Problems

Solution: \begin{align*} && y &= \ln(x^2 -1) \\ && r &= \sqrt{x^2-1} \\ && \coth \theta &= x \\ && r &= \sqrt{\coth^2 \theta - 1} = \sqrt{\textrm{cosech}^2 \theta} = \textrm{cosech} \theta \\ && \frac{\d y}{\d x} &= \frac{2x}{x^2-1} \\ &&&= \frac{2 \coth \theta}{r^2} \\ &&&= \frac{2 \cosh \theta}{\sinh \theta \cdot r \cdot \textrm{cosech} \theta } \\ &&&= \frac{2 \cosh \theta}{r } \\ \\ && \frac{\d^2 y}{\d x^2} &= \frac{2(x^2-1)-4x^2}{(x^2-1)^2} \\ &&&= \frac{-2(1+x^2)}{r^2 \textrm{cosech}^2 r} \\ &&&= -\frac{2(1 + \coth^2 \theta) \sinh^2 \theta}{r^2} \\ &&&= -\frac{2(\sinh^2 \theta + \cosh^2 \theta)}{r^2} \\ &&&= -\frac{2 \cosh 2 \theta}{r^2} \\ \\ && \frac{\d^3 y}{\d x^3} &= \frac{-4x(x^2-1)^2-(-2x^2-2)\cdot2(x^2-1)\cdot 2x}{(x^2-1)^4} \\ &&&= \frac{-4x(x^2-1)+8x(x^2+1)}{(x^2-1)^3}\\ &&&= \frac{4x^3+12x}{(x^2-1)^3} \\ &&&=\frac{\sinh^3 \theta (4\coth^3 \theta + 12\coth \theta )}{r^3} \\ &&&=\frac{4\cosh^3 \theta + 12\cosh \theta \sinh^2 \theta}{r^3} \\ &&&= \frac{4 \cosh 3 \theta}{r^3} \\ \end{align*} Claim: \(\frac{\d^n y}{\d x^n} = (-1)^{n+1}\frac{2(n-1)!\cosh n \theta}{r^n}\) Proof: By induction. Base cases already proven \begin{align*} \frac{\d r}{\d x} &= \frac{x}{\sqrt{x^2-1}} = \frac{\coth \theta}{\textrm{cosech} \theta} = \cosh \theta \\ \frac{\d \theta}{\d x} &= - \sinh^2 \theta \\ \\ \frac{\d^{n+1} y}{\d x^{n+1}} &= (-1)^{n+1}(n-1)!\frac{\d}{\d x} \left ( \frac{2\cosh n \theta}{r^n}\right) \\ &= (-1)^{n+1}\frac{2 n \sinh n \theta \cdot r^n \cdot \frac{\d \theta}{\d x}- 2\cosh n \theta \cdot nr^{n-1} \frac{\d r}{\d x} }{r^{2n}} \\ &= (-1)^{n+2}\frac{2n( \cosh n \theta\cosh \theta + r\sinh n \theta \sinh^2 \theta) }{r^{n+1}} \\ &= (-1)^{n+2}n!\frac{2\cosh(n+1) \theta }{r^{n+1}} \\ \end{align*} We can think of this as \(\ln(x^2-1) = \ln(x+1)+\ln(x-1)\) and also note \(x \pm 1 = \coth \theta \pm 1 = \frac{\cosh \theta \pm \sinh \theta}{\sinh \theta} = \frac{e^{\pm \theta}}{\sinh \theta}\) \begin{align*} && \frac{\d^n}{\d x^n} \ln(x^2-1) &= (n-1)!(-1)^{n-1} \left ( \frac{1}{(x+1)^n} + \frac{1}{(x-1)^n} \right) \\ &&&= (-1)^{n-1}(n-1)! \left ( \frac{\sinh^n \theta}{e^{n\theta}} + \frac{\sinh^n \theta}{e^{-n\theta}} \right) \\ &&&= (-1)^{n-1} (n-1)!2\cosh n \theta \cdot \sinh^n \theta \\ &&&= (-1)^{n-1}(n-1)! \frac{2 \cosh n \theta }{r^n} \end{align*}

