Let f be a differentiable function satisfying f(x) = 2√3_0^√3f!(^2/3x)d, x0, and f(1)=√3.

Question

Let $f$ be a differentiable function satisfying $f(x) = \dfrac2{\sqrt3}\displaystyle\int_0^{\sqrt3}f\!\left(\lambda^{2/3}x\right)d\lambda$, $x>0$, and $f(1)=\sqrt3$. If $y=f(x)$ passes through the point $(\alpha,6)$, then $\alpha$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 12

Accepted Answer

Differentiating both sides with respect to $x$:

$$f'(x) = \frac2{\sqrt3}\int_0^{\sqrt3}\lambda^{2/3}f'\!\left(\lambda^{2/3}x\right)d\lambda$$

Integrating by parts (treating $\lambda^{2/3}$ as arising from $d(\lambda^{2/3}x)/x$) and simplifying leads to

$$\sqrt3\,xf'(x) = \sqrt3f(x)-\frac{\sqrt3}2f(x) \implies xf'(x) = \frac{f(x)}2$$

Integrating: $\ln f = \dfrac12\ln x+\ln c$. Using $f(1)=\sqrt3$: $c=\sqrt3$, so $f(x)=\sqrt3\sqrt x$.

Since $(\alpha,6)$ lies on $y=f(x)$: $6=\sqrt3\sqrt\alpha \implies \sqrt\alpha = \dfrac6{\sqrt3}=2\sqrt3 \implies \alpha = 12$.

Solutions · 1