Problems

Solution:

1. \(\,\) \begin{align*} && f_1(n) &= n^2 + 6n + 11 \\ &&&= (n+3)^2 + 2 \\ &&&=F_1(n+3) \end{align*} Since \(n \mapsto n+3\) is a bijection on \(\mathbb{Z}\) both functions must have exactly the same range.
2. \(g_1(n) = n^2-2n+5 = (n-1)^2 + 4\). Since squares are always \(0, 1 \pmod{4}\) it's impossible for \(f_1\) and \(g_1\) to take the same value therefore the ranges have empty intersection.
3. \(\,\) \begin{align*} && f_2(n) &= n^2-2n - 6 \\ &&&= (n-1)^2-7 \\ && g_2(n) &= n^2-4n+2 \\ &&&= (n-2)^2 - 2 \end{align*} so suppose \(x^2 - 7 = y^2 - 2\) then \begin{align*} && x^2 - 7 &= y^2 -2 \\ \Rightarrow && 5 &= y^2 - x^2 \\ &&&= (y-x)(y+x) \end{align*} So we have cases: \(y-x = -5, y + x = -1 \Rightarrow y = -3\) and the output is \(7\) \(y-x=-1, y+x = -5 \Rightarrow y = -3\) same output \(y-x=1, y+x = 5 \Rightarrow y = 3\) same output \(y-x=5, y-x = 1 \Rightarrow y = 3\) same ouput.
4. \begin{align*} && 0 &\leq \frac12(p^2+q^2)+\frac12(p+q)^2 \\ &&&= p^2 + q^2 + pq \end{align*} Looking at \(f_3\) we see \begin{align*} && f_3(n) &= n^3 - 3n^2 + 7n \\ &&&= (n-1)^3 -3n + 7n +1 \\ &&&= (n-1)^3 +4(n-1) -3 \\ &&&= g_3(n-1) + 3 \end{align*} So suppose we have two values which are equal, ie \begin{align*} && x^3 + 4x -3 &= y^3 +4y -6 \\ \Rightarrow && 3 &= y^3-x^3+4y-4x \\ &&&= (y-x)(y^2+xy+x^2+4) \end{align*} Since \(x^2+xy+y^2 \geq 0\) then the right hand factor is always a positive integer bigger than \(3\) and in particular there will be no solutions and hence no integers in the intersection of the ranges.

Solution:

1. \((1+\sqrt{2})^2 = 3 + 2\sqrt{2}\) so \(\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}\)
2. We can always factorise any quartic in the form \((x^2+ax+b)(x^2+cx+d)\), since \(x^3\) has a coefficient of \(a+b\) we must have \(a = -b\), ie the form in the question. \begin{align*} && 0 &= (x^2+sx+p)(x^2-sx+q) \\ &&&= x^4+(p+q-s^2)x^2+s(q-p)x+pq \\ \Rightarrow && pq &= -2 \\ && s(q-p) &= 12 \\ && p+q-s^2 &= -10 \\ \\ && p+q &= s^2-10 \\ && (p+q)^2 &= (s^2-10)^2 \\ && (q-p)^2 &= \frac{12}{s^2} \\ \Rightarrow && (s^2-10)^2 &= \frac{12}{s^2} + 4pq \\ && (s^2-10)^2 &= \frac{144}{s^2} -8 \\ && 0 &= s^2(s^2-10)^2+8s^2-144 \\ &&&= s^6-20s^4+108s^2-144 \\ &&&= (s^2-2)(s^2-6)(s^2-12) \end{align*} Suppose \(s = \sqrt{2}\), and we have \begin{align*} && q-p &= 6\sqrt{2} \\ && p+q &= -8 \\ \Rightarrow && q &= 3\sqrt{2}-4 \\ && p &= -4-3\sqrt{2} \end{align*} Solving our quadratic equations, we have \begin{align*} && 0 &= x^2-\sqrt{2}x-4+3\sqrt{2} \\ \Rightarrow && x &= \frac{\sqrt{2}\pm \sqrt{2-4\cdot(-4+3\sqrt{2})}}{2} \\ &&&= \frac{\sqrt{2}\pm \sqrt{18-12\sqrt{2}}}{2} \\ &&&= \frac{\sqrt{2}\pm (2\sqrt{3}-\sqrt{6})}{2} \\ \\ && 0 &= x^2+\sqrt{2}x-3\sqrt{2}-4 \\ && x &= \frac{-\sqrt{2} \pm \sqrt{2-4\cdot(3\sqrt{3}-4)}}{2}\\ && &= \frac{-\sqrt{2} \pm \sqrt{18+12\sqrt{2}}}{2}\\ && &= \frac{-\sqrt{2} \pm (\sqrt{6}+2\sqrt{3})}{2}\\ \end{align*}

