Problems

Solution:

1. \(\,\) \begin{align*} && d &= u \cos \alpha t \\ && d \tan \beta &= u \sin \alpha t - \frac12 gt^2 \\ && &= d\tan \alpha - \frac1{2u^2} g d^2 \sec^2 \alpha \\ \Rightarrow && u^2 &= \frac{gd \sec^2 \alpha}{2(\tan \alpha + \tan \beta)} \\ &&&= \frac{gd t^2}{2(t+\tan \beta)} \\ && \frac{\d}{\d t} \left (u^2 \right) &= \frac{2gdt\cdot 2(t+\tan \beta) - gdt^2 \cdot 2}{4(t+\tan \beta)^2} \\ &&&= \frac{2gdt(2t-t+2\tan \beta)}{4(t+\tan \beta)^2} \\ &&&= \frac{gdt(t+2\tan \beta)}{2(t+\tan \beta)^2} \\ \end{align*} So either \(t = 0\) or \(t = -2 \tan \beta\) \begin{align*} && u^2 &= \frac{gd\cdot 4 \tan^2 \beta}{2(-2\tan \beta + \tan \beta)} \\ &&&= \frac{2gd \tan \beta}{-1} \\ &&&= gd (-2\tan \beta) \\ &&&= gd \tan \alpha \end{align*} \begin{align*} && d \tan \beta &= d \tan \alpha - \frac12 \frac{gd^2}{gd \tan \alpha \cdot \cos^2 \alpha } \\ \Rightarrow && \tan \beta&= \tan \alpha - \frac{1}{2\sin \alpha \cos \alpha} \\ &&&= \frac{2 \sin^2 \alpha - 1}{2 \sin \alpha \cos \alpha} \\ &&&= \frac{-\cos 2 \alpha}{\sin 2 \alpha} \\ &&&= -\cot 2 \alpha \\ &&&= \tan (2\alpha - 90^\circ) \\ \Rightarrow && \beta &= 2\alpha - 90^\circ \\ \Rightarrow && 2\alpha &= \beta + 90^\circ \end{align*}
2. Suppose the angle to the horizontal is \(\theta\), then \(\tan \theta = \frac{v_y}{v_x}\) so \begin{align*} && \tan \theta &= \frac{u \sin \alpha - gt}{u \cos \alpha} \\ &&&= \frac{u \sin \alpha - \frac12 g \frac{d}{u \cos \alpha}}{u \cos \alpha} \\ &&&= \frac{u^2\sin \alpha \cos \alpha - gd}{u^2 \cos^2 \alpha} \\ &&&= \frac{gd \tan \alpha \sin \alpha \cos \alpha- gd}{ gd \tan \alpha \cdot \cos^2 \alpha} \\ &&&= \frac{\tan \alpha \sin \alpha \cos \alpha - 1}{ \sin \alpha \cos \alpha} \\ &&&= \frac{\sin^2 \alpha - 1}{\sin \alpha \cos \alpha} \\ &&&= -\frac{\cos \alpha}{\sin \alpha}\\ &&&= - \cot \alpha = \tan (\alpha - 90^\circ)\\ \Rightarrow && \theta &= \alpha - 90^\circ \end{align*}

Solution:

1. \(\,\) \begin{align*} && \ln f(x) &= x \ln x \\ &&&> \ln x \quad (\text{if } 0 < x < 1)\\ \Rightarrow && f(x) &> x\quad\quad (\text{if } 0 < x < 1)\\ \Rightarrow && x^{f(x)} &< x^x \\ && g(x) &< f(x) \\ && 1&>f(x) \\ \Rightarrow && x &< x^{f(x)} = g(x) \end{align*}
2. \(\,\) \begin{align*} && f(x) &= e^{x \ln x} \\ \Rightarrow && f'(x) &= (\ln x + 1)e^{x \ln x} \\ \Rightarrow && f'(x) = 0 &\Leftrightarrow x = \frac1e \end{align*}
3. \(\,\) \begin{align*} && \lim_{x \to 0} f(x) &= \lim_{x \to 0} \exp \left ( x \ln x \right ) \\ &&&= \exp \left ( \lim_{x \to 0} \left ( x \ln x \right )\right) \\ &&&= \exp \left ( 0 \right) = 1\\ \\ && \lim_{x \to 0} g(x) &= \lim_{x \to 0} \exp \left ( f(x) \ln x\right) \\ &&&= \exp \left (\lim_{x \to 0} \ln x f(x)\right) \\ &&&= \exp \left (\lim_{x \to 0} \ln x \lim_{x \to 0}f(x)\right) \\ &&&= \exp \left (\lim_{x \to 0} \ln x\right) \\ &&&= 0 \end{align*}
4. \(y = x^{-1} + \ln x \Rightarrow y' = -x^{-2} + x^{-1}\) which has roots at \(x =1\), therefore the minimum value is \(1\). (We can see it's a minimum by considering \(x \to 0, x \to \infty\). So \begin{align*} && g'(x) &= x^{f(x)} \cdot (f'(x) \ln x + f(x) x^{-1})\\ &&&= x^{f(x)} \cdot f(x) \cdot ((1+\ln x) \ln x + x^{-1}) \\ &&&= x^{f(x)} \cdot f(x) \cdot (\ln x + x^{-1} + (\ln x)^2) \\ &&&\geq x^{f(x)} \cdot f(x) > 0 \end{align*}

