Problems

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Solution:

1. \begin{align*} \log_{10}5 &= \log_{10} 10 - \log_{10}2 \\ &= 1- \log_{10} 2 \\ &= 0.699\\ \\ \log_{10} 6 &= \log_{10} 2 + \log_{10} 3 \\ &= 0.301029996+0.477121255 \\ &= 0.778 \end{align*} \begin{align*} && 5 \times 10^{47} < 3^{100} < 6 \times 10^{47} \\ \Leftrightarrow && 47 + \log_{10} 5 < 100 \log_{10} 3 < \log_{10} 6 + 47 \\ \Leftrightarrow &&47.699< 47.71 < 47.778 \\ \end{align*} Which is true. Therefore the first digit of \(3^{100}\) is 5.
2. \(\log_{10} 2^{1000} = 1000 \log_{10} 2 = 301.02\cdots\). Therefore it starts with a \(1\). \(\log_{10}2^{10\, 000} = 10\,000 \log_{10} 2 = 3010.2\) therefore this also starts with a \(1\). \(\log_{10} 2^{100\, 000} = 100\,000 \log_{10} 2 = 30102.9996\) therefore it starts with a \(9\)

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Solution: The powers of \(x\) in the first bracket will be \(x^{20}, x^{14}, \cdots, x^{-10}\). The powers of \(x\) in the second bracket will be \(x^6, x^4, \cdots, x^{-6}\). Therefore we can achieve \(x^{-12}\) in only one way: \begin{array}{c|c|c|c|c} 1\text{st bracket} & 2\text{nd bracket} & 1\text{st coef} & 2\text{nd coef} & \text{prod} \\ \hline x^{-10} & x^{-2} & \binom{5}{5}(-1)^5 = -1 & \binom{6}{4}(-1)^4 = 15& -15 \\ \end{array} We can achieve \(x^2\) as follows: \begin{array}{c|c|c|c|c} 1\text{st bracket} & 2\text{nd bracket} & 1\text{st coef} & 2\text{nd coef} & \text{prod} \\ \hline x^{-4} & x^{6} & \binom{5}{4}(-1)^4 = 5 & \binom{6}{0}(-1)^0 = 1& 5 \\ x^{2} & x^{0} & \binom{5}{3}(-1)^3 = -10 & \binom{6}{3}(-1)^3 = -20 & 200 \\ x^{8} & x^{-6} & \binom{5}{2}(-1)^2 = 10 & \binom{6}{6}(-1)^6 = 1 & 10 \end{array} Therefore the coefficient is \(215\) \((x^2-1)(x^4+x^2+1) = x^6-1\), therefore \begin{align*} (x^2-1)^{11}(x^4+x^2+1)^5 &= (x^2-1)^6(x^6-1)^5 \\ &= x^6\left(x-\frac1x\right)^6(x^6-1)^6 \\ &= x^6\left(x-\frac1x\right)^6\left(x^2\left(x^4-\frac{1}{x^2}\right)\right)^5 \\ &= x^6\left(x-\frac1x\right)^6x^{10}\left(x^4-\frac{1}{x^2}\right)^5 \\ &= x^{16}\left(x-\frac1x\right)^6\left(x^4-\frac{1}{x^2}\right)^6 \\ \end{align*} Therefore the coefficient of \(x^4\) is the coefficient of \(x^{4-16} = x^{-12}\) in our original expression, ie \(-15\). Similarly, the coefficient of \(x^{38}\) is the coefficient of \(x^{38-16} = x^{22}\), which can only be achieved in one way: \begin{array}{c|c|c|c|c} 1\text{st bracket} & 2\text{nd bracket} & 1\text{st coef} & 2\text{nd coef} & \text{prod} \\ \hline x^{20} & x^{2} & \binom{5}{0}(-1)^0 = 1 & \binom{6}{2}(-1)^2 = 15& 15 \\ \end{array} Therefore the coefficient is \(15\)

