Problems

Solution:

1. 

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2. \(2 \textrm{Min}(x^2,2x) = 5x-3\) tells us either \(2x^2 = 5x-3 \Rightarrow 2x^2 - 5x +3 = 0 \Rightarrow (2x-3)(x-1) \Rightarrow x = 1, \frac32\) and \(0 \leq x \leq 2\) or \(4x = 5x-3 \Rightarrow x= 3\) and \(x < 0\) or \(2 > x\), therefore our solutions are \(x = 1, \frac32, 3\)
3. We have different cases based on \(x\) vs \(-2, 0, 2\), ie Case \(x \leq -2\): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&= 2x + x^3 \end{align*} So \(2x = 2x + x^3 \Rightarrow x^3 = 0\), but \(x \leq -2\) so no solutions. or \(6x = 2x + x^3 \Rightarrow 0 = x(x^2-4) \Rightarrow x = 0, 2, -2\) so \(x = -2\). Case \(-2 < x \leq 0\): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&= 2x + 4x \end{align*} So \(2x = 2x + 4x\) ie \(x = 0\) which is valid. Or \(6x = 2x + 4x\) ie valid for all values in \(-2 \leq x \leq 0\) Case \(0 < x \leq 2\): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&= x^2 + x^3 \end{align*} So \(2x = x^2 + x^3 \Rightarrow 0 = x(x^2+x-2) = x(x-1)(x+2)\) so \(x = 0, 1, -2\), but the range means \(x = 0\) or \(x = 1\) Or \(6x = x^2 + x^3 \Rightarrow 0 = x(x^2+x-6) = x(x-2)(x+3)\) so \(x = 0, 2, -3\), but the range means \(x = 0\) or \(x = 2\) Case \(2 \leq x \): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&=2x + 4x \end{align*} So \(2x = 2x + 4x \Rightarrow x = 0\) so no solutions. Or \(6x = 2x + 4x\) so a range of solutions. Therefore the final solutions for \(m = 2\) are \(x = 0, x = 1\) and for \(m = 2\) are \(x \in [-2,0] \cup [2, \infty)\)
4. \(\mathrm{Min}(x^2, x^3)\) switches when \(x = 1\), so we must consider both limits: \begin{align*} && \frac{\d y}{\d x}\vert_{x > 1} &= 4x - 5 \\ \\ && \frac{\d y}{\d x}\vert_{x < 1} &= 6x^2 - 5 \\ \end{align*} so when \(x = 1\) the sign of the derivative changes from positive to negative, hence a local maximum. The other local maxima and minima will be when \(x = \frac54\) or \(x = \pm \sqrt{5/6}\)
   

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Solution:

1. The horizontal position of the particle at time \(t\) is \(u \sin\alpha t\), so \(T = \frac{h \tan \beta}{u \sin \alpha}\) The vertical position of the particle at this time \(T\) satisifes: \begin{align*} && h &= u \cos\alpha \frac{h \tan \beta}{u \sin\alpha} - \frac12 g \left ( \frac{h \tan \beta}{u \sin\alpha} \right)^2 \\ &&&= h\cot \alpha \tan \beta - \frac{gh^2}{2u^2} \tan^2 \beta \cosec^2 \beta \\ \Rightarrow && 1 &= c \tan \beta - \frac{1}{k} \tan^2 \beta (1 + c^2) \\ \Rightarrow && k \cot^2 \beta &= kc\cot \beta -1-c^2 \\ \Rightarrow && 0 &= c^2 -ck \cot \beta + 1 + k \cot^2 \beta \end{align*}
   1. As a quadratic in \(c\) the sum of the roots is \(k \cot \beta\), therefore \(\cot \alpha_1 + \cot \alpha_2 = k \cot \beta\). We also have that \(\cot \alpha_1 \cot \alpha_2 = 1 + k \cot^2 \beta\), so \begin{align*} && \cot (\alpha_1 + \alpha_2) &= \frac{\cot \alpha_1 \cot \alpha_2-1}{\cot \alpha_1 + \cot \alpha_2} \\ &&&= \frac{1 + k \cot^2 \beta - 1}{k \cot \beta} \\ &&&= \cot \beta \\ \Rightarrow && \beta &= \alpha_1 + \alpha_2 \pmod{\pi} \end{align*} but since \(\alpha_i \in (0, \frac{\pi}{2})\) the equation must hold exactly.
   2. Since it has two real roots we must have \begin{align*} && 0 &<\Delta = k^2 \cot^2 \beta - 4 (1 + k \cot^2 \beta) \\ &&&= k^2 \cot^2 \beta-4k \cot^2 \beta -4 \\ &&&= \cot^2 \beta (k^2 - 4k - 4(\sec^2 \beta - 1)) \\ &&&= \cot^2 \beta ( (k-2)^2 -4\sec^2 \beta) \\ \Rightarrow && k &> 2 + 2\sec \beta = 2(1+\sec \beta) \end{align*}
2. The greatest height will satisfy \(v^2 = u^2 + 2as\) so \(0 = u^2 \cos^2 \alpha - 2gh_{max} \Rightarrow 4\sec^2 \alpha = \frac{2u^2}{gh_{max}} = k_{max}\), but this decreases with \(h\), so the smallest \(k\) can be is \(4\sec^2 \alpha\), ie \(k \geq 4 \sec^2 \alpha\)

