Problems

Solution:

1. Using the generalise binomial theorem \begin{align*} \left( 1+\frac k {100} \right)^{\frac12} &= 1 + \frac12 \frac{k}{100} + \frac{\tfrac12 \cdot \left ( -\tfrac12\right)}{2!} \left (\frac{k}{100} \right)^2 + \frac{\tfrac12 \cdot \left ( -\tfrac12\right)\cdot \left ( -\tfrac32\right)}{3!} \left (\frac{k}{100} \right)^3 + \cdots \\ &= 1 + \frac{1}{200}k - \frac{1}{80\,000}k^2 + \frac{1}{16\,000\,000}k^3 + \cdots \end{align*}
   1. If \(k = 8\), \begin{align*} && \left( 1+\frac 8 {100} \right)^{\frac12} &= 1 + \frac{1}{200}8 - \frac{1}{80\,000}8^2 + \frac{1}{16\,000\,000}8^3 + \cdots \\ \Rightarrow && \frac{6\sqrt{3}}{10} &\approx 1 + 0.04 - 0.0008 + 0.000032 \\ &&&= 1.039232\\ \Rightarrow && \sqrt{3} &\approx 1.73205 \, (5\, \text{d.p.}) \end{align*}
   2. If \(k = -4\), \begin{align*} && \left( 1-\frac 4 {100} \right)^{\frac12} &= 1 - \frac{1}{200}4 - \frac{1}{80\,000}4^2 - \frac{1}{16\,000\,000}4^3 + \cdots \\ \Rightarrow && \frac{4\sqrt{6}}{10} &\approx 1 -0.02-0.0002 -0.000004 \\ &&&= 0.979796\\ \Rightarrow && \sqrt{6} &\approx 2.44949\, (5\, \text{d.p.}) \end{align*}
2. \(\,\) \begin{align*} && \left( 1+\frac k {1000} \right)^{\!\frac13} &= 1 + \frac13 \frac{k}{1000} + \cdots \\ &&&= 1 + \frac{k}{3\,000} + \cdots \\ && 3 \times 7^3 &= 1029 \\ \Rightarrow && \left( 1+\frac {29} {1000} \right)^{\!\frac13} &\approx 1 + \frac{29}{3\,000} \\ \Rightarrow && \frac{7\sqrt[3]{3}}{10} &\approx \frac{3\,029}{3000} \\ \Rightarrow && \sqrt[3]{3} &= \frac{3\,029}{2\,100} \end{align*}

Solution:

1. \(\,\) \begin{align*} && y &= 2x^3-bx^2+cx \\ \Rightarrow && y' &= 6x^2-2bx+c \end{align*} We must have \(p, q\) are the roots of this equation, ie \(\frac13b = p+q, \frac16c = pq\)
2. The point of inflection will be at \(\frac{b}6\) The graph will have rotational symmetry of \(180^{\circ}\) about the point of inflection.
   

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3. \begin{align*} && m-n &= 2(p^3-q^3)-b(p^2-q^2)+c(p-q) \\ &&&= (p-q)(2(p^2+qp+q^2)-b(p+q)+c) \\ &&&= (p-q)(2(p^2+qp+q^2)-3(p+q)^2+6pq) \\ &&&= (p-q)(-p^2-q^2+2pq) \\ &&&= (q-p)^3 \end{align*}
4. The area of \(R\) is \begin{align*} A &= \frac12 bh \\ &= \frac12 (q-p)(m-n) = \frac12(q-p)^4 \end{align*} as required.

Solution: \begin{align*} && I &= \int \frac{1}{a^2+x^2} \d x\\ x = a \tan \theta, \d x =a \sec^2 \theta \d \theta &&&= \int \frac{1}{a^2+a^2\tan^2 x} a \sec^2 \theta \d \theta \\ &&&=\int \frac{\sec^2 \theta}{a \sec^2 \theta} \d \theta \\ &&&= \frac1a \theta + C \\ &&&= \frac1a \arctan \frac{x}{a} + C \end{align*}

1. 1. \(\,\) \begin{align*} && I &= \int_0^{\frac{1}{2}\pi} \frac {\cos x}{1+\sin^2 x} \d x \\ &&&= \left [ \arctan (\sin x) \right]_0^{\pi/2} \\ &&&= \arctan(1) - \arctan(0) = \frac{\pi}{4} \end{align*}
   2. \(\,\) \begin{align*} && t &= \tan \frac{x}{2} \\ \Rightarrow && \sin x &= \frac{2t}{1+t^2} \\ && \cos x &= \frac{1-t^2}{1+t^2} \\ && \d x &= \frac{2}{1+t^2} \d t \\ \Rightarrow && I &= \int_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x } \d x \\ &&&= \int_{t=0}^{t = 1} \frac{\frac{1-t^2}{1+t^2}}{1 + \left (\frac{2t}{1+t^2} \right)^2} \frac{2}{1+t^2} \d t \\ &&&= 2 \int_0^1 \frac{1-t^2}{(1+t^2)^2+(2t)^2} \d t\\ &&&= 2 \int_0^1 \frac{1-t^2}{1+6t^2+t^4} \d t\\ \end{align*} From which the conclusion follows
2. \(\,\) \begin{align*} && J &= \int_0^1 \frac {1-t^2}{1+14t^2+t^4} \, \d t \\ &&&= \int_0^1 \frac {\frac{1-t^2}{1+t^2}}{\frac{1+14t^2+t^4}{(1+t^2)^2}} \frac{1}{1+t^2} \, \d t \\ &&&= \int_0^1 \frac {\frac{1-t^2}{1+t^2}}{\frac{(t^2+1)^2+3(2t)^2}{(1+t^2)^2}} \frac{1}{1+t^2} \, \d t \\ &&&= \frac12\int_{x=0}^{x=\pi/2} \frac {\cos x}{1+3 \sin^2 x} \d x \\ &&&= \frac{1}{6}\left[ \sqrt{3} \arctan(\sin \sqrt{3}x)\right]_0^{\pi/2} \\ &&&= \frac16 \sqrt{3} \frac{\pi}{3} \\ &&&= \frac{\sqrt{3}\pi}{18} \end{align*}

