Problems

Solution:

1. The only way you wont be able to sell to them is if they are first, ie \(\frac1{m+1}\)
2. If \(n=2\), the the only way you fail to sell to them is if one comes first or they both appear before two people with pound coins, ie \(2\) or \(122\). These have probabilities \(\frac{2}{m+2}\) and \(\frac{m}{m+2} \cdot \frac{2}{m+1} \frac{1}{m} = \frac{2}{(m+1)(m+2)}\). Therefore the total probability you don't sell all the tickets is \(\frac{2}{m+2}\left ( 1 + \frac{1}{m+1} \right) = \frac{2}{m+2} \frac{m+2}{m+1} = \frac{2}{m+1}\). Therefore the probability you do sell all the tickets is \(1 - \frac{2}{m+1} = \frac{m-1}{m+1}\)
3. The only ways to fail when \(n=3\) are: \(2\), \(122\), or if all three \(2\)s appear before three \(1\)s. this can happen in \(11222\), \(12122\) These happen with probability: \begin{align*} 2: && \frac{3}{m+3} \\ 122: && \frac{m}{m+3} \cdot \frac{3}{m+2} \cdot \frac{2}{m+1} \\ 11222: && \frac{m(m-1) 6}{(m+3)(m+2)(m+1)m(m-1)} \\ 12122: && \frac{m(m-1) 6}{(m+3)(m+2)(m+1)m(m-1)} \\ \end{align*} Therefore the total probability is: \begin{align*} P &= \frac{1}{(m+3)(m+2)(m+1)} \left (3(m+2)(m+1)+6m + 12 \right) \\ &= \frac{1}{(m+3)(m+2)(m+1)} \left (3(m+1)(m+2) \right) \\ &= \frac{3}{m+1} \end{align*} and the result follows

Solution:

1. \(\,\) \begin{align*} && x &= (a-x)^{\frac12} \\ \Rightarrow && x^2 &= a - x \\ \Rightarrow && 0 &= x^2 + x - a \end{align*} This has a roots if \(\Delta = 1 + 4a \geq 0 \Rightarrow a \geq -\frac14\). These roots also need to be positive (since \(x \geq 0\)). Since \(f(0) = -a\) we have one positive root if \(a \geq 0\). If \(a \leq 0\) then since the roots are symmetric about \(x = -\frac12\), both roots are negative and there are no positive roots. Therefore we have on real solution if \(a \geq 0\) and non otherwise. \begin{align*} && x & = \left ( (1+a)x - a \right)^{\frac13} \\ \Leftrightarrow && x^ 3 &= (1+a)x - a \\ \Leftrightarrow && 0 &= x^3- (1+a)x + a \\ \Leftrightarrow && 0 &= (x-1)(x^2+x-a) \\ \end{align*} Since every solution to the first equation is a solution to the second, we have \(x = 1\) always works, and there is an additional two solutions if \(a > -\frac14\) and a single extra solution if \(a = -\frac14\). We can also repeat solutions if \(1\) is a root of \(x^2+x -a\), ie when \(a = 2\) Therefore: One solution if \(a < -\frac14\) Two solutions if \(a = -\frac14, 2\) Three solutions if \(a > -\frac14, a \neq 2\)
2. \(\,\) \begin{align*} && x &= (b+x)^{\frac12} \\ \Rightarrow && x^2 &= b + x \\ \Rightarrow && 0 &= x^2 - x - b \end{align*} This has a positive root if \(\frac14 - \frac12 - b \leq 0 \rightarrow b \geq \frac14\). It has two positive roots if \(b \geq 0\). Therefore two solutions if \(b > \frac14\) and one solution if \(b = \frac14\)

Solution:

+ - ↺

Solution:

1. \(\,\) \begin{align*} && I &= \int_0^1 \frac 1 {x^2 + (a+2)x +2a} \; \d x \\ &&&= \int_0^1 \frac{1}{(x+a)(x+2)} \d x\\ \end{align*} Case 1: \(a = 2\) \begin{align*} && I &= \int_0^1 \frac{1}{(x+2)^2} \d x \\ &&&= \left [ -(x+2)^{-1}\right]_0^1 = \frac12 - \frac13 = \frac16 \end{align*} Case 2: \(a \neq 2, a \not \in [0,1]\) \begin{align*} && I &=\frac{1}{a-2} \int_0^1 \left ( \frac{1}{x+2} - \frac{1}{x+a} \right) \d x \\ &&&= \frac{1}{a-2} \left [ \ln |x+2| - \ln |x + a|\right]_0^1 \\ &&&= \frac{1}{a-2} \left ( \ln \frac{3}{|1+a|} - \ln \frac{2}{|a|} \right) \\ &&&= \frac{1}{a-2} \ln \frac{3|a|}{2|a+1|} \end{align*} Case 3: \(a \in [0, 1]\), \(I\) does not converge
2. \(\,\) \begin{align*} && J &= \int _{1}^2\frac 1 {u^2 +au +a-1} \; \d u \\ &&&= \int_1^2 \frac{1}{(u+a-1)(u+1)} \d u \\ x = u-1:&&&= \int_0^1 \frac{1}{(x+a)(x+2)} \d x \end{align*} So it's the same as the previous integral

Solution:

1. Case \(b > 0\). Then \(bx+a\) is increasing and the greatest value is \(10b+a\), and the least value \(a-10b\) Case \(b=0\), then \(a\) is constant and the greatest and least value is \(a\) Case \(b < 0\), then \(bx+a\) is decreasing and the greatest value is \(-10b+a\) and the least value is \(10b+a\)
2. If \(c = 0\) we have the same cases as above. If \( c > 0\) the consider \(2cx+b\). if \(b-20c > 0\) then our function is increasing on our interval and the greatest value is \(100c+10b+a\) and the least value is \(100c-10b+a\) If \(20c+b < 0\) then our function is decreasing and that calculation is reversed. If neither of these are true, then the minimum will be when \(x = - \frac{b}{2c}\) and the max at one end point.

