Problems

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Solution: \begin{align*} y &= x^2-2bx+c \\ &= (x-b)^2+c-b^2 \end{align*} Therefore the turning point is at \((b,c-b^2)\)

Therefore there will be two distinct roots if \(c -b^2 < 0 \Rightarrow c < b^2\). For the equation to have two distinct positive roots it needs to have two distinct roots (ie the condition above) and \(y(0) = c\) needs to be positive, ie \(c > 0\). Therefore the curve has two distinct positive roots if \(0 < c < b^2\). The turning points on \(y = x^3-3b^2x+c\) will have \(0 = y' = 3x^2-3b^2 \Rightarrow x = \pm b\) Therefore for the cubic to have three distinct real root we must have a root between the turning points, \(y(-b) > 0 > y(b)\) \(b^3-3b^3+c = c-2b^3 < 0\) \((-b)^3+3b^3+c = c+2b^3 > 0\) ie \(-2b^3 < c < 2b^3\). The equation \(\l x - a \r^3 - 3b^2 \l x - a \r + c = 0\) will have 3 distinct roots if \(-2b^3 < c < 2b^3\), they will all be positive if the \(y(0) < 0\) and \(a+b > 0\) (ie the first turning point is in the first quadrant, ie \(-a^3+3b^2a+c < 0, a+b>0\). \begin{align*} 2x^3 - 9x^2 + 7x - 1 &= 2(x^3-\frac92x^2+\frac72 x-\frac12) \\ &= 2((x-\frac{3}{2})^3-\frac{27}{4}x+\frac{27}{8}+\frac72x-\frac12) \\ &= 2((x-\frac{3}{2})^3-\frac{13}{4}x+\frac{23}{8}) \\ &= 2((x-\frac{3}{2})^3-\frac{13}{4}(x-\frac{3}{2})-\frac{39}{8}+\frac{23}{8}) \\ &= 2((x-\frac{3}{2})^3-\frac{13}{4}(x-\frac{3}{2})-2) \\ \end{align*} Therefore in our notation \(a = \frac32, b = \sqrt{13/12}, c = 2\). Clearly \(a+b > 0\), so we need to check \(|c| < 2b^3\) which is clearly true as \(b > 1\). Finally we need to check \(y(0) = -1\), so all conditions are satisfied and there are 3 distinct roots.

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Solution:

1. If \(a > 0\) and \(b^2 < 4ac \Rightarrow \Delta < 0\) then \(f(x) = ax^2-bx+c > 0\) for all \(x\). Therefore \begin{align*} && 0 & < \int_0^1 (ax^2-bx+c) \d x\\ &&&= \frac13 a-\frac12b+c \\ \Rightarrow && 3b &< 2a+6c \end{align*}
2. Similar logic tells us this must also be always positive, therefore \begin{align*} && 0 &< \int_0^{\pi} (a \sin^2 x - b \sin x +c ) \d x\\ &&&= \frac{\pi}{2}a - 2b+\pi c \\ \Rightarrow && 4b &< (a+2c)\pi \end{align*}
3. Similar logic shows that \(f(x) = \frac{a}{x^2}-\frac{b}{x} +c>0\), therefore \begin{align*} && 0 &< \int_p^q \left (\frac{a}{x^2} - \frac{b}{x} + c\right) \d x \\ &&&=a\left (\frac{1}{p} - \frac{1}{q} \right) - b(\ln q - \ln p)+c(q-p) \\ \Rightarrow && b \ln (q/p) &,< a\left (\frac{1}{p} - \frac{1}{q} \right) +c(q-p) \end{align*}

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Solution: \begin{align*} && f'(x) &= \frac12 ax^{-1/2}-\frac12(x-b)^{-1/2} \\ \Rightarrow f'(x) = 0: && 0 &= \frac{a\sqrt{x-b}-\sqrt{x}}{\sqrt{x(x-b)}} \\ \Rightarrow && x &= a^2(x-b)\\ \Rightarrow && x &= \frac{a^2b}{a^2-1} \\ && f(x) &= a^2 \sqrt{\frac{b}{a^2-1}} - \sqrt{\frac{a^2b}{a^2-1}-b} \\ &&&= a^2 \sqrt{\frac{b}{a^2-1}} - \sqrt{\frac{b}{a^2-1}} \\ &&&= \sqrt{b(a^2-1)} \end{align*}

If \(c\) is below the turning point, ie \(c^2 < b(a^2-1)\) there is no solution. If \(c^2 = b(a^2-1)\) there is exactly one solution. If \(b(a^2-1) < c^2 < (f(b))^2 = a^2b\) then there are two solutions, otherwise there is exactly one solution.

