $\ln 2 - \ln (\frac{3}{2}) \geq \int_{2}^{3} \frac{f (x)}{x} d x$
Equality holds for $\ln 2 - \ln (\frac{3}{2}) \geq \int_{2}^{3} \frac{f (x)}{x} d x$