Problems

Solution:

1. 

   + - ↺
2. \(2 \textrm{Min}(x^2,2x) = 5x-3\) tells us either \(2x^2 = 5x-3 \Rightarrow 2x^2 - 5x +3 = 0 \Rightarrow (2x-3)(x-1) \Rightarrow x = 1, \frac32\) and \(0 \leq x \leq 2\) or \(4x = 5x-3 \Rightarrow x= 3\) and \(x < 0\) or \(2 > x\), therefore our solutions are \(x = 1, \frac32, 3\)
3. We have different cases based on \(x\) vs \(-2, 0, 2\), ie Case \(x \leq -2\): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&= 2x + x^3 \end{align*} So \(2x = 2x + x^3 \Rightarrow x^3 = 0\), but \(x \leq -2\) so no solutions. or \(6x = 2x + x^3 \Rightarrow 0 = x(x^2-4) \Rightarrow x = 0, 2, -2\) so \(x = -2\). Case \(-2 < x \leq 0\): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&= 2x + 4x \end{align*} So \(2x = 2x + 4x\) ie \(x = 0\) which is valid. Or \(6x = 2x + 4x\) ie valid for all values in \(-2 \leq x \leq 0\) Case \(0 < x \leq 2\): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&= x^2 + x^3 \end{align*} So \(2x = x^2 + x^3 \Rightarrow 0 = x(x^2+x-2) = x(x-1)(x+2)\) so \(x = 0, 1, -2\), but the range means \(x = 0\) or \(x = 1\) Or \(6x = x^2 + x^3 \Rightarrow 0 = x(x^2+x-6) = x(x-2)(x+3)\) so \(x = 0, 2, -3\), but the range means \(x = 0\) or \(x = 2\) Case \(2 \leq x \): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&=2x + 4x \end{align*} So \(2x = 2x + 4x \Rightarrow x = 0\) so no solutions. Or \(6x = 2x + 4x\) so a range of solutions. Therefore the final solutions for \(m = 2\) are \(x = 0, x = 1\) and for \(m = 2\) are \(x \in [-2,0] \cup [2, \infty)\)
4. \(\mathrm{Min}(x^2, x^3)\) switches when \(x = 1\), so we must consider both limits: \begin{align*} && \frac{\d y}{\d x}\vert_{x > 1} &= 4x - 5 \\ \\ && \frac{\d y}{\d x}\vert_{x < 1} &= 6x^2 - 5 \\ \end{align*} so when \(x = 1\) the sign of the derivative changes from positive to negative, hence a local maximum. The other local maxima and minima will be when \(x = \frac54\) or \(x = \pm \sqrt{5/6}\)
   

   + - ↺

Solution:

1. 

   + - ↺
2. \begin{enumerate}
3. since \(\frac{\ln x}{x}\) is decreasing on \((e, \infty)\) we must have that \(\frac{\ln 3}{3} > \frac{\ln \pi}{\pi} \Rightarrow e^\pi > \pi^3\)
4. similarly, since \(\frac{\ln x}{x}\) is increasing on \((0, e)\) we must have that \(\frac{\ln \sqrt{5}}{\sqrt{5}} < \frac{\ln 9/4}{9/4} \Rightarrow \left(\frac{9}{4}\right)^{\sqrt{5}} > \sqrt{5}^{\frac{9}{4}}\)
5. Since \(2^4 = 4^2\) notice also that:
   