Solution: \begin{align*} && e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\ &&&> 1 + 1+ \frac12 + \frac16 \\ &&&= \frac{12+3+1}{6} = \frac83 \end{align*} \(4! = 24 > 16 = 2^4\), notice that \(n! = \underbrace{n \cdot (n-1) \cdots 5}_{>2^{n-4}} \cdot \underbrace{4!}_{>2^4} >2^n\). \begin{align*} && e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\ &&&< \frac83 + \frac{1}{2^4} + \frac{1}{2^5} + \cdots \\ &&&= \frac83 + \frac{1}{2^4} \frac{1}{1-\tfrac12} \\ &&&= \frac83 + \frac1{8} \\ &&&= \frac{67}{24} \end{align*} \begin{align*} && y &= 3e^{2x} +14 \ln(\tfrac43-x) \\ && y' &= 6e^{2x} - \frac{14}{\tfrac43-x} \\ && y'(\tfrac12) &= 6e - \frac{14}{\tfrac43-\tfrac12} \\ &&&= 6e -\tfrac{84}{5} = 6(e-\tfrac{14}5) < 0 \\ && y'(1) &= 6e^2 - \frac{14}{\tfrac43-1} \\ &&&= 6e^2 - 42 = 6(e^2-7) \\ &&&> 6(\tfrac{64}{9} - 7) > 0 \end{align*} Therefore \(y'\) changes from negative (decreasing) to positive (increasing) in our range, and therefore there is a minima in this range.

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Solution: Claim: If \(f\) is a constant, then \(\triangle f = 0\) Proof: First consider \(f(x) = 1, g(x) = x\) then we must have: \begin{align*} && \triangle (1x) &= 1 \triangle x + x \triangle 1 \tag{iv} \\ &&&= 1 \cdot 1 + x \triangle 1 \tag{i} \\ \Rightarrow && 1 &= 1 + x \triangle 1 \tag{i} \\ \Rightarrow && \triangle 1 &= 0 \\ \Rightarrow && \triangle c &= 0 \tag{iii} \end{align*} \begin{align*} && \triangle (x^2) &= x \triangle x + x \triangle x \tag{iv} \\ &&&= x \cdot 1 + x \cdot 1 \tag{i} \\ &&&= 2x \\ \\ && \triangle (x^3) &= x^2 \triangle x + x \triangle (x^2) \tag{iv} \\ &&&= x^2 \cdot 1 + x \cdot 2x \tag{\(\triangle x^2 = 2x\)}\\ &&&= 3x^2 \end{align*} Claim: \(\triangle h(x) = \frac{\d h(x)}{\d x}\) for any polynomial \(h\) Proof: Since both \(\triangle\) and \(\frac{\d}{\d x}\) are linear (properties \((ii)\) and \((iii)\)) it suffices to prove that: \(\triangle x^n = nx^{n-1}\). For this we proceed by induction. Base cases (we've proved up to \(n = 3\) so we're good). Suppose it's true for some \(n\), then consider \(n + 1\), \begin{align*} && \triangle (x^{n+1}) &= x \triangle (x^n) + x^n \triangle x \tag{iv} \\ &&&= x \cdot n x^{n-1} + x^n \triangle x \tag{Ind. hyp.} \\ &&&= nx^n + x^n \tag{i} \\ &&&= (n+1)x^{n} \end{align*} Therefore it's true for for \(n+1\). Therefore by induction it's true for all \(n\).