Solution: The tangent to the curve \(y = f(x)\) at \(x = a\) has the equation \(\frac{y-f(a)}{x-a} = f'(a) = g(a)+(a-p)g'(a)\). This passes through \((p,0)\) iff \begin{align*} && \frac{-f(a)}{p-a} &= g(a)+(a-p)g'(a) \\ \Leftrightarrow && -f(a) &= (p-a)g(a) -(a-p)^2g'(a) \\ \Leftrightarrow && -f(a) &= -f(a) -(a-p)^2g'(a) \\ \Leftrightarrow && 0 &= g'(a) \\ \end{align*}

1. In this case \(g(x) = A(x-q)(x-r) = A(x^2-(q+r)x+qr)\) and so we must have that \(g'(a) = 0\), ie \(A(2a-(q+r)) = 0 \Rightarrow 2a = q+r\) The gradient is \(g(a) +(a-p)g'(a) = g(a) = A(a-q)(a-r)\)
2. By the same reasoning, but with \(g(x) = A(x-p)(x-q)\) we have the gradient is \(A(c-p)(c-r)\). This is parallel iff \begin{align*} && (c-p)(c-r) &= (a-q)(a-r) \end{align*} The tangent at \(x = q\) is \(\frac{y-0}{x-q} = A(q-p)(q-r)\) or \( y = A(q-p)(q-r)(x-q)\)

Solution:

1. If \(f(a) = a\) then the sequence is constant, ie \(a = p+a^2-pa \Rightarrow 0 = (a-p)(a-1)\). Therefore \(a = 1, p\) If there sequence has period \(2\) then there must be a solution to \(f(f(x)) = x\), ie \begin{align*} && x &= p+(f(x)-p)f(x) \\ &&&= p+(p+(x-p)x-p)(p+(x-p)x) \\ &&&= p + (x-p)x(p+(x-p)x) \\ &&&= p+(x^2-px)(x^2-px+p) \\ \Rightarrow && 0 &= x^4-2px^3+(p+p^2)x^2-(p^2+1)x+p \\ &&&= (x-1)(x-p)(x^2-(p-1)x+1) \end{align*} The first two roots (\(x = 1, p\)) are constant sequences, so we need the second quadratic to have a root, ie \((p-1)^2-4 \geq 0 \Rightarrow p \geq 3 , p \leq -1\). We also need this root not to be \(1\) or \(p\), ie \(1-(p-1)+1 = 3-p \neq 0\) and \(p^2-(p-1)p + 1 = 1+p \neq 0\) so \(p \neq -1, 3\). Therefore \(p > 3\) or \(p < -1\).
2. There exists a constant sequence iff there is a solution to \(f(x) = x\), ie \begin{align*} && x &= f(x) \\ &&&= q + (x-p)x \\ \Leftrightarrow && 0 &= x^2-(p+1)x + q \tag{has a solution} \\ \end{align*} But if it doesn't have a solution, clearly the RHS is always larger, and if it does have a solution then there is some point where the inequality doesn't hold. Suppose \(f(x) > x\) for all \(x\) then \(f(f(x)) > f(x) > x\) therefore there is no value where \(f(f(x)) = x\) which is required for any sequence of period 2. No, consider \(p = q = 0\) so \(f(x) = x^2\) then there cannot be a period \(2\) sequence by the first part, but also clearly \(u_n = 1\) is a valid constant sequence.

