Problems

Solution:

1. \(\,\) \begin{align*} && y&=\frac{x^2+x\sin\theta+1}{x^2+x\cos\theta+1} \\ \Leftrightarrow && 0 &= x^2(y-1) + x(y \cos \theta - \sin \theta) + y-1 \\ \Leftrightarrow && 0 &\leq \Delta = (y\cos \theta - \sin \theta)^2 - 4(y-1)^2 \\ \Leftrightarrow && (y\cos \theta - \sin \theta)^2 &\geq 4(y-1)^2 \end{align*} [Assuming that \(y \neq 1\), if \(y = 1\) then the RHS is \(0\) and it is automatically satisfied]. Notice that \((y\cos \theta - \sin \theta)^2 \leq (y^2+1)(\cos^2 \theta + \sin^2 \theta)\) by Cauchy-Schwarz, so \(y^2 + 1 \geq 4(y-1)^2\). \begin{align*} && y^2 + 1 &\geq 4(y-1)^2 \\ \Leftrightarrow && 0 &\geq 3y^2-8y+3 \\ \text{c.v.} && y&= \frac{8 \pm \sqrt{64-4\cdot3 \cdot 3}}{6} \\ &&&= \frac{4 \pm \sqrt{16-9}}{3} = \frac{4 \pm \sqrt{7}}3 \end{align*} so \(\frac{4-\sqrt{7}}3 \leq y \leq \frac{4+\sqrt7}3\).
2. If \(y = \frac{4+\sqrt7}3\) then \(y - 1 = \frac{1+\sqrt7}3\) and since \(y^2+1 = 4(y-1)^2\) taking square roots we obtain \(\sqrt{y^2+1} = 2(y-1)\). Since equality must hold in our C-S identity, we must have \(\langle y, -1 \rangle\) parallel to \( \langle \cos \theta , \sin \theta \rangle\), ie \(\tan \theta = -\frac{3}{4+\sqrt{7}}\) and \begin{align*} && x & = \frac{-(y \cos \theta - \sin \theta) \pm \sqrt{\Delta}}{2(y-1)} \\ &&&= \frac{\pm2(y-1)}{2(y-1)} \\ &&&= \pm1 \end{align*}

Solution:

1. \begin{align*} && 0 &= x + 3\sqrt{x} - \frac12 \\ \sqrt{x} = y: && 0&= y^2 + 3y - \frac12 \\ \Rightarrow && y &= \frac{-3\pm\sqrt{3^2+2}}{2} \\ &&&= \frac{-3 \pm \sqrt{11}}{2} \\ y > 0: && x &= \left ( \frac{\sqrt{11}-3}{2} \right)^2 \end{align*}
2. • \begin{align*} && 0 &= x + 10\sqrt{x+2} - 22 \\ y = \sqrt{x+2}: && 0 &= y^2 - 2 + 10y - 22 \\ &&&= y^2 + 10y - 24 \\ &&&= (y-2)(y+12) \\ \Rightarrow && y &= 2, -12 \\ y > 0: && x &= 2 \end{align*}
   • Let \(y = \sqrt{2x^2-8x-3}\), so \begin{align*} && 0 &= x^2 - 4x +\sqrt{2x^2-8x-3} - 9 \\ && 0 &= \frac{y^2+3}{2} + y - 9 \\ &&&= \frac12 y^2 +y - \frac{15}{2} \\ &&&= \frac12 (y-3)(y+5) \\ \Rightarrow && y &= 3,-5 \\ y > 0: && 9 &= 2x^2-8x-3 \\ \Rightarrow && 0 &= 2x^2-8x-12 \\ &&&= 2(x^2-4x-6) \\ \Rightarrow && x &= 2 \pm \sqrt{10} \end{align*}

Solution:

1. \(\,\) \begin{align*} && x &= (a-x)^{\frac12} \\ \Rightarrow && x^2 &= a - x \\ \Rightarrow && 0 &= x^2 + x - a \end{align*} This has a roots if \(\Delta = 1 + 4a \geq 0 \Rightarrow a \geq -\frac14\). These roots also need to be positive (since \(x \geq 0\)). Since \(f(0) = -a\) we have one positive root if \(a \geq 0\). If \(a \leq 0\) then since the roots are symmetric about \(x = -\frac12\), both roots are negative and there are no positive roots. Therefore we have on real solution if \(a \geq 0\) and non otherwise. \begin{align*} && x & = \left ( (1+a)x - a \right)^{\frac13} \\ \Leftrightarrow && x^ 3 &= (1+a)x - a \\ \Leftrightarrow && 0 &= x^3- (1+a)x + a \\ \Leftrightarrow && 0 &= (x-1)(x^2+x-a) \\ \end{align*} Since every solution to the first equation is a solution to the second, we have \(x = 1\) always works, and there is an additional two solutions if \(a > -\frac14\) and a single extra solution if \(a = -\frac14\). We can also repeat solutions if \(1\) is a root of \(x^2+x -a\), ie when \(a = 2\) Therefore: One solution if \(a < -\frac14\) Two solutions if \(a = -\frac14, 2\) Three solutions if \(a > -\frac14, a \neq 2\)
2. \(\,\) \begin{align*} && x &= (b+x)^{\frac12} \\ \Rightarrow && x^2 &= b + x \\ \Rightarrow && 0 &= x^2 - x - b \end{align*} This has a positive root if \(\frac14 - \frac12 - b \leq 0 \rightarrow b \geq \frac14\). It has two positive roots if \(b \geq 0\). Therefore two solutions if \(b > \frac14\) and one solution if \(b = \frac14\)

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: \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:

1. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} + \sqrt{x} -2x-1 \\ \Rightarrow && 2x+1 &= \sqrt{3x^2+1} + \sqrt{x} \\ \Rightarrow && 4x^2+4x+1 &= 3x^2+1+x+2\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2(x^2+6x+9) &= 4x(3x^2+1) \\ \Rightarrow && 0 &= x^4-6x^3+9x^2-4x \\ &&&= x(x-4)(x-1)^2 \end{align*} So clearly we have \(x = 0, x = 1, x = 4\). \(x = 0\) works, \(x = 1\) works, \(x = 4\) works.
2. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} - 2\sqrt{x} +x-1 \\ \Rightarrow && 1-x &= \sqrt{3x^2+1}-2\sqrt{x} \\ \Rightarrow && x^2-2x+1 &= 3x^2+1+4x-4\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)} \\ \Rightarrow && 0 &= x(x-4)(x-1)^2 \end{align*} Again we must check \(x = 0, x = 1, x = 4\). \(x = 0,1\) work, but \(x = 4\) is not a solution.
3. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} - 2\sqrt{x} -x+1 \\ \Rightarrow && x-1 &= \sqrt{3x^2+1} - 2\sqrt{x} \\ \Rightarrow && x^2-2x+1 &= 3x^2+1+4x-4\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)}\\ \Rightarrow && 0 &= x(x-4)(x-1)^2 \end{align*} So again, we need to check \(x = 0, 1, 4\). \(x = 0, 4\) work, but \(x = 1\) fails.

Solution: There are two cases to consider by they are equivalent to \(x^2 \pm \alpha x + 2 < 0\), which has no solution solutions if \(\Delta < 0\), ie if \(\alpha^2 - 4\cdot1\cdot2 < 0 \Leftrightarrow |\alpha| < 2\sqrt{2}\). If \(\alpha = 3\), we have \begin{align*} && 0 & > x^2-3x+2 \\ &&&= (x-2)(x-1) \\ \Rightarrow && x & \in (1,2) \\ \\ && 0 &> x^2+3x+2 \\ &&& = (x+2)(x+1) \\ \Rightarrow && x &\in (-2,-1) \end{align*} Both cases work here, so \(x \in (-2, -1) \cup (1,2)\). \begin{align*} && 0 &> x^2 \pm \alpha x + 2 \\ &&&= (x \pm \tfrac{\alpha}{2})^2 -\frac{\alpha^2-8}{4} \end{align*} The potential intervals therefore are \((\frac{\alpha -\sqrt{\alpha^2-8}}{2}, \frac{\alpha +\sqrt{\alpha^2-8}}{2})\) and \((\frac{-\alpha -\sqrt{\alpha^2-8}}{2}, \frac{-\alpha +\sqrt{\alpha^2-8}}{2})\). Neither of these intervals overlap with \(0\), since \(\alpha^2 > \alpha^2-8\), and their lengths are both \(\sqrt{\alpha^2-8}\), therefore \(S = 2\sqrt{\alpha^2-8} < 2\alpha\)