1. \(c^2 = 16\), \(2 \cdot (3^2-1) = 16\), so we should have exactly one solution at \(x = \frac{3^2 \cdot 2}{3^2 -1 } = \frac{9}{4}\)
2. \(c^2 = 25\) and \(3 \cdot (3^2 - 1) = 24, 3 \cdot (3^2) = 27\), so we look for two solutions. \begin{align*} && 5 & = 3 \sqrt{x} - \sqrt{x-3} \\ \Rightarrow && 25 &= 9x+x-3-6\sqrt{x(x-3)} \\ \Rightarrow && 3\sqrt{x(x-3)} &= 5x-14 \\ \Rightarrow && 9x(x-3) &= 25x^2-140x+196 \\ \Rightarrow && 0 &= 16x^2-113x+196 \\ &&&= (x-4)(16x-49) \\ \Rightarrow && x &= 4, \frac{49}{16} \end{align*}

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Solution:

Since there are two turning points the derivative (a quadratic) has two distinct real roots. \begin{align*} && f'(x) &= 3x^2-2(p+q+r)x+(pq+qr+rp) \\ && 0 &< \Delta = 4(p+q+r)^2 - 4\cdot 3(pq+qr+rp) \\ \Rightarrow && (p+q+r)^2 &> 3(pq+qr+rp) \end{align*} If \(g^2 > 4h\) then \(p(x) = (x^2+gx+h)(x-k)\) has at least 2 real roots (possibly one repeated, and in particular it has two turning point, ie \begin{align*} && p'(x) &= (2x+g)(x-k)+(x^2+gx+h) \\ &&&= 3x^2+(2g-2k)x + (h-kg) \\ && 0 &< \Delta = 4(g-k)^2 - 4\cdot 3 (h-gk) \\ \Rightarrow && (g-k)^2 &> 3(h-gk) \end{align*} Pick \(g = h = 1\) and \(k = 1000\) then \((-999)^2 > 0 > 3(1-1000)\) so it is sufficient but not necessary.

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Solution: If \(p \geq 0\) then the derivative is \(x^2+p \geq 0\) and in particular the function is increasing. Therefore it will have exactly \(1\) real root (as for very large negative \(x\) it is negative, and vice-versa fo positive \(x\)). \begin{align*} && \frac{\d y}{\d x} &= \frac{\dot{y}}{\dot{x}} \\ &&&= \frac{2a}{2at} \\ &&&= \frac{1}{t} \\ \text{eq of normal} && \frac{k-2at}{h-at^2} &= -t \\ \Rightarrow && k-2at &= at^3-th \\ && 0 &= at^3+(2a-h)t-k \end{align*} Since \(a > 0\) this is the same constraint as the first part, in particular \(2a-h \geq 0 \Leftrightarrow 2a \geq h\). If exactly two normals can be drawn to \(C\) we must have that our equation has a repeated root, ie \begin{align*} && 0 &= at^3+(2a-h)t-k\\ && 0 &= 3at^2+2a-h\\ \Rightarrow && 0 &= 3at^3+ 3(2a-h)t-3k \\ && 0 &= 3at^3+(2a-h)t \\ \Rightarrow && 0 &= 2(2a-h)t-3k \\ \Rightarrow && t &= \frac{3k}{2(2a-h)} \\ \Rightarrow && 0 &= 3a \left (\frac{3k}{2(2a-h)} \right)^2+2a-h \\ && 0 &= 27ak^2+4(2a-h)^3 \end{align*}

