Problems

Solution:

+ - ↺

1. Since the area not visible is increasing as \(p\) increases, we would like \(p\) to be as large as possible, ie \(p = 4\).
2. Similarly, he can see the most when \(p =0\)

Solution:

1. \begin{align*} && g(x) &= \frac{2x^3+3}{x^4+4} \\ \Rightarrow && g'(x) &= \frac{6x^2(x^4+4) - (2x^3+3)(4x^3)}{(x^4+4)^2} \\ &&&= \frac{-2x^6-12x^3+24x^2}{(x^4+4)} \\ &&&= \frac{-2x^2(x^4+6x-12)}{(x^4+4)} \end{align*} So \(g'(x)\) is not \(0\) for \(x = \pm 1\). We can also note that \(g(-1) = \frac1{5} \neq 1\) Even if the other turning point was \(1\), we would also need to check the behaviour as \(x \to \pm \infty\). We can also note that \(g(-1) = \frac{1}{5} < \frac34\) so the conclusion is also not true.
2. There are several errors. \[ \int (1-x)^{-3} \d x = \underbrace{\frac{1}{4}}_{\text{correct constant}}(1-x)^{-4} + \underbrace{C}_{\text{constant of integration}} \] We cannot integrate through the asymptote at \(1\). There is a sense in which we could argue \(\displaystyle \int_{-1}^3 (1-x)^{-3} \d x = 0\), specifically using Cauchy principal value \begin{align*} \mathrm {p.v.} \int_{-1}^3 (1-x)^{-3} &=\lim_{\epsilon \to 0} \left [ \int_{-1}^{1-\epsilon} (1-x)^{-3} \d x+ \int_{1+\epsilon}^{3} (1-x)^{-3} \d x\right] \\ &=\lim_{\epsilon \to 0} \left [ \left[ \frac14 (1-x)^{-4}\right]_{-1}^{1-\epsilon}+ \left[ \frac14 (1-x)^{-4}\right]_{1+\epsilon}^3\right] \\ &=\lim_{\epsilon \to 0} \left [ \frac14 \epsilon^{-4}-\frac14 \frac1{2^4} + \frac14 \frac1{2^4} - \frac14 \epsilon^{-4} \right] \\ &= \lim_{\epsilon \to 0} 0 \\ &= 0 \end{align*} However, in many normal ways of treating this integral it would be undefined.

Solution: \[ f(x) = 2|x(x-1)| + |(x-1)(x+1)|-2|x(x+1)| \] Therefore the absolute value terms will change behaviour at \(x = -1, 0, 1\). Then \begin{align*} f(x) &= \begin{cases} 2(x^2-x)+(x^2-1)-2(x^2+x) & x \leq -1 \\ 2(x^2-x)-(x^2-1)+2(x^2+x) & -1 < x \leq 0 \\ -2(x^2-x)-(x^2-1)-2(x^2+x) & 0 < x \leq 1 \\ 2(x^2-x)+(x^2-1)-2(x^2+x) & 1 < x\end{cases} \\ &= \begin{cases} x^2-4x-1 & x \leq -1 \\ 3x^2+1& -1 < x \leq 0 \\ -5x^2+1& 0 < x \leq 1 \\ x^2-4x-1 & 1 < x\end{cases} \\ \\ f'(x) &= \begin{cases} 2x-4 & x <-1 \\ 6x & -1 < x < 0 \\ -10x & 0 < x < 1 \\ 2x-4 & 1 < x\end{cases} \\ \end{align*} Therefore \(f'(x) = 0 \Rightarrow x = 0, 2\) and so we should check all the turning points. Therefore the minimum is \(x = 2, y = -5\), maximum is \(x = -2, y = 11\) (assuming the range is actually \(|x| < 2\). There is a point of inflection at \(x = 0, y = 1\).

+ - ↺

Solution:

1. \begin{align*} \int_{1}^{3}\frac{1}{6x^{2}+19x+15}\,\mathrm{d}x &= \int_1^3 \frac1{(2x+3)(3x+5)} \d x \\ &= \int_1^3 \l \frac{2}{2x+3} - \frac{3}{3x+5} \r \d x \\ &= \left [\ln(2x+3) - \ln(3x+5) \right ]_1^3 \\ &= \l \ln9 - \ln14 \r - \l \ln 5 - \ln 8 \r \\ &= \ln \frac{72}{70} \\ &= \ln \frac{36}{35} \end{align*}
2. 

