Problems
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2007 Paper 1 Q14
The discrete random variable \(X\) has a Poisson distribution with mean \(\lambda\).
- Sketch the graph \(y=\l x+1 \r \e^{-x}\), stating the coordinates of the turning point and the points of intersection with the axes. It is known that \(\P(X \ge 2) = 1-p\), where \(p\) is a given number in the range \(0 < p <1\). Show that this information determines a unique value (which you should not attempt to find) of \(\lambda\).
- It is known (instead) that \(\P \l X = 1 \r = q\), where \(q\) is a given number in the range \(0 < q <1\). Show that this information determines a unique value of \(\lambda\) (which you should find) for exactly one value of \(q\) (which you should also find).
- It is known (instead) that
\(\P \l X = 1 \, \vert \, X \le 2 \r = r\),
where \(r\) is a given number in the range \(0
2007 Paper 2 Q1
In this question, you are not required to justify the accuracy of the approximations.
- Write down the binomial expansion of \(\displaystyle \left( 1+\frac k {100} \right)^{\!\frac12}\)in ascending powers of \(k\), up to and including the \(k^3\) term.
- Use the value \(k=8\) to find an approximation to five decimal places for \(\sqrt{3}\,\).
- By choosing a suitable integer value of \(k\), find an approximation to five decimal places for \(\sqrt6\,\).
- By considering the first two terms of the binomial expansion of \(\displaystyle \left( 1+\frac k {1000} \right)^{\!\frac13}\), show that \(\dfrac{3029}{2100}\) is an approximation to \(\sqrt[3]{3}\).
Solution:
- 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*}
- 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*}
- 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*}
- \(\,\) \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*}
2007 Paper 2 Q2
A curve has equation \(y=2x^3-bx^2+cx\). It has a maximum point at \((p,m)\) and a minimum point at \((q,n)\) where \(p>0\) and \(n>0\). Let \(R\) be the region enclosed by the curve, the line \(x=p\) and the line \(y=n\).
- Express \(b\) and \(c\) in terms of \(p\) and \(q\).
- Sketch the curve. Mark on your sketch the point of inflection and shade the region \(R\). Describe the symmetry of the curve.
- Show that \(m-n=(q-p)^3\).
- Show that the area of \(R\) is \(\frac12 (q-p)^4\).
Solution:
- \(\,\) \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 \(\frac13b = p+q, \frac16c = pq\)
- The point of inflection will be at \(\frac{b}6\)
The graph will have rotational symmetry of \(180^{\circ}\) about the point of inflection.
- \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*}
- The area of \(R\) is \begin{align*} A &= \frac12 bh \\ &= \frac12 (q-p)(m-n) = \frac12(q-p)^4 \end{align*} as required.
2007 Paper 2 Q3
By writing \(x=a\tan\theta\), show that, for \(a\ne0\), $\displaystyle \int \frac 1 {a^2+x^2}\, \d x =\frac 1 a \arctan \frac x a + \text{constant}\,$.
- Let $\displaystyle I=\int_0^{\frac{1}{2}\pi}
\frac {\cos x}{1+\sin^2 x} \, \d x\,$.
- Evaluate \(I\).
- Use the substitution \(t=\tan \frac12 x\) to show that \(\displaystyle \int_0^1 \frac {1-t^2}{1+6t^2+t^4} \, \d t = \tfrac12 I\,\).
- Evaluate \(\displaystyle \int_0^1 \frac {1-t^2}{1+14t^2+t^4} \, \d t \,\).
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*}
- \(\,\) \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*}
- \(\,\) \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
- \(\,\) \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*}
2007 Paper 2 Q4
Given that \(\cos A\), \(\cos B\) and \(\beta\) are non-zero, show that the equation \[ \alpha \sin(A-B) + \beta \cos(A+B) = \gamma \sin(A+B) \] reduces to the form \[ (\tan A-m)(\tan B-n)=0\,, \] where \(m\) and \(n\) are independent of \(A\) and \(B\), if and only if \(\alpha^2=\beta^2+\gamma^2\). Determine all values of \(x\), in the range \(0\le x <2\pi\), for which:
- $2\sin(x-\frac14\pi) + \sqrt 3 \cos(x+\frac14\pi) = \sin(x+\frac14\pi)\, \(;
- \)2\sin(x-\frac16\pi) + \sqrt 3 \cos(x+\frac16\pi) = \sin(x+\frac16\pi)\, \(;
- \)2\sin(x+\frac13\pi) + \sqrt 3 \cos(3x) = \sin(3x)\, $.
2007 Paper 2 Q5
In this question, \(\f^2(x)\) denotes \(\f(\f(x))\), \(\f^3(x)\) denotes \(\f( \f (\f(x)))\,\), and so on.
- The function \(\f\) is defined, for \(x\ne \pm 1/ \sqrt3\,\), by $$ \f(x) = \ds \frac{x+\sqrt3} {1-\sqrt3\, x }\,. $$ Find by direct calculation \(\f^2(x) \) and \(\f^3(x)\), and determine \(\f^{2007}(x)\,\).