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Solution:

\begin{align*} \int_0^5 [x]\;\d x &= 0 \cdot 1 + 1 \cdot 1 + 2 \cdot 1 + 3 \cdot 1 + 4 \cdot 1 \\ &= 10 \end{align*} \begin{align*} \int_{0}^{\ln n}[\e^{x}]\, \d x &= \sum_{k=1}^{n-1} \int_{\ln k}^{\ln (k+1)}[\e^{x}]\, \d x \\ &= \sum_{k=1}^{n-1} k \l \ln (k+1) - \ln (k) \r\\ &= \sum_{k=1}^{n-1} \l( (k+1) \l \ln (k+1) - \ln (k) \r - \ln(k+1) \r \\ &= \sum_{k=1}^{n-1} (k+1) \ln (k+1) - \sum_{k=1}^{n-1} k \ln (k) - \sum_{k=1}^{n-1} \ln (k+1) \\ &= n \ln n - 1 \ln 1 - \sum_{k=1}^{n-1} \ln (k+1) \\ &= n \ln n - \ln \l \prod_{k=1}^{n-1} (k+1)\r \\ &= n \ln n - \ln (n!) \end{align*}

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Solution:

1. \(x^2 + 1 \geq 2x \Rightarrow \frac{1}{x^2+1} \leq \frac1{2x} \Rightarrow \frac{x^6}{(x^2+1)^4}\leq \frac{x^6}{16x^2} = \frac1{16}x^2 \leq \frac1{16}\) with equality when \(x = 1\)
2. \(\,\) \begin{align*} && RHS &= \frac{\d \ }{\d x} \left( \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3}\right) +\frac{Dx^6}{(x^2+1)^4} \\ &&&= \frac{(5Ax^4+3Bx^2+C)(x^2+1)^3-6x(x^2+1)^2(Ax^5+Bx^3+Cx)}{(x^2+1)^6}+\frac{Dx^6}{(x^2+1)^4} \\ &&&= \frac{(5Ax^4+3Bx^2+C)(x^2+1)-6x(Ax^5+Bx^3+Cx)+Dx^6}{(x^2+1)^4} \\ &&&= \frac{(D-A) x^6 + (5A-3B) x^4 + (3B-5C)x^2 + C}{(x^2+1)^4} \\ \Rightarrow && C &= 1 \\ && 3B &= 5 \quad\quad\quad\Rightarrow B = \frac53 \\ && 5A &= 3B = 5\Rightarrow A = 1 \\ && D &= A \quad\quad \Rightarrow D = 1 \end{align*}
3. So \begin{align*} && I &= \int_{0}^{1}\frac{1}{(x^{2}+1)^{4}}\, \d x \\ &&&= \int_{0}^{1}\frac{\d \ }{\d x} \left( \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3}\right) +\frac{Dx^6}{(x^2+1)^4} \, \d x \\ &&&= \left [ \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3} \right]_0^1 + \int_{0}^{1}\frac{Dx^6}{(x^2+1)^4} \, \d x \\ &&& \leq \frac{A+B+C}{8} + \frac1{16} \\ &&&= \frac{2+\frac53}{8} + \frac1{16} = \frac{11}{24} + \frac{1}{16} \\ && I &\geq \left [ \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3} \right]_0^1 = \frac{11}{24} \end{align*}

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Solution: \begin{align*} && (x-y+2)(x+y-1) &= (x-y)(x+y)-(x-y)+2(x+y)-2 \\ &&&= x^2-y^2+x+3y-2 \end{align*}

\begin{align*} x^2-4y^2 +3x-2y+2 &= (x - 2 y + 1) (x + 2 y + 2) \end{align*} Consider the point \(x = 0, y = \frac32\), then \(\frac92 - \frac94 = \frac94 > 2\) and \(-4\cdot\frac94-2\cdot \frac32 = -12 < -2\) so this is an example of a point in both regions