+ - ↺

Solution: \((x, y) = (f(\theta)\cos(\theta), f(\theta)\sin(\theta))\) so \begin{align*} \frac{dy}{d\theta} &= -f(\theta)\sin(\theta) + f'(\theta)\cos(\theta) \\ \frac{dx}{d\theta} &= f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) \\ \frac{dy}{dx} &= \frac{-f(\theta)\sin(\theta) + f'(\theta)\cos(\theta)}{f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) } \\ &= \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \end{align*} If the curves meet at right angles then the product of their gradients is \(-1\), ie \begin{align*} \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \cdot \frac{-g(\theta)\tan(\theta) + g'(\theta)}{g(\theta) + g'(\theta)\tan(\theta) } &= -1 \\ f(\theta)g(\theta)\tan^2 \theta - f(\theta)g'(\theta)\tan \theta - f'(\theta)g(\theta)\tan \theta + f'(\theta)g'(\theta) &= \\ \quad - \l f(\theta)g(\theta) + f(\theta)g'(\theta)\tan(\theta) + f'(\theta)g(\theta)\tan(\theta) + f'(\theta)g'(\theta)\tan^2 \theta \r \\ \tan^2\theta \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r + f'(\theta)g'(\theta) + f(\theta)g(\theta) &= 0 \\ (\tan^2\theta + 1) \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r &= 0 \\ f(\theta)g(\theta) + f'(\theta)g'(\theta) &= 0 \end{align*} \(g(\theta) = a(1+\sin\theta), g'(\theta) = a\cos\theta\) Therefore \(f'(\theta)a\cos \theta+f(\theta)a(1+\sin(\theta)) = 0\) \begin{align*} && \frac{f'(\theta)}{f(\theta)} &= -\sec(\theta) - \tan(\theta) \\ \Rightarrow && \ln(f(\theta)) &= -\ln |\tan(\theta) + \sec(\theta)| + \ln |\cos(\theta)| + C \\ \Rightarrow && f(\theta) &= A \frac{\cos \theta}{\tan \theta + \sec \theta} \\ &&&= A \frac{\cos^2 \theta}{\sin \theta + 1} \\ &&&= A \frac{1-\sin^2 \theta}{\sin \theta + 1} \\ &&&= A (1-\sin \theta) \end{align*} When \(\theta = -\frac12 \pi, r = 4\), so \(A = 2\).

+ - ↺

Solution: \begin{align*} && y &= (ax^2+bx+c)\,\ln \big( x+\sqrt{1+x^2}\big) +\big(dx+e\big)\sqrt{1+x^2} \\ && y' &= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + (ax^2+bx+c) \frac{1}{x + \sqrt{1+x^2}} \cdot \left(1 + \frac{x}{\sqrt{1+x^2}} \right) + d\sqrt{1+x^2} + \frac{x(dx+e)}{\sqrt{1+x^2}} \\ &&&= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + \frac{1}{\sqrt{1+x^2}} \left ( (ax^2+bx+c) + d(1+x^2) + x(dx+e) \right) \\ &&&= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + \frac{1}{\sqrt{1+x^2}} \left ( (a+2d)x^2+(b+e)x+(d+c) \right) \\ \end{align*}

