Problems

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Solution:

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   The area under the blue curve is \(1-\cos b\). The area under the green line is \(\frac12 b \sin b\) The area under the red line is \(\frac12 b^2\) Therefore \(\frac12 b \sin b < 1- \cos b < \frac12 b^2 \Rightarrow 1- \frac12 b^2 < \cos b < 1 - \frac12 b \sin b\)
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   \begin{align*} &&\text{Area under blue curve}: &= \int_0^1 a^x \d x\\ &&&= \left [ \frac{1}{\ln a}e^{x \ln a} \right]_0^1 \\ &&&= \frac{a-1}{\ln a} \\ \\ &&\text{Area under green line}: &=\frac12 \cdot 1 \cdot (a + 1)\\ &&&= \frac{a+1}{2} \\ \\ &&\text{Area under tangent}: &=\frac12 \cdot 1 \cdot (1+\ln a + 1)\\ &&&= \frac{\ln a+2}{2} \\ \\ \Rightarrow && \frac{a+1}{2} & > \frac{a-1}{\ln a} \\ \Rightarrow && \ln a& > \frac{2(a-1)}{a+1} \\ \\ \Rightarrow && \frac{a-1}{\ln a} &> \frac{\ln a +2}{2} \\ \Rightarrow && 2(a-1) -2\ln a - (\ln a)^2 &> 0 \\ \Rightarrow && \ln a & < -1 + \sqrt{2a-1} \end{align*}

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Solution: \begin{align*} && \frac{\d y}{\d x} &= -\frac1{2x^2} \\ \Rightarrow && \frac{y - \frac{1}{2p}}{x - p} &= - \frac{1}{2p^2} \\ \Rightarrow && y - \frac1{2p} &= -\frac{1}{2p^2}x +\frac1{2p} \\ \Rightarrow && y +\frac{1}{2p^2}x &= \frac1p \\ \Rightarrow && 2p^2 y + x &= 2p\\ \Rightarrow && 2q^2 y + x &= 2q \\ \Rightarrow && (p^2-q^2)y &= p-q \\ \Rightarrow && y &= \frac{1}{p+q} \\ && x &= \frac{2pq}{p+q} \end{align*} \begin{align*} \text{normal}: && \frac{y-\frac1{2p}}{x-p} &= 2p^2 \\ \Rightarrow && y - \frac1{2p} &= 2p^2x - 2p^3 \\ \Rightarrow && 2py -4p^3x &= 1-4p^4 \\ \Rightarrow && 2qy -4q^3x &= 1-4q^4 \\ pq = \tfrac12: && y - 2p^2 x &= q-2p^3 \\ && y - 2q^2 x &= p-2q^3 \\ \Rightarrow && (2q^2-2p^2)x &= q-p +2q^3-2p^3 \\ &&&= (q-p)(q+p+2q^2+1+2p^2) \\ \Rightarrow && x &= \frac{1+2(p^2+q^2)+1}{2(p+q)} \\ &&&= \frac{1+2(p^2+q^2+2pq-1)+1}{2(p+q)} \\ &&&= p+q\\ && y &= 2p^2 \left ( p+q \right) + q - 2p^3 \\ &&&= p+q \end{align*} So \(N(p+q, p+q)\) and \(T\left (\frac{1}{p+q}, \frac{1}{p+q} \right)\), so both points lie on \(y = x\). \[ OT \cdot ON = \frac{\sqrt{2}}{p+q} \cdot (p+q)\sqrt{2} = 2 \] which is clearly constant.