   + - ↺
   from the graph we must have the green area between \(1\) and \(2\) mapping to the (higher) green area between \(3\) and \(4\). Therefore \((x+2)^x > x^{x+2}\) for \(1 < x < 2\)
6. \begin{align*} && 9^{\sqrt 2} & \stackrel{?}{>} \sqrt{2}^9 \\ \Leftrightarrow && (3^2)^{\sqrt2} &\stackrel{?}{>} (\sqrt{2}^3)^3 \\ \Leftrightarrow && 3^{2 \sqrt2} &\stackrel{?}{>} (2\sqrt2)^3 \end{align*} Since \(e^2 < 8 < 9\Rightarrow e < 2\sqrt2 < 3\) therefore: \begin{align*} && \frac{\ln 2 \sqrt2}{2 \sqrt 2} &> \frac{\ln 3}{3} \\ \Leftrightarrow && (2 \sqrt{2})^3 &> 3^{2 \sqrt{2}} \\ \Leftrightarrow && \sqrt{2}^9 &> 9^{\sqrt 2} \\ \end{align*}
7. \begin{align*} && 8^{\sqrt[3]{3}} & \stackrel{?}{>} \sqrt[3]{3}^8 \\ \Leftrightarrow && 2^{3 \sqrt[3] 3} & \stackrel{?}{>} (\sqrt[3]{3}^4)^2 \\ \Leftrightarrow && 2^{3 \sqrt[3] 3} & \stackrel{?}{>} (3\sqrt[3]{3})^2 \\ \end{align*} Since \(3\sqrt[3]{3} > 4\) we have \begin{align*} && \frac{\ln (3 \sqrt[3]3)}{3 \sqrt[3]3} &< \frac{\ln 4}{4} \\ &&&= \frac{\ln 2}{2}\\ \Rightarrow && (3 \sqrt[3]{3})^2 &< 2^{3 \sqrt[3]{3}} \\ \Rightarrow && \sqrt[3]3^8 &< 8^{\sqrt[3]3} \end{align*}

Solution:

1. Suppose \(d_r = p_r - p_{r-1}\) then \begin{align*} d_r &= p_r - p_{r-1} \\ &= \mathbb{P}(X = r) - \mathbb{P}(X = r-1) \\ &= e^{-\lambda} \left ( \frac{\lambda^r}{r!} - \frac{\lambda^{r-1}}{(r-1)!} \right) \\ &= e^{-\lambda} \frac{\lambda^{r-1}}{(r-1)!} \left ( \frac{\lambda}{r} - 1\right) \end{align*} Therefore \(d_r > 0 \Leftrightarrow \lambda > r\)ie, \(p_r\) is increasing while \(r < \lambda\) and reaches a (unique) maximum when \(r = \lfloor \lambda \rfloor\).
2. Let \(dd_r = d_r - d_{r-1}\), so: \begin{align*} dd_r &= d_r - d_{r-1} \\ &= p_r - 2p_{r-1} + p_{r-2} \\ &= e^{-\lambda} \frac{\lambda^{r-2}}{r!} \left ( \lambda^2 - 2 \lambda r + r(r-1)\right ) \end{align*} Therefore \(dd_r < 0 \Leftrightarrow \lambda^2 - 2\lambda r +r(r-1) < 0 \Leftrightarrow r^2 -(1+2\lambda)r + \lambda^2 < 0\), but this has roots \(r = \frac{(1+2\lambda) \pm \sqrt{(1+2\lambda)^2-4\lambda^2}}{2} = \lambda + \frac12 \pm \sqrt{\lambda + \frac14}\). Therefore \(d_r\) is decreasing when \(r \in \left (\lambda + \frac12 -\sqrt{\lambda + \frac14},\lambda + \frac12 + \sqrt{\lambda + \frac14} \right)\), therefore the possible minimums are \(d_1\) and \(d_k\) where \(k < \lambda + \frac{1}{2} + \sqrt{\lambda + \frac{1}{4}} < k + 1\). \(d_1 = e^{-\lambda}(\lambda - 1)\), \(d_k = e^{-\lambda} \frac{\lambda^{k-1}}{(k-1)!}(\frac{\lambda}{k}-1)\)
3. If the maximum value of \(d_r\) is \(r = 1\) then \(d_r\) must be decreasing, ie considering \(dd_2\) we have \(\lambda^2 -4\lambda + 2< 0 \Leftrightarrow 2 - \sqrt{2} < \lambda < 2 + \sqrt{2}\). It must also be the case that it doesn't get beaten as \(\lambda \to \infty\). In this case \(d_r \to 0\), so we need \(d_1 > 0\), ie \(\lambda > 1\). Therefore \(1 < \lambda < 2 + \sqrt{2}\)
4. 