1. We want \(a = 0, b = 1, d = 0, e = -1, c =0\), so \begin{align*} I &= \int \ln \big( x+\sqrt{1+x^2}\,\big) \,\d x \\ &= x\ln(x+\sqrt{1+x^2})-\sqrt{1+x^2}+C \end{align*}
2. We want \(a = b =0, e = 0, d = \frac12, c = \frac12\), so \begin{align*} I &= \int \sqrt{1+x^2}\, \d x \\ &= \frac12\ln(x+\sqrt{1+x^2}) + \frac12x\sqrt{1+x^2}+C \end{align*}
3. We want \(a = \frac12, b = 0, d = -\frac14, e = 0, c = \frac14\), so \begin{align*} I &= \int x \ln (x+\sqrt{1+x^2}) \, \d x \\ &= \left (\frac12 x^2+\frac14 \right)\ln(x+\sqrt{1+x^2}) -\frac14x\sqrt{1+x^2}+C \end{align*}

Solution:

1. Let \(\displaystyle y = \frac z {(1+z^2)^{\frac12}}\) then \(\frac{d y}{d x} = \frac{(1+z^2)^{\frac12} - z^2(1+z^2)^{-\frac12}}{1+z^2} = \frac{(1+z^2)-z^2}{(1+z^2)^\frac32} = \frac{1}{(1+z^2)^\frac32}\)
2. \(\kappa = \frac {f''(x)}{\big({1+ (f'(x))^2\big)^{\frac32}}}\) then \begin{align*} && \int \kappa \, dx &= \int \frac{f''(x)}{( 1 + (f'(x))^2)^{\frac32}} \, dx \\ && \kappa x &= \frac{f'(x)}{(1 + (f'(x))^2)^\frac12} + C \\ \Rightarrow && (\kappa x-C)^2 &= \frac{f'(x)^2}{1 + (f'(x))^2} \\ \Rightarrow && f'(x)^2((\kappa x - C)^2 - 1) &= -(\kappa x-C)^2 \\ \Rightarrow && f'(x) &= \frac{\kappa x - C}{\sqrt{1-(\kappa x - C)^2 }} \\ \Rightarrow && f(x) &= \frac{1}{\kappa} \sqrt{1 - (\kappa x - C)^2} \\ \Rightarrow && (\kappa y)^2 + (\kappa x - C)^2 &= 1 \\ \Rightarrow && y^2 + (x - C')^2 &= \frac{1}{\kappa^2} \end{align*} Therefore all the curves are circles and \(\kappa\) is the reciprocal of the radius.

Solution:

1. \(\,\) \begin{align*} && f'_n(x) &= 0 + 1 + \frac{2x}{2!} + \frac{3x^2}{3!} + \cdots + \frac{nx^{n-1}}{n!} \\ &&&= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{n-1}}{(n-1)!} \\ &&&= f_{n-1}(x) \end{align*}
2. Claim: \(f_n(x) > 0\) for all \(x > 0\) Proof: (By induction) Base case: (\(n = 1\)) \(f_1(x) = 1 + x > 1\) therefore \(f_1(x) > 0\) Suppose it's true for \(n = k\), then consider \(f_{k+1}\), if we differentiate it, we find it is increasing on \((0, \infty)\) by our inductive hypothesis. But then \(f_{k+1}(0) = 1 > 0\). Therefore \(f_{k+1}(x) > 0\) as well. Therefore by the principle of mathematical induction we are done. Since \(f_n(x) > 0\) for non-negative \(x\), if \(a\) is a root it must be negative.
3. Suppose \(f_n(a) = f_n(b) = 0\) then \(f'_n(a) = -\frac{a^n}{n!}\) and \(f'_n(b) = -\frac{b^n}{n!}\), but then \(f_n'(a) f_n'(b) = \frac{(-a)^n(-b)^n}{(n!)^2} > 0\) since \(a < 0, b < 0\). \(_n'(a) f_n'(b)\) is positive, the two gradients must have the same sign (and not be zero). Therefore if they are both increasing, at some point the curve must cross the axis in between. Therefore there is some root \(c\) between \(a\) and \(b\). But then there is also a root between \(c\) and \(a\) and \(c\) and \(b\), and very quickly we find more than \(n\) roots which is not possivel. Therefore there must be at most \(1\) root. If \(n\) is odd there must be exactly one root, since \(f_n\) changes sign as \(x \to -\infty\) vs \(x = 0\). If \(n\) is even then there can't be any roots, since if it crossed the \(x\)-axis there would be two roots (not possible) and it cannot touch the axis, since \(f'_n(a) \neq 0\) unless \(a = 0\), and we know \(a < 0\)