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Solution:

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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 \(b > a+2\):
   

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   If \(b = a+2\):
   

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Solution:

1. 1. \(\,\)
      

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2. Suppose \(m = 12\) \begin{align*} && y &= 4-x^2 \\ && 12x &= k-y^2 \\ \Rightarrow && 12 x&=k-(4-x^2)^2 \\ &&&= k-16+8x^2-x^4 \\ \Rightarrow && 0 &= x^4- 8x^2+12x+16-k \end{align*} When the curves touch, we will have repeated root, ie \(a\) is a root of \(4x^3-16x+12 \Rightarrow a^3-4a+3 =0\). \begin{align*} &&0 &= a^3-4a+3 \\ &&&= (a-1)(a^2+a-3) \\ \Rightarrow &&a &= 1, \frac{-1 \pm \sqrt{13}}{2} \end{align*} \begin{align*} && 0 &= a^4-8a^2+12a+16-k \\ \Rightarrow && k &= a(a^3-8a+12)+16 \\ &&&= a(4a-3-8a+12)+16 \\ &&&= -4a^2+9a+16 \\ \\ \Rightarrow && a = 1& \quad k = 21 \\ && k &= -4(3-a)+9a+16 = 13a+4\\ && a = \frac{-1-\sqrt{13}}2& \quad k = \frac{-5 - 13\sqrt{13}}{2} < 0 \\ && a = \frac{-1+\sqrt{13}}2& \quad k = \frac{-5 + 13\sqrt{13}}{2} \\ \end{align*} So we have type (a), and (d).

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Solution: \begin{align*} && 0 &= z^2 + pz + 1\\ &&&= (x+iy)^2 + (x+iy)p + 1 \\ &&& = (x^2-y^2+px+1) + (2xy+py)i \\ \Rightarrow && 0 &= x^2 - y^2 + px + 1 \\ && 0 &= (2x+p)y \\ \Rightarrow && p &= -2x \\ \text{ or } && y &= 0 \\ \Rightarrow && p &= -(x^2+1)/x \end{align*} Therefore as \(p\) varies with either have \(y = 0\) and \(x\) taking any real value except \(0\) ie the real axis minus the origin. Or \(p = -2x\) and \(-y^2-x^2+1 = 0 \Rightarrow x^2 + y^2 = 1\) which is a circle. Suppose \(pz^2 + z + 1 = 0\) \begin{align*} && 0 &= pz^2 + z +1\\ &&&= p(x+iy)^2 + (x+iy) + 1\\ &&&= (px^2-py^2+x+1) + (2xyp + y) i \\ \Rightarrow && 0 &= (2xp+1)y \\ \Rightarrow && y & = 0, p = \frac{-(x+1)}{x^2}, x \neq 0 \\ \text{ or } && p &= -\frac{1}{2x}\\ \Rightarrow && 0 &= -\frac{1}{2}x + \frac{y^2}{2x} + x + 1 \\ &&&= \frac{y^2 - x^2 +2x^2 + 2x}{2x} \\ &&&= \frac{(x+1)^2+y^2-1}{2x} \end{align*} So we either have the real axis (except \(0\)) or a circle radius \(1\) centre \((-1, 0)\) (excluding \(x = 0\)).

Suppose \(pz^2 + p^2 z + 2 = 0\) then \begin{align*} && 0 &= p(x+iy)^2 + p^2(x+iy) + 2 \\ &&&= (p(x^2-y^2) + p^2x + 2) + (2xyp + p^2y)i \\ \Rightarrow && 0 &= py(2x+p) \\ \Rightarrow && y &= 0, \Delta = x^4-8x \\ \Rightarrow && x &\in (-\infty, 0) \cup [2, \infty) \\ \text{ or } && p &= -2x \\ && 0 &= (-2x)(x^2-y^2) + 4x^3+2 \\ &&&= 2x^3+2xy^2+2 \\ \Rightarrow && 0 &= x^3+xy^2+1 \end{align*}

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Solution:

