Problems

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Solution:

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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 \(x = \pm \sqrt{5/6}\)
   

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Solution: When \((\ln x)^2 = 1\) we have \(f(x) = \frac{1}{x\ln x}(1 - 1^2)^2 = 0\) \(f'(x) = \frac{2(1 - (\ln x)^2) \cdot (-2 \ln x ) \cdot \frac1x \cdot (x \ln x) - (\ln x +1)(1-(\ln x)^2)^2}{(x\ln x)^2} = \frac{2\cdot 0 \cdot (-2 \ln x ) \cdot \frac1x \cdot (x \ln x) - (\ln x +1) \cdot 0}{(x\ln x)^2} = 0\)

1. First consider \(0 < x < 1\), so \begin{align*} && F(x) &= \int_{1/e}^x f(t) \d t \\ &&&= \int_{1/e}^x \frac{1}{t\ln t} \left(1 - (\ln t)^2 \right)^2 \d t \\ u = \ln t, \d u = \frac1t \d t: &&&= \int_{u=-1}^{u=\ln x} \frac{1}{u}(1-u^2)^2 \d u \\ &&&= \int_{-1}^{\ln x} \left ( u^3 - 2u+\frac1u \right) \d u \\ &&&= \left [ \frac{u^4}{4} - u^2+ \ln |u| \right]_{-1}^{\ln x} \\ &&&= \frac{(\ln x)^4}{4} -(\ln x)^2 + \ln |\ln x| - \frac14+1 \end{align*} Now consider \(x > 1\) \begin{align*} && F(x) &= \int_{e}^x f(t) \d t \\ &&&= \int_{e}^x \frac{1}{t\ln t} \left(1 - (\ln t)^2 \right)^2 \d t \\ u = \ln t, \d u = \frac1t \d t: &&&= \int_{u=1}^{u=\ln x} \frac{1}{u}(1-u^2)^2 \d u \\ &&&= \int_{1}^{\ln x} \left ( u^3 - 2u+\frac1u \right) \d u \\ &&&= \left [ \frac{u^4}{4} - u^2+ \ln |u| \right]_{1}^{\ln x} \\ &&&= \frac{(\ln x)^4}{4} -(\ln x)^2 + \ln| \ln x| - \frac14+1 \end{align*} Notice that \begin{align*} F(x^{-1}) &= \frac{(\ln x^{-1})^4}{4} -(\ln x^{-1})^2 + \ln| \ln x^{-1}| - \frac14+1 \\ &= \frac{(-\ln x)^4}{4} -(-\ln x)^2 + \ln| -\ln x| - \frac14+1 \\ &= \frac{(\ln x)^4}{4} -(\ln x)^2 + \ln| \ln x| - \frac14+1 \\ &= F(x) \end{align*}
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Solution:

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Solution: \(f'(x) = 6ax-18x^2\), therefore \(f\) has turning points at \(0\) and \(\frac{a}3\) (ie decreasing for \(x \leq 0\) and \(x \geq \frac{a}{3}\) and increasing otherwise). Therefore possible maxima are \(f(-\tfrac13), f(\frac{a}{3}), f(1)\) where we consider \(\frac{a}{3}\) if \(a \leq 3\) and \(1\) otherwise. \(f(-\frac13) = \frac{a}{3} + \frac{2}{9} = \frac{3a+2}{9}\) \(f(\frac{a}{3}) = \frac{a^3}{3} - \frac{2a^3}{9} = \frac{a^3}{9}\) \(f(1) = 3(a-2)\) Comparing \(\frac{a^3}{9}\) to \(\frac{3a+2}{9}\) we have a double root at \(a = -1\) and a single root at \(a = 2\), therefore \(\frac{a^3}9\) is larger if \(a \geq 2\) Comparing \(3(a-2)\) to \(\frac{3a+2}9\) we have a cross-over at \(a = \frac{7}3\). Therefore we have: \begin{align*} M(a) &= \begin{cases} \frac{3a+2}{9} & 0 \leq a \leq 2 \\ \frac{a^3}{9} & 2 \leq a \leq 3 \\ 3(a-2) & 3 \leq a \end{cases} \end{align*}