Solution:

1. \begin{align*} && \int \frac {x^3-2}{(x+1)^2}\, \e ^x \d x &= \frac{\P(x)}{Q(x)}\,\e^x + \text{constant} \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{x^3-2}{(x+1)^2}e^x &= \frac{P'(x)Q(x)-Q'(x)P(x)}{Q(x)^2}e^x + \frac{P(x)}{Q(x)}e^x \\ \Rightarrow && \frac{x^3-2}{(x+1)^2} &= \frac{(P(x)+P'(x))Q(x)-Q'(x)P(x)}{Q(x)^2} \\ \Rightarrow && Q(x)^2(x^3-2) &= ((P(x)+P'(x))Q(x)-Q'(x)P(x))(x+1)^2 \\ \Rightarrow && Q(-1) &= 0 \\ \Rightarrow && x+1 &\mid Q(x) \end{align*} We have \(\frac{x^3-2}{(x+1)^2}\) has degree \(1\) (plus some remainder term). Therefore \begin{align*} 1 &= \deg \l (P(x)+P'(x))Q(x)-Q'(x)P(x)\r - 2 \deg Q(x) \\ &= \deg P(x) + \deg Q(x) - 2 \deg Q(x) \\ &= \deg P(x) - \deg Q(x) \end{align*} as required. Suppose \(Q(x) = x+1, P(x) = ax^2+bx+c\) then \begin{align*} && \frac{x^3-2}{(x+1)^2} &= \frac{(P(x)+P'(x))(x+1)-P(x)}{(x+1)^2} \\ \Rightarrow && x^3-2 &= (P(x)+P'(x))(x+1) - P(x) \\ \Rightarrow && x^3-2 &= (ax^2+bx+c+2ax+b)(x+1) - (ax^2+bx+c) \\ &&&= a x^3+ x^2 (2 a + b) + x (2 a + b + c)+b \\ \Rightarrow && a &= 1 \\ && b &= -2 \\ && c &= 0 \end{align*} So \(P(x) = x^2-2x\)
2. \begin{align*} && \int \frac1{x+1}e^x \d x &= \frac{P(x)}{Q(x)}e^x + c \\ \Rightarrow && \frac{1}{x+1} e^x &= \frac{P'(x)Q(x)-Q'(x)P(x)}{Q(x)^2}e^x + \frac{P(x)}{Q(x)}e^x \\ \Rightarrow && \frac{1}{x+1} &= \frac{(P(x)+P'(x))Q(x)-Q'(x)P(x)}{Q(x)^2} \end{align*} Therefore \(Q(-1) = 0\) and so \(x +1 \mid Q(x)\). Considering degrees, we must have that \(P(x)\) has degree \(1\) less than \(Q(x)\). Consider also the number of factors of \(x+1\) in the numerator and denominator. Since \(P(x)\) and \(Q(x)\) have no common factors, the \(Q(x)\) could have \(q\) factors and \(P(x)\) must have none. The denominator therefore has \(2q\) factors and the numerator must have \(q-1\) factors (coming from \(Q'(x)\)), we must have \(2q = (q-1) + 1\), but that implies \(q = 0\). Contradiction! \end{align*}

Solution: The position of the hands are \(\begin{pmatrix} a\sin(-t) \\ a \cos(-t) \end{pmatrix}\) and \(\begin{pmatrix} b\sin(-60t) \\ b \cos(-60t) \end{pmatrix}\), the distance between the hands is \begin{align*} x &= \sqrt{\left ( a \sin t - b \sin 60t\right)^2+\left ( a \cos t - b \cos 60t\right)^2} \\ &= \sqrt{a^2+b^2-2ab\left (\sin t \sin 60t+\cos t \cos 60t \right)} \\ &= \sqrt{a^2+b^2-2ab \cos(59t)} = \sqrt{a^2+b^2-2ab \cos \theta} \\ \\ \frac{\d x}{\d \theta} &= \frac{ab \sin \theta}{ \sqrt{a^2+b^2-2ab \cos \theta}} \\ \frac{\d^2 x}{\d \theta^2} &= \frac{ab \cos \theta\sqrt{a^2+b^2-2ab \cos \theta} - \frac{a^2b^2 \sin^2 \theta}{\sqrt{a^2+b^2-2ab \cos \theta}} }{a^2+b^2-2ab \cos \theta} \\ &= \frac{ab \cos \theta(a^2+b^2-2ab \cos \theta) - a^2b^2 \sin^2 \theta }{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{ab \cos \theta(a^2+b^2-2ab \cos \theta) - a^2b^2(1-\cos^2 \theta)}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{ab(a^2+b^2) \cos \theta-a^2b^2 \cos \theta- a^2b^2}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{-ab(a\cos \theta -b)(b \cos \theta - a)}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ \end{align*} So the rate of increase is largest when \(\cos \theta = \frac{a}{b}\) (since \(\frac{b}{a}\) is impossible. Therefore when \(x = \sqrt{a^2+b^2-2ab \frac{a}{b}} = \sqrt{a^2+b^2-2a^2} = \sqrt{b^2-a^2}\) If \(b = 2a\) then \(\cos \theta = \frac{a}{2a} = \frac12 = \frac{\pi}{3} = 60^\circ\) The relative speed of the hands is \(5.5^\circ\) per minute, so \(\frac{60}{5.5} = \frac{120}{11} \approx 11\) but clearly also less than since \(121 = 11^2\).