Solution: Case 1: \(x \leq -1\) \begin{align*} && -1-x+x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && -x-2 &= x + 2 \\ \Leftrightarrow && x = -2 \end{align*} Case \(-1 < x \leq 0\): \begin{align*} && x+1+x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && x&= x + 2 \\ \end{align*} No solutions Case \(0 < x \leq 1\): \begin{align*} && x+1-x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && -x&= x + 2 \\ \end{align*} No solutions Case \(1 < x \leq 2\): \begin{align*} && x+1-x+3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && 5x-6&= x + 2 \\ \Leftrightarrow && x = 2 \end{align*} Case \(2 < x\): \begin{align*} && x+1-x+3(x-1)-2(x-2) &= x + 2 \\ \Leftrightarrow && x+2&= x + 2 \\ \end{align*} Therefore the solutions are \(x \in \{-2\} \cup [2, \infty)\)

Solution: First suppose \(|a| < 1\), then \begin{align*} && I &= \int_0^1 \frac{1}{x^2+2ax+1} \d x \\ &&&= \int_0^1 \frac{1}{(x+a)^2 +1-a^2} \d x \\ &&&= \int_0^1 \frac{1}{(x+a)^2 +(\sqrt{1-a^2})^2} \d x \tag{\(1-a^2 > 0\)}\\ &&&= \left [\frac{1}{\sqrt{1-a^2}} \tan^{-1} \frac{x+a}{\sqrt{1-a^2}} \right]_0^1 \\ &&&= \frac{1}{\sqrt{1-a^2}} \left ( \tan^{-1} \frac{a+1}{\sqrt{1-a^2}} - \tan^{-1} \frac{a}{\sqrt{1-a^2}} \right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \left ( \frac{\frac{a+1}{\sqrt{1-a^2}}-\frac{a}{\sqrt{1-a^2}}}{1+\frac{(a+1)a}{1-a^2}} \right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \left ( \frac{\frac{1}{\sqrt{1-a^2}}}{\frac{1+a}{1-a^2}} \right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \left ( \frac{1-a}{\sqrt{1-a^2}}\right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \sqrt { \frac{1-a}{1+a}} \\ \end{align*} Second, suppose \(|a| > 1\), then \begin{align*} && I &= \int_0^1 \frac{1}{x^2+2ax+1} \d x \\ &&&= \int_0^1 \frac{1}{(x+a)^2-(a^2-1)} \d x \\ &&&= \int_0^1 \frac{1}{(x+a-\sqrt{a^2-1})(x+a+\sqrt{a^2-1})} \d x \tag{\(a^2-1 > 0\)} \\ &&&= \frac{1}{2\sqrt{a^2-1}}\int_0^1 \left ( \frac{1}{x+a-\sqrt{a^2-1}} - \frac{1}{x+a+\sqrt{a^2-1}} \right) \d x \\ &&&= \frac{1}{2\sqrt{a^2-1}} \left [ \ln |x+a-\sqrt{a^2-1}|- \ln |x+a+\sqrt{a^2-1}| \right]_0^1 \\ &&&= \frac{1}{2\sqrt{a^2-1}} \left ( \ln |1+a-\sqrt{a^2-1}| - \ln|1+a+\sqrt{a^2-1}| - \ln|a-\sqrt{a^2-1}| +\ln|a + \sqrt{a^2-1}| \right) \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln | \frac{(1+a-\sqrt{a^2-1})(a+\sqrt{a^2-1})}{(1+a+\sqrt{a^2-1})(a-\sqrt{a^2-1})}|\\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |\frac{a+a^2-(a^2-1) +\sqrt{a^2-1}}{1+a-\sqrt{a^2-1}}| \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |\frac{(1+a +\sqrt{a^2-1})^2}{(1+a)^2-(a^2-1)}| \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |\frac{1+2a+a^2+a^2-1+2(1+a)\sqrt{a^2-1}}{2+2a}| \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |a+\sqrt{a^2-1}| \\ \end{align*}

Solution: \begin{align*} x^4 - 2x^3+x^2 &= x^2(x^2-2x+1) \\ &= x^2(x-1)^2 > 0 \end{align*} Since \(x\)and \(x-1\) can't both be zero, and square cannot be negative. \begin{align*} x^2 - 8x+17 &= (x-4)^2 +1 \geq 1 > 0 \end{align*} If \(x \leq 2\) then \(x^4 - 2x^3+x^2 > 0\) and \(17-8x \geq 1\) so \(x^4-2x^3+x^2-8x+17 > 0\) If \(x > 2\) then \(x^4-2x^3 = x^3(x-2) \geq 0\) and \(x^2-8x+17 > 0\) so \(x^4-2x^3+x^2-8x+17 > 0\), so for all real numbers our polynomial is positive and therefore cannot have any roots. Note that: \(x^4-x^3+x^2 = x^2(x^2-x+1) > 0\) and \(x^2-4x+4 =(x-2)^2 \geq 0\) If \(x < 1\) then \(x^4-x^3+x^2 > 0\) and \(4(1-x) > 0\) so \(x^4-x^3+x^2-4x+4 > 0\). If \(x > 1\) then \(x^4-x^3 = x^3(x-1) > 0\) and \(x^2-4x+4 \geq 0\) therefore \(x^4-x^3+x^2-4x+4 > 0\). Therefore \(x^4-x^3+x^2-4x+4 > 0\) for all real \(x\) and hence there are no real roots.