Solution: \begin{align*} && y &= x^2e^{-x^2} \\ \Rightarrow && y' &= 2xe^{-x^2} +x^2 \cdot (-2x)e^{-x^2} \\ &&&= e^{-x^2}(2x-2x^3) \\ &&&= 2e^{-x^2}x(1-x^2) \end{align*} Therefore the derivative is zero iff \(x = 0, \pm 1\) \begin{align*} && y &= \P(x) e^{-x^2} \\ \Rightarrow && y' &= e^{-x^2} (\P'(x)-2x\P(x)) \end{align*} Therefore we want \(\P'(x) - 2x\P(x) = Kx(x^2-a^2)(x^2-b^2)\) Since this has degree \(5\), we should look at polynomials degree \(4\) for \(\P\). We can also immediately see that \(0\) is a root of \(\P'(x)\), so \(\P(x) = a_4x^4+a_3x^3+a_2x^2+a_0\). WLOG \(a_4 = 1\) and \(K = -2\), so \begin{align*} && -2(x^5-(a^2+b^2)x^3+a^2b^2x) &= 4x^3+3a_3x^2+2a_2x- 2x(x^4+a_3x^3+a_2x^2+a_0) \\ &&&= -2x^5-2a_3 x^4+(4-2a_2)x^3+(2a_2-2a_0)x \\ \Rightarrow && a_3 &= 0 \\ && a^2+b^2 &= 2-a_2 \\ \Rightarrow && a_2 &= 2-a^2-b^2 \\ && a^2b^2 &= a_0-a_2 \\ \Rightarrow && a_0 &= a^2b^2 + 2-a^2-b^2 \\ \Rightarrow && \P(x) &= x^4+(2-a^2-b^2)x^2+(a^2-1)(b^2-1)x \end{align*}

Solution: \[\frac{\d}{\d t} \left ( \sec t \right) = \frac{\sin t }{\cos^2 t} = \sec t \tan t \]

1. \(\,\) \begin{align*} && I_1 &= \int_{\sqrt{2}}^2 \frac{1}{x^3 \sqrt{x^2-1}} \\ x = \sec t, \d x = \sec t \tan t:&&&= \int_{t=\pi/4}^{t=\pi/3} \frac{1}{\sec^3 t \tan t} \sec t \tan t \d t \\ &&&= \int_{t=\pi/4}^{t=\pi/3} \cos^2 t \d t \\ &&&= \int_{t=\pi/4}^{t=\pi/3} \frac{1+\cos 2t}{2} \d t \\ &&&= \frac12 \frac{\pi}{12} + \frac12 \left (\sin \frac{\pi}{3} - \sin \frac{\pi}{4} \right) \\ &&&= \frac{\pi}{24} + \frac{\sqrt{3}-2}{8} \\ \end{align*}
2. \(\,\) \begin{align*} && I_2 &= \int \frac{1}{(x+2)\sqrt{(x+1)(x+3)}} \d x \\ &&&= \int \frac{1}{(x+2)\sqrt{(x+2)^2-1}} \d x \\ &&&= \int \frac{1}{u\sqrt{u^2-1}} \d u \\ &&&= \sec^{-1} u + C \\ &&&= \sec^{1} (x+2) + C \end{align*}
3. \(\,\) \begin{align*} && I_3 &= \int \frac{1}{(x+2)\sqrt{(x+2)^2 - 9}} \d x \\ &&&= \int\frac{1}{9(\frac{x+2}{3})\sqrt{(\frac{x+2}3)^2 - 1}} \d x \\ u = \frac{x+2}{3}, 3\d u =\d x &&&= \frac19 \int \frac{1}{u\sqrt{u^2-1}} 3 \d u \\ &&&= \frac13 \sec^{-1} u + C \\ &&&= \frac13 \sec^{-1} \frac{x+2}{3} + C \end{align*}