Solution:

1. \(\,\) \begin{align*} && y &= 9x^2 - 12x \cos \theta + 4 \\ &&&= (3x-2\cos \theta)^2+4-4\cos^2 \theta \\ &&&= (3x-2\cos \theta)^2 + 4 \sin^2 \theta \end{align*} Therefore the minimum is \(4\sin^2 \theta\) when \(x = \frac23 \cos \theta\). \begin{align*} && y &= 12x^2 \sin \theta - 9x^4 \\ &&&=4\sin^2 \theta -(3x^2-2\sin\theta)^2 \end{align*} Therefore the maximum is \(4\sin^2 \theta\) when \(x^2 = \frac23\sin \theta\) Therefore \begin{align*} && 0 &= 9x^4 + (9 - 12 \sin \theta)x^2 - 12x \cos \theta + 4 \\ && \underbrace{-9x^4+12x^2\sin \theta}_{\leq 4\sin^2 \theta } &= \underbrace{9x^2 - 12x \cos \theta + 4 }_{\geq 4 \sin^2 \theta} \end{align*} Therefore the equality cases must be achieved in both cases, ie \(x = \frac23 \cos \theta\) and \(x^2 = \frac23 \sin \theta\) \begin{align*} && x^2 &= \frac49\cos^2 \theta \\ &&&= \frac49(1-\sin^2 \theta) \\ &&&= \frac49(1-\frac94 x^2) \\ \Rightarrow && 2x^2 &= \frac49 \\ \Rightarrow && x &= \pm \frac{\sqrt{2}}3\\ \Rightarrow && \cos \theta &=\pm \frac32 \frac{\sqrt{2}}3 \\ &&&= \pm \frac{1}{\sqrt{2}} \\ \Rightarrow && \theta &= \frac{\pi}{4}, \frac{3\pi}{4} \\ \Rightarrow && (x, \theta) &= \left (\frac{\sqrt{2}}{3}, \frac{\pi}{4} \right), \left (-\frac{\sqrt{2}}{3}, \frac{3\pi}{4} \right) \end{align*}
2. Sketching we obtain, noticing we can find the turning point by: \begin{align*} && \frac{x^2}{x-\theta} &= \lambda \\ \Leftrightarrow && x^2 - \lambda x +\theta \lambda &= 0 \\ \Leftrightarrow && 0 &\leq \Delta = \lambda^2 -4\lambda \theta \\ \Leftrightarrow && \lambda &\geq 4 \theta, \lambda \leq 0 \end{align*}
   

   + - ↺
   Notice that \(\sin^2 \theta \cos^2 x \leq 1\) and \(1 + cos^2 \theta \sin^2 x \geq 1\) and therefore we must have the inequality desired. \begin{align*} && \underbrace{\frac{x^2}{4\theta(x - \theta)}}_{\geq 1 \text{ or } \leq 0} &= \underbrace{\frac{\sin^2 \theta \cos^2 x}{1 + \cos^2 \theta \sin^2 x}}_{\in [0,1]} \\ \text{both}=0: && x = 0 &, \sin \theta = 0 \\ \text{both}=1: && x = 2\theta &, \sin^2 \theta = 1,\cos^2 x = 1 \\ && 1 &= \cos^2 2 \theta \\ &&&= (1-2 \sin^2 \theta)^2 \\ &&&= 1 \\ \Rightarrow && (x, \theta) &= \left(\frac{\pi}{2}, \pi\right) \end{align*}

Solution:

