Multiple Choice Question (MCQ)

Given the function y=\sin x-\cos x+\sin x\cos x, x\in[0,\dfrac{\pi}{2} ], then the maximum value of the function is

  1. ×

    -1

  2. ×

    \dfrac{1}{2}

  3. 1

  4. ×

    2

Collected in the board: Trigonometry

Steven Zheng posted 2 months ago


Answer

  1. y=\sin x-\cos x+\sin x\cos x

    =\sin x-\cos x -\dfrac{(\sin x-\cos x)^2-1}{2}

    Let t=\sin x-\cos x

    =\sqrt{2}(\dfrac{\sqrt{2} }{2}\sin x-\dfrac{\sqrt{2} }{2} \cos x)

    =\sqrt{2}\sin(x-\dfrac{\pi}{4})

    -\sqrt{2} < t <\sqrt{2}

    Then,

    y=-\dfrac{1}{2} t^2+t+\dfrac{1}{2}

    =-\dfrac{1}{2} (t^2-2t-1)

    =-\dfrac{1}{2}(t+1)^2 +1

    Therefore,

    When t=-1, t_{max}=1

Steven Zheng posted 1 month ago

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