- Show that \(\f^n(x) = \tan(\theta + \frac 13 n\pi)\), where \(x=\tan\theta\) and \(n\) is any positive integer.
- The function \(\g(t)\) is defined, for \(\vert t\vert\le1\) by \(\g(t) = \frac {\sqrt3}2 t + \frac 12 \sqrt {1-t^2}\,\). Find an expression for \(\g^n(t)\) for any positive integer \(n\).
2007 Paper 2 Q6
- Differentiate \(\ln\big (x+\sqrt{3+x^2}\,\big)\) and \(x\sqrt{3+x^2}\) and simplify your answers. Hence find \(\int \! \sqrt{3+x^2}\, \d x\).
- Find the two solutions of the differential equation \[ 3\left(\frac{\d y}{\d x}\right)^{\!2} + 2 x \frac {\d y}{\d x} =1 \] that satisfy \(y=0\) when \(x=1\).
2007 Paper 2 Q7
A function \(\f(x)\) is said to be concave on some interval if \(\f''(x)<0\) in that interval. Show that \(\sin x\) is concave for \(0< x < \pi\) and that \(\ln x\) is concave for \(x > 0\). Let \(\f(x)\) be concave on a given interval and let \(x_1\), \(x_2\), \(\ldots\), \(x_n\) lie in the interval. Jensen's inequality states that \[ \frac1 n \sum_{k=1}^n\f(x_k) \le \f \bigg (\frac1 n \sum_{k=1}^n x_k\bigg) \] and that equality holds if and only if \(x_1=x_2= \cdots =x_n\). You may use this result without proving it.
- Given that \(A\), \(B\) and \(C\) are angles of a triangle, show that \[ \sin A + \sin B + \sin C \le \frac{3\sqrt3}2 \,. \]
- By choosing a suitable function \(\f\), prove that
\[
\sqrt[n]{t_1t_2\cdots t_n}\; \le \; \frac{t_1+t_2+\cdots+t_n}n
\]
for any positive integer \(n\) and for any positive numbers \(t_1\), \(t_2\), \(\ldots\), \(t_n\).
Hence:
- show that \(x^4+y^4+z^4 +16 \ge 8xyz\), where \(x\), \(y\) and \(z\) are any positive numbers;
- find the minimum value of \(x^5+y^5+z^5 -5xyz\), where \(x\), \(y\) and \(z\) are any positive numbers.
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 \(x > 0\)
- 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*}
- 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*}
- 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*}
- 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 \(-2\)
2007 Paper 2 Q8
The points \(B\) and \(C\) have position vectors \(\mathbf{b}\) and \(\mathbf{c}\), respectively, relative to the origin \(A\), and \(A\), \(B\) and \(C\) are not collinear.
- The point \(X\) has position vector \(s \mathbf{b}+t\mathbf{c}\). Describe the locus of \(X\) when \(s+t=1\).
- The point \(P\) has position vector \(\beta \mathbf{b}+\gamma\mathbf{c}\), where \(\beta\) and \(\gamma\) are non-zero, and \(\beta+\gamma\ne1\). The line \(AP\) cuts the line \(BC\) at \(D\). Show that \(BD:DC=\gamma:\beta\).
- The line \(BP\) cuts the line \(CA\) at \(E\), and the line \(CP\) cuts the line \(AB\) at \(F\). Show that \[ \frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA}=1\,. \]
Solution:
- \(X\) lies on the line including \(B\) and \(C\).
- 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*}
- 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 \[\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = \frac{\beta}{\gamma+\beta} \frac{\gamma}{\beta} \frac{\beta+\gamma}{\gamma} = 1 \]
2007 Paper 2 Q9
A solid right circular cone, of mass \(M\), has semi-vertical angle \(\alpha\) and smooth surfaces. It stands with its base on a smooth horizontal table. A particle of mass \(m\) is projected so that it strikes the curved surface of the cone at speed \(u\). The coefficient of restitution between the particle and the cone is \(e\). The impact has no rotational effect on the cone and the cone has no vertical velocity after the impact.
- The particle strikes the cone in the direction of the normal at the point of impact. Explain why the trajectory of the particle immediately after the impact is parallel to the normal to the surface of the cone. Find an expression, in terms of \(M\), \(m\), \(\alpha\), \(e\) and \(u\), for the speed at which the cone slides along the table immediately after impact.
- If instead the particle falls vertically onto the cone, show that the speed \(w\) at which the cone slides along the table immediately after impact is given by \[ w= \frac{mu(1+e)\sin\alpha\cos\alpha}{M+m\cos^2\alpha}\,. \] Show also that the value of \(\alpha\) for which \(w\) is greatest is given by \[ \cos \alpha = \sqrt{ \frac{M}{2M+m}}\ . \]