Problems

---

Solution:

1. \(\,\) \begin{align*} && y &= 9x^2 - 12x \cos \theta + 4 \\ &&&= (3x-2\cos \theta)^2+4-4\cos^2 \theta \\ &&&= (3x-2\cos \theta)^2 + 4 \sin^2 \theta \end{align*} Therefore the minimum is \(4\sin^2 \theta\) when \(x = \frac23 \cos \theta\). \begin{align*} && y &= 12x^2 \sin \theta - 9x^4 \\ &&&=4\sin^2 \theta -(3x^2-2\sin\theta)^2 \end{align*} Therefore the maximum is \(4\sin^2 \theta\) when \(x^2 = \frac23\sin \theta\) Therefore \begin{align*} && 0 &= 9x^4 + (9 - 12 \sin \theta)x^2 - 12x \cos \theta + 4 \\ && \underbrace{-9x^4+12x^2\sin \theta}_{\leq 4\sin^2 \theta } &= \underbrace{9x^2 - 12x \cos \theta + 4 }_{\geq 4 \sin^2 \theta} \end{align*} Therefore the equality cases must be achieved in both cases, ie \(x = \frac23 \cos \theta\) and \(x^2 = \frac23 \sin \theta\) \begin{align*} && x^2 &= \frac49\cos^2 \theta \\ &&&= \frac49(1-\sin^2 \theta) \\ &&&= \frac49(1-\frac94 x^2) \\ \Rightarrow && 2x^2 &= \frac49 \\ \Rightarrow && x &= \pm \frac{\sqrt{2}}3\\ \Rightarrow && \cos \theta &=\pm \frac32 \frac{\sqrt{2}}3 \\ &&&= \pm \frac{1}{\sqrt{2}} \\ \Rightarrow && \theta &= \frac{\pi}{4}, \frac{3\pi}{4} \\ \Rightarrow && (x, \theta) &= \left (\frac{\sqrt{2}}{3}, \frac{\pi}{4} \right), \left (-\frac{\sqrt{2}}{3}, \frac{3\pi}{4} \right) \end{align*}
2. Sketching we obtain, noticing we can find the turning point by: \begin{align*} && \frac{x^2}{x-\theta} &= \lambda \\ \Leftrightarrow && x^2 - \lambda x +\theta \lambda &= 0 \\ \Leftrightarrow && 0 &\leq \Delta = \lambda^2 -4\lambda \theta \\ \Leftrightarrow && \lambda &\geq 4 \theta, \lambda \leq 0 \end{align*}
   

   + - ↺
   Notice that \(\sin^2 \theta \cos^2 x \leq 1\) and \(1 + cos^2 \theta \sin^2 x \geq 1\) and therefore we must have the inequality desired. \begin{align*} && \underbrace{\frac{x^2}{4\theta(x - \theta)}}_{\geq 1 \text{ or } \leq 0} &= \underbrace{\frac{\sin^2 \theta \cos^2 x}{1 + \cos^2 \theta \sin^2 x}}_{\in [0,1]} \\ \text{both}=0: && x = 0 &, \sin \theta = 0 \\ \text{both}=1: && x = 2\theta &, \sin^2 \theta = 1,\cos^2 x = 1 \\ && 1 &= \cos^2 2 \theta \\ &&&= (1-2 \sin^2 \theta)^2 \\ &&&= 1 \\ \Rightarrow && (x, \theta) &= \left(\frac{\pi}{2}, \pi\right) \end{align*}

---

Solution:

1. \begin{align*} && f(\beta) &= \beta - \frac1{\beta}-\frac1{\beta^2} \\ \Rightarrow && f'(\beta) &= 1 +\frac{1}{\beta^2}+\frac{2}{\beta^3} \\ \Rightarrow && 0 &= f'(\beta) \\ &&&= 1 + \frac1{\beta^2} + \frac{2}{\beta^3} \\ \Rightarrow && 0 &= \beta^3 + \beta + 2 \\ &&&= (\beta+1)(\beta^2-\beta+2) \end{align*} Therefore the only stationary point is at \(\beta = -1, f(-1) = -1\)
   

