Problems

Solution:

1. 

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   Since \(y(0) < 0\) and \(y(\pm \infty) > 0\) we must cross the axis twice. Therefore there are two distinct real roots. It is not necessary, for example \((x-2)(x-3)\) has distinct real roots by the constant term is \(6 > 0\)
2. For \(x^2+bx+c=0\) to have distinct, positive real roots we need \(\Delta > 0\) and \(\frac{-b -\sqrt{\Delta}}{2a} > 0\) where \(\Delta = b^2-4ac\), ie \(b < 0\) and \(b^2 > \Delta = b^2-4ac\) or \(4ac > 0\). Therefore we need \(b^2-4ac > 0, b < 0, 4ac > 0\)
3. Since \(q < 0\) at least one of the roots is positive. The gradient is \(3x^2+p > 0\) therefore there is exactly one positive root. If \(p < 0\) then there are turning points when \(3x^2+p = 0\) ie \(x = \pm \sqrt{\frac{-p}{3}}\). If the first turning point is above the \(x\)-axis then there will be 3 roots. If it is on the \(x\)-axis then 2, otherwise only 1. \begin{align*} y &= \left (-\sqrt{\frac{-p}{3}}\right)^3 + p\left (-\sqrt{\frac{-p}{3}}\right)+q \\ &= \sqrt{\frac{-p}{3}} \left (p - \frac{p}{3} \right) + q \\ &= \frac{2}{3} \sqrt{\frac{-p}{3}}p +q \\ \end{align*} Therefore it is positive if \(-\frac{4}{27}p^3 >q^2\) ie if \(4p^3+27q^2 < 0\)

Solution:

1. Suppose \(a^2-2b^2=1\) then \begin{align*} (3a+4b)^2-2(2a+3b)^2 &= 9a^2+24ab+16b^2-2\cdot(4a^2+12ab+9b^2) \\ &=a^2-2b^2 \\ &= 1 \end{align*} Therefore \((3a+4b,2a+3b)\) also lies on the curve.
2. Suppose \(Ma^2-Nb^2 = 1\) then \begin{align*} M(5a+6b)^2-N(4a+5b)^2 &= M\cdot(25a^2+60ab+36b^2) - N\cdot(16a^2+40ab+25b^2) \\ &= (25M-16N)a^2+20\cdot(3M-2N)ab+(36M-25N)b^2 \end{align*} Therefore we need \(3M = 2N\) so the smallest possible value would have to be \(M = 2, N = 3\), which does work
3. Consider \(x + \sqrt{3}y\), then consider \((x+\sqrt{3}y)(2+\sqrt{3}) = (2x+3y)+(x+2y)\sqrt{3}\). Notice that \((x+\sqrt{3}y)(x-\sqrt{3}y) = 1\) and \((2+\sqrt{3})(2-\sqrt{3}) = 1\) so \(((2x+3y)+(x+2y)\sqrt{3})((2x+3y)-(x+2y)\sqrt{3}) = 1\), so we can take \(P=2,Q=3,R=1,S=2\)

Solution:

1. First notice that \(\frac{1}{5+\sqrt{24}} = \frac{5-\sqrt{24}}{25-24} = 5 - \sqrt{24}\), hence \begin{align*} && ( 5 + \sqrt {24})^4 + \frac{1 }{(5 + \sqrt {24})^4} &= ( 5 + \sqrt {24})^4 + ( 5 - \sqrt {24})^4 \\ \end{align*} where clearly all terms including \(\sqrt{24}\) will cancel out, therefore it is an integer. \begin{align*} && 5 + \sqrt{24} &< 5 + 5 = 10 \\ \Rightarrow && \frac{1}{5+\sqrt{24}}& > \frac{1}{10} = 0.1 \\ && 2(5 + \sqrt{24}) &=10 + \sqrt{96} > 19 \\ \Rightarrow && \frac{1}{5+\sqrt{24}} & < \frac{2}{19} < \frac{2}{18} = \frac19 = 0.11111\ldots < 0.11 \end{align*} Therefore, \(10^{-4} < (5+\sqrt{24})^4 < 0.11^{-4} = 0.00014641\) \begin{align*} && (5+\sqrt{24})^4 + (5-\sqrt{24})^4 &= 2(5^4+6\cdot5^2\cdot24+24^2) \\ &&&= 2\cdot (625 + 3600+576) \\ &&&= 9602 \\ \Rightarrow && (5+\sqrt{24})^4 &= 9602 - \epsilon, \epsilon \in (0.0001, 0.00014641) \\ \Rightarrow && (5+\sqrt{24})^4 &\in (9601.999854, ,9601.9999) \\ \Rightarrow && (5+\sqrt{24})^4 &= 9601.9998 \, (4 \text{ d.p.}) \end{align*}
2. Notice that \((N+\sqrt{N^2-1})^{k}+(N-\sqrt{N^2-1})^{k}\) is an integer for the same reason as before (sum of conjugates). Notice also that \(\frac{1}{N+\sqrt{N^2-1}} = N - \sqrt{N^2-1}\) and that so it sufficies to show that \begin{align*} && N + \sqrt{N^2-1} &> 2N-\tfrac12 \\ \Leftrightarrow && \sqrt{N^2-1} &> N - \tfrac12 \\ \Leftrightarrow && N^2-1 &> N^2-N+1\\ \Leftrightarrow && N &> \tfrac32\\ \end{align*} Which is true since \(N > 1\) and \(N\) is an integer.

