Question #17:
This is a geometric series. If \(x\ne1\), then the partial sum has the
following closed form: \(\displaystyle 1 + x + x^2 + \cdots +
x^n=\frac{x^n-1}{x-1}\). So the limit exists (and is equal to
\(\displaystyle \frac{1}{1-x}\) if \(-1\lt x\lt 1\)), otherwise the
limit does not exist. Alternatively, you can use the ratio test or the
root test to show that it converges for \(|x|<1\) and diverges for
\(|x|>1\), and then check \(x=1\) and \(x=-1\) directly.
Answer: (c)
Question #18:
The series \(\displaystyle {1 \over 1^p} + {1 \over 2^p} + {1 \over
3^p} + \cdots\) converges if \(p>1\) and diverges if \(p\le1\). This
can be checked by the integral test.
Answer: (a)
Question #19:
\(\displaystyle \frac{1}{1+x}=1-x+x^2-x^3+\cdots\). This is a
geometric series.
Answer: (b)
Question #20:
The series \(\displaystyle 1 - {x^2 \over 2!} + {x^4 \over 4!} - {x^ 6
\over 6!} + \cdots\) is the Taylor series for \(\cos(x)\) around
0.
Answer: (d)
Question #21:
\(x = 2 \cos t\) and \(y = 1 + \sin t\).
\(\displaystyle \left(\frac{x(t)}{2}\right)^2 + (y(t)-1)^2
= \cos^2(t)+\sin^2(t) = 1\).
\(\displaystyle \frac{x^2}{4}+(y-1)^2=1\) is an ellipse centered at (0,1)
and with semi-axes of radius 2 and 1.
Answer: (b)
Question #22:
\(\bigl(x(t),y(t)\bigr) = (1+t^2,1-t^3)\).
\(\bigl(x'(t),y'(t)\bigr) = (2t,-3t^2)\).
\(\bigl(x'(1),y'(1)\bigr) = (2,-3)\), a vector of length \(\sqrt{13}\).
Answer: (c)
Question #23:
\(\displaystyle {dy \over dx} + y = 1\).
\(\displaystyle \frac{1}{1-y}\frac{dy}{dx}=1\).
\(-\ln(1-y)=\ln(x)+c\).
\(\displaystyle \frac{1}{1-y}=Ce^x\), where \(C=e^c\).
Substitute \(x=0\) and \(y=0\) to get \(1=C\),
so \(\displaystyle \frac{1}{1-y}=e^x\)
\(\displaystyle 1-y=\frac{1}{e^x}\).
\(\displaystyle y=1-\frac{1}{e^x}\).
Substitute \(x=1\) to get \(\displaystyle
y(1)=1-\frac{1}{e}=\frac{e-1}{e}\).
Answer: (c)
Question #24:
The general solution to \(y''+y=0\) is \(y=c_1 \cos x + c_2 \sin x\),
and a specific solution to \(y''+y=e^x\)
is \(\displaystyle y=\frac{1}{2}e^x\),
so the general solution to \(y''+y=e^x\) is
\(\displaystyle y=c_1 \cos x + c_2 \sin x+\frac{1}{2}e^x\).
Answer: (d)