   + - ↺
   \begin{align*} && g(\beta) &= \beta + \frac3{\beta}-\frac1{\beta^2} \\ \Rightarrow && g'(\beta) &= 1 -\frac{3}{\beta^2}+\frac{2}{\beta^3} \\ \Rightarrow && 0 &= f'(\beta) \\ &&&= 1 - \frac3{\beta^2} + \frac{2}{\beta^3} \\ \Rightarrow && 0 &= \beta^3 - 3\beta + 2 \\ &&&= (\beta-1)^2(\beta+2) \end{align*} Therefore there are stationary points at \(\beta=1,f(1) = 3, \beta=-2, f(-2) = \frac14\)
   

   + - ↺
2. Let \(u,v\) be the roots of \(x^2 + \alpha x + \beta = 0\), then since \((x-u)(x-v) = 0\) we must have \(\alpha = -(u+v), \beta = uv\). Therefore: \begin{align*} && u+v +\frac{1}{uv} &= -\alpha + \frac{1}{\beta} \\ && \frac1u+\frac1v + uv &= \frac{u+v}{uv} + uv \\ &&&= -\frac{\alpha}{\beta} + \beta \end{align*} Given \(u+v + \frac 1 {uv} = -1\), ie \(-\alpha + \frac{1}{\beta} = -1\). Since the roots are real, we must also have that \(\alpha^2 - 4\beta \geq 0\), so \begin{align*} && -\alpha + \frac1\beta &= -1 \\ \Rightarrow && \alpha &= 1 +\frac1\beta \\ \Rightarrow && -\frac{\alpha}{\beta}+\beta &= -\frac{1}{\beta} \l1+\frac1{\beta}\r + \beta \\ &&&=\beta - \frac{1}{\beta}-\frac{1}{\beta^2} \end{align*} So we want to maximise \(f(\beta)\) subject to \(\alpha ^2 - 4\beta \geq 0\) \begin{align*} && 0 &\leq \alpha^2 -4 \beta \\ &&&= \l 1 + \frac1{\beta} \r^2 - 4\beta \\ &&&= 1+ \frac2{\beta} + \frac1{\beta^2} - 4\beta \\ \Leftrightarrow && 0 &\leq -4\beta^3+\beta^2 + 2\beta + 1 \\ &&&=-(\beta-1)(4\beta^2+3\beta+1)\\ \Leftrightarrow && \beta &\leq 1 \end{align*} But we know \(f(\beta) \leq -1\) on \((-\infty,1]\) so we're done.
3. Given that \(-\alpha + \frac{1}{\beta} = 3\) we have \begin{align*} && -\alpha + \frac1\beta &= 3 \\ \Rightarrow && \alpha &= -3 +\frac1\beta \\ \Rightarrow && -\frac{\alpha}{\beta}+\beta &= -\frac{1}{\beta} \l-3+\frac1{\beta}\r + \beta \\ &&&=\beta + \frac{3}{\beta}-\frac{1}{\beta^2} \end{align*} which we want to maximise, subject to: \begin{align*} && 0 &\leq \alpha^2 -4 \beta \\ &&&= \l -3 + \frac1{\beta} \r^2 - 4\beta \\ &&&= 9- \frac6{\beta} + \frac1{\beta^2} - 4\beta \\ \Leftrightarrow && 0 &\leq -4\beta^3+9\beta^2 - 6\beta + 1 \\ &&&=-(\beta-1)^2(4\beta-1)\\ \Leftrightarrow && \beta &\leq \frac14 \end{align*} Therefore the maximum will either be \(f(-2) = \frac14\) or \(f(\frac14) = -\frac{15}4\). Therefore the maximum is \(\frac14\)

---

Solution:

1. \(\,\)
   

   + - ↺
2. \(\,\)
   

   + - ↺
3. \(\,\)
   

   + - ↺
4. Note that \(\frac{x+x^4}{1+x^2+x^4}\) is continuous, and nicely behaved on \((-\infty, \infty)\) so we can see that \(\lim_{x \to \infty} h_1(x) = \pi + \frac{\pi}{4} = \frac{5\pi}{4}\). \(\frac{4x-x^3}{1-x^4}\) on the other hand has asymptotes at \(\pm 1\). So as as \(x \to 1\), \(h_1(x) \to \pi + \frac{\pi}{2} = \frac{3\pi}{2}\). Then as \(x \to \infty\) we increase by another \(\frac{\pi}{2}\), so \(\lim_{x \to \infty} h_2(x) = 2\pi\)