Note that the line \(y = x\) is the tangent \((0,0)\) and \(y = \sin x \) is always below it. For any other line through the origin with gradient \(0 < k < 1\) it must start below \(y = \sin x\), but finish above it at \(x = \pi\). It also can only cross once due the the convexity of \(\sin\) in this interval. \begin{align*} && I &= \int_0^\pi | \sin x -kx | \d x \\ &&&= \int_0^\alpha (\sin x -k x) \d x + \int_{\alpha}^\pi (kx - \sin x) \d x \\ &&&= \left [ -\cos x - \frac{kx^2}{2} \right]_0^{\alpha} + \left [ \cos x +\frac{kx^2}{2} \right]_{\alpha}^\pi \\ &&&= -\cos \alpha - \frac{k \alpha^2}{2} +1+(-1)+\frac{k\pi^2}{2} - \cos \alpha - \frac{k\alpha^2}{2} \\ &&&= -2\cos \alpha - k\left (\alpha^2 - \frac{\pi^2}{2} \right) \\ &&&= -2\cos \alpha - \frac{\sin \alpha}{\alpha}\left (\alpha^2 - \frac{\pi^2}{2} \right) \\ &&&= \frac{\pi^2 \sin \alpha}{\alpha} - \alpha \sin \alpha - 2\cos \alpha \end{align*} \begin{align*} && \frac{\d I}{\d \alpha} &= \frac{\pi^2(\alpha \cos \alpha - \sin \alpha)}{2\alpha^2} + 2 \sin \alpha - \sin \alpha - \alpha \cos \alpha \\ &&&= \frac{-2\alpha^3 \cos \alpha + 2\alpha \sin\alpha + \pi^2 \alpha \cos \alpha - \pi^2 \sin \alpha}{2\alpha^2} \\ \\ &&&= \left ( \alpha \cos \alpha - \sin \alpha\right) \left ( \frac{\pi^2}{2\alpha^2}-1 \right)\ \end{align*} Therefore \(I' = 0\) if \(\tan \alpha = \alpha\) or \(\alpha = \frac{\pi}{\sqrt{2}}\). Since \(\tan \alpha = \alpha\) only at \(\alpha = 0\) (between \(0 \leq \alpha < \pi\) (by considering the tangent), we must have a unique turning point when \(\alpha = \frac{\pi}{\sqrt{2}}\). Note that \(I(\frac{\pi}{\sqrt{2}}) = \frac{\pi^2 \sqrt{2} \sin \alpha}{2\pi} - \frac{\pi}{\sqrt{2}} \sin \alpha - 2\cos \frac{\pi}{\sqrt{2}}=- 2\cos \frac{\pi}{\sqrt{2}}\). Notice that \(I(0) = \frac{\pi^2}2 - 2 > 2\) and \(I(\pi) = 2\), but \(-2\cos \frac{\pi}{\sqrt{2}} < 2\) so we must be a at a minimum

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Solution:

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   Clearly the only solution is \(x = 1\)
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   There is clearly only one solution, with \(x \approx -2\) \begin{align*} && 4(1-x) &= 6+2\sqrt{9-x^2} \\ && -2x-1 &=\sqrt{9-x^2} \\ \Rightarrow && 4x^2+4x+1 &= 9-x^2 \\ \Rightarrow && 0 &= 5x^2+4x-8 \\ &&x&= \frac{-2\pm 2\sqrt{11}}{5} \\ \Rightarrow && x &= -\left ( \frac{2+2\sqrt{11}}{5} \right) \end{align*}

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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.
   