   + - ↺

Solution:

1. 1. 

      + - ↺
   2. If \(a = 1\) then \(u_1 = f(a) = 7-2 = 5\), \(u_2 = f(5) = -3\), \(u_3 = f(-3) = 7-6 = 1\). Therefore it must be the case that \(f(f(f(x))) = x\) for \(x = 1, 5, -3\). Similarly, if \(a = -\tfrac79\) then \(u_1 = f(-\tfrac79) = \tfrac{49}{9}\), \(u_2 = f(\tfrac{49}{9}) = -\tfrac{35}{9}\) and \(u_3 = f(-\tfrac{35}{9}) = -\tfrac79\). Therefore we must also have roots \(x = -\tfrac79, \tfrac{49}{9}, -\tfrac{35}9\). We also have the roots \(x = -7, \tfrac73\) from the first part so we have found all \(8\) roots.
   3. We need \(f(f(f(x))) = 1\) but \(f(f(x)) \neq -3, f(x) \neq 5, x \neq 1\). Suppose \(f(y) = 1 \Rightarrow 7-2|y| = 1 \Rightarrow y = \pm 3\). So \(y = 3\), ie \(f(f(x)) = 3\). Suppose \(f(z) = 3 \Rightarrow 7-2|z| = 3 \Rightarrow z = \pm 2\). Finally we need \(f(x) = \pm 2\), so say \(7-2|x| = 2 \Rightarrow x = \tfrac52\), so we have the sequence \(\tfrac52, 2, 3, 1, 5, -3, 1, \cdots\)as required.

Solution:

1. Suppose \begin{align*} && \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} \\ \Rightarrow && \binom{x_1}{y_1} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \binom{x_2}{y_2} \end{align*} Therefore if \(\frac{x_1^2}9+\frac{y_1^2}{4} = 1\) we must have \begin{align*} \frac{(\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2 }{9} + \frac{(-\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2}{4} = 1 \end{align*} but this is precisely the statement that \((x_1, y_1)\) is on \(G_1\) is equivalent to \((x_2,y_2)\) being on the \(G_2\). Since the point \((x_2,y_2)\) is a \(45^{\circ}\) rotation of \((x_1,y_1)\) anticlockwise about the origin, this means \(G_2\) is a \(45^{\circ}\) anticlockwise rotation of \(G_1\).
2. 1. \begin{align*} && \begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} -0.6 x + 0.8y \\ 0.8x + 0.6y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} -1.6 x + 0.8y \\ 0.8x -0.4y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \Rightarrow && y &=2 x \end{align*}
   2. 

      + - ↺
3. Consider the transformation \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) which is a shear, leaving the \(x\)-axis invariant. Then we must have:
   

   + - ↺
   Since the shear leaves lines of the form \(y = k\) invariant, the points where \(\frac{\d y}{\d x} = 0\) must also map to points where this is true, ie \((\tfrac{\pi}{2}, 1), (\tfrac{3\pi}{2}, -1)\) map to points \((\tfrac{\pi}{2}+2,1), (\tfrac{3\pi}{2} -2,-1)\) where the tangent is horizontal. The line \(x = c\) map back to lines \(\begin{pmatrix} 1 & -2 \\ 0 & 1\end{pmatrix} \begin{pmatrix} c \\ t\end{pmatrix} = \begin{pmatrix}c - 2t \\ t \end{pmatrix}\), ie \(y = -\frac12 x- \frac{c}{2}\). Therefore we are interested in points on the original curve where the gradient is \(-\frac12\), ie \((\frac{2\pi}{3}, \frac{\sqrt{3}}{2}), (\frac{4\pi}{3}, -\frac{\sqrt{3}}{2})\), these map to \((\frac{2\pi}{3}+\sqrt{3},\frac{\sqrt{3}}{2}), (\frac{4\pi}{3}-\sqrt{3}, -\frac{\sqrt{3}}{2})\)

Solution:

1. \begin{align*} \sqrt{x^2+1} &= |x|\sqrt{1+\frac{1}{x^2}} \\ &=|x| \left (1 + \frac12 \frac{1}{x^2} + \cdots \right) & \text{if } \left (\frac{1}{x^2} < 1 \right) \\ &= |x| + \frac12 \frac{1}{|x|} + \cdots \\ &\approx |x| + \frac{1}{2|x|} \end{align*}
   

   + - ↺
2. 1. \begin{align*} && \tan( g(x) + g(-x)) &= \tan \left ( \tan^{-1}(\sqrt{x^2+1}-x) + \tan^{-1}(\sqrt{x^2+1}+x) \right) \\ &&&= \frac{\sqrt{x^2+1}-x+\sqrt{x^2+1}+x}{1-1} \\ \Rightarrow && g(x) + g(-x) &\in \left \{\cdots, -\frac{\pi}{2}, \frac{\pi}{2}, \cdots \right\} \end{align*} But \(g(x), g(-x) > 0\) and \(g(x), g(-x) \in (-\frac{\pi}{2}, \frac{\pi}{2})\), therefore it must be \(\frac{\pi}{2}\).
   2. \begin{align*} && \tan(2(k(x) + k(-x))) &= \tan(\tan^{-1}x + \tan^{-1}(-x)) \\ &&&= 0 \\ \Rightarrow && k(x)+k(-x) &\in \left \{\cdots, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \cdots \right\} \\ \end{align*} But \(k(x) \in (-\frac{\pi}{4}, \frac{\pi}{4})\), therefore \(k(x) + k(-x) = 0\).
   3. Let \(t = \tan k(x)\). \begin{align*} && \tan \left ( \tan^{-1} \frac{1}{x} \right) &= \frac{2 \tan\left ( \frac12 \tan^{-1} \frac1x \right)}{ 1-\tan^2\left ( \frac12 \tan^{-1} \frac1x \right)} \\ \Rightarrow && \frac1x &= \frac{2t}{1-t^2} \\ \Rightarrow && 1-t^2 &= 2tx \\ \Rightarrow && 0 &= t^2+2tx - 1 \\ \Rightarrow && 0 &= (t+x)^2 - 1-x^2 \\ \Rightarrow && t &= -x \pm \sqrt{1+x^2} \end{align*} Since \(t > 0\), \(t = \sqrt{1+x^2}-x = f(x) = \tan g(x)\)
   4. 

      + - ↺
   5. \begin{align*} \int_0^1 k(x) \d x &= \int_0^1 \frac12 \tan^{-1} \left ( \frac1x \right) \d x \\ &= \left [ \frac{x}{2} \tan^{-1}\left ( \frac1x \right) \right]_0^1 - \int_0^1 \frac{x}{2} \frac{-1/x^2}{1+1/x^2} \d x \\ &= \left [ \frac{x}{2} \tan^{-1}\left ( \frac1x \right) \right]_0^1 + \frac14 \int_0^1 \frac{2x}{1+x^2} \d x \\ &= \frac12 \frac{\pi}{4} + \frac14 \ln(2) \\ &= \frac{\pi + \ln 4}{8}\end{align*} Therefore \(\displaystyle \int_{-1}^0 g(x) \d x = -\frac{\pi + \ln 4}{8}\)

Solution:

1. 

