Problems

---

Solution:

1. \(L_a : \frac{y}{x-a} = \frac{1-a-0}{0-a} = \frac{a-1}{a} \Rightarrow ay+(1-a)x = a(1-a)\) \begin{align*} && ay + (1-a)x &= a(1-a) \\ && by + (1-b)x &= b(1-b) \\ \Rightarrow && aby + b(1-a)x &= ba(1-a) \\ && aby + a(1-b)x &= ab(1-b) \\ \Rightarrow && (b-a)x &= ab(b-a) \\ \Rightarrow && x &= ab \\ && y &= \frac{a-1}{a} \cdot a(b-1) \\ &&&= (1-a)(1-b) \end{align*}
2. As \(a \to b\), \(x \to a^2, y \to 1-2a+a^2 =(1-a)^2 = (1-\sqrt{x})^2\)
3. \(\frac{\d y}{\d x} = 2(1-\sqrt{x})\cdot \left (-\tfrac12 \frac{1}{\sqrt{x}} \right) = 1 - \frac{1}{\sqrt{x}}\). Therefore the tangent when \(x = c^2, y = (1-c)^2\) is \begin{align*} && \frac{y-(1-c)^2}{x-c^2} &= 1 - \frac{1}{c} \\ \Rightarrow && cy + (1-c)x &= c(c-1)+c(1-c)^2 \\ &&&= c(1-c) \end{align*} Which is an equation of the form \(L_c\)

---

Solution:

1. \(\,\)
   

   + - ↺
   Note that we have a right angled triangle, with the sides in a ratio of \(m\). So if our target length is \(x\) we have \(x^2 + (mx)^2 = c^2\) and so \(x = c(m^2+1)^{-\frac12}\)
2. The distance from the origin to \(L\) is \(a = c(m^2+1)^{-\frac12}\) so \begin{align*} && a^2(m^2+1) &= c^2 \\ && \frac{c-y(x)}{0-x} &= y' \\ \Rightarrow && c-y &= -xy' \\ \Rightarrow && a^2((y')^2+1) &= (y-xy')^2 \\ \\ && 2a^2y'y'' &= 2(y-xy')(y'-xy''-y') \\ &&&= 2(xy'-y)xy'' \\ \Rightarrow && y'' &= 0 \\ \text{ or } && 2a^2y' &= 2(xy'-y)x \end{align*} If \(y'' = 0\) then \(y = mx + c\) and the result follows immediately. \begin{align*} && 0 &= (a^2-x^2)y' + yx \\ \Rightarrow &&\frac1{y} y' &= -\frac{x}{a^2-x^2} \\ \Rightarrow && \ln y &= \frac12\ln (a^2-x^2) + K \\ \Rightarrow && y^2 &= M(a^2-x^2) \\ \Rightarrow && x^2 + y^2 &= a^2 \end{align*} Where in the last step we know the tangents from an ellipse are not all equidistant to the origin.
3. 

   + - ↺

---

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 \(\cos x = -1 \Rightarrow x = \pi\)
   

   + - ↺
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:

1. \(\,\)
   

   + - ↺
   \((a,0)\), \((b,0)\), \((b,\f(b))\) and \((a, \f(a))\) forms a rectangle.
2. There are no solutions if \(a < c < b\):
   

   + - ↺
   There is one solution if \(a=c\) or \(a = b\)
   

   + - ↺
   And there are two solution if \(c \not \in [a,b]\)
   

   + - ↺
   There is exactly one solution unless....
   

   + - ↺
   ... there are infinitely many solutions when the gradients line up perfectly, ie when \(a+b=c+d\)
   

   + - ↺

---

Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(T > t + \delta t | T > t) &= \frac{\mathbb{P}(T < t + \delta t)}{\mathbb{P}(T > t )} \\ &&&= \frac{\int_t^{t+\delta t} f(s) \d s}{1-F(t)} \\ &&&\approx \frac{f(t)\delta t}{1-F(t)} \\ &&&= h(t) \delta t \end{align*}
2. If \(F(t) = t/a\) then \(f(t) = 1/a\) and \(h(t) = \frac{1/a}{1-t/a} = \frac{1}{a-t}\)
   