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Solution: \begin{align*} && x+ c &= f(x) \\ \Rightarrow && (x+c)(x^2-1) &= x(x-2)(x-a) \\ \Rightarrow && x^3 + cx^2-x-c &= x^3-(2+a)x^2+2ax \\ \Rightarrow && 0 &= (c+2+a)x^2-(1+2a)x-c \\ && 0 &\leq \Delta = (1+2a)^2 + 4(2+c+a)c \\ &&&= 4c^2+(4a+8)c + (1+2a)^2 \\ && \Delta_c &= 16(a+2)^2-16(1+2a)^2 \\ &&&= 16(1-a)(3a+3) \\ &&&= 48(1-a^2) \end{align*} Therefore if \(|a| \geq 1\) we must have \(\Delta_c \leq 0\) which means \(\Delta \geq 0\) and so there are always solutions. If \(|a| < 1\) there are values for \(c\) where \(\Delta < 0\) and there would be no solutions. \begin{align*} && y &= \frac{x^3-(2+a)x^2+2ax}{x^2-1} \\ &&&= \frac{(x^2-1)(x-(2+a))+(2a+1)x-(2+a)}{x^2-1} \\ &&&= x - (2+a) + \frac{(2a+1)x-(2+a)}{x^2-1} \end{align*} therefore the oblique asymptote has equation \(y = x - (2+a)\)

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Solution:

1. \(\,\) \begin{align*} && 0 &= nx^2 + 2\sqrt{3n^2+50}x + 2n + 15 \\ && 0 &> \Delta = 4(3n^2+5) - 4\cdot n \cdot (2n + 15) \\ \Leftrightarrow && 0 &> n^2-15n+5\\ \text{cv}: && n &= \frac{15 \pm \sqrt{225 - 20}}{2} \\ &&&\approx \frac{15\pm14.x}{2}\\ \Leftrightarrow &&n &\in \{1, 2, \cdots, 14\} \end{align*}
2. \(\,\) \begin{align*} && 0 &> \Delta = 4(pn^2+q) - 4\cdot n \cdot (rn+s) \\ \Leftrightarrow && 0&>(p-r)n^2-sn+q \end{align*} Which is always true if \(r > p\) and \(s^2 < 4q(p-r)\)
3. \(\,\) \begin{align*} && 0 &= x^2 + 2\sqrt{p+q}x+ r+s \\ && 0 &\leq \Delta = 4(p+q) - 4(r+s) \\ && 0 &\leq 1 + s^2/8 - s \\ \text{c.v}: && s &= \frac{1 \pm \sqrt{1-4 \cdot \frac{1}{8}}}{2\cdot\frac18} \\ &&&= 4 \pm 4\sqrt{\frac12} \\ &&&= 4 \pm 2\sqrt{2} \\ \Rightarrow && s \leq 4 - 2\sqrt{2} &\text{ or } s \geq 4 + 2\sqrt{2} \end{align*}

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Solution: Suppose it is fired with angle to the horizontal \(\alpha\), then \begin{align*} \rightarrow: && x &= u\cos \alpha \cdot t \\ \uparrow: && h &= u \sin \alpha \cdot t - \frac12 g t^2 \\ \Rightarrow && u\cos \alpha &= \frac{x}{t} \\ && u \sin \alpha &= \frac{h + \frac12 gt^2}{t} \\ \Rightarrow && u^2 &= \frac{x^2}{t^2} + \frac{(h + \frac12 gt^2)^2}{t^2} \\ \Rightarrow && 0 &= x^2+h^2-u^2t^2+ght^2+\tfrac14 g^2 t^4 \\ &&&= \tfrac14 g^2 t^4 - (u^2 - gh)t^2 + h^2 + x^2 \end{align*} For a distance \(x\) to be achievable there must be a root to this quadratic in \(t^2\), ie \begin{align*} && 0 &\leq \Delta = (u^2-gh)^2 - 4 \cdot \tfrac14 g^2 (h^2 + x^2) \\ \Rightarrow && x^2 &\leq \frac{(u^2-gh)^2}{g^2} - h^2 \\ &&&= \frac{u^4+g^2h^2 - 2ghu^2-g^2h^2}{g^2} \\ &&&= \frac{u^2(u^2-2gh)}{g^2} \\ \Rightarrow && x &\leq \frac{u\sqrt{u^2-2gh}}{g} \end{align*} This is achieved when \begin{align*} && t^2 &= \frac{u^2-gh}{\tfrac12g^2}\\ &&&= \frac{2(u^2-gh)}{g^2} \\ \Rightarrow && \cos \alpha &= \frac{u\sqrt{u^2-2gh}}{g} \cdot \frac{g}{\sqrt{2(u^2-gh)}} \frac{1}{u} \\ &&&= \frac{1}{\sqrt{2}} \end{align*} ie when \(\alpha = \frac{\pi}{4}\)

