Question

Требуется решить задачу с пикчи

Answer 1

Ответ:   $\boldsymbol{\alpha =-\dfrac{\pi +1}{\pi }\approx -1,32}$   .

Производная дроби равна  $\bf \Big(\dfrac{u}{v}\Big)'=\dfrac{u'v-uv'}{v^2}$   .

$f(x)=\pi \cdot \dfrac{sinx}{x^2+\alpha \pi ^2}\ \ ,\ \ \ f'(\pi )=\dfrac{1}{\pi }\\\\\\f'(x)=\pi \cdot \dfrac{cosx\cdot (x^2+\alpha \pi ^2)-sinx\cdot 2x}{(x^2+\alpha \pi ^2)^2}\\\\\\f'(\pi )=\dfrac{cos\pi \cdot (\pi ^2+\alpha \pi ^2)-sin\pi \cdot 2\pi }{(\pi ^2+\alpha \pi ^2)^2}=\dfrac{-1\cdot (1+\alpha )\cdot \pi ^2-0}{(1+\alpha )^2\cdot \pi ^4}=-\dfrac{1}{(1+\alpha )\cdot \pi ^2}$

$-\dfrac{1}{(1+\alpha )\cdot \pi ^2}=\dfrac{1}{\pi }\ \ \ \Rightarrow \ \ \ (1+\alpha )\cdot \pi ^2=-\pi \ \ ,\ \ \ (1+\alpha )\cdot \pi =-1\ ,\\\\\\1+\alpha =-\dfrac{1}{\pi }\ \ ,\ \ \ \alpha =-1-\dfrac{1}{\pi }\ \ \ ,\ \ \ \boldsymbol{\alpha =-\dfrac{\pi +1}{\pi }}\approx -1,32$