Question text 9Marks a) The Maclaurin series of [math: f(x)=2cos(x)]f(x) = 2\cos(x) is [math: f(x)=]Answer 1[input][math: +]Answer 2[input][math: x+]Answer 3[input][math: x2+⋯]x^2+\cdotsb) The Maclaurin series of [math: g(x)=6ln(1−x)]g(x) = 6\ln(1-x) is [math: g(x)=]Answer 4[input][math: +]Answer 5[input][math: x+]Answer 6[input][math: x2+⋯]x^2+\cdotsc) The Maclaurin series of [math: h(x)=2cos(3x)+6ln(1−x2)]h(x) = 2\cos(3x) + 6\ln(1-x^2) is[math: h(x)=]Answer 7[input][math: +]Answer 8[input][math: x+]Answer 9[input][math: x2+⋯]x^2+\cdotsNotes Report question issue Question 7 Notes多项填空题
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Question text 9Marks a) The Maclaurin series of [math: f(x)=2cos(x)]f(x) = 2\cos(x) is [math: f(x)=]Answer 1[input][math: +]Answer 2[input][math: x+]Answer 3[input][math: x2+⋯]x^2+\cdotsb) The Maclaurin series of [math: g(x)=6ln(1−x)]g(x) = 6\ln(1-x) is [math: g(x)=]Answer 4[input][math: +]Answer 5[input][math: x+]Answer 6[input][math: x2+⋯]x^2+\cdotsc) The Maclaurin series of [math: h(x)=2cos(3x)+6ln(1−x2)]h(x) = 2\cos(3x) + 6\ln(1-x^2) is[math: h(x)=]Answer 7[input][math: +]Answer 8[input][math: x+]Answer 9[input][math: x2+⋯]x^2+\cdotsPlease answer all parts of the question.Notes Report question issue Question 7 Notes
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