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Solution:

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2. Notice that there are no solutions when \(x < 0\) since \(f(x) \geq 1\) in that region. Suppose \(x = n + \epsilon, 0 < \epsilon < 1\), then \(f(x) = \frac{n}{n+\epsilon}\), ie \(12n = 7n + 7 \epsilon \Rightarrow 5 n = 7\epsilon \Rightarrow \epsilon = \frac{5}{7}n \Rightarrow n < \frac75\), so \(n = 1 ,\epsilon = \frac57, x = \frac{12}5\). \begin{align*} && \frac{17}{24} &= f(x) \\ \Rightarrow && 17n + 17 \epsilon &= 24 n \\ \Rightarrow && 17 \epsilon &= 7 n \\ \Rightarrow && n &< \frac{17}{7} \\ \Rightarrow && n &= 1, 2 \\ \Rightarrow && x &= \frac{24}{17}, \frac{48}{17} \end{align*}. For \(f(x) = \frac{4}{3}\) we notice that \(x < 0\), so let \(x = -n +\epsilon\), ie \begin{align*} && \frac43 &= f(x) \\ \Rightarrow && \frac43 &= \frac{-n}{-n+\epsilon} \\ \Rightarrow && 4\epsilon &= n \\ \Rightarrow && n &= 1,2,3 \\ \Rightarrow && x &= -\frac{5}{4}, -\frac{3}{2}, -\frac{9}{4} \end{align*}
3. \begin{align*} && \frac9{10} &= f(x) \\ \Rightarrow && 9n + 9 \epsilon &= 10 n \\ \Rightarrow && 9 \epsilon &= n \\ \Rightarrow && n < 9 \end{align} So largest will be when \(n = 8, \epsilon = \frac{8}{9}\), ie \(\frac{80}{9}\)

If \(c < 1\) \begin{align*} && c &= \frac{k}{k + \epsilon} \\ \Rightarrow && \frac{c}{1-c} \epsilon &= k \end{align*} For this to have exactly \(n\) solutions, we need \(n < \frac{c}{1-c} \leq n+1\). If \(c > 1\) \begin{align*} && c &= \frac{-k}{-k+\epsilon} \\ \Rightarrow && c \epsilon &= (c-1) k \\ \Rightarrow && \frac{c}{c-1} \epsilon &= k \end{align*} Therefore for there to be exactly \(n\) solutions we need \(n < \frac{c}{c-1} \leq n+1\)

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Solution: First notice that \begin{align*} && 1 &= \int_{-a}^2 f(x) \d x \\ &&&= 2ka + k\pi \\ \Rightarrow && k &= (\pi + 2a)^{-1} \end{align*}

1. \(\,\) \begin{align*} && \E[X] &= \int_{-a}^2 x f(x) \d x\\ &&&= \int_{-a}^0 2kx \d x + k\int_0^{2} x\sqrt{4-x^2} \d x\\ &&&= \left [kx^2 \right]_{-a}^0 +k \left [-\frac13(4-x^2)^{\frac32} \right]_0^2 \\ &&&= -ka^2 + \frac83k \\ &&&= \frac{\frac83-a^2}{\pi + 2a} \end{align*}
2. Consider \(\mathbb{P}(X < 0)\) then \(d < 0 \Leftrightarrow \mathbb{P}(X < 0) > \frac{9}{10}\) \begin{align*} && \frac{9}{10} &< \mathbb{P}(X < 0) \\ &&&= \int_{-a}^0 2k \d x \\ &&&= \frac{2a}{\pi+2a} \\ \Leftrightarrow && 9\pi &< 2a \\ \\ && \frac{9}{10} &= \int_{-a}^d 2k \d x \\ &&&= \frac{2(d+a)}{\pi + 2a} \\ \Rightarrow && 9\pi &= 2a + 20d \\ \Rightarrow && d &= \frac{2a-9\pi}{20} \end{align*}
3. Suppose \(d=\sqrt 2\) then \begin{align*} && \frac1{10} &= \int_{\sqrt{2}}^2 f(x) \d x \\ &&&= \int_{\sqrt{2}}^2 k\sqrt{4-x^2} \d x \\ &&&= k\left [ 2 \sin^{-1} \tfrac12 x + \tfrac12 x \sqrt{4-x^2}\right]_{\sqrt{2}}^2 \\ &&&= k\left (\pi -\frac{\pi}{2} - 1 \right) \\ \Rightarrow && \pi + 2a &= 5\pi - 10 \\ \Rightarrow && a &= 2\pi-5 \end{align*}