Solution: Consider \(\begin{pmatrix} a \\ b \\ c \end{pmatrix}\), \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\), then we know that \begin{align*} && \begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} &= \sqrt{a^2+b^2+c^2} \sqrt{x^2+y^2+z^2} \cos \theta \\ \Rightarrow && (ax+by+cz)^2 &= (a^2+b^2+c^2)(x^2+y^2+z^2) \cos^2 \theta \\ &&&\leq (a^2+b^2+c^2)(x^2+y^2+z^2) \end{align*} For equality to hold, we must have that the vectors are parallel, ie \(\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)

1. By applying our inequality from the first part with \(a=1, b = 2, c=2\) we have \((x+2y+2z)^2 \leq (1+2^2+2^2)(x^2+y^2+z^2) = 9(x^2+y^2+z^2)\)
2. Since \begin{align*} && (p^2+(2q)^2+(3r)^2)\left (8^2 +4^2+1^2 \right) &\geq (8p+8q+3r)^2 \\ \Leftrightarrow && 729 \cdot 81 &\geq 243^2 \\ &&3^6 \cdot 3^4 &\geq 3^{10} \end{align*} Therefore we must be in the equality case, ie \(p = 8\lambda, 2q = 4\lambda, 3r = \lambda\) as well as \(64\lambda + 16\lambda +\lambda = 243 \Rightarrow 81\lambda = 243 \Rightarrow \lambda = 3\) so we have \[ (p,q,r) = \left (24, 6, 1 \right) \]

Solution: \begin{align*} &&2f(x) &\equiv f(x) + f(x) \\ &&&\equiv f(x+x) \\ &&&\equiv f(2x) \\ \\ \Rightarrow && 2f(0) &= f(0) \\ \Rightarrow && f(0) &= 0 \\ && f''(0) &= \lim_{h \to 0} \frac{f(2h)-2f(0)+f(-2h)}{h^2} \\ &&&= \lim_{h \to 0} \frac{f(2h)+f(-2h)}{h^2} \\ &&&= \lim_{h \to 0} \frac{f(0)}{h^2} \\ &&&= 0 \\ \Rightarrow && f''(0) &= 0 \end{align*} If \(f(x)\) satisfies the equation, then \(f'(x)\) satisfies the equation. In particular this means that \(f^{(n)}(0) = 0\) for all \(n \geq 2\). Therefore the only non-zero term in the Maclaurin series is \(x^1\). Therefore \(f(x) = cx\)

1. Suppose \(g(x)g(y) \equiv g(x+y)\), then if \(G(x) = \ln g(x)\) we must have \(G(x)+G(y) \equiv G(x+y)\), ie \(G(x) = cx \Rightarrow g(x) = e^{cx}\)
2. Suppose \(h(x)+h(y) \equiv h(xy)\), then if \(h(e^u) = H(u)\) we must have that \(H(u)+H(v) \equiv h(e^u) + h(e^v) \equiv h(e^{u+v}) \equiv H(u+v)\).Therefore \(H(u) = cu\), ie \(h(e^u) = cu\) or \(h(x) = h(e^{\ln x}) = c \ln x\).
3. Finally if \(t(x) + t(y) \equiv t(z)\), the considering \(T(w) = t(\tan w)\) then \(T(x) + T(y) \equiv t(\tan x) + t(\tan y) \equiv t( \frac{\tan x + \tan y}{1- \tan x \tan y}) \equiv t (\tan (x+y)) \equiv T(x+y)\). Therefore \(T(x) = cx\) Therefore \(t(\tan w) = c w \Rightarrow t(x) = c \tan^{-1} x\)

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:

1. \(\,\) \begin{align*} && 1 &= \frac{x}{x-a} + \frac{x}{x-b} \\ \Leftrightarrow && (x-a)(x-b) &= x(2x-a-b) \\ \Leftrightarrow && 0 &= x^2-ab \end{align*} Therefore if \(a,b\) are both positive or both negative, \(ab > 0\) and there are two distinct solutions \(x = \pm \sqrt{ab}\)
2. \(\,\) \begin{align*} && 1+c &= \frac{x}{x-a} + \frac{x}{x-b} \\ \Leftrightarrow && (1+c)(x-a)(x-b) &= x(2x-a-b) \\ \Leftrightarrow && 0 &= (c-1)x^2-c(a+b)x+ab(1+c) \\ \\ && 0 &= \Delta = c^2(a+b)^2 - 4 \cdot(c-1)\cdot ab(1+c) \\ &&&= c^2(a+b)^2-4ab(c^2-1) \\ &&&= c^2 ((a+b)^2-4ab)+4ab \\ &&&= c^2(a-b)^2+4ab \\ \Rightarrow && c^2 &= -\frac{4ab}{(a-b)^2} \\ &&&= -\frac{(a+b)^2-(a-b)^2}{(a-b)^2} \\ &&&= 1 - \left ( \frac{a+b}{a-b} \right)^2 \end{align*} Note that \(c^2 \geq 0\) and \(1-x^2 \leq 1\) so \(0 \leq c^2 \leq 1\). \(c^2 = 0 \Rightarrow ab = 0\), but this is not possible since \(a,b \neq 0\), therefore \(0 < c^2 \leq 1\)

Solution:

1. \(f(12) = f(2^2 \cdot 3) = 12 \cdot (1-\frac12)\cdot(1-\frac13) = 12 \cdot \frac12 \cdot \frac 23 = 4\) \(f(180) = f(2^2 \cdot 3^2 \cdot 5) = 180 \cdot \frac12 \cdot \frac23 \cdot \frac 45 = 12 \cdot 4 = 48\) Suppose \(N\) has prime decomposition \(p_1^{a_1} \cdots p_k^{a_k}\), then \(f(N) = p_1^{a_1} \cdots p_k^{a_k} (1 - \frac{1}{p_1})\cdots ( 1- \frac{1}{p_k}) = p_1^{a_1-1} \cdots p_k^{a_k-1}(p_1-1) \cdots (p_k-1)\) which is clearly an integer.
2. \(f(2) = 1, f(4) = 2 \neq f(2) \cdot f(2)\) If \(p, q\) are distinct primes then \(f(p) = p \cdot \frac{p-1}{p} = p-1\) and \(f(q) = q-1\). \(f(pq) = pq \frac{p-1}{p} \cdot \frac{q-1}{q} = (p-1)(q-1) = f(p)f(q)\) \(f(12) = 4 = 2\cdot 2 = f(4) \cdot f(3)\)
3. \(f(p^m) = p^{m-1} (p-1)\) \(146410 = 10 \cdot 14641 = 10 \cdot 11^4\). Therefore \(p = 11, m = 5\)

Solution: \begin{align*} && \tan \left (\arctan\left(\frac1 {a+b}\right)+ \arctan\left(\frac1 {a+c}\right) \right) &= \frac{\frac1{a+b}+\frac1{a+c}}{1-\frac{1}{(a+b)(a+c)}} \\ &&&= \frac{a+c+a+b}{(a+b)(a+c)-1} \\ &&&= \frac{2a+b+c}{a^2+ab+ac+bc-1} \\ &&&= \frac{2a+b+c}{2a^2+ab+ac} \\ &&&= \frac{1}{a} \\ &&&= \tan \arctan \frac1a\\ \Rightarrow && \arctan\left(\frac1 {a+b}\right)+ \arctan\left(\frac1 {a+c}\right) &= \arctan \frac{1}{a} + n \pi \end{align*} Since \(\arctan x \in (-\frac{\pi}{2}, \frac{\pi}{2})\) the LHS \(\in (0, \pi)\) so \(n = 0\). \begin{align*} a=p+q, b = s, c = t:&& \arctan \! \!\left(\!\frac1 {p+q+s}\!\right) + \arctan \! \!\left(\!\frac 1{p+q+t}\!\right) &= \arctan \left ( \frac{1}{p+q} \right) \\ a=p+r, b= u, c = v && \arctan \! \!\left(\!\frac 1 {p+r+u}\!\right) + \arctan \! \!\left(\!\frac1 {p+r+v}\!\right) &= \arctan \! \!\left(\!\frac1 {p+r}\!\right) \\ a = p, b = q, c = r:&& \arctan \left ( \frac{1}{p+q} \right) +\arctan \! \!\left(\!\frac1 {p+r}\!\right) &= \arctan \left ( \frac1p \right) \end{align*} and the result follows. Taking \(p = 7\) we need to solve \[ \begin{cases} q+s &= 6 \\ q+t &= 14 \\ r+u &= 75 \\ r+v &= 180 \end{cases} \] also satisfying \(qr = 50\) etc, so say \(q = 1, r = 50, s = 5, v=25\)

Solution: \begin{align*} u = x^m, \d u = m x^{m-1} && \int \frac{m}{x f(x^m)} \d x &= \int \frac{m x^{m-1}}{uf(u)} \d x \\ &&&= \int \frac{1}{u f(u)} \d u \\ &&&= F(u) + c \\ &&&= F(x^m) + c \end{align*}