1. \(\,\) \begin{align*} && I_1 &= \int_{-\infty}^{\infty} \frac{1}{x^2+2ax+b} \d x \\ &&&= \int_{-\infty}^{\infty} \frac{1}{b-a^2 +(x+a)^2} \d x \\ &&&= \left [ \frac{1}{\sqrt{b-a^2}} \tan^{-1} \frac{x+a}{\sqrt{b-a^2}} \right]_{-\infty}^{\infty} \\ &&&= \frac{\pi}{\sqrt{b-a^2}} \end{align*}
2. \(\,\) Here is the corrected LaTeX code for the second part, maintaining your exact styling and notation.
3. \(\,\) \begin{align*} && I_{n} &= \int_{-\infty}^{\infty} \frac{1}{(x^2+2ax+b)^{n}} \d x \\ &&&= \left[ \frac{x}{(x^2+2ax+b)^n} \right]_{-\infty}^{\infty} - \int_{-\infty}^\infty x \cdot \frac{-n(2x+2a)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 0 + n \int_{-\infty}^\infty \frac{2x^2+2ax}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= n \int_{-\infty}^\infty \frac{2(x^2+2ax+b) - (2ax+2b)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{2ax+2b}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{a(2x+2a) + 2(b-a^2)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{a(2x+2a)}{(x^2+2ax+b)^{n+1}} \d x - 2n(b-a^2) I_{n+1} \\ &&&= 2n I_n - n \left[ \frac{-a}{n(x^2+2ax+b)^n} \right]_{-\infty}^\infty - 2n(b-a^2) I{n+1} \\ &&&= 2n I_n - 0 - 2n(b-a^2) I_{n+1} \\ \Rightarrow && 2n(b-a^2)I_{n+1} &= (2n-1)I_n \end{align*}
4. \(\,\) \begin{align*} && I_{n+1} &= \frac{2n-1}{2n(b-a^2)} I_n \\ &&&= \frac{2n-1}{2n(b-a^2)} \cdot \frac{2n-3}{2(n-1)(b-a^2)} I_{n-1} \\ &&&= \frac{2n-1}{2n(b-a^2)} \cdot \frac{2n-3}{2(n-1)(b-a^2)} \cdots I_{1} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1}{2n \cdot 2(n-1) \cdots 2 (b-a^2)^n} \frac{\pi}{\sqrt{b-a^2}} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1}{2^n n!} \frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1 \cdot 2n \cdot 2(n-1) \cdots 2}{2^{2n} n!n!} \frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{(2n)!}{2^{2n}n!n!}\frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{\pi}{2^{2n}(b-a^2)^{n+\frac12}} \binom{2n}{n} \\ \Rightarrow && I_n &= \frac{\pi}{2^{2n-2}(b-a^2)^{n-\frac12}} \binom{2n-2}{n-1} \\ \end{align*}

Solution:

1. \begin{align*} && I - I_1 &= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) \d x - \int_0^1 \left ( y' + y \tan x \right)^2 \d x\\ &&&= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) - \left ( y' + y \tan x \right)^2 \d x\\ &&&= \int_0^1 \left (-y^2-2yy' \tan x - y^2 \tan^2 x \right) \d x\\ &&&= \int_0^1 \left (-2yy' \tan x - y^2(1+ \tan^2 x )\right) \d x\\ &&&= \int_0^1 \left (-2yy' \tan x - y^2 \sec^2 x\right) \d x\\ &&&= \int_0^1 -\frac{\d}{\d x} \left (y^2 \tan x \right) \d x\\ &&&= \left [-y^2 \tan x \right]_0^1 \\ &&&= 0 \\ \\ \Rightarrow && I &= I_1 = \int_0^1 \left ( y' + y \tan x \right)^2 d x \geq 0 \end{align*} The only way \(I_0 = 0\) is is \(y' + y \tan x =0\), so \begin{align*} && \frac{\d y}{\d x} &= - y \tan x \\ \Rightarrow && \int \frac{1}{y} &= \int -\tan x \d x \\ \Rightarrow && \ln |y| &= \ln |\cos x| + C \\ \Rightarrow && y &= A \cos x \\ \Rightarrow && A &= 0 \Rightarrow y = 0 \end{align*}
2. Let \(J_1 = \int_0^1 (y'+ay\tan ax)^2 \d x\), then \begin{align*} && J-J_1 &= \int_0^1 \left ( \left ( y' \right)^2 - a^2y^2 \right) - \left ( y' + ya \tan ax \right)^2 \d x\\ &&&= \int_0^1 \left (-a^2y^2-2yy' a \tan a x-y^2a^2 \tan^2 ax \right) \d x \\ &&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2(1+\tan^2 ax) \right) \d x \\ &&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2\sec^2 ax \right) \d x \\ &&&= \left [ - a y^2 \tan a x \right]_0^1 = 0 \end{align*} This is true if \(a < \frac{\pi}{2}\), since otherwise we might care about the order of the zero for \(y\) at \(x = 1\). Consider \(y = \cos \frac{\pi}{2} x\), then \(y' = -\frac{\pi}{2} \sin^2\frac{\pi}{2} x\) and \begin{align*} && \int_0^1 \frac{\pi^2}{4} \left (\sin^2 \frac{\pi}{2}x - \cos^2 \frac{\pi}{2} x \right) \d x &= -\frac{\pi^2}{4} \int_0^1 \cos(\pi x) \d x \\ &&&= 0 \end{align*}