   + - ↺
   When \(q = 0\) the roots are \(-1, 0, 1\) There can be \(0, 2, 3, 4\) roots. There will be no roots if \(q > -\min (x^{1760} - x^{220})\) since the whole graph will be above the axis. There will be \(2\) roots if \(q = -\min (x^{1760} - x^{220})\) or \(q > 0\) There will be \(4\) roots if \(0 > q > -\min (x^{1760} - x^{220})\). There will be \(3\) roots if \(q =0\)

Solution:

1. 

   + - ↺
2. 

   + - ↺
3. 

   + - ↺
4. 

   + - ↺
5. 

   + - ↺
6. 

   + - ↺
7. 

   + - ↺
8. 

   + - ↺

+ - ↺

+ - ↺

+ - ↺

Solution:

+ - ↺

Solution: \(g'(x) = f'(x) - \frac{f'(x)}{(f(x))^2} = f'(x) \l 1 - \frac{1}{(f(x))^2} \r\) \(g''(x) = f''(x) - \frac{f''(x)f(x)^2-f'(x)\cdot 2f(x) f'(x)}{(f(x))^4} = f''(x) + \frac{f''(x)f(x)-2(f'(x))^2}{(f(x))^3}\) \begin{align*} f(x) &=4+\cos2x+2\sin x \\ f'(x) &=-2\sin2x+2\cos x \\ f''(x) &= -4\cos2x-2\sin x \end{align*} Therefore, since the stationary points of \(g\), ie points where \(g'(x) = 0\) are where \(f'(x) = 0\) or \(f(x) = \pm 1\) we should look at \begin{align*} && 0 &= f'(x) \\ && 0 &= 2 \cos x - 2 \sin 2x \\ &&&= 2 \cos x - 4 \sin x \cos x \\ &&&= 2\cos x (1 - 2 \sin x) \\ \Rightarrow && x &= \frac{\pi}2, \frac{3\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6} \end{align*} \begin{align*} && 1 &= f(x) \\ && 1 &= 4 + \cos 2x + 2 \sin x \\ \Rightarrow && \cos 2x = -1,& \sin x = -1 \\ \Rightarrow && x &= \frac{3\pi}{2} \end{align*} which we were already checking. For each of these points we have: \begin{array}{c|c|c|c||c} x & f(x) & f'(x) & f''(x) & g''(x) \\ \hline \frac{\pi}{2} & 5 & 0 & 2 & > 0\\ \frac{3\pi}{2} & 1 & 0 & 6 &> 0\\ \frac{\pi}{6} & 5.5 & 0 & -3 & < 0 \\ \frac{5\pi}{6} & 5.5 & 0 & -3 & < 0\\ \end{array} Therefore \(\frac{\pi}{2}, \frac{3\pi}{2}\) are minimums and \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\) are maxima.

Solution:

+ - ↺

Solution: Let \(f(x) = e^{ax} \cos bx\) then, \(f'(x) = ae^{ax} \cos bx - be^{ax} \sin bx = e^{ax} \l a\cos bx - b \sin bx \r\). Therefore the stationary points are where \(f'(x) = 0 \Leftrightarrow \tan bx = \frac{b}a\), ie \(x = \tan^{-1} \frac{a}{b} + \frac{n}{b} \pi, n \in \mathbb{Z}\). \begin{align*} f(\tan^{-1} \frac{a}{b} + \frac{n}{b} \pi) &= e^{a \tan^{-1} \frac{a}{b} + \frac{an}{b} \pi} \cos \l b \tan^{-1} \frac{a}{b} +n \pi\r \\ &= e^{a \tan^{-1} \frac{a}{b}} \cos \l b \tan^{-1} \frac{a}{b}\r \cdot e^{\frac{an}{b} \pi}(-1)^n \\ &= e^{a \tan^{-1} \frac{a}{b}} \cos \l b \tan^{-1} \frac{a}{b}\r \cdot (-e^{\frac{a}{b} \pi})^n \\ \end{align*} showing the form the desired geometric progression.

+ - ↺