Tìm điều kiện của m để phương trình 2^sin^2x có nghiệm.

* Hướng dẫn giải

Đáp án D

$$\begin{gathered} 2^{{\sin }^2\mathrm{x}}+3^{{\cos }^2\mathrm{x}}=\mathrm{m}.3^{{\sin }^2\mathrm{x}}(*) \\ \mathrm{Đặt} {\sin }^2\mathrm{x}=\mathrm{t} (0\le \mathrm{t}\le 1) \\ \Rightarrow 2^{\mathrm{t}}+3^{1-\mathrm{t}}=\mathrm{m}.3^{\mathrm{t}} \\ \iff {\left(\frac{2}{3}\right)}^{\mathrm{t}}+3^{1-2\mathrm{t}}=\mathrm{m} \\ \mathrm{Đặt} \mathrm{f}\left(\mathrm{t}\right)={\left(\frac{2}{3}\right)}^{\mathrm{t}}+3^{1-2\mathrm{t}} \\ \mathrm{f}\text{'}\left(\mathrm{t}\right)={\left(\frac{2}{3}\right)}^{\mathrm{t}}\ln \left(\frac{2}{3}\right)-2.3^{1-2\mathrm{t}}\ln 3<0 \mathrm{với}\forall \mathrm{t}\in [0;1] \\ \Rightarrow \mathrm{Hàm} \mathrm{f}\left(\mathrm{t}\right) \mathrm{NB} \mathrm{trên} \mathrm{đoạn} [0;1] \\ \mathrm{Ta} \mathrm{có} \mathrm{f}\left(0\right)=4 \mathrm{và} \mathrm{f}\left(1\right)=1 \\ \mathrm{KL}: \mathrm{PT} (*) \mathrm{có} \mathrm{nghiệm} \mathrm{khi} 1\le \mathrm{m}\le 4 \end{gathered}$$