Solution: \begin{align*} && \sin(r + \tfrac12)x - \sin(r - \tfrac12) x &= \sin rx \cos \tfrac12x + \cos r x\sin\tfrac12x - \sin r x \cos \tfrac12 x + \cos rx \sin \tfrac12 x \\ &&&= 2\cos r x \sin\tfrac12 x \\ \\ && S &= \cos x + \cos 2x + \cdots + \cos n x \\ && 2\sin \tfrac12 x S &= \sin(1 + \tfrac12)x - \sin \tfrac12 x + \\ &&&\quad+ \sin(2+\tfrac12)x - \sin(2- \tfrac12)x + \\ &&&\quad+ \sin(3+\tfrac12)x - \sin(3 - \tfrac12)x + \\ &&& \quad + \cdots + \\ &&&\quad + \sin(n+\tfrac12)x - \sin(n-\tfrac12)x \\ &&&=\sin(n+\tfrac12)x - \sin\tfrac12 x \\ \Rightarrow && S &= \frac{\sin(n+\tfrac12)x - \sin\tfrac12 x}{2 \sin \tfrac12 x} \end{align*}

1. \(\,\) \begin{align*} && S_2(x) &= \sin x + \tfrac12 \sin 2 x \\ && S'_2(x) &= \cos x + \cos 2x \\ &&&= \cos x + 2\cos^2 x - 1 \\ &&&= (2\cos x -1)(\cos x + 1) \\ \end{align*} Therefore the turning points are \(\cos x= \frac12 \Rightarrow x = \frac{\pi}{3}\) and \(\cos x = -1 \Rightarrow x = \pi\)
   

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2. Suppose \(S_n(x)\) has a stationary point at \(x_0\), then $$ therefore \begin{align*} &&0 &= S_n'(x_0) \\ &&&= \cos x_0 + \cos 2x_0 + \cdots + \cos n x_0 \\ &&&= \frac{\sin(n+\tfrac12)x_0 - \sin \tfrac12x_0}{2 \sin \tfrac12 x_0} \\ \Rightarrow &&\sin\tfrac12 x_0&= \sin nx_0 \cos \tfrac12 x_0 + \cos nx_0 \sin \tfrac12x_0 \\ \Rightarrow && \sin nx_0 &= (1-\cos nx_0)\tan \tfrac12 x_0 \end{align*} Therefore \(S_n(x_0) -S_{n-1}(x_0) = \tfrac1n \sin n x_0 = \tfrac1n \underbrace{(1-\cos nx_0)}_{\geq 0}\underbrace{\tan\tfrac12 x_0}_{\geq 0} \geq 0\). Therefore if \(S_{n-1}(x) > 0\) for all \(x\) on \(0 < x < \pi\) then since \(S_n(x) > S_{n-1}(x)\) at the turning points and since they agree at the end points, it must be larger at all points inbetween.
3. Notice that \(S_1(x) = \sin x \geq 0\) for all \(x \in [0,1]\) and by our previous argument we can show \(S_n > S_{n-1}\) inside the interval and equal on the boundary we must have \(S_n(x) \geq 0\) for \(x \in [0, \pi]\)

