Science:Math Exam Resources/Courses/MATH101/April 2016/Question 02 (a)

MATH101 April 2016

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q7  •  Q8 (a)  •  Q8 (b)  •  Q9 (a)  •  Q9 (b)  •  Q10 (a)  •  Q10 (b)  •  Q11 (a)  •  Q11 (b)  •  

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Question 02 (a)

Only one of the following statements is always true; determine which one is true. (Assume $f(x)$ and $g(x)$ are continuous functions.) A: If $\displaystyle \lim _{n\rightarrow \infty }a_{n}=0$, then $\textstyle \sum _{n=1}^{\infty }a_{n}$ converges. B: If $f(x)$ is an odd function, then $\displaystyle \int _{-3}^{-2}f(x)\,dx=\int _{2}^{3}f(x)\,dx$. C: If $\displaystyle \lim _{n\rightarrow \infty }a_{n}=0$, then $\sum _{n=1}^{\infty }(-1)^{n}a_{n}$ converges. D: If $g(x)\geq f(x)\geq 0$ for all $x$ and $\displaystyle \int _{1}^{\infty }\!\!g(x)\,dx$ converges, then $\displaystyle \int _{1}^{\infty }\!\!f(x)\,dx$ converges. E: $\displaystyle {\int _{0}^{7}f(x^{2})\,x\,dx}={\int _{0}^{7}f(u)\,{\frac {du}{2}}}.$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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Hint
For A and C, consider the fact that $\sum _{i=1}^{\infty }{\frac {1}{n}}$ diverges. For B, try to compute the integrals for some odd function. (for example, $f(x)=x$ ) For D, recall the comparison test. For E, try the substitution $u=x^{2}$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. A: False. A counter example is $a_{n}={\frac {1}{n}}$. B: False. Try $f(x)=x$. Then $\int _{-3}^{-2}xdx=-{\frac {5}{2}}\neq {\frac {5}{2}}=\int _{2}^{3}xdx.$ In fact, using substitution $u=-x$, $\int _{-3}^{-2}xdx=\int _{3}^{2}f(-u)(-du)=-\int _{2}^{3}(-f(-u))du=-\int _{2}^{3}f(u)du=-\int _{2}^{3}f(x)dx$. Note that $-f(-u)=f(u)$ because $f$ is an odd function. C: False. A counter example is $a_{n}=(-1)^{n}{\frac {1}{n}}$. D: True. Since $\int _{1}^{\infty }g(x)dx$ converges, this integral has certain value(number), say $M$. Then the assumption on $g(x)\geq f(x)\geq 0$ implies that $0=\int _{1}^{\infty }0dx\leq \int _{1}^{\infty }f(x)dx\leq \int _{1}^{\infty }g(x)dx=M$ by the comparison property of integral. Therefore, $0\leq \int _{1}^{\infty }f(x)dx\leq M$, so that $\int _{1}^{\infty }f(x)dx$ also has a certain value between 0 and M. i.e., convergent. E: False. A counter example is $f(x)=1$. $\int _{0}^{7}f(u){\frac {du}{2}}=\int _{0}^{7}{\frac {1}{2}}du={\frac {7}{2}}$, while $\int _{0}^{7}f(x^{2})xdx=\int _{0}^{7}xdx={\frac {49}{2}}$. In fact, using substitution $u=x^{2}$, $\int _{0}^{7}f(u){\frac {du}{2}}=\int _{0}^{\sqrt {7}}f(x^{2})xdx$. The answer is D.