tìm các giá trị m nguyên để : sin^3 x + 3sin^2x + 4sinx +2-m= căn3 m có nghiệm

Đáp án:

\(0 \le m \le - 5 + 5\sqrt 3 \)

Giải thích các bước giải:

$$\eqalign{ & {\sin ^3}x + 3{\sin ^2}x + 4\sin x + 2 - m = \sqrt 3 m \cr & \Leftrightarrow {\sin ^3}x + 3{\sin ^2}x + 4\sin x + 2 = m\left( {\sqrt 3 + 1} \right) \cr & Dat\,\,t = \sin x\,\,\left( { - 1 \le t \le 1} \right) \cr & PT:\,\,{t^3} + 3{t^2} + 4t + 2 = m\left( {\sqrt 3 + 1} \right) \cr & \Leftrightarrow \left( {t + 1} \right)\left( {{t^2} + 2t + 2} \right) = m\left( {\sqrt 3 + 1} \right) \cr & \Leftrightarrow \left( {t + 1} \right)\left[ {{{\left( {t + 1} \right)}^2} + 1} \right] = m\left( {\sqrt 3 + 1} \right) \cr & \Leftrightarrow {\left( {t + 1} \right)^3} + \left( {t + 1} \right) = m\left( {\sqrt 3 + 1} \right) \cr & Dat\,\,u = t + 1\,\,\left( {0 \le u \le 2} \right) \cr & PT:\,\,{u^3} + u = m\left( {\sqrt 3 + 1} \right)\,\,\left( * \right) \cr & Xet\,\,ham\,\,so\,\,f\left( u \right) = {u^3} + u \cr & H = {{f\left( {{u_2}} \right) - f\left( {{u_1}} \right)} \over {{u_2} - {u_1}}} \cr & = {{u_2^3 + {u_2} - u_1^3 - {u_1}} \over {{u_2} - {u_1}}} \cr & = {{\left( {{u_2} - {u_1}} \right)\left( {u_2^2 + {u_2}{u_1} + u_1^2} \right) + \left( {{u_2} - {u_1}} \right)} \over {{u_2} - {u_1}}} \cr & = u_2^2 + {u_2}{u_1} + u_1^2 + 1 > 0\,\,\forall {u_1},\,\,{u_2} \cr & \Rightarrow Ham\,\,so\,\,DB\,\,tren\,\,\left( {0;2} \right) \cr & Ta\,\,co:\,\,0 \le u \le 2 \cr & \Rightarrow f\left( 0 \right) \le f\left( u \right) \le f\left( 2 \right) \cr & \Leftrightarrow 0 \le f\left( u \right) \le 10 \cr & Vay\,\,pt\,\,\left( * \right)\,\,co\,\,nghiem \cr & \Leftrightarrow 0 \le m\left( {\sqrt 3 + 1} \right) \le 10 \cr & \Leftrightarrow 0 \le m \le {{10} \over {\sqrt 3 + 1}} = - 5 + 5\sqrt 3 \cr} $$