Problems

Solution:

1. \(\,\) \begin{align*} && 0 & \leq x^3 - 4x^2 - x + 4 \\ &&&= (x-1)(x^2-3x-4) \\ &&&= (x-1)(x-4)(x+1) \\ \Leftrightarrow && x &\in [-1, 1] \cup [4, \infty) \end{align*}
2. \(\,\) \begin{align*} && 0 &= x^{3}-4x^{2}y-xy^{2}+4y^{3} \\ && 0 &= (x-y)(x-4y)(x+y) \end{align*} Therefore the lines are \(y = x, 4y = x, y=-x\).
3. 

   + - ↺
   (quickest way to see this is to check the \(x\) or \(y\)-axis)

Solution:

1. \begin{align*} && S + T &= \int \frac{\cos x + \sin x }{\cos x + \sin x} \d x \\ &&&= \int \d x \\ &&&= x + C \\ && S - T &= \int \frac{\cos x - \sin x}{\cos x + \sin x} \d x \\ &&&= \ln( \cos x + \sin x) + C \\ \Rightarrow && 2S &= x + \ln(\cos x + \sin x) + C \\ \Rightarrow && S &= \frac12 \left ( x + \ln(\cos x + \sin x) \right) + C \\ \Rightarrow && 2T &= x - \ln(\cos x + \sin x) + C \\ \Rightarrow && T &= \frac12 \left ( x - \ln(\cos x + \sin x) \right) + C \end{align*}
2. \begin{align*} && I &= \int_{1/4}^{1/2} (1-4x)\sqrt{\frac1x-1} \d x \\ x = \sin^2 t, \d x = 2 \sin t \cos t \d t: &&&= \int_{\pi/6}^{\pi/4} (1-4\sin^2 t) \sqrt{\frac{1-\sin^2 t}{\sin^2 t}} 2 \sin t \cos t \d t\\ &&&=\int_{\pi/6}^{\pi/4} (1-4\sin^2 t)\frac{\cos t}{\sin t} 2 \sin t \cos t \d t \\ &&&= \int_{\pi/6}^{\pi/4} (1-4\sin^2 t) 2 \cos^2 t \d t \\ &&&= \int_{\pi/6}^{\pi/4} \left ( 2\cos^2t - 8 \sin^2t \cos^2 t \right) \d t \\ &&&= \int_{\pi/6}^{\pi/4} \left ( 1+\cos2t - 2 \sin^2 2t \right) \d t \\ &&&= \int_{\pi/6}^{\pi/4} \left ( 1+\cos2t +(\cos 4t-1)\right) \d t \\ &&&= \left[\frac12 \sin 2t + \frac14 \sin 4t \right]_{\pi/6}^{\pi/4} \\ &&&= \left ( \frac12 \right) - \left (\frac12 \frac{\sqrt{3}}{2} + \frac14 \frac{\sqrt{3}}{2} \right) \\ &&&= \frac{4-3\sqrt{3}}{8} \end{align*}

Solution:

1. \(\,\) \begin{align*} && \sum_{r=1}^n \left (f(r) - f(r-1) \right) &= \cancel{f(1)} - f(0) + \cdots\\ &&&\quad\, \cancel{\f(2)}-\cancel{f(1)} + \cdots \\ &&&\quad\, \cancel{f(3)}-\cancel{f(2)} + \cdots \\ &&&\quad\, +\cdots + \cdots \\ &&&\quad\, \cancel{f(n-1)}-\cancel{f(n-2)} + \cdots \\ &&&\quad\, f(n)-\cancel{f(n-1)} \\ &&&=f(n) - f(0) \end{align*}
2. If \(f(r) = r^2(r+1)^2\) then \begin{align*} && f(r) - f(r-1) &= r^2(r+1)^2 - (r-1)r^2 \\ &&&= r^2((r+1)^2-(r-1)^2) \\ &&&=4r^3 \end{align*} Therefore \begin{align*} && \sum_{r=1}^n 4r^3 &= n^2(n+1)^2-0 \\ \Rightarrow && \sum_{r=1}^n r^3 &= \frac{n^2(n+1)^2}{4} \\ \end{align*}
3. So \begin{align*} && \sum_{r=1}^{2n+1} (-1)^{r-1}r^3 &= \sum_{r=1}^{2n+1} r^3 - 2 \sum_{r=1}^n (2r)^3 \\ &&&= \frac14(2n+1)^2(2n+2)^2 - 8 \frac{n^2(n+1)^2}{4} \\ &&&= (2n+1)^2(n+1)^2 - 2n^2(n+1)^2 \\ &&&= (n+1)^2(4n^2+4n+1 - 2n^2) \\ &&&= (n+1)^2(2n^2+4n+1) \end{align*}