Solution: \begin{questionparts} \item The tangent to \(y = \ln x\) is \begin{align*} && \frac{y - \ln x_1}{x - x_1} &= \frac{1}{x_1} \\ \Rightarrow && \frac{x_1y -x_1 \ln x_1}{ x- x_1} &= 1 \\ \Rightarrow && x_1 y - x_1 \ln x_1 &= x - x_1 \end{align*} So to run through the origin, we need \(\ln x_1 = 1 \Rightarrow x_1 = e\) so the line will be \(y = \frac1{e} x\) If \(ma = \ln a \Rightarrow m = \frac{\ln a}{a} = \frac{\ln b}{b} \Rightarrow b \ln a = a \ln b \Rightarrow a^b = b^a\). \item

Solution:

1. \(\,\) \begin{align*} && f(x) &= f(1-x) \\ \Rightarrow && f'(x) &= -f'(1-x) \\ \Rightarrow && f'(\tfrac12) &= -f'(\tfrac12) \\ \Rightarrow && f'(\tfrac12) &= 0 \\ \\ && f(x) &= f(\tfrac1x) \\ \Rightarrow && f'(x) &= f'(\tfrac1x) \cdot \frac{-1}{x^2} \\ \Rightarrow && f'(-1) &= -f'(-1) \\ \Rightarrow && f'(-1) &= 0 \\ \\ && f'(2) &= -\frac{1}{4}f'(\tfrac12) \\ &&&= 0 \end{align*}
2. Suppose \begin{align*} && f(x) &= \frac{(x^2-x+1)^3}{(x^2-x)^2} \\ && f(1/x) &= \frac{(x^{-2}-x^{-1}+1)^3}{(x^{-2}-x^{-1})^2} \\ &&&= \frac{(1-x+x^2)^3/x^6}{((x-x^2)^2/x^6} \\ &&&= f(x) \\ \\ && f(1-x) &= \frac{((1-x)^2-(1-x)+1)^3}{((1-x)^2-(1-x))^2} \\ &&&= \frac{(1-x+x^2)^3}{(x^2-x)^2} = f(x) \end{align*}
   

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3. Clearly \(x = -1\) is a root of \(f(x) = \frac{27}{4}\), so we must also have \(x=2\) and \(x = \frac12\), therefore \(f(x) > \frac{27}{4}\) if \(x \in \mathbb{R} \setminus \{-1, 2, \tfrac12, 0, 1 \}\). Clearly \(x = 3\) and \(x = -2\) are solutions so we also have: \(\frac13, -\frac12, \frac32, \frac23\) and these must be all solutions so we must have: \(f(x) > \frac{343}{36} \Leftrightarrow x \in (-\infty, -2) \cup (-\frac12, 0) \cup (0, \frac13) \cup (\frac23, 1) \cup (1, \frac32) \cup (3, \infty)\)

Solution:

1. First, not that the point must be ony the curve:
   

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   Since otherwise it's pretty clear we could make the area of the rectangle larger by moving the point onto the curve. Therefore \(A = x(f(x)-f(t))\). To maximise this we need \(xf'(x) + f(x)-f(t) = 0\), ie \(x_0f'(x_0) + f(x_0) = f(t)\)
2. Suppose \(\displaystyle \g(t) =\frac 1t \int_0^t \f(x) \d x\) then \begin{align*} && \g(t) &=\frac 1t \int_0^t \f(x) \d x\\ \Rightarrow && tg(t) &= \int_0^t \f(x) \d x \\ \Rightarrow && tg'(t) +g(t) &= f(t) \\ \Rightarrow && tg'(t) &= f(t) - g(t) \end{align*}
   

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   Clearly the blue area + green area is larger than the green area. So \(\displaystyle \int_0^t (f(x) - f(t))\d x > A_0(t)\). Notice that \(f(t) = \frac1{t} \int_0^t f(t) \d x \) so \(-t^2g'(t) = \int_0^t f(x) \d x > A_0(t)\)
3. Not that if \(f(x) = \dfrac{1}{1+x}\), the \(f'(x) = -\frac{1}{(1+x)^2}\) and so \begin{align*} && -\frac{x_0}{(1+x_0)^2} + \frac{1}{1+x_0} &= \frac{1}{1+t} \\ && \frac{1}{(1+x_0)^2} &= \frac{1}{1+t} \\ \Rightarrow && x_0 &= \sqrt{1+t} - 1 \\ && A_0(t) &= (\sqrt{1+t} - 1) \left ( \frac{1}{\sqrt{1+t}} - \frac{1}{t+1} \right) \\ &&&= 1 - \frac{1}{\sqrt{1+t}} - \frac{1}{\sqrt{1+t}} + \frac{1}{1+t} \\ &&&= \frac{2+t}{1+t} - \frac{2}{\sqrt{1+t}} \\ && g(t) &= \frac{1}{t} \int_0^t \frac{1}{1+x} \d x \\ &&&= \frac{\ln(1+t)}{t} \\ \Rightarrow && g'(t) &= \frac{\frac{t}{1+t} - \ln(1+t)}{t^2} \\ \Rightarrow && -t^2g(t) &= \ln(1+t) - \frac{t}{1+t} \\ \Rightarrow && \ln(1+t) - \frac{t}{1+t} &> \frac{2+t}{1+t} - \frac{2}{\sqrt{1+t}} \\ \Rightarrow && \ln \sqrt{1+t} & > 1 - \frac{1}{\sqrt{1+t}} \end{align*}

