Đáp án:
`5/{64}`
Giải thích các bước giải:
`M=sin^4x-cos^2x+5/ 4 cos2x`
`M=(1-cos^2x)^2-cos^2x+5/ 4 .(2cos^2x-1)`
`M=(1-2cos^2 x+cos^4x)-cos^2x+5/ 2 cos^2x -5/ 4`
`M=cos^4x -1/ 2 cos^2x -1/ 4`
`M=(cos^2 x-2. cos^2x . 1/ 4 +1/{16})-1/{16}-1/ 4`
`M=(cos^2 x -1/ 4)^2 - 5/{16}`
Với mọi `π/4\le x\le {2π}/3` ta có:
`\qquad 0\le |cosx|\le \sqrt{2}/2`
`=>0\le cos^2x \le 1/ 2`
`=>-1/ 4 \le cos^2 x -1/ 4 \le 1/ 4`
`=>0\le |cos^2 x-1/ 4|\le 1/ 4`
`=>0\le (cos^2 x-1/ 4)^2 \le 1/{16}`
`=> -5/{16}\le (cos^2 x-1/ 4)^2- 5/{16}\le 1/{16}-5/{16}`
`=>-5/{16}\le A\le -1/ 4`
`=>M_{min}=-5/{16}`
`\qquad M_{max}=-1/4`
`=>M_{min}.M_{max}={-5}/{16} . {-1}/ 4=5/{64}`
Vậy `M_{min}.M_{max}=5/{64}`
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`M_{max}=-1/ 4` khi $\left[\begin{array}{l}x=\dfrac{π}{4}\\x=\dfrac{π}{2}\end{array}\right.$
`M_{min}=-5/{16}` khi `cos^2x=1/ 4`
`<=>cosx=±1/ 2<=>`$\left[\begin{array}{l}x=\dfrac{π}{3}\\x=\dfrac{2π}{3}\end{array}\right.$
