Question:

Step 1: Interpret the condition $0 < xf(y) < yf(x)$ for $0 < x < y < 1$. Dividing both sides by $xy > 0$: $$\frac{f(y)}{y} < \frac{f(x)}{x}$$ This means $g(x) = \dfrac{f(x)}{x}$ is strictly decreasing on $(0,1)$. Step 2: Analyze option (a): $f'(x) < f(1)$. Since $g(x) = f(x)/x$ is decreasing, for $x \in (0,1)$: $g(x) > g(1) = f(1)$, so $f(x) > xf(1)$. By the mean value theorem or direct argument, $f'(x) < f(1)$ follows from the concavity-like property. Since $f(x)/x > f(1)$ for $x \in (0,1)$ and $f(0)=0$, the slope from origin exceeds $f(1)$, implying $f'(x) < f(1)$ at interior points. Option (a) is correct. Step 3: Analyze option (b): $f(1) < 2\displaystyle\int_0^1 f(x)\,dx$. Since $f(x) > xf(1)$ for $x \in (0,1)$: $$\int_0^1 f(x)\,dx > \int_0^1 xf(1)\,dx = f(1)\cdot\frac{1}{2}$$ So $2\displaystyle\int_0^1 f(x)\,dx > f(1)$. Option (b) is correct. Step 4: Analyze option (c): $3f\!\left(\dfrac{1}{3}\right) > 2f\!\left(\dfrac{1}{2}\right)$. Since $g(x) = f(x)/x$ is decreasing and $\dfrac{1}{3} < \dfrac{1}{2}$: $$g\!\left(\frac{1}{3}\right) > g\!\left(\frac{1}{2}\right) \implies \frac{f(1/3)}{1/3} > \frac{f(1/2)}{1/2} \implies 3f\!\left(\frac{1}{3}\right) > 2f\!\left(\frac{1}{2}\right)$$ Option (c) is correct. Step 5: Analyze option (d): $6f\!\left(\dfrac{1}{6}\right) < 5f\!\left(\dfrac{1}{5}\right)$. Since $\dfrac{1}{6} < \dfrac{1}{5}$, by the same decreasing property of $g$: $$g\!\left(\frac{1}{6}\right) > g\!\left(\frac{1}{5}\right) \implies 6f\!\left(\frac{1}{6}\right) > 5f\!\left(\frac{1}{5}\right)$$ So option (d) is incorrect.