Question

1) Para que valores de m existam x tal que senx = (m-1)/(m-2)

2) Para que valores de m existam x tal que senx = 2m -5?

Answer 1

$\large\boxed{\begin{array}{l}\rm 1)~\sf sen(x)\exists\Longleftrightarrow-1\leqslant sen(x)\leqslant1\\\sf -1\leqslant\dfrac{m-1}{m-2}\leqslant1\implies\begin{cases}\sf\dfrac{m-1}{m-2}\geqslant-1\\\\\sf\dfrac{m-1}{m-2}\leqslant1\end{cases}\end{array}}$

$\boxed{\begin{array}{l}\sf\dfrac{m-1}{m-2}\geqslant-1\\\\\sf\dfrac{m-1}{m-2}+1\geqslant0\\\\\sf\dfrac{m-1+m-2}{m-2}\geqslant0\\\\\sf\dfrac{2m-3}{m-2}\geqslant0\\\sf f(m)=2m-3\\\sf f(m)\geqslant0~para~m\geqslant\dfrac{3}{2}\\\sf f(m)<0~para~m<\dfrac{3}{2}\\\sf g(m)=m-2\\\sf g(m)>0~para~m>2\\\sf g(m)<0~para~g<2\\\underline{\rm montando~o~quadro~sinal~e}\\\underline{\rm assinalando~a~resposta~temos}\\\sf S_1=\bigg\{m\in\mathbb{R}/m\leqslant\dfrac{3}{2}~ou~m>2\bigg\}\end{array}}$

$\boxed{\begin{array}{l}\sf\dfrac{m-1}{m-2}\leqslant1\\\\\sf \dfrac{m-1}{m-2}-1\leqslant0\\\\\sf\dfrac{m-1-m+2}{m-2}\leqslant0\\\sf\dfrac{1}{m-2}\leqslant0\\\sf \\\sf h(m)=1\implies h(m)>0~\forall m\in\mathbb{R}\\\sf g(m)=m+2\\\sf g(m)>0~para~m>2\\\sf g(m)<0~para~m<2\\\sf S_2=\{m\in\mathbb{R}/m<2\}\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm fazendo~a~intersecc_{\!\!,}\tilde ao~das~soluc_{\!\!,}\tilde oes~temos:}\\\sf S=\bigg\{m\in\mathbb{R}/m\leqslant\dfrac{3}{2}\bigg\}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm 2)~\sf sen(x)\exists\Longleftrightarrow -1\leqslant sen(x)\leqslant1\\\sf-1\leqslant 2m-5\leqslant1\\\sf -1+5\leqslant2m\leqslant1+5\\\sf 4\leqslant2m\leqslant6\\\sf\dfrac{4}{2}\leqslant m\leqslant\dfrac{6}{2}\\\sf 2\leqslant m\leqslant3\end{array}}$