Solution:

1. \(\,\) \begin{align*} && E(x) &= \left ( \frac{\d y}{\d x} \right)^2 + \frac12 y^4 \\ \Rightarrow && E'(x) &= 2 \frac{\d y}{\d x} \frac{\d^2 y}{\d x^2} + 2y^3 \frac{\d y}{\d x} \\ &&&= 2\frac{\d y}{\d x} \left ( \frac{\d^2 y}{\d x^2} + y^3 \right) \\ &&&= 0 \end{align*} Therefore \(E\) is constant. \(E(0) = \frac12\) and \begin{align*} &&\frac12 &= \left ( \frac{\d y}{\d x} \right)^2 + \frac12 y^4 \\ &&&\geq \frac12 y^4 \\ \Rightarrow && |y| &\leq 1 \end{align*}
2. \(\,\) \begin{align*} && E(x) &= \left ( \frac{\d v}{\d x} \right)^2 + 2 \cosh v \\ \Rightarrow && E'(x) &= 2 \frac{\d v}{\d x}\frac{\d^2 v}{\d^2 x} + 2 \sinh v \frac{\d v}{\d x} \\ &&&= 2 \frac{\d v}{\d x} \left ( \frac{\d^2 v}{\d^2 x} + \sinh v\right) \\ &&&= 2 \frac{\d v}{\d x} \left ( -x \frac{\d v}{\d x}\right) \\ &&&= -2x \left ( \frac{\d v}{\d x} \right)^2 \leq 0 \tag{\(x \geq 0\)} \\ \\ && E(0) &= 0^2 + 2 \cosh \ln 3 \\ &&&= 3 + \frac13 = \frac{10}{3} \\ \Rightarrow && \frac{10}{3} &\geq E(x) \\ &&&= \left ( \frac{\d v}{\d x} \right)^2 + 2 \cosh v \\ &&&\geq 2 \cosh v \\ \Rightarrow && \cosh v &\leq \frac53 \end{align*}
3. \(\,\) \begin{align*} && E(x) &= \left ( \frac{\d w}{\d x} \right)^2 + 2(w \sinh w + \cosh w) \\ && E'(x) &= 2 \frac{\d w}{\d x}\frac{\d^2 w}{\d^2 x} + 2(w \cosh w + 2 \sinh w) \frac{\d w}{\d x} \\ &&&= 2 \frac{\d w}{\d x} \left ( \frac{\d^2 w}{\d^2 x}+(w \cosh w + 2 \sinh w)\right) \\ &&&= -2 \left ( \frac{\d w}{\d x} \right)^2 \left (\underbrace{5\cosh x - 4 \sinh x -3}_{\geq0} \right) \\ &&&\leq 0 \\ && E(0) &= \frac12 + 2 = \frac52 \\ \Rightarrow && \frac52 &\geq E(x) \\ &&&=\underbrace{ \left ( \frac{\d w}{\d x} \right)^2}_{\geq0} + 2(\underbrace{w \sinh w}_{\geq 0} + \cosh w) \\ &&&\geq2\cosh w \\ \Rightarrow && \cosh w &\leq \frac54 \end{align*}