1. \begin{align*} && \int \frac{1}{u(u-1)} \d u &= \int \left ( \frac{1}{u-1}-\frac{1}{u} \right ) \d u \\ &&&= \ln \left ( \frac{u-1}{u} \right) + c \\ &&&= \ln \left ( 1 - \frac{1}{u} \right) + c \\ && \int \frac{1}{x^n - x} \d x &= \int \frac{1}{x (x^{n-1}-1)} \d x \\ f(u) = u - 1: && &= \frac{1}{n-1} \ln \left ( 1 - \frac{1}{x^{n-1}} \right) + c \end{align*}
2. \begin{align*} v = \sqrt{u+1}, \d v = \tfrac12 (u+1)^{-1/2} \d u && \int \frac{1}{u\sqrt{u+1}} \d u &= \int \frac{1}{(v^2-1)} (u+1)^{-1/2} \d u \\ &&&= \int \frac{2}{v^2-1} \d v \\ &&&=\ln \frac{1-v}{1+v} + c \\ &&&= \ln \left (\frac{1-\sqrt{u+1}}{1+\sqrt{u+1}} \right)+ c \\ f(u) = \sqrt{x+1}:&& \int \frac{1}{\sqrt{x^n + x^2}} \d x &= \int \frac{1}{x\sqrt{x^{n-2}+1}} \d x \\ &&&= \frac{1}{n-2} \ln \left ( \frac{1-\sqrt{x^{n-2}+1}}{1+\sqrt{x^{n-2}+1}} \right)+c \end{align*}

Solution: Suppose \begin{align*} && \frac{1}{a+b} &= \frac{1}{a} + \frac{1}{b} \\ \Rightarrow && ab &= b(a+b)+a(a+b) \\ &&&= (a+b)^2 \\ \Rightarrow && 0 &= a^2+ab + b^2 \\ &&&= \tfrac12 (a^2+(a+b)^2+b^2) \end{align*} Which clearly has no solution for non-zero \(a,b\). Suppose \begin{align*} && \frac{1}{a+b+c} &= \frac1a + \frac1b+\frac1c \\ \Leftrightarrow && abc &= (a+b+c)(bc+ca+ab) \\ \Leftrightarrow && 0 &= (a+b+c)(bc+ca+ab) - abc \\ &&&= (a+b)(b+c)(c+a) \end{align*} Therefore it is true iff \(a+b = 0\) or \(b+c=0\) or \(c+a =0\) as required.

Solution: First note that \(10k - k^2 = 25 - (5-k)^2\) which is clearly bounded above by \(25\). The smallest it can be is when \(|5-k|\) is as large as possible, ie when \(k =0\) and we get a lower bound of \(0\). For \((k-1)k(k+1)\) notice this is the product of \(3\) consecutive integers, and therefore must be divisible by \(3\). (In fact, it's divisible by six, since \(\binom{k+1}{3}\) is the number of ways to choose \(3\) objects from \(k+1\). Let \(N = 100a + 10b + c\) where \(0 \leq a,b,c < 10\) and \(1 \leq a\). \(S = a + b^2 + c^3\) we want to find \begin{align*} && 100a +10b + c &= a + b^2 + c^3 \\ \Rightarrow && 0 &= \underbrace{99a}_{3 \mid 99 } + 10b - b^2 -\underbrace{c(c+1)(c-1)}_{3 \mid c(c+1)(c-1)} \\ \end{align*} Therefore \(3 \mid 10b - b^2 = b(10-b)\). Therefore \(3 \mid b\) or \(3 \mid 10-b\) so \(b = 0, 3, 6, 1, 4, 7\) We also have \(99a \geq 99\) and \(10b-b^2 \in [0, 25]\) so we need \(c^3-c \geq 99\), so \(c \geq 5\) Case \(c = 5\), Then \(c^3-c = 120\) so \(a = 1\) and \(10b-b^2 = 21\) so \(b= 3, 7\) \(N = 135, 175\) Case \(c = 6\), so \(c^3 - c = 210\) so \(a = 2\) and \(25-(5-k)^2 = 12\) so no solutions. Case \(c = 7\), so \(7^3 - 7 = 336\) so \(a = 3\) and \(25-(5-k)^2 = 39\) so no solutions. Case \(c = 8\) so \(8^3-8 = 504\) so \(a = 5\) and \(25-(5-k)^2 = 9\), so \(b = 1, 9\) and \(N = 518, 598\) Case \(c = 9\) so \(9^3 - 9 = 720\), so \(a = 7\) and \(25-(5-k)^2 = 27\) so no solutions. Therefore all the solutions are \(N = 135, 175, 518, 598\)