   + - ↺
3. \(\,\) \begin{align*} && \frac{F'}{1-F} &= \frac{1}{t} \\ \Rightarrow && -\ln (1-F) &= \ln t + C\\ \Rightarrow && 1-F &= \frac{A}{t} \\ && F &= 1 - \frac{A}{t} \\ F(a) = 0: && F &= 1 - \frac{a}{t} \\ && f(t) &= \frac{a}{t^2} \end{align*}
4. (\(\Rightarrow\)) \begin{align*} && \frac{F'}{1-F} &= k \\ \Rightarrow && -\ln(1-F) &= kt+C \\ \Rightarrow && 1-F &= Ae^{-kt} \\ F(b) = 0: && 1 &= Ae^{-kb} \\ \Rightarrow && 1-F &= e^{-k(t-b)}\\ \Rightarrow && f &= ke^{-k(t-b)} \\ \end{align*} (\(\Leftarrow\)) \(f(t) = ke^{-k(t-b)} \Rightarrow F(t) = 1-e^{-k(t-b)}\) and the result is clear.
5. \(\,\) \begin{align*} && \frac{F'}{1-F} &= \left ( \frac{\lambda}{\theta^{\lambda}} \right) t^{\lambda-1} \\ \Rightarrow && -\ln(1-F) &= \left ( \frac{t}{\theta} \right)^{\lambda} +C\\ \Rightarrow && F &= 1-A\exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \\ F(0) = 0: && 0 &= 1-A \\ \Rightarrow && F &= 1 - \exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \\ \Rightarrow && f &= \lambda t^{\lambda -1} \theta^{-\lambda} \exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \end{align*}

---

Solution:

1. 

   + - ↺
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 \(n < \frac{c}{c-1} \leq n+1\)

---

Solution:

1. \(k = 0\), we have \(x^2 + y^2 + 3x + y = 0\), ie \((x+\tfrac32)^2+(y+\tfrac12)^2 = \frac{10}{4}\).
   

   + - ↺
2. \(3x^2 + 3y^2 +10xy = (3x+y)(x+3y)\) so \(x^2 + y^2 + \tfrac{10}3xy + 3x+y = (3x+y)(\frac{x+3y}{3}+1) = 0\) so we have the line pair \(3x +y =0\), \(x+3y + 3 = 0\)
   

   + - ↺
3. If \(k = 2\) then \((x+y)^2 + (x+y)+2x = 0\). If \(\theta = 45^\circ\) then \( X = \frac1{\sqrt{2}}(x+y), Y = \frac{1}{\sqrt{2}}(y-x)\), ie \(x+y = \sqrt{2}X\) and \(x = \frac{1}{\sqrt2}(X-Y)\), so our equation is: \begin{align*} 0 &= 2X^2 + \sqrt{2}X + \sqrt{2}(X-Y) \\ &= (\sqrt{2}X + 1)^2 - 1 - \sqrt{2} Y \end{align*} which would be a parabola with line of symmetry \(X = -\frac{1}{\sqrt{2}}\). However, we are actually looking at that parabola rotated by \(45^\circ\) anticlockwise.
   

   + - ↺

---

Solution:

1. \begin{align*} cb(x) &= c(b(x)) \\ &= 2 \ln x \quad (x > 0) \\ ab(x) &= (b(x))^2 \\ &= (\ln x)^2 \quad (x > 0) \\ da(x) &= \sqrt{a(x)} \\ &= \sqrt{x^2} \\ &= |x| \quad (-\infty < x < \infty) \\ ad(x) &= (d(x))^2 \\ &= (\sqrt{x})^2 \\ &= x \quad (x \geq 0) \end{align*} The domains are specified above. The ranges are \(\mathbb{R}, \mathbb{R}_{\geq 0}, \mathbb{R}_{\geq 0}, \mathbb{R}_{\geq 0}\) respectively.
2. \begin{align*} fg(x) &= \sqrt{g(x)^2-1} \quad (|g(x)| \geq 1) \\ &= \sqrt{x^2+1-1} \\ &= |x| \end{align*} So \(fg: \mathbb{R} \to \mathbb{R}_{\geq 0}\). \begin{align*} gf(x) &= \sqrt{f(x)^2 + 1} \\ &= \sqrt{\left ( \sqrt{x^2-1} \right)^2+1} \quad (|x| \geq 1) \\ &= |x| \end{align*} So \(gf: \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}\)
3. • 

     + - ↺
   • 

     + - ↺
   \begin{align*} kh(x) &= h(x) - \sqrt{h(x)^2 -1} \quad (|h(x)| \geq 1)\\ &= x + \sqrt{x^2+1} - \sqrt{(x + \sqrt{x^2+1})^2 - 1} \\ &= x + \sqrt{x^2+1} - \sqrt{x^2 + x^2 - 2x} \quad (x \geq 1) \\ &= x + \sqrt{x^2+1} - \sqrt{2x^2-2x} \quad (x \geq 1) \end{align*} This has domain \(x \geq 1\) and range, \((0, 1]\)

---

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. First notice that this cubic has leading first term \(1\) and three real roots, so it must have the shape:
   

   + - ↺
   With the \(x\)-axis running somewhere between the dashed lines. Since \(c < 0\), the \(y\)-axis must meet the curve below the \(x\)-axis, ie somewhere on the blue section of this curve:
   

   + - ↺
   Therefore there will be either \(1\) (if it meets it in the \(\cup\) area) or \(3\) (if it meets it on the far left) positive roots.
2. First notice that if \(c > 0\) we cannot have three positive real roots since the function would need to pass \(0\) between \(0\) and \(-\infty\). Secondly, notice both turning points must be larger than zero, ie \begin{align*} && 0 &= 3x^2 + 6ax + 3b \\ \Leftrightarrow && 0 &= (x+a)^2 + b - a^2 \end{align*} has both roots larger than zero, (and it needs to have two roots, so \(a^2 > b\) and \(-a > 0\), ie \(a < 0\). If \(b < 0\), then just looking at \(x^2+2ax+b\) we can see that it is \(<0\) at \(0\) and one of the roots will be negative, therefore \(c < 0\), \(a^2 > b > 0\) and \(a < 0\)
3. Since \(c > 0\) we can see that at least one root is negative.
   

