Đáp án:
$${1 \over 2}\left( {A + C - \sqrt {{B^2} + {{\left( {A - C} \right)}^2}} } \right) \le M \le {1 \over 2}\left( {A + C + {B^2} + {{\left( {A - C} \right)}^2}} \right)$$
Giải thích các bước giải:
$$\eqalign{
& M = A{\sin ^2}x + B\sin x\cos x + C{\cos ^2}x \cr
& = A{{1 - \cos 2x} \over 2} + {B \over 2}\sin 2x + C{{1 + \cos 2x} \over 2} \cr
& = {1 \over 2}\left( {A - A\cos 2x + B\sin 2x + C + C\cos 2x} \right) \cr
& = {1 \over 2}\left( {B\sin 2x - \left( {A - C} \right)\cos 2x + A + C} \right) \cr
& Ta\,\,co:\,\,\left| {B\sin 2x - \left( {A - C} \right)\cos 2x} \right| \le \sqrt {{B^2} + {{\left( {A - C} \right)}^2}} \cr
& \Rightarrow - \sqrt {{B^2} + {{\left( {A - C} \right)}^2}} \le B\sin 2x - \left( {A - C} \right)\cos 2x \le {B^2} + {\left( {A - C} \right)^2} \cr
& \Rightarrow A + C - \sqrt {{B^2} + {{\left( {A - C} \right)}^2}} \le B\sin 2x - \left( {A - C} \right)\cos 2x + A + C \le A + C + {B^2} + {\left( {A - C} \right)^2} \cr
& \Rightarrow {1 \over 2}\left( {A + C - \sqrt {{B^2} + {{\left( {A - C} \right)}^2}} } \right) \le M \le {1 \over 2}\left( {A + C + {B^2} + {{\left( {A - C} \right)}^2}} \right) \cr} $$
