اگر تابع $f(x)=\left\{ \begin{align} & \frac{۲a}{\sqrt{x}}+۳x\,\,\,\,\,\,\,\,\,\,;x\ge ۱ \\ & b{{x}^{۲}}+۶\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;x \lt ۱ \\ \end{align} \right.$ در $x=۱$ مشتقپذیر باشد، حاصل $\frac{a}{b}$ کدام است؟
شرط پیوستگی :$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=f(1)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)$ $\Rightarrow 2a+3=b+6\Rightarrow 2a-b=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$ $f'(x)=\left\{ \begin{align} & 3-\frac{a}{\sqrt{{{x}^{3}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,;x\ge 1 \\ & 2bx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;x \lt 1 \\ \end{align} \right.$ شرط مشتقپذیری :$f_{+}^{'}(1)=f_{-}^{'}(1)$ $\Rightarrow 3-a=2b\Rightarrow a+2b=3\,\,\,\,\,\,\,\,\,(2)$ $\xrightarrow{(1),(2)}a=\frac{9}{5},b=\frac{3}{5}\Rightarrow \frac{a}{b}=3$