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Solution: \begin{align*} && x &= \frac{1}{t^2-1} \\ && t &= \sqrt{\frac{x+1}{x}}\\ \Rightarrow && \frac{\d x}{\d t} &= \frac{-2t}{(t^2-1)^2} \\ \Rightarrow && I &= \int \frac{1}{\sqrt{x(x+1)}} \d x \\ &&&= \int \frac{1}{\sqrt{\frac1{t^2-1} \frac{t^2}{t^2-1}}} \cdot \frac{-2 t}{(t^2-1)^2} \d t \\ &&&= \int \frac{t^2-1}{t} \frac{-2t}{(t^2-1)^2} \d t \\ &&&= -\int \frac{2}{t^2-1} \d t \\ &&&= - \frac{2}{2 \cdot 1} \ln \left | \frac{t-1}{t+1} \right| +C \\ &&&= \ln \left | \frac{t+1}{t-1} \right| + C \\ &&&= \ln \left | \frac{\sqrt{\frac{x+1}{x}}+1}{\sqrt{\frac{x+1}{x}}-1} \right| + C \\ &&&= \ln \left | \frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}-\sqrt{x}} \right| + C \\ &&&= 2\ln \left | \sqrt{x+1}+\sqrt{x}\right| + C \\ &&&= 2\ln \left ( \sqrt{x+1}+\sqrt{x}\right) + C \\ \end{align*} \begin{align*} && V&= \pi \int_{1/8}^{9/16} y^2 \d x \\ &&&= \pi \int_{1/8}^{9/16} \left ( \frac1x + \frac{1}{x+1} - \frac{2}{\sqrt{x(x+1)}}\right) \d x \\ &&&= \pi \left [ \ln x + \ln (x+1) - 4 \ln(\sqrt{x+1} + \sqrt{x}) \right]_{1/8}^{9/16} \\ &&&= \pi \left ( \ln \frac{9}{16} + \ln \frac{25}{16} - 4 \ln \left ( \frac54 + \frac34\right) \right) +\\ &&&\quad -\pi \left ( \ln \frac{1}{8} + \ln \frac{9}{8} - 4 \ln \left ( \frac1{2\sqrt{2}} + \frac3{2\sqrt{2}}\right) \right) \\ &&&= \pi \left ( 2 \ln 3 - 8 \ln 2 + 2 \ln 5 - 4\ln2 \right) - \pi \left ( -6 \ln 2 + 2\ln 3 - 2\ln 2\right) \\ &&&= \pi (2 \ln 5 - 4 \ln 2 ) \\ &&&= 2 \pi \ln \tfrac54 \end{align*}

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Solution: \begin{align*} && {\bf r} &=\left( \e^{t}\cos t \right) {\bf i}+ \left(\e^t \sin t\right) {\bf j} \\ \Rightarrow && \dot{\bf r} &= \left( \e^{t}\cos t -\e^t \sin t\right) {\bf i}+ \left(\e^t \sin t+\e^t \cos t\right) {\bf j} \\ \Rightarrow && \mathbf{r}\cdot\dot{ \mathbf{r}} &= e^{2t}(\cos^2 t - \sin t \cos t) + e^{2t}(\sin^2 t+ \sin t \cos t) \\ &&&= e^{2t} (\cos^2 t + \sin ^2 t)\\ &&&= e^{2t} \\ \\ && | {\bf r}| &= e^{t} \\ && |{\bf \dot{r}}| &= e^t \sqrt{(\cos t - \sin t)^2 + (\sin t + \cos t)^2} \\ &&&= e^t \sqrt{2 \cos^2 t + 2 \sin^2 t} \\ &&&= \sqrt{2} e^t \\ \\ \Rightarrow && \frac{\mathbf{r}\cdot\dot{ \mathbf{r}}}{ |{\bf {r}}| |{\bf \dot{r}}|} &= \frac{e^{2t}}{\sqrt{2}e^te^t} \\ &&&= \frac{1}{\sqrt{2}} \end{align*} Therefore the angle between the velocity and displacement is \(\frac{\pi}{4}\). \begin{align*} && \ddot{\bf{r}} &= \left( \e^{t}(\cos t - \sin t) - \e^t (\sin t + \cos t)\right) {\bf i}+ \left(\e^t (\sin t + \cos t) + \e^t(\cos t - \sin t)\right) {\bf j} \\ &&&= \left ( -2\e^{t} \sin t \right) {\bf i}+ \left ( 2\e^{t} \cos t \right) {\bf j} \\ \Rightarrow && {\bf r} \cdot \ddot{\bf{r}} &= 2e^{2t} \left ( -\sin t \cos t + \sin t \cos t \right) \\ &&&= 0 \end{align*} Therefore the acceleration is perpendicular.