Solution: \begin{align*} && b &= c - ma \\ \Rightarrow && c &= b+ma \\ \Rightarrow && y &= m(a-x)+b \\ \Rightarrow && q &= ma+b \\ && p &= \frac{ma+b}{m} \\ \\ && d &= p+q \\ &&&= a + \frac{b}{m} + ma + b \\ \Rightarrow && d' &= -bm^{-2}+a \\ \Rightarrow && m &= \sqrt{b/a} \\ \\ \Rightarrow &&d &= a + \sqrt{ba}+\sqrt{ba} + b \\ &&&= (\sqrt{a}+\sqrt{b})^2 \\ \\ && |PQ|^2 &= p^2 + q^2 \\ &&&= a^2 + \frac{2ab}{m} + \frac{b^2}{m^2} + m^2a^2 + 2mab + b^2 \\ &&&= a^2+b^2 + \frac{b^2}{m^2} + \frac{2ab}{m}+ 2abm + a^2m^2 \\ && \frac{\d}{\d m}&= -2b^2m^{-3}-2abm^{-2}+2ab + 2a^2m \\ && 0 &=2a^2m^4+2abm^3-2abm-2b^2 \\ &&&= 2(am^3-b)(am+b) \\ \Rightarrow && m &= \sqrt[3]{\frac{b}{a}} \\ \\ &&|PQ|^2 &= \left[ a^{1/3}(a^{2/3} + b^{2/3}) \right]^2 + \left[ b^{1/3}(a^{2/3} + b^{2/3}) \right]^2 \\ &&&= a^{2/3}(a^{2/3} + b^{2/3})^2 + b^{2/3}(a^{2/3} + b^{2/3})^2 \\ &&&= (a^{2/3} + b^{2/3})^2 \cdot (a^{2/3} + b^{2/3}) \\ &&&= (a^{2/3} + b^{2/3})^3 \\ \Rightarrow && |PQ| &= (a^{2/3} + b^{2/3})^{3/2} \end{align*} We can also do this with AM-GM instead: \begin{align*} && d &= a + b + \frac{b}{m} + am \\ &&&\geq a+b + 2 \sqrt{\frac{b}{m} \cdot am} \\ &&&= a+2\sqrt{ab}+b \\ \\ && |PQ|^2 &= a^2+b^2 + \frac{b^2}{m^2} + \frac{2ab}{m}+ 2abm + a^2m^2 \\ &&&= a^2+b^2 + \frac{b^2}{m} + abm + abm + a^2m^2 + \frac{ab}{m} + \frac{ab}{m} \\ &&&= a^2+b^2 + 3\sqrt[3]{ \frac{b^2}{m} \cdot abm \cdot abm} + 3 \sqrt[3]{ a^2m^2 \cdot \frac{ab}{m} \cdot \frac{ab}{m} } \\ &&&= a^2 + 3b^{4/3}a^{2/3}+3b^{2/3}a^{4/3}+b^2 \\ &&&= (a^{2/3}+b^{2/3})^3 \end{align*}

Solution:

1. \(\,\)
   

   + - ↺
   1. \(n = 0\) if \(b > 9\)
   2. \(n = 1\) is not possible, since by symmetry if \(x\) is a root, so is \(-x\), and \(0\) can never be the only root.
   3. \(n = 2\) if \(b < 0\) or \(b = 9\)
   4. \(n = 3\) if \(b = 0\)
   5. \(n = 4\) if \(0 < b < 9\)
2. \(\,\) \begin{align*} && y' &= 4x^3-12x+a \\ && y'' &= 12x^2-12 \\ \Rightarrow && x &= \pm 1 \\ \Rightarrow && 0 &= 4(\pm 1) - 12 (\pm 1) + a \\ &&&= a \mp 8 \\ \Rightarrow && a &= \pm 8 \end{align*} When \(a = 8\), we have \(y = x^4-6x^2+8x\) and \begin{align*} &&y' &= 4x^3-12x+8 \\ &&&= 4(x^3-3x+2) \\ &&&= 4(x-1)^2(x+2) \\ \Rightarrow && y(1) &= 3\\ && y(-2) &= -24 \end{align*}
   

   + - ↺
   Therefore there are no solutions if \(b > 24\), one solution if \(b = 24\) and two solutions otherwise. Similarly, if \(a = -8\), we have \(y = x^4 - 6x^2-8x\) \begin{align*} && y' &= 4x^3-12x-8 \\ &&&= 4(x^3-3x-2) \\ &&&= 4(x-2)(x+1)^2 \end{align*} So we have stationary points at \(x = 2\) and \(x = -1\) (which is also a inflection point) and at \(x = 2\) \(y = -24\), so we have the same story: there are no solutions if \(b > 24\), one solution if \(b = 24\) and two solutions otherwise.
3. \(\,\)
   

   + - ↺