   + - ↺
   By symmetry, the centre of mass will lie on the \(y\) axis. Notice that a single slice (when revolved around the \(y\)-axis) has volume \(y \cdot \pi \cdot ((x+ \delta x)^2 - x^2) = 2 \pi x y \delta x\), and COM at height \(\frac12 y\) so we can conclude: \[ \overline{y} \sum_{\delta x} 2 \pi x y \delta x = \sum_{\delta x} \pi xy^2 \delta x\] \begin{align*} && \overline{y} \int_0^r 2xy \d x &= \int_0^r y^2 x \d x \\ \Rightarrow && \overline{y} 2\int_0^r \left (r - \frac{x^n}{r^{n-1}} \right)x \d x &= \int_0^r \left (r - \frac{x^n}{r^{n-1}} \right)^2 x \d x \\ \Rightarrow && \overline{y} \left [r \frac{x^2}{2} - \frac{1}{r^{n-1}} \frac{x^{n+2}}{n+2} \right]_0^r &= \left [r^2 \frac{x^2}{2} - \frac{2}{r^{n-2}} \frac{x^{n+2}}{n+2} + \frac{1}{r^{2n-2}} \frac{x^{2n+2}}{2n+2} \right]_0^r \\ \Rightarrow && 2\overline{y} \left (\frac{r^3}{2} - \frac{r^3}{n+2} \right) &= \left (\frac12 r^4 - \frac{2}{n+2}r^4 + \frac{1}{2n+2}r^4 \right) \\ \Rightarrow && \overline{y}r^3 \frac{n}{(n+2)} &= r^4\frac{(n+1)(n+2)-2\cdot2\cdot(n+1)+(n+2)}{2(n+1)(n+2)} \\ \Rightarrow && \overline{y} \frac{n}{(n+2)} &= r \left ( \frac{n^2}{2(n+1)(n+2)} \right) \\ \Rightarrow && \overline{y} &= \frac{nr}{2(n+1)} \\ &&&= r \left (1 -p^n \right) \end{align*} as required.
2. \begin{align*} && r^{n-1}y &= r^n - x^n \\ \frac{\d}{\d x}: && r^{n-1} \frac{\d y}{\d x} &= -n x^{n-1} \\ && \frac{\d y}{\d x} &= -np^{n-1} \end{align*} Therefor the normal has the equation: \begin{align*} && \frac{y-r(1-p^n)}{x-rp} &= \frac{1}{np^{n-1}} \\ \Rightarrow && Y &= \frac{-rp}{np^{n-1}} + r(1-p^n) \\ &&&= r \left (1 - p^n - \frac{1}{np^{n-2}} \right) \end{align*} If \(n = 4\) then \begin{align*} && Y &= r\left (1 - p^4 - \frac{1}{4p^{2}} \right) \\ \Rightarrow && \frac{\d Y}{\d p} &= r \left (-4p^3 + \frac{1}{2p^3} \right) \end{align*} Therefore there is a stationary point if \(p^6 = \frac18 \Rightarrow p =2^{-1/2}\). Clearly this will be a maximum (sketch or second derivative) therefore, \(Y = r(1 - \frac14 - \frac{2}{4}) = \frac14 r\)
3. The centre of mass of this shape can be found using this table: \begin{array}{|c|c|c|} \hline \text{} & \overline{y} & \text{mass} \\ \hline r^3y = r^4 - x^4 & \frac{2r}{5} & \frac{4\pi r^3}{6} = \frac23 \pi r^3\\ ry = -(r^2 - x^2) & -\frac{r}{3}& \frac{2 \pi r^3}{4}=\frac12\pi r^3 \\ \text{combined} & \frac{(\frac25 \cdot \frac23-\frac13 \cdot \frac12)r^4}{\frac76 r^3} = \frac3{35}r & \frac76 \pi r^3\\ \hline \end{array} Normals to the surface through points on the upper surface will meet the \(y\)-axis between \((-\infty, \frac14 r)\), and since \(p = 0 \to -\infty\) and \(p = 1 \to -\frac14 r\), so normals will pass through \((0, \frac3{35}r)\) from two different points. Normals to the surface through points on the lower surface will go through \(-r(1 - p^2 - \frac12) =- r(\frac12 -p^2)\) which ranges monotonically from \(\frac12 r \to -\frac12 r\) so there will be one point where the normal goes through \(\frac3{35}r\). Therefore there are three angles where the vector \(\overrightarrow{AB}\) is not vertical but the normal to the surfaces runs through the centre of mass (ie the the solid can rest in equilibrium)

Solution:

1. \(\,\)
   