Solution:

1. \begin{align*} && \frac12 &= \mathbb{P}(X \leq x_m) \\ \Leftrightarrow && \frac12 &= \mathbb{P}(e^X \leq e^{x_m} = y_m) \end{align*} Therefore the median is \(y_m = e^{x_m}\)
2. \begin{align*} && \mathbb{P}(Y \leq y) &= \mathbb{P}(e^X \leq y) \\ &&&= \mathbb{P}(X \leq \ln y) \\ &&&= F(\ln y) \\ \Rightarrow && f_Y(y) &= f(\ln y)/y \\ \\ && f'_Y(y) &= \frac{f'(\ln y) - f(\ln y)}{y^2} \end{align*} Therefore since the mode satisfies \(f'_Y = 0\) we must have \(f'(\ln \lambda ) = f(\ln \lambda)\)
3. This is the integral of the pdf of \(N(\mu + \sigma^2, \sigma^2)\) and therefore is clearly \(1\). \begin{align*} && \E[Y] &= \int_{-\infty}^{\infty} e^x \cdot \frac{1}{\sqrt{2\pi \sigma^2}} e^{-(x-\mu)^2/(2\sigma^2)} \d x \\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (x - (x-\mu)^2/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp ((2x \sigma^2- (x-\mu)^2)/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu-\sigma^2)^2+2\mu \sigma^2-\sigma^4)/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu+\sigma^2)^2)/(2\sigma^2)+\mu +\frac12\sigma^2) \d x\\ &&&= \e^{\mu +\frac12\sigma^2}\frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu-\sigma^2)^2)/(2\sigma^2)) \d x\\ &&&= \e^{\mu +\frac12\sigma^2} \end{align*}
4. Notice that \(y_m = e^\mu < e^{\mu + \tfrac12 \sigma^2} = \E[Y]\), so it suffices to prove that \(\lambda < e^{\mu}\) Notice that \(f'(x) - f(x) = f(x)[-(x-\mu)/\sigma^2 - 1]\) and therefore \(\ln y - \mu = -\sigma^2\) so \(\lambda = e^{\mu - \sigma^2}\) which is clearly less than \(e^{\mu}\) as required.

Solution: \(a\left(x-y+2\right)^{2} +b\left(cx+y\right)^{2}+d\, \equiv (a + bc^2)x^2 + (a+b)y^2 + (-2a+2bc)xy + (4a)x+(-4ay) + 4a+d\) so we want to solve \[ \begin{cases} a + bc^2 &= 5 \\ a+b &= 2 \\ 2bc - 2a &= -6 \\ 4a &= 4 \\ -4a &= 4 \\ 4a+d &= -9 \end{cases} \Rightarrow a = 1, b = 1, c = -2, d = -13 \] Therefore we have: \((x-y+2)^2 + (2x+y)^2-13 = 0\) and our simultaneous equations will be: \[ \begin{cases} (x-y+2)^2 + (-2x+y)^2 &= 13 \\ 2(x-y+2)^2 + (-2x+y)^2 &= 22 \end{cases} \] which are simultaneous equations in \((x-y+2)^2\) and \((-2x+y)^2\) which solve to \((x-y+2)^2 = 9, (-2x+y)^2 = 4 \) so we need to solve \(4\) sets of simultaneous equations: \begin{align*} &\begin{cases} x - y + 2 &= 3 \\ -2x + y &= 2 \end{cases} &&\Rightarrow (x,y) = (-3, -4) \\ &\begin{cases} x - y + 2 &= -3 \\ -2x + y &= 2 \end{cases} &&\Rightarrow (x,y) = (3, 8) \\ &\begin{cases} x - y + 2 &= 3 \\ -2x + y &= -2 \end{cases} &&\Rightarrow (x,y) = (1, 0) \\ &\begin{cases} x - y + 2 &= -3 \\ -2x + y &= -2 \end{cases} &&\Rightarrow (x,y) = (7, 12) \\ \end{align*} So \((x,y) = (-3, -4), (3, 8), (1, 0), (7,12)\)