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Solution: \begin{align*} && f(x) &= a x-\frac{x^{3}}{1+x^{2}} \\ \Rightarrow && f'(x) &= a - \frac{3x^2(1+x^2)-x^3 \cdot 2 x}{(1+x^2)^2} \\ &&&= a - \frac{-x^4+3x^2}{(1+x^2)^2} \\ &&&= a - \frac{-t^2+3t}{(1+t)^2} \\ &&&= \frac{a+2at+at^2-t^2-3t}{(1+t)^2} \\ &&&= \frac{(a-1)t^2+(2a-3)t+a}{(1+t)^2} \\ \\ && 0 \leq \Delta &= (2a-3)^2 - 4 \cdot (a-1) \cdot a \\ &&&= 4a^2-12a+9 - 4a^2+4a \\ &&&= -8a + 9 \\ \Leftrightarrow && a &\geq 9/8 \end{align*} Therefore if \(a \geq 9/8\) the numerator is always non-negative and \(f'(x) \geq 0\)

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Solution: \begin{align*} && \int_{-1}^1 |x e^x |\d x &= \int_{-1}^0 |xe^x| \d x + \int_0^1 |xe^x| \d x \\ &&&= \int_{-1}^0 -xe^x \d x + \int_0^1 x \e^x \d x \\ &&&= -\int_{-1}^0 xe^x \d x + \int_0^1 x \e^x \d x \\ \\ && \int xe^x \d x &= xe^x - \int e^x \d x \\ &&&= xe^x - e^x \\ \\ \Rightarrow && \int_{-1}^1 |x e^x |\d x &= \left [ xe^x - e^x \right]_0^{-1}+ \left [ xe^x - e^x \right]_0^{1} \\ &&&= -e^{-1}-e^{-1} +e^{0} + e^1 - e^1 +e^0 \\ &&&= 2-2e^{-1} \end{align*}

1. \(\,\) \begin{align*} && I &= \int_0^4 | x^3-2x^2-x+2| \d x \\ &&&= \int_0^4 |(x-2)(x-1)(x+1)| \d x\\ &&&= \int_0^1( x^3-2x^2-x+2) \d x- \int_1^2 ( x^3-2x^2-x+2) \d x + \int_2^4 ( x^3-2x^2-x+2) \d x \\ &&&= \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_0^1 - \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_1^2 + \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_2^4 \\ &&&= 2 \left ( \frac14 - \frac23 -\frac12 + 2\right) - 2 \left ( \frac14 2^4 - \frac23 2^3 -\frac12 2^2 + 2 \cdot 2\right)+ \left ( \frac14 4^4 - \frac23 4^3 -\frac12 4^2 + 2 \cdot 4\right) \\ &&&= \frac{133}{6} \end{align*}
2. \(\,\) \begin{align*} && J &= \int_{-\pi}^\pi | \sin x + \cos x | \d x \\ &&&= \int_{-\pi}^{\pi} | \sqrt{2} \sin(x + \tfrac{\pi}{4})| \d x \\ &&&= 2\sqrt{2}\int_0^\pi \sin x \d x \\ &&&= 4\sqrt{2} \end{align*}

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Solution: \begin{align*} && s_x &= V \cos \theta t + \frac12at^2 \\ && s_y &= V \sin \theta t - \frac12 gt^2 \\ \Rightarrow && T &= \frac{2V \sin \theta}g \\ \Rightarrow && s_{max} &= \frac{2V^2 \sin \theta \cos \theta}{g} + \frac12a \frac{4V^2 \sin^2 \theta}{g^2} \\ &&&= (g \sin 2 \theta+2a\sin^2 \theta)\frac{V^2}{g^2} \\ && \frac{\d s_{max}}{\d \theta} &= (2g \cos 2 \theta +4 a \cos \theta \sin \theta)\frac{V^2}{g^2} \\ &&&= (2g \cos 2\theta + 2a \sin2 \theta) \frac{V^2}{g^2} \\ \Rightarrow && \tan 2\theta &= -\frac{a}{g} \\ \Rightarrow && 2 \theta &\in (\frac{\pi}2, \pi) \\ \Rightarrow && \tan \left (\frac{\pi}{2} - 2 \theta\right) &=-\frac{a}{g} \\ \Rightarrow && \frac{\pi}{2} - 2 \theta&\approx -\frac{a}{g} \\ \Rightarrow && \theta &\approx \frac{\pi}{4} + \frac{a}{2g} \\ \\ && s_{max} & \approx \left (g \sin \left (\frac{\pi}{2} + \frac{a}{g} \right)+2a\sin^2 \left ( \frac{\pi}{4} + \frac{a}{2g}\right)\right)\frac{V^2}{g^2} \\ &&&\approx \left (g \cdot 1+2a\left( \frac{1}{\sqrt{2}}(\frac{a}{2g}+1)\right)^2\right)\frac{V^2}{g^2} \\ &&&\approx \left (g+a\left(1+\frac{a}{g}\right)^2\right)\frac{V^2}{g^2} \\ &&&\approx \left (g+a\right)\frac{V^2}{g^2} \\ &&&= \frac{V^2}{g} + \frac{V^2a}{g} \end{align*}