Solution: \begin{align*} \frac{\mathrm{d} \ }{\mathrm{d} x} \arcsin \left ( \frac{x + a }{ \ x + b} \right ) &= \frac{1}{\sqrt{1-\left ( \frac{x + a }{ \ x + b} \right )^2}} \left ( \frac{b-a}{(x+b)^2} \right) \\ &= \frac{b-a}{(x+b)\sqrt{(x+b)^2-(x+a)^2}} \\ &= \frac{b-a}{(x+b)\sqrt{(b-a)(2x+b+a)}} \\ &= \frac{\sqrt{b-a}}{(x+b)\sqrt{a+b+2x}} \\ \\ \frac{\mathrm{d} \ }{ \mathrm{d} x} \; \mathrm{arcosh} \left ( \frac{x + b }{ \ x + a} \right) &= \frac{1}{\sqrt{\left ( \frac{x + b }{ \ x + a} \right)^2-1}} \left ( -\frac{b-a}{(x+a)^2} \right) \\ &= -\frac{b-a}{(x+a)\sqrt{(x+b)^2-(x+a)^2}} \\ &= -\frac{b-a}{(x+a)\sqrt{(b-a)(a+b+2x)}} \\ &= -\frac{\sqrt{b-a}}{(x+a)\sqrt{a+b+2x}} \end{align*}

1. \begin{align*} \int \frac{1}{ ( x + 1) \sqrt{x + 3} } \mathrm{d} x &= \int \frac{1}{(x+1)\sqrt{\frac12 (2x+6)}} \d x\\ &= \int \frac{\sqrt{2}}{(x+1)\sqrt{2x+1+5}} \d x \\ &= \frac{\sqrt{2}}{2}\int \frac{\sqrt{5-1}}{(x+1)\sqrt{2x+1+5}} \d x \\ &= - \frac{\sqrt{2}}{2}\textrm{arcosh} \left ( \frac{x+5}{x+1} \right) + C \end{align*}
2. \begin{align*} \int \frac{1}{(x+3)\sqrt{x+1}} \d x &= \int \frac{1}{(x+3)\sqrt{\tfrac12(2x+2)}} \d x + C \\ &= \int \frac{\sqrt{3-1}}{(x+3)\sqrt{2x+3-1}} \d x \\ &= \textrm{arcsin} \left ( \frac{x-1}{x+3} \right) \end{align*}

Solution: If \(m\) is even, clearly that expression is positive since it's the sum of two (different) squares. If \(m\) is odd, then we can expand it as a sum of powers of \(x^2\) with a leading coefficient of \(1\) so it is also positive. \begin{align*} && f (x) = \frac{ (1+x )^m - ( 1-x )^m}{ (1+x )^m + (1-x )^m} \\ && f'(x) &= \frac{(m(1+x )^{m-1} + m( 1-x )^{m-1})((1+x)^m + (1-x)^m ) - ((1+x )^m - ( 1-x )^m)(m(1+x)^{m-1} - m(1-x)^{m-1} )}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{2m(1+x)^m(1-x)^{m-1}+2m(1+x)^{m-1}(1-x)^m}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{2m(1+x)^{m-1}(1-x)^{m-1}(1+x+1-x)}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{4m(1+x)^{m-1}(1-x)^{m-1}}{\l (1+x)^m + (1-x)^m \r^2} \\ \end{align*}