Solution: \begin{align*} && f(x) &= \sin x \\ \Rightarrow && f''(x) &= -\sin x \end{align*} which is clearly negative on \((0,\pi)\) since \(\sin\) is positive on this interval. \begin{align*} && f(x) &= \ln x \\ \Rightarrow && f''(x) &= -1/x^2 \end{align*} which is clearly negative for \(x > 0\)

1. Since \(A,B,C\) are angles in a triangle, we must have \(0 < A,B,C< \pi\) and so we can apply Jensen with \(f = \sin\) to obtain: \begin{align*} &&\frac13( \sin A + \sin B + \sin C) &\leq \sin \left ( \frac{A+B+C}{3}\right) \\ &&&= \sin \frac{\pi}{3} = \frac{\sqrt{3}}2 \\ \Rightarrow && \sin A + \sin B + \sin C &\leq\frac{3\sqrt{3}}2 \end{align*}
2. Suppose \(f(x) = \ln x\), then applying Jensen on the positive numbers \(t_1, \ldots, t_n\) we obtain \begin{align*} && \frac1n \left ( \sum_{i=1}^n \ln t_n \right) &\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \frac1n \ln\left (\prod_{i=1} t_n\right)&\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \ln\left (\left (\prod_{i=1} t_n\right)^{1/n}\right)&\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \left (\prod_{i=1} t_n\right)^{1/n}&\leq\frac1n\sum_{i=1}^n t_n \\ \Rightarrow && \sqrt[n]{t_1t_2 \cdots t_n}&\leq\frac1n(t_1 + t_2 + \cdots + t_n) \tag{AM-GM}\\ \end{align*}
   1. Applying AM-GM with \(t_1 = x^4, t_2 = y^4, t_3 = z^4, t_4 = 2^4\) we have \begin{align*} && \frac{x^4+y^4+z^4+16}{4} & \geq \sqrt[4]{x^4y^4z^42^4} \\ \Rightarrow && x^4+y^4+z^4+16 &\geq 8xyz \end{align*}
   2. Applying AM-GM with \(t_1 = x^5, t_2 = y^5, t_3 = z^5, t_4 = 1^5, t_5 = 1^5\) we have \begin{align*} && \frac{x^5+y^5+z^5+1+1}{5} & \geq \sqrt[5]{x^5y^5z^5} \\ \Rightarrow && x^5+y^5+z^5+2 &\geq 5xyz \\ \Rightarrow && x^5+y^5+z^5 - 5xyz &\geq -2 \end{align*} Therefore the minimum is \(-2\)

Solution:

1. \(X\) lies on the line including \(B\) and \(C\).
2. points on the line \(AP\) have the form \(\lambda(\beta \mathbf{b}+\gamma\mathbf{c})\), and the point \(D\) will be the point where \(\lambda\beta + \lambda \gamma = 1\). \begin{align*} && \frac{|BD|}{|DC|} &= \frac{|\mathbf{b} -\lambda(\beta \mathbf{b}+\gamma\mathbf{c})| }{|\lambda(\beta \mathbf{b}+\gamma\mathbf{c})- \mathbf{c}|} \\ &&&= \frac{|(1-\lambda \beta)\mathbf{b} - \lambda \gamma \mathbf{c}|}{|\lambda \beta \mathbf{b}+(\lambda \gamma -1)\mathbf{c}|}\\ &&&= \frac{|\lambda \gamma\mathbf{b} - \lambda \gamma \mathbf{c}|}{|\lambda \beta \mathbf{b}-(\lambda \beta)\mathbf{c}|} \\ &&&= \frac{\gamma}{\beta} \end{align*}
3. The line \(BP\) is \(\mathbf{b} + \mu(\beta \mathbf{b}+\gamma\mathbf{c})\) and will meet \(CA\) when \(1+\mu\beta = 0\), ie \(\mu = -\frac{1}{\beta}\), therefore \(E\) is \(-\frac{\gamma}{\beta}\mathbf{c}\), and so \(\frac{|CE|}{|EA|} = \frac{1+\gamma/\beta}{\gamma/\beta} = \frac{\beta+\gamma}{\gamma}\). Similarly, \(F\) is \(-\frac{\beta}{\gamma}\mathbf{b}\) and \(\frac{|AF|}{|FB|} = \frac{\beta/\gamma}{1+\frac{\beta}{\gamma}} = \frac{\beta}{\gamma+\beta}\), and so \[\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = \frac{\beta}{\gamma+\beta} \frac{\gamma}{\beta} \frac{\beta+\gamma}{\gamma} = 1 \]