Solution:

1. Notice that \(P = \begin{pmatrix} u \cos \alpha t\\ u \sin \alpha t - \frac12 g t^2 \end{pmatrix}\), so \begin{align*} && |OP|^2 &= u^2 \cos^2 \alpha t^2 + \left (u \sin \alpha t - \frac12 g t^2 \right)^2 \\ &&&= u^2 \cos^2 \alpha t^2 +u^2 \sin^2 \alpha t^2 - u \sin \alpha g t^3 +\frac14 g^2 t^4 \\ &&&= u^2 t^2 -u\sin \alpha g t^3 + \frac14g^2t^4 \\ && \frac{\d |OP|^2}{\d t} &= 2u^2 t - 3u \sin \alpha g t^2+g^2 t^3 \\ &&&= t \left (2u^2 - 3u \sin \alpha (gt)+(gt)^2\right) \\ && \Delta &= 9u^2 \sin^2 \alpha -4 \cdot 2u^2 \cdot 1 \\ &&&= u^2 (9\sin^2 \alpha -8) \\ \end{align*} Therefore if \(\sin \alpha < \frac{2\sqrt{2}}3\) the discriminant is negative, the quadratic factor is always positive and the distance \(|OP|\) is always increasing. Similarly, if \(\sin \alpha > \frac{2 \sqrt{2}}3\) then the derivative has a root. This means somewhere on its (possibly extended) trajectory \(OP\) is decreasing. This must be before it lands, since if it were after it 'landed' then both the \(x\) and \(y\) distances are increasing, therefore it cannot occur after it 'lands'.
2. Note that \(Q = \begin{pmatrix} -v t \\0 \end{pmatrix}\) \begin{align*} && |QP|^2 &= (u \cos \alpha t+vt)^2 + \left (u \sin \alpha t - \frac12 g t^2 \right)^2 \\ &&&= u^2 \cos^2 \alpha t^2+2u\cos \alpha v t^2 + v^2 t^2 +u^2 \sin^2 \alpha t^2 - u \sin \alpha g t^3 +\frac14 g^2 t^4 \\ &&&= (u^2+2u v \cos \alpha+v^2) t^2 - u \sin \alpha g t^3 + \frac14 g^2 t^4 \\ \\ \Rightarrow && \frac{\d |QP|^2}{\d t} &= 2(u^2+u v \cos \alpha+v^2) t - 3u \sin \alpha g t^2 + g^2 t^3 \\ &&&= t \left ( 2(u^2+2u v \cos \alpha+v^2) - 3u \sin \alpha (g t) + (g t)^2\right) \\ && \Delta &= 9u^2 \sin^2 \alpha - 8(u^2+2u v \cos \alpha+v^2) \\ &&&= (9 \sin^2 \alpha -8)u^2 - 16v \cos \alpha u - 8v^2 \\ &&&= \left (( \sin \alpha-2\sqrt{2}\cos \alpha)u-2\sqrt{2} v \right) \left ( ( \sin \alpha+2\sqrt{2}\cos \alpha)u+2\sqrt{2} v \right) \end{align*} Since the second bracket is clearly positive, the first bracket must be negative (for \(\Delta < 0\) and our derivative to be positive), ie \(2\sqrt{2} v > ( \sin \alpha-2\sqrt{2}\cos \alpha)u\)

Solution:

1. \(\,\)
   

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   Notice that \(1+r+r^2\) ranges from \(\frac34\) to \(3\) over \((-1,1)\) therefore \(\frac34 \leq p < 3\) attaining its minimum but not its maximum. If \(p > 1\) we know we must be on the right branch and hence we can determine \(r\) and \(S\) uniquely. If \(\frac34 < p < 1\) then we must have two possible values for \(r\), satisfying \(r_i^2 + r+1 = p\) and so our two possible values for \(S_i = \frac{1}{1-r_i}\) or \(r_i = 1-\frac{1}{S_i}\) and so \begin{align*} && p &= \left ( 1 - \frac{1}{S} \right)^2 + 1 - \frac1S + 1 \\ \Rightarrow && pS^2 &= (S-1)^2 + S^2-S + S^2 \\ \Rightarrow && 0 &= (3-p)S^2 -3S + 1 \end{align*}
2. \(\,\)
   

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   So \(\frac23 \leq q < 6\) and \(r\) and hence \(T\) is uniquely determined if \(2 \leq q < 6\). if \(\frac23

Solution:

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1. \(\,\) \begin{align*} \bf{COE}: && \frac12 m \cdot 5ag &= mg\cdot 2a + \frac12 m v^2 \\ \Rightarrow && v^2 &= ag \\ && v &= \sqrt{ag} \end{align*} If it just clears the second wall, we must have: \begin{align*} && 0 &= \sqrt{ag} \sin \theta t - \frac12 gt^2 \\ \Rightarrow && t &= \frac{2\sqrt{ag}\sin \theta}{g} \\ && a &= \sqrt{ag} \cos \theta t \\ &&&=\sqrt{ag} \cos \theta \frac{2\sqrt{ag}\sin \theta}{g} \\ &&&= a \sin 2 \theta \\ \Rightarrow && \theta &= 45^{\circ} \end{align*} Imagine firing the particle backwards from the top of the wall at \(45^\circ\) then \begin{align*} && -2a &= \sqrt{ag}\cdot \left ( -\frac1{\sqrt{2}} \right) t - \frac12 g t^2 \\ \Rightarrow && 0 &= gt^2+\sqrt{2ag} t -4a \\ &&&= (\sqrt{g}t -\sqrt{2} \sqrt{a})(\sqrt{g}t +2\sqrt{2} \sqrt{a}) \\ \Rightarrow && t &= \sqrt{\frac{2a}{g}} \\ \Rightarrow && s &= \left ( -\frac1{\sqrt{2}} \right) \sqrt{ag} \sqrt{\frac{2a}{g}} \\ &&&= -a \end{align*} Therefore the \(A\) is \(a\) from the wall.
2. When it passes over the first wall, \begin{align*} \bf{COE}: && \frac52amg &= (2a+h)mg + \frac12 m v^2 \\ \Rightarrow && v^2 &= (a-2h)g \end{align*} Now imagine firing a particle with this speed in any direction. The question is asking whether we can ever travel \(2a\) without descending more than \(h\). \begin{align*} && a &= \sqrt{(a-2h)g} \cos \beta t \\ \Rightarrow && t &= \frac{a}{\sqrt{(a-2h)g} \cos \beta}\\ && -h &= \sqrt{(a-2h)g} \sin \beta t - \frac12 g t^2 \\ &&&= a \tan \beta - \frac12 \frac{a^2}{(a-2h)} \sec^2 \beta \\ &&&= a \tan \beta - \frac{a^2}{2(a-2h)}(1+ \tan^2 \beta )\\ \Rightarrow && 0 &= \frac{a^2}{2(a-2h)} \tan^2 \beta-a \tan \beta + \frac{a^2-2ah+4h^2}{2(a-2h)} \\ && \Delta &= a^2 - \frac{a^2}{a-2h} \frac{a^2-2ah+4h^2}{a-2h} \\ &&&= \frac{a^2}{(a-2h)^2}\left ( a^2-4ah+4h^2-a^2+2ah-4h^2\right) \\ &&&= \frac{a^2}{(a-2h)^2}\left ( -2ah\right) < 0 \\ \end{align*} So there are no solutions if \(h > 0\)

Solution:

1. Notice that \(u^2 = v+2wz\), so \(w,z\) are roots of the quadratic \(t^2 -ut+\frac{u^2-v}{2}\). Therefore they are both real if \(u^2 \geq 2(u^2-v) \Rightarrow 2v \geq u^2\).
2. \begin{align*} && w+z &= u \\ && w^2+z^2 &= u^2 - \tfrac23 \\ && w^3+z^3 &= \lambda(u-1) \\ \\ && wz &= \frac{u^2 - (u^2-\tfrac23)}{2} = \tfrac13\\ \\ && (w+z)(w^2+z^2) &= w^3+z^3+wz(w+z) \\ &&u(u^2-\tfrac23)&= \lambda(u-1)+\frac13u \\ \Rightarrow && u^3-u&= \lambda (u-1) \\ \Rightarrow && 0 &= (u-1)(u(u+1) - \lambda) \\ \Rightarrow && 0 &= (u-1)(u^2+u - \lambda) \end{align*} Therefore there will be at most 3 values for \(u\), unless \(1\) is a root of \(u^2+u-\lambda\), ie \(\lambda = 2\). Suppose \(u = 1\), then we have: \(w+z = 1, wz = 1/3 \Rightarrow w,z = \frac{-1 \pm \sqrt{-1/3}}{2}\) which are clearly complex.

Solution: If \(\p(x)\) and \(\q(x)\) are polynomials of degree \(m\) and \(n\), \(\p(\q(x))\) has degree \(mn\).