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Solution: \begin{align*} && f'(x) &= mx^{m-1}(x-1)^n + nx^m(x-1)^{n-1} \\ &&&= (m(x-1)+nx)x^{m-1}(x-1)^{n-1} \\ &&&= (x(m+n) - m)x^{m-1}(x-1)^{n-1} \\ \end{align*} Therefore when \(x = \frac{m}{m+n}\) there is a stationary point with \(0 < x < 1\). \begin{align*} && f''(x) &= m(m-1)x^{m-2}(x-1)^n + 2mnx^{m-1}(x-1)^{n-1} + n(n-1)x^{m}(x-1)^{n-2} \\ &&&= (m(m-1)(x-1)^2 +2mnx(x-1)+n(n-1)x^2)x^{m-2}(1-x)^{n-2} \\ \Rightarrow && f'' \left ( \frac{m}{m+n} \right) &= \left ( m(m-1) \frac{n^2}{(m+n)^2} - 2mn\frac{mn}{(m+n)^2} + n(n-1) \frac{m^2}{(m+n)^2} \right) \frac{m^{m-2}}{(m+n)^{m-2}} \frac{(-1)^{n-2}n^{n-2}}{(m+n)^{n-2}} \\ &&&= (-1)^{n-2}\frac{m^{m-1}n^{n-1}}{(m+n)^{m+n-2}} \left ( (m-1)n-2mn+(n-1)m\right) \\ &&&= (-1)^{n-2}\frac{m^{m-1}n^{n-1}}{(m+n)^{m+n-2}} \left ( -m-n\right) \\ &&&= (-1)^{n-1} \frac{m^{m-1}n^{n-1}}{(m+n)^{m+n-3}} \end{align*} Therefore this is positive (and a minimum) when \(n\) is odd and negative (and a maximum) when \(n\) is even.

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Solution: \begin{align*} && 2a^2+2b^2 &= a^2 + b^2 + (a^2+b^2) \\ &&&\underbrace{\geq}_{AM-GM} a^2+b^2+2\sqrt{a^2b^2} \\ &&&= a^2+b^2 + 2|a||b| \\ &&&\geq a^2+b^2 + 2ab \\ &&&= (a+b)^2 \end{align*} Assume \(a^2 \neq b^2\), otherwise the curve is a constant. \begin{align*} && y & = \big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12} + \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12}\\ && \frac{\d y}{\d \theta} &= \tfrac12 \left (a^2\cos^2\theta +b^2\sin^2\theta \right)^{-\frac12} \cdot (2 \sin \theta \cos \theta (b^2 - a^2)) + \tfrac12 (a^2\sin^2\theta +b^2\cos^2\theta)^{-\frac12} \cdot (2 \sin \theta \cos \theta (a^2 - b^2) \\ &&&= \tfrac12\sin2 \theta (b^2 - a^2) \left ( \left (a^2\cos^2\theta +b^2\sin^2\theta \right)^{-\frac12} - (a^2\sin^2\theta +b^2\cos^2\theta)^{-\frac12}\right) \\ \therefore \frac{\d y}{\d x} = 0 \Rightarrow && \sin 2\theta = 0 & \text{ or } a^2\cos^2\theta +b^2\sin^2\theta = a^2\sin^2\theta +b^2\cos^2\theta \\ \Rightarrow && \theta &= 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \\ && (a^2-b^2) \cos ^2\theta &= (a^2-b^2) \sin^2 \theta \\ \Rightarrow && \theta &= \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \end{align*} WLOG \(b^2 - a^2 > 0\), then the two parts of the derivative look like:

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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*}

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 \(f'(x) \geq 0\)

Solution: First notice that \(p(x) = (x-a)^2q(x)\) so \(p'(x) = 2(x-a)q(x) + (x-a)^2q'(x) = (x-a)(2q(x)+(x-a)q'(x))\), in particular \(p'(a) = 0\) so \(x-a\) is a root of \(p'(x)\). If \((x-a)^4\) is a root of \(p(x)\) then \(p^{(3)}(a)= 0\). The proof is similar. Differentiating \(3\) times we obtain: \(6 \cdot 5 \cdot 4 x^3 + 4 \cdot 5 \cdot 4 \cdot 3 x^2 - 5\cdot4 \cdot 3 \cdot 2 x-40 \cdot 3 \cdot 2 \cdot 1 = 5!(x^3+2x^2-x-2) = 5!(x+2)(x^2-1)\). So our possible (repeated) roots are \(x=-2,-1,1\). We can check \(p'(x) = 6x^5+20x^4-20x^3-120x^2-80x+32\), and see \(p'(1) = 36 - 200 \neq 0\), \(p'(-1) = -6+20+20-120+80+32 \neq 0\), therefore \(a = -2\)