   + - ↺
   ie the \(y\)-axis passes through an orange section of this curve. What now matters is where the larger turning point is. Considering \(x^2 + 2ax + b\), we notice that \(ab < 0\) means that \((x-\alpha)(x-\beta)\) we must have \((\alpha + \beta)\alpha \beta > 0\) which isn't possible if both roots are negative. Therefore the \(y\)-axis passes through the orange \(\cap\) and there are \(2\) positive real roots.
4. If we take \(a = 1, b = -1, c = 1\) then we have \(x^3 + 3x^2-3x+1\). This has turning points when \(x^2+2x-1 = 0\), ie \(x = -1 \pm \sqrt{2}\) Notice that \begin{align*} && y(-1\pm \sqrt2) &= (-1 \pm \sqrt{2})^3 + 3(-1 \pm \sqrt{2})^2-3(-1 \pm \sqrt{2}) + 1 \\ &&&= (-1\pm \sqrt{2}) \cdot (3 \mp 2\sqrt2) + 3(3 \mp \sqrt2) -3(-1\pm \sqrt2) + 1 \\ &&&= (-7 \pm 5 \sqrt2) + (9 \mp 3\sqrt2) +(3 \mp 3\sqrt2) + 1 \\ &&&= 24 \mp 16\sqrt2 = 8(3 \mp 2 \sqrt2) >0 \end{align*} ie both turning points are above zero and hence only one real root

---

Solution: \begin{align*} && y &= bx-ba \\ && 1 &= x^2 + y^2 \\ \Rightarrow && 1 &= x^2 + b^2(x-a)^2 \\ \Rightarrow && 0 &= (1+b^2)x^2-2ab^2x+b^2a^2-1 \end{align*} This will have roots which sum to \(\frac{2ab^2}{1+b^2}\), therefore \(M = \left ( \frac{ab^2}{1+b^2}, \frac{ab^3}{1+b^2}-ba \right)=\left ( \frac{ab^2}{1+b^2}, \frac{-ba}{1+b^2} \right)\)

1. Since \(b\) is fixed so is \(\frac{b}{1+b^2} = t\) and all the points are \((bta, -ta)\), ie \(x = -by\). If \(a \in [-1,1]\) we are ranging on the points \((bt, -t)\) to \((-bt, t)\) which is a distance of \begin{align*} && d &= \sqrt{(bt+bt)^2+(-2t)^2} \\ &&&= \sqrt{4(b^2+1)t^2} \\ &&&=2 \sqrt{(b^2+1)\frac{b^2}{(b^2+1)^2}} \\ &&&= \frac{2b}{\sqrt{b^2+1}} \end{align*}
   

   + - ↺
2. • If \(a\) is fixed we have \(\left ( \frac{ab^2}{1+b^2}, -\frac{ba}{1+b^2} \right)\) \begin{align*} && \frac{x}{y} &= - b \\ \Rightarrow && y &= \frac{a\frac{x}{y}}{1 + \frac{x^2}{y^2}} \\ \Rightarrow && y^2 \left ( 1 + \frac{x^2}{y^2} \right) &= ax \\ \Rightarrow && x^2-ax + y^2 &= 0 \\ \Rightarrow && \left (x - \frac{a}{2} \right)^2 + y^2 &= \frac{a^2}{4} \end{align*} Therefore we will end up with a circle centre \((\tfrac{a}{2}, 0)\) going through the origin.
     

     + - ↺
   • If \(ab = 1\), we have \(\left ( \frac{b}{1+b^2}, -\frac{1}{1+b^2} \right)\) \begin{align*} && \frac{x}{y} &= -b \\ \Rightarrow && y &= -\frac{1}{1+\frac{x^2}{y^2}} \\ \Rightarrow && y + \frac{x^2}{y} &= - 1 \\ \Rightarrow && y^2 +y+ x^2 &= 0 \\ \Rightarrow && \left ( y + \frac12 \right)^2 + x^2 &= \frac14 \end{align*}
     

     + - ↺

---

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*}
   

   + - ↺
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:
   

   + - ↺
   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*}
   

   + - ↺
   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. \(\,\)
   

   + - ↺
   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. \(\,\)
   

   + - ↺