Đáp án:
T=-5
Giải thích các bước giải:
\(\begin{array}{l}
y' = \dfrac{{ - 2.2\cos x.\left( { - \sin x} \right)}}{{{{\left( {2{{\cos }^2}x + 3} \right)}^2}}} = \dfrac{{2.2\sin x\cos x}}{{{{\left( {2{{\cos }^2}x + 3} \right)}^2}}}\\
= \dfrac{{2.\sin 2x}}{{{{\left( {2{{\cos }^2}x + 3} \right)}^2}}}\\
y' = 0\\
\to 2.\sin 2x = 0\\
\to 2x = k\pi \left( {k \in Z} \right)\\
\to x = \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\\
Do:x \in \left[ { - \dfrac{\pi }{3};\dfrac{\pi }{6}} \right]\\
\to x = 0\\
Có:y\left( 0 \right) = \dfrac{1}{{2{{\cos }^2}0 + 3}} = \dfrac{1}{5}\\
y\left( { - \dfrac{\pi }{3}} \right) = \dfrac{1}{{2{{\cos }^2}\left( { - \dfrac{\pi }{3}} \right) + 3}} = \dfrac{2}{7}\\
y\left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{{2{{\cos }^2}\left( {\dfrac{\pi }{6}} \right) + 3}} = \dfrac{2}{9}\\
\to Max = y\left( { - \dfrac{\pi }{3}} \right) = \dfrac{2}{7}\\
\to a = 2;b = 7\\
\to T = a - b = 2 - 7 = - 5
\end{array}\)