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Solution: \(X \sim U(-1,1)\) \begin{align*} \E[X^2] &= \int_{-1}^1 \frac12 x^2 \, dx \\ &= \frac{1}{6} \left [ x^3 \right]_{-1}^1 \\ &= \frac{1}{3} \end{align*} \begin{align*} \E[X^4] &= \int_{-1}^1 \frac12 x^4 \, dx \\ &= \frac{1}{10} \left [ x^5 \right]_{-1}^1 \\ &= \frac{1}{5} \end{align*} \begin{align*} \var[X^2] &=\E[X^4] - \E[X^2]^2 \\ &= \frac{1}{5} - \frac{1}{9} \\ &= \frac{4}{45} \end{align*} \begin{align*} \E(Z^2) &= \E(Y^2 - 2XY+Z^2) \\ &= \E(Y^2) - 2\E(X)\E(Y)+\E(Z^2) \\ &= \frac{1}{3} - 0 + \frac{1}{3} \\ &= \frac{2}{3} \end{align*} \begin{align*} \E[Z^4] &= \E[Y^4 -4Y^3X+6Y^2X^2-4YX^3+X^4] \\ &= \E[Y^4]-4\E[Y^3]\E[X]+6\E[Y^2]\E[X^2]-4\E[Y]\E[X^3]+\E[X^4] \\ &= \frac{1}{5}+6 \frac{1}{3} \frac13 + \frac{1}{5} \\ &= \frac{2}{5} + \frac{2}{3} \\ &= \frac{16}{15} \end{align*} \begin{align*} \var(Z^2) &= \E(Z^4) - \E(Z^2) \\ &= \frac{16}{15} - \frac{4}{9} \\ &= \frac{28}{45} \\ &= 7 \var(X^2) \end{align*}

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Solution: Notice that \(\frac{1}{N} = \frac{1}{N+1} + \frac{1}{N(N+1)}\), so any unit fraction can be expressed as the sum of two distinct unit fractions. \begin{align*} && \frac{1}N &= \frac1a + \frac1b \\ \Leftrightarrow && ab&= Nb+Na \\ \Leftrightarrow && 0 &= (a-N)(b-N)-N^2 \\ \Leftrightarrow && N^2 &= (a-N)(b-N) \end{align*} If \(N\) is prime then the only factors of \(N^2\) are \(1,N\) and \(N^2\). if \(a-N = b-N = N\) then \(a=b\) and we don't have distinct fractions. Therefore \(a-N = 1\) and \(b-N = N^2\) and we obtain the decomposition earlier (and it must be the only solution). \begin{align*} && \frac2N &= \frac1a+\frac1b \\ \Leftrightarrow && 2ab &= Nb+Na \\ \Leftrightarrow && 4ab &= 2Na+2Nb \\ \Leftrightarrow && N^2 &= (2a-N)(2b-N) \end{align*} Therefore for \(a,b\) to be distinct we must have \(2a = N+1\) and \(2b = N+N^2\) as the only possible factorisation. Both of the right hand sides are even so we can write \[ \frac{1}{N} = \frac{1}{\frac{N+1}{2}} + \frac{1}{\frac{N(N+1)}{2}} \] and this is unique