Question

Answer 1

\(P=sinx-\left(1-sin^2x\right)+\dfrac{1}{2}\)

\(P=sin^2x+sinx-\dfrac{1}{2}=\left(sinx+\dfrac{1}{2}\right)^2-\dfrac{3}{4}\ge-\dfrac{3}{4}\)

\(\Rightarrow m=-\dfrac{3}{4}\)

\(P=sin^2x+sinx-2+\dfrac{3}{2}=\left(sinx-1\right)\left(sinx+2\right)+\dfrac{3}{2}\le\dfrac{3}{2}\)

\(\Rightarrow M=\dfrac{3}{2}\)

\(\Rightarrow Q=A\)