\(Q\) has position $\mathbf{r}' = \left( \e^{t-T}\cos (t-T) \right) {\bf i}+ \left(\e^{t-T} \sin (t-T)\right) {\bf j}\( for \)t > T$. \begin{align*} && {\bf r' \cdot r} &= e^{2t-T} \left (\cos t \cos (t-T) + \sin t \sin(t - T) \right) \\ &&&= e^{2t-T} \cos (t - (t-T)) \\ &&&= e^{2t-T} \cos T \\ \\ && |{\bf r'}- {\bf r} |^2 &= |{\bf r}|^2 + |{\bf r}'|^2 - 2 {\bf r' \cdot r} \\ &&&= e^{2t} + e^{2(t-T)} - 2e^{2t-T} \cos T \\ &&&= e^{2t} \left (1 - 2e^{-T} \cos T + e^{-2T} \right) \\ \Rightarrow && |{\bf r'}- {\bf r} | &= e^{t} \sqrt{1 - 2e^{-T} \cos T + e^{-2T} } \end{align*} as required

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Solution:

The curves touch when \(\sin x = 1\) ie \(x = \frac{4n+1}{2} \pi\). therefore \begin{align*} && I &= \int e^{-x} \sin x \d x \\ &&&= -e^{-x} \sin x + \int e^{-x} \cos x \d x \\ &&&= -e^{-x} \sin x -e^{-x} \cos x - I \\ && I &= -\frac{e^{-x}}2 ( \sin x + \cos x)\\ \\ && A_n &= \int_{\frac{4n+3}{2}\pi}^{\frac{4n-1}{2}\pi} e^{-x} (1-\sin x) \d x \\ &&&= \left [ -e^{-x} - \frac{e^{-x}}2(\sin x + \cos x) \right]_{\frac{4n+1}{2}\pi}^{\frac{4n+5}{2}\pi} \\ &&&= -\frac12\exp \left ( -\frac{4n+1}{2}\pi\right)+ \frac12\exp \left ( -\frac{4n-3}{2}\pi\right) \\ &&&= \frac12 (e^{2 \pi}-1) \left ( -\frac{4n+1}{2}\pi\right) \\ \\ && \sum_{n=1}^\infty A_n &= \frac12(e^{2\pi}-1)e^{-\pi/2} \sum_{n=1}^\infty e^{-2n\pi} \\ &&&= \frac12(e^{2\pi}-1)e^{-\pi/2} \frac{e^{-2\pi}}{1-e^{-2\pi}} \\ &&&= \frac12e^{-\pi/2} \end{align*}

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Solution: The curve \(x^4 + y^4 = u\) has lines of symmetry:

• \(y = 0\)
• \(x = 0\)
• \(y = x\)
• \(y = -x\)

The curve \(xy = v\) has lines of symmetry:

• \(y = x\)
• \(y = -x\)

The points are \(A = (\alpha, \beta), B = (\beta, \alpha), C = (-\alpha, -\beta), D = (-\beta, -\alpha)\) \(AD\) has gradient \(\frac{\beta+\alpha}{\alpha+\beta} = 1\), \(BC\) has the same gradient. \(AB\) has gradient \(\frac{\alpha-\beta}{\beta-\alpha} = -1\), as does \(CD\). Therefore it has two sets of perpendicular and parallel sides, hence a rectangle. The area is \(|AD||AB| = \sqrt{2(\alpha+\beta)^2}\sqrt{2(\alpha-\beta)^2} = 2(\alpha^2-\beta^2)\) The squared area is \(4(\alpha^4+\beta^4 - 2 \alpha^2\beta^2) = 4(u - 2v^2)\) ie the area is \(2\sqrt{u-2v^2}\) When \(u = 81, v = 4\) we have the area is \(2 \sqrt{81 - 2 \cdot 16} = 14\) as required.