Solution: \begin{align*} && (\cos \theta + i \sin \theta)^n &= \cos n \theta + i \sin n \theta \\ n = 5: && \cos 5 \theta + i \sin 5 \theta &= (\cos \theta + i \sin \theta)^5 \\ \textrm{Im}: && \sin 5 \theta &= \binom{5}{1}\cos^4 \theta \sin \theta + \binom{5}{3} \cos^2 \theta (- \sin^3 \theta) + \binom{5}{5} \sin^5 \theta \\ &&&= \sin \theta (5\cos^4 \theta-10\cos^2 \theta \sin^2 \theta+\sin^4 \theta) \\ &&&= \sin \theta (5\cos^4 \theta-10\cos^2 \theta (1-\cos^2 \theta)+(1-\cos^2 \theta)^2) \\ &&&= \sin \theta((5+10+1)\cos^4 \theta +(-10-2)\cos^2 \theta + 1) \\ &&&= \sin \theta(16\cos^4 \theta -12\cos^2 \theta + 1) \\ \end{align*} Suppose \(\theta= \frac{\pi}{5}\), then \(\sin 5 \theta = 0, \sin \theta \neq 0\), therefore if \(c = \cos \theta\) we must have \begin{align*} && 0 &= 16c^4-12c^2+1 \\ \Rightarrow && c^2 &= \frac{3 \pm \sqrt{5}}{8} \\ &&&= \frac{6\pm 2\sqrt{5}}{16} \\ &&&= \frac{(1 \pm \sqrt{5})^2}{16} \\ \Rightarrow && c &= \pm \frac{1 \pm \sqrt{5}}{4} \end{align*} Since \(c > 0\) we either have \(\cos \frac15 \pi = \frac{1+\sqrt{5}}4\) or \(\cos \frac15 \pi = \frac{\sqrt{5}-1}4\), however \(\sqrt{5}-1 < 1.5\) and so \(\frac{\sqrt{5}-1}{4} < \frac12 = \cos \frac13 \pi\) we must have \(\cos \frac15 \pi = \frac{1+\sqrt{5}}4\)

Solution: \begin{align*} f(x) & =nx-\binom{n}{2}\frac{x^{2}}{2}+\binom{n}{3}\frac{x^{3}}{3}-\cdots+(-1)^{r+1}\binom{n}{r}\frac{x^{r}}{r}+\cdots+(-1)^{n+1}\frac{x^{n}}{n} \\ f'(x) &= n - \binom{n}{2} x + \binom{n}{3}x^2 - \cdots (-1)^{r+1} \binom{n}{r} + \cdots + (-1)^{n+1} x^{n-1} \\ &= \frac{1-(1-x)^n}{x} \end{align*} Therefore, since \(\displaystyle f(x) = \int_0^xf'(t)\,dt\) \begin{align*} f(x) &= \int_0^x \frac{1 - (1-t)^n}{t} \, dt \\ &= \int_{1}^{1-x} \frac{1-y^n}{1-y} (-1)\, dy \tag{Let \(y = 1-t, \frac{dy}{dt} = -1\)} \\ &= \boxed{\int_{1-x}^1 \frac{1-y^n}{1-y} dy} \\ &= \int_{1-x}^1 \l 1 + y + y^2 + \cdots + y^{n-1} \r \, dy \\ &= \left [ y + \frac{y^2}{2} + \frac{y^3}{3} + \cdots + \frac{y^n}{n} \right]_{1-x}^1 \\ \end{align*} So when \(x = 1, 1-x = 0\) so we exactly have the sum required.

Solution:

1. \(,\) \begin{align*} u = 1/y, \frac{\d u}{\d x} = -1/y^2 y': && -\frac{\d u}{\d x} + u &= e^{2x} \\ \Rightarrow && \frac{\d}{\d x} \left (e^{-x}u \right) = -e^{x} \\ \Rightarrow && e^{-x}u &= -e^x+C \\ \Rightarrow && u &= Ce^x-e^{2x} \\ \Rightarrow && y &= \frac{1}{Ce^x-e^{2x}} \end{align*}
2. \(\,\) \begin{align*} u = 1/y^2, u' = -2/y^3 y': && -2u'+u &= e^{2x} \\ \Rightarrow && \frac{\d}{\d x} \left (e^{-1/2x}u\right) &= -\frac12e^{\frac32x} \\ \Rightarrow && e^{-\frac12x}u &= -\frac13e^{\frac32x}+C \\ \Rightarrow && u &= -\frac13e^{2x}+Ce^{\frac12x} \\ \Rightarrow && y &= \left ( \frac{1}{Ce^{\frac12x}-\frac13e^{2x}}\right)^{\frac12} \end{align*}

Solution: The position of the particle is projected at angle \(\theta\) is \((x,y) = (v \cos \theta t, v \sin \theta t - \frac12 g t^2)\), ie \(t = \frac{x}{v \cos \theta}\), \begin{align*} && y &= x\tan \theta -\frac12 g \frac{x^2}{v^2} \sec^2 \theta \\ && y &= x \tan \theta -\frac{1}{4h} (1+\tan^2 \theta) x^2 \\ && 0 &= \frac{1}{4h} x^2\tan^2 \theta - x \tan \theta + \frac{x^2}{4h} +y \\ \Delta \geq 0: && 0 &\leq \Delta = x^2-4\frac{x^2}{4h}\left (\frac{x^2}{4h}+y \right) \\ &&&=1-\frac{1}{4h^2}(x^2+4hy) \\ \Rightarrow && x^2+4hy &\leq 4h^2 \\ \Rightarrow && x^2 &\leq 4h(h-y) \end{align*} We are assuming that there are no forces acting other than gravity (eg air resistance)