Questions
ENG1005 - MUM S2 2025 Practice Exam
Multiple fill-in-the-blank
Question texta) The Maclaurin series of [math: ex2]e^{x^2} is [math: ex2=]e^{x^2} =Answer 1 Question 9[input] [math: +] Answer 2 Question 9[input][math: x] [math: +] Answer 3 Question 9[input][math: x2]x^2 [math: +] Answer 4 Question 9[input][math: x3]x^3 [math: +…]+\ldotsb) The Maclaurin series of [math: ∫0x6ln(1+t)sin(t)dt]\int_0^x 6\ln(1+t)\sin(t) dt is [math: ∫0x6ln(1+t)sin(t)dt=]\int_0^x 6\ln(1+t)\sin(t) dt =Answer 5 Question 9[input] [math: +] Answer 6 Question 9[input][math: x] [math: +] Answer 7 Question 9[input][math: x2]x^2 [math: +] Answer 8 Question 9[input][math: x3]x^3 [math: +…]+\ldots
View Explanation
Verified Answer
Please login to view
Step-by-Step Analysis
The task contains two separate parts (a and b) with multiple blanks to fill in the Maclaurin (Taylor at 0) series coefficients.
Part a) The Maclaurin series of e^{x^2} is asked. To analyze the options:
- The function e^{x^2} is the exponential of x^2. Its Taylor expansion around x=0 only contains even powers of x, since e^{t} = 1 + t + t^2/2! + t^3/3! + ... and substituting t = x^2 yields terms x^{2k} for k=0,1,2,... . Therefore, the coefficient of x^1 (the x-term) must be 0, because there is no x^1 term in the series. This means the place where you would write the coefficient for x^1 should reflect a value of 0.
- The constant term (x^0) is e^{0} = 1, so the coefficient of x^0 is 1.
- The x^2 term comes from the k=1 term: (x^2)/1!, g......Login to view full explanationLog in for full answers
We've collected over 50,000 authentic exam questions and detailed explanations from around the globe. Log in now and get instant access to the answers!
Similar Questions
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
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
Question texta) The Maclaurin series of [math: ex2]e^{x^2} is [math: ex2=]e^{x^2} =Answer 1 Question 9[input] [math: +] Answer 2 Question 9[input][math: x] [math: +] Answer 3 Question 9[input][math: x2]x^2 [math: +] Answer 4 Question 9[input][math: x3]x^3 [math: +…]+\ldotsb) The Maclaurin series of [math: ∫0x6ln(1+t)sin(t)dt]\int_0^x 6\ln(1+t)\sin(t) dt is [math: ∫0x6ln(1+t)sin(t)dt=]\int_0^x 6\ln(1+t)\sin(t) dt =Answer 5 Question 9[input] [math: +] Answer 6 Question 9[input][math: x] [math: +] Answer 7 Question 9[input][math: x2]x^2 [math: +] Answer 8 Question 9[input][math: x3]x^3 [math: +…]+\ldotsPlease answer all parts of the question.
What are the constants in the first four terms of the Maclaurin series of \(\cos(x^2)\)? Input in the form \((a_0,a_1,a_2,a_3)\).
More Practical Tools for Students Powered by AI Study Helper
Making Your Study Simpler
Join us and instantly unlock extensive past papers & exclusive solutions to get a head start on your studies!