   + - ↺
2. 1. Suppose \(y=k\) meets the curve \(C_2\) then \begin{align*} && \tfrac1{16} &= (k^2+(x-1)^2-1)(k^2+(x+1)^2-1) \\ &&&= (k^2+x^2-2x)(k^2+x^2+2x) \\ &&&= k^4+2k^2x^2+x^4-4x^2 \\ &&&= x^4+(2k^2-4)x^2+k^4 \\ \Rightarrow && 0 &= x^4+(2k^2-4)x^2+(k^4-\tfrac1{16}) \\ \\ && \Delta &= 4(k^2-2)^2 - 4 \cdot 1 \cdot (k^4-\tfrac1{16}) \\ &&&= 4(k^4-4k^2+4 - k^4 +\tfrac{1}{16}) \\ &&&= 16(1+\tfrac{1}{64} - k^2) \\ &&&= 16(\tfrac{65}{64} - k^2) \end{align*} Therefore if \(|k| < \frac{\sqrt{65}}{8}\) there are \(4\) intersections. If \(|k| = \frac{\sqrt{65}}{8}\) there are \(2\) intersections, otherwise there are \(0\).
   2. The greatest possible \(y\) value is \( \frac{\sqrt{65}}{8}\) and at this point \(x^2 = \frac{2(2-\frac{65}{64}}{2} =1 - \frac{1}{64} < 1\) so they are close to the \(y\)-axis.
   3. The regions where \(y^2+(x-1)^2-1 < 0\) and \(y^2+(x+1)^2-1 < 0\) is the interior of the two circles from the first part. However, since they don't overlap they can never both be negative. Therefore in our equation both are positive and therefore \(C_2\) is entirely outside \(C_1\)
   4. \(\,\)
      

      + - ↺

Solution:

1. \(\,\)
   

   + - ↺
   \begin{align*} && y & = x^{1/x} = \exp \left ( \tfrac1x \ln x \right ) \\ \Rightarrow && y' &= \exp \left ( \tfrac1x \ln x \right ) \cdot \left ( \frac{1}{x^2} - \frac{\ln x}{x^2} \right ) \\ &&&= \frac{\exp \left ( \tfrac1x \ln x \right ) }{x^2}(1 - \ln x) \\ y' =0: && x &= e \end{align*} Therefore the largest integer values will be \(2\) or \(3\). Comparing \((2^{\frac12})^6 = 8 < 9 = (3^{1/3})^6\) we see the maximum value of \(n^{1/n}\) where \(n\) is an integer is \(\sqrt[3]{3}\)
2. The number of tests is \(r\) plus however many groups fail times \(k\). The probability of a group failing is \(g = 1-(1-p)^k\) and the number of failing groups is \(\sim B(r, g)\) so the expected number of additional groups is \(rg\) and the expected total number of tests is \[ \frac{N}{k} + N(1-(1-p)^k) \]
3. \(\,\) \begin{align*} && N &\geq E = \frac{N}{k} + N(1-(1-p)^k) \\ \Rightarrow && 1 &\geq \frac{1}{k} + 1-(1-p)^k \\ \Rightarrow && (1-p)^k &\geq \frac1k \\ \Rightarrow && k \ln(1-p) &\geq - \ln k \\ \Rightarrow && \ln(1 - p) &\geq -\frac{1}{k} \ln k \geq -\frac13 \ln 3 \\ \Rightarrow && 1-p &\geq \frac{1}{\sqrt[3]{3}} \\ \Rightarrow && p &\leq 1-\frac{1}{\sqrt[3]{3}} \end{align*} (taking \(k=3\)) Claim: \(1 - \frac{1}{\sqrt[3]{3}} > \frac14\) Proof: This is equivalent to \(\sqrt[3]{3} > \frac43\) or \(81 > 4^3 = 64\) which is clearly true.
4. If \(pk\) is small then \((1-p)^k \approx 1 - pk\) and so we obtain \(N \left ( \frac1k +pk \right)\) as required. If \(p = 0.01\) and \(k = 10\) then \(\frac{1}{10} + 0.01 \cdot 10 = 0.2\) so the expected number of tests is \(\sim 20\%\) of \(N\)