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Solution: \begin{align*} && 0 &= z^4 +5z^3 +4z^2 -5z +1 \\ &&0 &= z^2 + z^{-2} + 5(z-z^{-1}) + 4 \\ &&&= (z-z^{-1})^2+2+5(z-z^{-1})+4 \\ &&&= w^2 + 5w + 6 \\ &&&= (w+3)(w+2) \\ \Rightarrow && 0 &= z-z^{-1}+3 \\ \Rightarrow && 0 &= z^2+3z-1 \\ \Rightarrow && z &= \frac{-3 \pm \sqrt{3^2+4}}{2} = \frac{-3 \pm \sqrt{13}}{2} \\ \Rightarrow && 0 &= z-z^{-1}+2 \\ \Rightarrow && 0 &= z^2+2z-1 \\ \Rightarrow && z &= \frac{-2 \pm \sqrt{2^2+4}}{2} = - 1 \pm \sqrt{2} \\ \end{align*} \begin{align*} &&0 &= 2z^8 -3z^7-12z^6 +12z^5 +22z^4-12z^3 -12 z^2 +3z +2 \\ && 0 &= 2(z^4+z^{-4}) - 3(z^3-z^{-3})-12(z^2+z^{-2})+12(z-z^{-1})+22 \\ &&&= 2\left ((z-z^{-1})^4+4(z^2+z^{-2})-6\right)-3 \left ((z-z^{-1})^3+3(z-z^{-1}) \right)-12 \left ((z-z^{-1})^2+2 \right)+12(z-z^{-1})+22 \\ &&&= 2(w^4+4(w^2+2)-6)-3w^3-9w-12w^2-24+12w+22 \\ &&&= 2 w^4-3w^3-4w^2+3w+2 \\ \Rightarrow && 0 &= 2(w^2+w^{-2})-3(w-w^{-1})-4 \\ &&&= 2((w-w^{-1})^2+2)-3(w-w^{-1})-4 \\ &&&= 2x^2-3x \\ &&&= x(2x-3) \\ \Rightarrow && 0 &= w -w^{-1} \\ \Rightarrow && w &= \pm 1 \\ \Rightarrow && \pm 1 &= z-z^{-1} \\ \Rightarrow && 0 &= z^2 \mp z-1 \\ \Rightarrow && z &= \frac{\pm 1 \pm \sqrt{5}}{2} \\ \Rightarrow && \frac32 &= w-w^{-1} \\ \Rightarrow && 0 &= 2w^2-3w -2 \\ &&&= (2w+1)(w-2) \\ \Rightarrow && 2 &= z-z^{-1} \\ \Rightarrow && 0 &= z^2-2z-1 \\ \Rightarrow && z &= 1 \pm \sqrt{2} \\ \Rightarrow && -\frac12 &= z-z^{-1} \\ \Rightarrow && 0 &= 2z^2+z-2 \\ \Rightarrow && z &= \frac{-1 \pm \sqrt{17}}{4} \\ \Rightarrow && z &\in \left \{ \frac{\pm 1 \pm \sqrt{5}}{2}, 1 \pm \sqrt{2}, \frac{-1 \pm \sqrt{17}}{4} \right \} \end{align*}

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*}

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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*}

Solution:

1. \(\,\)
   

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   \begin{align*} && 1+2x-x^2 = 2/x \\ \Rightarrow && 0 &= x^3-2x^2-x+2 \\ &&&= (x+1)(x^2-3x+2) \\ &&&= (x+1)(x-1)(x-2) \end{align*} Therefore the inequality is satisfied on \((1,2)\) and \((-1,0)\)
2. \(\,\)
   

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   \begin{align*} && \sqrt{3x+10} &= 2+\sqrt{x+4} \\ && 3x+10 &= x+8 + 4\sqrt{x+4} \\ && 16(x+4) &= 4(x+1)^2 \\ && 4x+16 &= x^2+2x+1 \\ \Rightarrow && 0 &= x^2-2x-15 \\ &&&= (x-5)(x+3) \end{align*} Therefore \(x > 5\)

Solution: \begin{align*} && (x-y+2)(x+y-1) &= (x-y)(x+y)-(x-y)+2(x+y)-2 \\ &&&= x^2-y^2+x+3y-2 \end{align*}

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