1. Suppose \(\p(\p(\p(x)))- 3 \p(x)= -2x\), and suppose \(p(x)\) has degree \(n = \geq 2\), then \(\p(\p(\p(x)))\) has degree \(n^3\) and so the left hand side has degree higher than \(1\) and the right hand side is degree \(1\). Therefore \(\p(x)\) is degree \(1\) or \(0\). If \(p(x) = c\) then \(c^3-3c = -2x\) but the LHS doesn't depend on \(x\) which is also a contradiction. Therefore \(\p(x)\) is degree \(1\). Suppose \(\p(x) = ax+b\) then: \begin{align*} && -2x &= \p(\p(\p(x))) - 3\p(x) \\ &&&= \p(\p(ax+b)) - 3(ax+b) \\ &&&= \p(a(ax+b)+b) - 3ax -3b \\ &&&= a(a^2x+ab+b) + b - 3ax - 3b \\ &&&= (a^3-3a)x + b(a^2+a-2) \\ \Rightarrow &&& \begin{cases} a^3-3a&=-2 \\ b(a^2+a-2) &= 0\end{cases} \\ \Rightarrow &&& \begin{cases} a^3-3a+2 = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow &&& \begin{cases} (a-1)(a^2+a-2) = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow && (a,b) &= (1, b), (-2,b) \end{align*}
2. Suppose \(2\p(\p(x)) +3 [\p(x)]^2 -4\p(x) =x^4\) and let \(\deg \p(x) = n\), then LHS has degree \(\max(n^2,2n,n)\) and the right hand side has degree \(4\). Therefore \(\p(x)\) must have degree \(2\). Let \(\p(x) = ax^2 + bx + c\), then, considering the coefficient of \(x^4\) in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) we will have \(2a^3+3a^2=1 \Rightarrow 2a^3+3a^2-1 = (a+1)^2(2a-1) \Rightarrow a = -1, a=\frac12\). Consider the coefficient of \(x^3\) in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) we have \(4a^2b+6ab = 0 \Rightarrow 2ab(2a+3) = 0\) Since \(a = -1, \frac12\) this means \(b = 0\). Consider the constant coefficient in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) (using \(b = 0\)). \(2ac^2+c+3c^2-4c = 0 \Rightarrow c(2ac+3c-3) = 0\). Therefore \(c = 0\) or \(a = -1, c = 3, a = \frac12, c = \frac34\), so our possible polynomials are: \(\p(x) = -x^2, \frac12x^2, -x^2+3, \frac12x^2+\frac34\)

Solution: \begin{align*} && s &= ut \\ \Rightarrow && a &= u \cos \alpha t\\ \Rightarrow && t &= \frac{a}{u \cos \alpha}\\ && s &= ut+ \frac12at^2 \\ \Rightarrow && -h &< -u\sin \alpha \frac{a}{u \cos \alpha}-\frac12 g \left (\frac{a}{u \cos \alpha} \right)^2 \\ &&&= -a \tan \alpha-\frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha \\ \Rightarrow && \frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha &< h -a\tan \alpha \\ \Rightarrow &&\frac{1}{u^2} &< \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \end{align*} \begin{align*} && s &= ut + \frac12a t^2 \\ \Rightarrow && 2h &= u\sin \alpha t + \frac12 gt^2 \\ \Rightarrow && t &= \frac{-u\sin \alpha \pm \sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\ && t &= \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\ && s &= ut \\ \Rightarrow && b &> u \cos \alpha t \\ \Rightarrow && \frac{b}{u \cos \alpha} &> \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g} \\ \Rightarrow && \sqrt{u^2 \sin^2 \alpha+4hg} &< \frac{bg}{u \cos \alpha} + u \sin \alpha \\ \end{align*} \begin{align*} \Rightarrow && u^2 \sin^2 \alpha+4hg &< \frac{b^2g^2}{u^2 \cos^2 \alpha} +u^2 \sin^2 \alpha + 2bg \tan \alpha \\ \Rightarrow && 4hg - 2bg \tan \alpha &< \frac{b^2g^2}{u^2 \cos^2 \alpha} \\ &&&< \frac{b^2g^2}{\cos^2 \alpha} \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \\ &&&= \frac{2b^2g(h-a\tan \alpha)}{a^2} \\ \Rightarrow && \tan \alpha \left (\frac{2b^2g}{a} - 2bg \right) &< \frac{2b^2gh}{a^2} - 4hg \\ \Leftrightarrow && \tan \alpha \left (\frac{2b^2g- 2abg}{a} \right) &< \frac{2b^2gh- 4hga^2}{a^2} \\ \Leftrightarrow && \tan \alpha \left (\frac{2bg(b- a)}{a} \right) &< \frac{2hg(b^2- 2a^2)}{a^2} \\ \Rightarrow && \tan \alpha &< \frac{h(b^2-2a^2)}{ab(b-a)} \end{align*}

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:

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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 \(4p^3+27q^2 < 0\)