Solution: \begin{align*} && t &= x+ \sqrt{x^2+2bx+c} \\ && \frac{\d t}{\d x} &= 1 + \frac{x+b}{\sqrt{x^2+2bx+c}} \\ &&&= \frac{x + \sqrt{x^2+2bx+c} + b}{\sqrt{x^2+2bx+c}} \\ &&&= \frac{t+b}{t-x} \\ \Rightarrow && \frac{\d x}{\d t} &= \frac{t-x}{t+b} \\ \\ && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= \int \frac{1}{t-x} \frac{t-x}{t+b} \d t \\ &&&= \int \frac{1}{t+b} \d t \\ &&&= \ln (t + b) +C \\ &&&= \ln \left (x + \sqrt{x^2+2bx+c} + b \right) + C \end{align*} If \(b^2 = c\) then we have \(x^2+2bx+b^2 = (x+b)^2\) so \(\sqrt{x^2+2bx+c^2} = x+b\) (if \(x+b>0\)), so \begin{align*} && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= \int \frac{1}{x+b} \d x\\ &&&= \ln (x + b) + C \\ &&&= \ln(x+b) + \ln 2 + C' \\ &&&= \ln (2(x+b)) + C' \\ &&&= \ln \left(x + b + \sqrt{(x+b)^2} \right)+C'\\ &&&= \ln \left(x + b + \sqrt{x^2+2bx+c} \right)+C'\\ \end{align*} If \(x+b < 0\) then the antiderivative is \(\ln 0\). \begin{align*} && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= -\int \frac{1}{x+b} \d x\\ &&&= -\ln |x + b| + C \\ \end{align*} which are clearly different.

Solution: \begin{align*} && f(g(x)) &= (g(x))^2 + p(g(x)) + q \\ &&&= (x^2+rx+s)^2 + p(x^2+rx+s) + q \\ &&&= x^4 + 2rx^3 + (2s+r^2+p)x^2 +(2rs+pr)x + (s^2+ps+q) \end{align*} So we need \(2r=a ,2s+r^2+p = b, r(2s+p) = c\). (We have full control over \(d\) since we can always chance \(q\) only affecting \(d\). \begin{align*} && r &= \frac{a}{2} \\ && b-r^2 & =rc \\ && b - \frac{a^2}{4} & =\frac{ac}{2} \\ \Rightarrow && 4b-a^2&= 2ac \end{align*} Clearly this condition is necessary. It is sufficient since if it is true the equations are solveable. \((x^2+vx+w)^2 = x^4 + 2vx^3 + (2vw+v^2)x^2+2vw x + w^2\). We don't care about the constant term since we can control this with \(k\), so we just need to check \(4(2vw+v^2) - (2v)^2 = 8wv\) so this does satisfy the condition. The reverse is also clear. \begin{align*} && 0 &= x^4-4x^3+10x^2-12x+4 \\ &&&= (x^2-2x+3)^2-5 \\ \Rightarrow && 0 &= x^2 - 2x+3 \pm \sqrt{5} \\ && x &= \frac{2 \pm \sqrt{4 - 4(3 \pm \sqrt{5})}}{2} \\ &&&= 1 \pm \sqrt{\mp \sqrt{5} -2} \\ &&& = 1 \pm \sqrt{\sqrt{5}-2}, 1 \pm i\sqrt{\sqrt{5}+2} \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:

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