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Solution:

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2. Since \(f(x)\) is continuous, \(a -bx\) joins \((2k, \ln k)\) and \((4k ,0)\). ie it has a gradient of \(\frac{-\ln k}{2k}\) and is zero at \(4k\), hence \(\displaystyle b = -\frac{\ln k}{2k}, a = 2\ln k\). The \(3\) sections have areas \(\int_1^k \ln x \d x = k \ln k -k +1\), \(k \ln k, k \ln k\). Therefore \begin{align*} &&1&= 3k\ln k - k +1 \\ \Rightarrow &&0 &= k(3\ln k - 1) \\ \Rightarrow &&\ln k &= \frac13 \\ \Rightarrow &&k &= e^{1/3} \\ && a &= \frac23 \\ && b&= -\frac16e^{-1/3} \end{align*}
3. Clearly \(1 > k \ln k > \frac{1}{3}\), therefore the median must lie between \(k\) and \(2k\). So we need, \(\frac12\) to be the area of the rectangle + the triangle, ie: \begin{align*} && \frac12 &= k \ln k + (2k-M) \ln k \\ &&&= \frac13 k + \frac13 (2k - M) \\ \Rightarrow && M &= 3k - \frac32 \\ \Rightarrow && M &= 3e^{1/3} - \frac32 \end{align*}

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Solution:

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   \begin{align*} && \int_0^a \sqrt{\boldsymbol{[}x\boldsymbol{]}} \d x &= \sum_{r=0}^{a-1} \int_{x=r}^{x=r+1} \sqrt{\boldsymbol{[}x\boldsymbol{]}} \d x \\ &&&= \sum_{r=0}^{a-1} \int_r^{r+1} \sqrt{r} \d x \\ &&&= \sum_{r=0}^{a-1} \sqrt{r} \\ \end{align*}
2. \(\,\) \begin{align*} && \int^{a}_0 2^{\boldsymbol {[} x \boldsymbol{]}} \d x &= \sum_{r=0}^{a-1} \int_{x=r}^{x=r+1} 2^{\boldsymbol {[} x \boldsymbol{]}} \d x \\ &&&= \sum_{r=0}^{a-1} \int_{x=r}^{x=r+1} 2^{r} \d x \\ &&&= \sum_{r=0}^{a-1} 2^{r}\\ &&&= 2^{a}-1 \end{align*}
3. \(\,\) \begin{align*} && \int^{a}_0 2^{\boldsymbol {[} x \boldsymbol{]}} \d x &= \int_0^{\boldsymbol {[} a \boldsymbol{]}} 2^{\boldsymbol {[} x \boldsymbol{]}} \d x + \int_{\boldsymbol {[} a \boldsymbol{]}}^a 2^{\boldsymbol {[} x \boldsymbol{]}} \d x \\ &&&= 2^{ \boldsymbol {[} a \boldsymbol{]}}-1 + (a-\boldsymbol {[} a \boldsymbol{]})2^{\boldsymbol {[} a \boldsymbol{]}} \\ &&&= (a-\boldsymbol {[} a \boldsymbol{]}+1)2^{\boldsymbol {[} a \boldsymbol{]}} -1 \end{align*}