An alternative way to think about the last two parts is to consider \(h\) as giving the (continuous) argument (shifted by \(\pi\)) of \((1-t^2)+it\) (blue), \((1+t^2+t^4)+i(t+t^4)\) (orange) or \((1-t^4)+i(4t-t^3)\) (green). We can see the orange line never wraps around the origin, so the argument is always easy to find. The blue does one full circuit, from \(-\pi\) to \(\pi\) (or \(0\) to \(2\pi\) in our world. And the green line also does a full \(2\pi\) loop.

---

Solution:

1. \(\,\) \begin{align*} && \int \ln (2- x) \d x &= \left [ (x-2) \ln (2- x) \right] - \int -\frac{x-2}{2-x} \d x \\ && \int \ln (2- x) \d x &= \left [ (x-2) \ln (2- x) \right] + \int 1 \d x \\ &&&= -(2-x) \ln (2-x) +(2-x) + C \end{align*}
2. \(\,\)
   

   + - ↺
3. \begin{align*} && \text{Area} &= \int_0^{\sqrt{3}} \ln | x^2 - 4 | \d x \\ &&&= \int_0^\sqrt{3} \ln(4-x^2) \d x \\ &&&= \int_0^\sqrt{3} \left ( \ln(2-x) + \ln (2+x) \right) \d x \\ &&&= \left [ -(2-x) \ln (2-x) +(2-x) +(2+x)\ln(2+x)-(2+x)\right]_0^{\sqrt{3}} \\ &&&= \left ( -(2-\sqrt{3}) \ln (2-\sqrt{3}) +(2-\sqrt{3}) +(2+\sqrt{3})\ln(2+\sqrt{3})-(2+\sqrt{3}) \right) - \\ &&&\quad \quad \left (- 2\ln (2)+2 +2\ln(2)-2 \right) \\ &&&=\left ( -(2-\sqrt{3}) \ln \left ( \frac{1}{2+\sqrt{3}} \right) -2\sqrt{3} +(2+\sqrt{3})\ln(2+\sqrt{3}) \right) \\ &&&= 4\ln(2 + \sqrt{3}) - 2 \sqrt{3} \end{align*}
4. 

   + - ↺
   \begin{align*} && \text{Area} &= 2 \left ( \int_0^\sqrt{3} \ln (4-x^2) \d x - \lim_{t \to 2}\int_{\sqrt{3}}^t \ln(4-x^2) \d x \right) \\ &&&= 8\ln(2 + \sqrt{3})-4\sqrt{3} - 2 \lim_{t \to 2}\int_{\sqrt{3}}^t \left ( \ln (2-x) + \ln (2+x) \right) \d x \\ &&&= 8\ln(2 + \sqrt{3})-4\sqrt{3} - 2 \lim_{t \to 2} \left [ -(2-x) \ln (2-x) +(2-x) +(2+x)\ln(2+x)-(2+x)\right]_{\sqrt{3}}^{t} \\ &&&= 16 \ln(2+\sqrt{3})-8\sqrt{3} - 2 \lim_{t \to 2} \left(-(2-t) \ln (2-t) +(2-t) +(2+t)\ln(2+t)-(2+x) \right) \\ &&&= 16 \ln(2+\sqrt{3})-8\sqrt{3} -2(4 \ln4-4) \\ &&&= 16 \ln(2 + \sqrt{3}) - 16 \ln 2 +8(1-\sqrt{3}) \end{align*}

---

Solution:

1. \begin{align*} cb(x) &= c(b(x)) \\ &= 2 \ln x \quad (x > 0) \\ ab(x) &= (b(x))^2 \\ &= (\ln x)^2 \quad (x > 0) \\ da(x) &= \sqrt{a(x)} \\ &= \sqrt{x^2} \\ &= |x| \quad (-\infty < x < \infty) \\ ad(x) &= (d(x))^2 \\ &= (\sqrt{x})^2 \\ &= x \quad (x \geq 0) \end{align*} The domains are specified above. The ranges are \(\mathbb{R}, \mathbb{R}_{\geq 0}, \mathbb{R}_{\geq 0}, \mathbb{R}_{\geq 0}\) respectively.
2. \begin{align*} fg(x) &= \sqrt{g(x)^2-1} \quad (|g(x)| \geq 1) \\ &= \sqrt{x^2+1-1} \\ &= |x| \end{align*} So \(fg: \mathbb{R} \to \mathbb{R}_{\geq 0}\). \begin{align*} gf(x) &= \sqrt{f(x)^2 + 1} \\ &= \sqrt{\left ( \sqrt{x^2-1} \right)^2+1} \quad (|x| \geq 1) \\ &= |x| \end{align*} So \(gf: \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}\)
3. • 

     + - ↺
   • 

     + - ↺
   \begin{align*} kh(x) &= h(x) - \sqrt{h(x)^2 -1} \quad (|h(x)| \geq 1)\\ &= x + \sqrt{x^2+1} - \sqrt{(x + \sqrt{x^2+1})^2 - 1} \\ &= x + \sqrt{x^2+1} - \sqrt{x^2 + x^2 - 2x} \quad (x \geq 1) \\ &= x + \sqrt{x^2+1} - \sqrt{2x^2-2x} \quad (x \geq 1) \end{align*} This has domain \(x \geq 1\) and range, \((0, 1]\)

---

Solution:

1. \(\,\)
   

   + - ↺
2. \(\,\) \begin{align*} && \frac{\d}{\d x} \left ( \frac{1}{g(x)} \right) &= \frac{\d }{\d x} \left ( \big( (x-a)^2-1 \big) \big( (x-b)^2 -1\big)\right) \\ &&&= ((x-a)^2-1)(2(x-b))+((x-b)^2-1)(2(x-a)) \\ &&&= 2(2x-a-b)(x^2-(a+b)x+ab-1) \\ \Rightarrow && x &= \frac{a+b}{2}, \frac{a+b \pm \sqrt{(a+b)^2-4ab+4}}{2} \\ &&&= \frac{a+b}{2}, \frac{a+b \pm \sqrt{(a-b)^2+4}}{2} \end{align*} If \(b > a+2\):
   

   + - ↺
   If \(b = a+2\):
   

   + - ↺

---

Solution: \begin{align*} && y &= \left ( \frac{x-a}{x-b} \right )e^x \\ &&y'& = \left ( \frac{x-a}{x-b} \right )e^x + \left ( \frac{(x-b)-(x-a)}{(x-b)^2}\right )e^x \\ &&&= \left ( \frac{(x-b)(x-a) +a-b}{(x-b)^2} \right)e^x \\ &&&= \left ( \frac{x^2-(a+b)x+a-b+ab}{(x-b)^2} \right)e^x \\ && 0 &< \Delta = (a+b)^2 - 4 \cdot 1 \cdot (a-b+ab) \\ &&&= a^2+2ab+b^2-4a+4b-4ab \\ &&&= a^2-2ab+b^2-4a+4b\\ &&&= (a-b)^2-4(a-b) \\ &&&= (a-b)(a-b-4) \\ \end{align*} Considered as a quadratic in \(a-b\) we can see \(a-b < 0\) or \(a-b > 4\)

1. If \(a = 0, b = \frac12\), we have \(x^2-\frac12x -\frac12 = 0 \Rightarrow (2x+1)(x-1) = 0 \Rightarrow x = -\frac12, x=1\). The asymptote is at \(x = \frac12\) so they are on either side.
   

   + - ↺
2. \(\,\)
   

   + - ↺

---

Solution:

1. \begin{align*} 1 &= \int_{1/4}^{1/2} \frac{\lambda}{x\ln x} \, dx \\ &= \lambda\left [ \ln |\ln x| \right ]_{1/4}^{1/2} \\ &= \lambda \l \ln |-\ln 2| - \ln |-2 \ln 2| \r \\ &= \lambda (-\ln 2) \end{align*} So \(\lambda = -\frac{1}{\ln 2} = \frac{1}{\ln \frac12}\)
2. \begin{align*} 1 &= \int_{a}^{b} \frac{1}{x\ln x} \, dx \\ &= \left [ \ln |\ln x| \right ]_{a}^{b} \\ &= \l \ln \ln b - \ln \ln a \r \\ &= \ln \l \frac{\ln b}{\ln a} \r \\ \end{align*} So \(b = e^{a}\)
3. If \(\lambda = 1, a = e, b = e^e\), then \begin{align*} \P(\e^{3/2}\le X \le \e^2) &= \int_{e^{3/2}}^{e^2} \frac{1}{x \ln x} \, dx \\ &= \left [ \ln \ln x \right]_{e^{3/2}}^{e^2} \\ &= \ln 2 - \ln \frac{3}{2} \\ &= \ln \frac{4}{3} \\ &= \ln \l 1 + \frac{1}{3} \r \\ &\approx \frac{1}{3} - \frac{1}{2 \cdot 3^2} + \frac{1}{3 \cdot 3^3} - \frac{1}{4 \cdot 3^4} \\ &= \frac{31}{108} \end{align*}
4. Note that \(2 > e^{\frac12} > 1\) so \(a = e^{\frac12}, b = e^{\frac{e}2}\). Since \(3 > e \Rightarrow e^{3/2} > e^{\frac{e}{2}}\) this probability is out of range, therefore \(\P(\e^{3/2}\le X \le \e^2) = 0\)

---

Solution: \begin{align*} && y &= \frac{2x(x^2-5)}{x^2-4} \\ &&&= 2x(x^2-5)(-\tfrac14)(1-\tfrac14x^2)^{-1} \\ &&&= \tfrac52x + \cdots \\ &&&= \frac{2x(x^2-4)-2x}{x^2-4} \\ &&&= 2x - \frac{2x}{x^2-4} \end{align*}

1. We are looking for the intersections of \(y = \frac23(x+3)\) and \(y = f(x)\)
   

   + - ↺
   Therefore 3 real roots.
2. We are looking for intersections of \(y = \frac12(5x-2)\) and \(y = f(x)\)
   

   + - ↺
   so one solution.
3. We are looking for intersections of \(y = f(x)^2\) and \(y = x^2+1\), or \(y = \sqrt{x^2+1}\) and \(y = f(x)\) where \(f(x) \geq 0\)
   

   + - ↺
   So \(3\) solutions.

---

Solution: \begin{align*} && {\frac{\d y} {\d x}} &= \frac{x^3y^2}{(1 + x^2)^{5/2}} \\ \Rightarrow &&\int \frac{1}{y^2} \d y &= \int \frac{x^3}{(1+x^2)^{5/2}} \d x \\ \Rightarrow && -\frac1y &= \int \frac{x^3+x-x}{(1+x^2)^{5/2}} \d x \\ &&&= \int \left ( \frac{x}{(1+x^2)^{3/2}}-\frac{x}{(1+x^2)^{5/2}} \right) \d x \\ &&&= \frac{-1}{(1+x^2)^{1/2}} + \frac{1}{3(1+x^2)^{3/2}} + C \\ &&&= \frac{1-3(1+x^2)}{3(1+x^2)^{3/2}} + C \\ &&&= \frac{-3x^2-2}{3(1+x^2)^{3/2}} + C \\ (x,y) = (0,1): &&-1 &= -\frac23 + C \\ \Rightarrow && C &= -\frac13 \\ \Rightarrow && \frac1y &= \frac{3x^2+2}{3(1+x^2)^{3/2}} + \frac13 \end{align*} \begin{align*} y &= \frac{1}{\frac13 +\frac{3x^2+2}{3(1+x^2)^{3/2}} } \\ &= \frac{3}{1+ \frac{3x^2+2}{3(1+x^2)^{3/2}}} \\ &= \frac{3}{1+ \frac{3}{x} + \cdots} \\ &\approx 3 - \frac{9}{x} \end{align*}

\begin{align*} && \frac {\d z} {\d x} &= \frac{x^3z^3}{2(1 + x^2)^{5/2}} \\ \Rightarrow && \int \frac{1}{z^3} \d z &= \int \frac{x^3}{2(1+x^2)^{5/2}} \\ && -\frac{1}{2z^2} &= -\frac{3x^2+2}{3(1+x^2)^{3/2}} - C \\ (x,z) = (0, \pm 1): && \frac{1}{2} &= \frac{2}{3} + C \\ \Rightarrow && C &= -\frac16 \\ \Rightarrow && \frac{1}{z^2} &= \frac{6x^2+4}{3(1+x^2)^{3/2}} - \frac13 \end{align*} So as \(x \to \infty\) \(z \sim \pm (3 + \frac{2}{x} + \cdots)\) and so:

---

Solution:

1. Clearly \(x = 0\) and \(x = 4\) are vertical asymptotes. Notice that \(\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r}\) tends to \(0\) as \(x \to \infty\). Therefore the oblique asymptote is \(y = x-4\).
2. \begin{align*} && 0 &= \frac{x^2(x-4)^2-4^2(2x+1)^2}{x^2(x-4)} \\ &&&= \frac{(x(x-4)-4(2x+1))(x(x-4)+4(2x+1))}{x^2(x-4)} \\ &&&= \frac{(x^2-12x-4)(x^2+4x+4)}{x^2(x-4)}\\ &&&= \frac{(x^2-12x-4)(x+2)^2}{x^2(x-4)} \end{align*} Therefore \(f(x) = 0\) has a double root at \(x = -2\). Notice it also has roots at \(6 \pm 2\sqrt{10}\)
3. 

   + - ↺

---

Solution: If \(m\) is even, clearly that expression is positive since it's the sum of two (different) squares. If \(m\) is odd, then we can expand it as a sum of powers of \(x^2\) with a leading coefficient of \(1\) so it is also positive. \begin{align*} && f (x) = \frac{ (1+x )^m - ( 1-x )^m}{ (1+x )^m + (1-x )^m} \\ && f'(x) &= \frac{(m(1+x )^{m-1} + m( 1-x )^{m-1})((1+x)^m + (1-x)^m ) - ((1+x )^m - ( 1-x )^m)(m(1+x)^{m-1} - m(1-x)^{m-1} )}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{2m(1+x)^m(1-x)^{m-1}+2m(1+x)^{m-1}(1-x)^m}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{2m(1+x)^{m-1}(1-x)^{m-1}(1+x+1-x)}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{4m(1+x)^{m-1}(1-x)^{m-1}}{\l (1+x)^m + (1-x)^m \r^2} \\ \end{align*}

---

Solution: \begin{align*} && x+ c &= f(x) \\ \Rightarrow && (x+c)(x^2-1) &= x(x-2)(x-a) \\ \Rightarrow && x^3 + cx^2-x-c &= x^3-(2+a)x^2+2ax \\ \Rightarrow && 0 &= (c+2+a)x^2-(1+2a)x-c \\ && 0 &\leq \Delta = (1+2a)^2 + 4(2+c+a)c \\ &&&= 4c^2+(4a+8)c + (1+2a)^2 \\ && \Delta_c &= 16(a+2)^2-16(1+2a)^2 \\ &&&= 16(1-a)(3a+3) \\ &&&= 48(1-a^2) \end{align*} Therefore if \(|a| \geq 1\) we must have \(\Delta_c \leq 0\) which means \(\Delta \geq 0\) and so there are always solutions. If \(|a| < 1\) there are values for \(c\) where \(\Delta < 0\) and there would be no solutions. \begin{align*} && y &= \frac{x^3-(2+a)x^2+2ax}{x^2-1} \\ &&&= \frac{(x^2-1)(x-(2+a))+(2a+1)x-(2+a)}{x^2-1} \\ &&&= x - (2+a) + \frac{(2a+1)x-(2+a)}{x^2-1} \end{align*} therefore the oblique asymptote has equation \(y = x - (2+a)\)

1. 

   + - ↺
2. 

   + - ↺

---

Solution: \begin{align*} && y &= \frac x {\sqrt{x^2-2x+a}} \\ && y' &= \frac{\sqrt{x^2-2x+a} - \frac{x(x-1)}{\sqrt{x^2-2x+a}}}{x^2-2x+a} \\ &&&= \frac{-x+a}{(x^2-2x+a)^{3/2}} \end{align*} Since the denominator is always positive, the only stationary point is when \(x = a\)