Question 9: Given that $f(x)$ is an odd function defined on $R$, and when $x<0$, $f(x)$ is monotonically increasing, $f(-1)=0$. Let $\varphi(x)=\sin ^{2} x+m \cdot \cos x-2 m$, set $M=\left\{m \mid\right.$ for $\left.\forall x \in\left[0, \frac{\pi}{2}\right], \varphi(x)<0\right\}, N=\{m \mid$ for $\left.\forall \mathrm{x} \in\left[0, \frac{\pi}{2}\right], \mathrm{f}(\varphi(\mathrm{x}))<0\right\}$, find $\mathrm{M} \cap \mathrm{N}$.
Question 9, Answer: Since $f(x)$ is an odd function defined on $R$, and when $x>0$, $f(x)$ is also monotonically increasing, and $f(1)=0$.

According to the definition: $M \cap N=\left\{m \mid\right.$ for $\forall x \in\left[0, \frac{\pi}{2}\right], \varphi(x)>0$
$\Leftrightarrow(2-\cos x) m>2-\cos ^{2} x$
$\Leftrightarrow m>\max _{x \in\left[0, \frac{\pi}{2}\right]} \frac{2-\cos ^{2} x}{2-\cos x}=\max _{t \in[0,1]}\left\{4-\left[(2-t)+\frac{2}{2-t}\right]\right\}=4-2 \sqrt{2}$
In summary, we know $M \cap N=(4-2 \sqrt{2},+\infty)$.