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Solution: \[ f(x) = 2|x(x-1)| + |(x-1)(x+1)|-2|x(x+1)| \] Therefore the absolute value terms will change behaviour at \(x = -1, 0, 1\). Then \begin{align*} f(x) &= \begin{cases} 2(x^2-x)+(x^2-1)-2(x^2+x) & x \leq -1 \\ 2(x^2-x)-(x^2-1)+2(x^2+x) & -1 < x \leq 0 \\ -2(x^2-x)-(x^2-1)-2(x^2+x) & 0 < x \leq 1 \\ 2(x^2-x)+(x^2-1)-2(x^2+x) & 1 < x\end{cases} \\ &= \begin{cases} x^2-4x-1 & x \leq -1 \\ 3x^2+1& -1 < x \leq 0 \\ -5x^2+1& 0 < x \leq 1 \\ x^2-4x-1 & 1 < x\end{cases} \\ \\ f'(x) &= \begin{cases} 2x-4 & x <-1 \\ 6x & -1 < x < 0 \\ -10x & 0 < x < 1 \\ 2x-4 & 1 < x\end{cases} \\ \end{align*} Therefore \(f'(x) = 0 \Rightarrow x = 0, 2\) and so we should check all the turning points. Therefore the minimum is \(x = 2, y = -5\), maximum is \(x = -2, y = 11\) (assuming the range is actually \(|x| < 2\). There is a point of inflection at \(x = 0, y = 1\).

\(f\) is continuous everywhere as a sum of continuous functions. \(f\) is not differentiable at \(x = -1, 1\) Suppose \begin{align*} &&y &=x^2-4x-1 \\ &&&= (x-2)^2 -5 \\ \Rightarrow &&x &= 2\pm \sqrt{y+5} \\ \\ && y &= 3x^2+1 \\ \Rightarrow && x &= \pm \sqrt{\frac{y-1}{3}} \\ \\ && y &= -5x^2+1 \\ \Rightarrow && x &=\pm \sqrt{\frac{1-y}{5}} \\ \\ \Rightarrow && f^{-1}(y) &= \begin{cases} 2 - \sqrt{y+5} & y > 4 \\ -\sqrt{\frac{y-1}{3}} & 1 < y < 4 \\ \sqrt{\frac{1-y}{5}} & -4 < y < 1 \\ 2 + \sqrt{y+5} & y < -4 \end{cases} \end{align*}

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Solution: Using Newton's second law in the form, \(\F(x) = m \ddot{x}\). Our two different differential equations can be solved as follows: When \(x \geq 0\) \(-\mu^2x = \ddot{x} \Rightarrow x = A\sin \mu t + B \cos \mu t\) when \(x \geq 0\). And when \(x < 0\) \(-\kappa \dot{x} = \ddot{x} \Rightarrow \dot{x} = Ce^{-\kappa t} \Rightarrow x = De^{-\kappa t} + E\) when \(x < 0\) Following the trajectory of the particle: At \(t = 0, x = 0, \dot{x} = v_0 > 0\), so \(x = \frac{v_0}{\mu} \sin \mu t\) until \(t = \frac{\pi}{\mu}\). When \(t = \frac{\pi}{\mu}\) the particle will head into the negative \(x\)-axis with velocity \(-v_0\). At which point our initial conditions for our differential equations give us that \(De^{-\frac{\pi\kappa}{\mu}} + E = 0, -\kappa De^{-\frac{\pi\kappa}{\mu}} = -v_0 \Rightarrow De^{-\frac{\pi\kappa}{\mu}} = \frac{v_0}{\kappa}, E = -\frac{v_0}{\kappa}\). To summarise: \[ x(t) = \begin{cases} \frac{v_0}{\mu} \sin \mu t & 0 \leq t \leq \frac{\pi}{\mu} \\ -\frac{v_0}{\kappa} \l 1-e^{-\kappa(t-\frac{\pi}{\mu})}\r & t > \frac{\pi}{\mu}\end{cases}\]