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Solution: \(p = a+b+c, q = ab+bc+ca, r = abc\)

1. Suppose two roots are equal to each other, this means that one of the roots is also a root of the derivative. ie \begin{align*} && 0 &= x^3+qx - r \\ && 0 &= 3x^2+q \end{align*} have a common root, but this root must satisfy \(x^2 = -\frac{q}{3}\). Then \begin{align*} &&0 &= x^3 + qx - r \\ &&&= x^3 -3x^3 - r \\ &&&= -2x^3 -r \\ \Rightarrow && r^2 &= 4x^6 \\ &&&= 4 \left ( -\frac{q}{3}\right)^3 \\ \Rightarrow && 0 &= 27r^2+4q^3 \end{align*}
2. Consider \(x = z + \frac{p}{3}\), then the equation is: \begin{align*} x^{3}-px^{2}+qx-r &= (z + \frac{p}{3})^3 - p(z + \frac{p}{3})^2 + q(z + \frac{p}{3}) - r \\ &= z^3 + pz^2 + \frac{p^2}{3}z + \frac{p^3}{27} - \\ &\quad -pz^2-\frac{2p^2}{3}z-\frac{p^3}{9} + \\ &\quad\quad qz + \frac{pq}{3} - r \\ &= z^3+\left (\frac{p^2}{3}-\frac{2p^2}{3}+q \right)z + \left (\frac{p^3}{27}-\frac{p^3}{9}+\frac{pq}{3}-r \right) \\ &= z^3+\left (-\frac{p^2}{3}+q \right)z + \left (-\frac{2p^3}{27}+\frac{pq}{3}-r \right) \\ \end{align*} Since this equation must also have repeated roots we must have: \begin{align*} 4\left (-\frac{p^2}{3}+q \right)^3 + 27 \left (-\frac{2p^3}{27}+\frac{pq}{3}-r \right)^2 = 0 \end{align*} which is exactly our desired result

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Solution: For \(h(x)\) to be written in this form \(b^2=4ac\). Suppose \(f(x) = 3x^2+4x\), \(g(x) = x^2-2\). then, \begin{align*} && f(x) + \lambda g(x) &= (3+\lambda)x^2+4x - 2 \lambda \\ \Rightarrow && 0 &= 16 + 8(3+\lambda) \lambda \\ \Rightarrow && 0 &= 2+ 3 \lambda + \lambda^2 \\ &&&= (\lambda +1)(\lambda + 2) \\ \Rightarrow && \lambda &= -1 , -2 \\ \end{align*} \begin{align*} && f(x) - g(x) &= 2(x+1)^2 \\ && f(x) -2g(x) &= (x+2)^2 \\ \Rightarrow && g(x) &= 2(x+1)^2 - (x+2)^2 \\ && f(x) &= 4(x+1)^2 - (x+2)^2 \end{align*} Suppose \(f(x) = 3x^2+4x, g(x) = x^2 + \alpha\), then \begin{align*} && f(x) + \lambda g(x) &= (3+\lambda)x^2+4x+\lambda \alpha \\ \Rightarrow && 0 &= 16 -2\lambda \alpha(\lambda + 3) \\ && 0 &= \alpha \lambda^2 +3\lambda-8 \\ \Rightarrow && 0 &= 9 +32 \alpha \\ \Rightarrow && \alpha &= -\frac{9}{32} \end{align*}

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Solution:

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\((i)\) \((v)\) \((vii)\)