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    Question
    SLPaper 2

    Consider the curves y=x2sin⁡xy=x^2 \sin xy=x2sinx and y=−1−1+4(x+2)2y=-1 - \sqrt{1+4(x+2)^2}y=−1−1+4(x+2)2​ for −π≤x≤0-\pi \leq x \leq 0−π≤x≤0.

    1.

    Find the xxx-coordinates of the points of intersection of the two curves.

    [3]
    Verified
    Solution

    • Attempts to solve x2sin⁡x=−1−1+4(x+2)2x^2 \sin x = -1 - \sqrt{1+4(x+2)^2}x2sinx=−1−1+4(x+2)2​ M1

    • x=−2.76x = -2.76x=−2.76 A1

    • x=−1.54x = -1.54x=−1.54 A1

    Award A1A0 if additional solutions outside the domain are given.

    3 marks total

    2.

    Find the area, A{A}A, of the region enclosed by the two curves.

    [4]
    Verified
    Solution
    • A=∫−2.762...−1.537...(−1−1+4(x+2)2−x2sin⁡x) dxA = \int_{-2.762...}^{-1.537...} (-1-\sqrt{1+4(x+2)^2}-x^2\sin x)\, dxA=∫−2.762...−1.537...​(−1−1+4(x+2)2​−x2sinx)dx (or equivalent) M1A1

    Award M1 for attempting to form an integrand involving "top curve" − "bottom curve".

    • A=1.47A = 1.47A=1.47 A2

    4 marks total

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