Đáp án: $M=\dfrac1{2\pi+1}$
Giải thích các bước giải:
Ta có :
$f'(x)=(\dfrac{x}{\cos x})'=\dfrac{x'\cos \left(x\right)-\left(\cos \left(x\right)\right)'\:x}{\left(\cos \left(x\right)\right)^2}=\dfrac{1\cdot \cos \left(x\right)-\left(-\sin \left(x\right)\right)x}{\left(\cos \left(x\right)\right)^2}=\dfrac{\cos x+x\sin x}{\cos^2x}$
$\to f'(\pi)=\dfrac{\cos\pi+\pi\sin\pi}{\cos^2\pi}=-1$
Lại có:
$g'(x)=((2x+1)\sin x)'=\left(2x+1\right)'\sin \left(x\right)+\left(\sin \left(x\right)\right)'\left(2x+1\right)=2\sin \left(x\right)+\cos \left(x\right)\left(2x+1\right)$
$\to g'(\pi)=2\sin \left(\pi\right)+\cos \left(\pi\right)\left(2\cdot\pi+1\right)=-(2\pi+1)$
$\to M=\dfrac{-1}{-(2\pi+1)}=\dfrac1{2\pi+1}$
