Problems

If \(m\) is a positive integer, show that \(\l 1+x \r^m + \l 1-x \r^m \ne 0\) for any real \(x\,\). The function \(\f\) is defined by \[ \f (x) = \frac{ (1+x )^m - ( 1-x )^m}{ (1+x )^m + (1-x )^m} \;. \] Find and simplify an expression for \(\f'(x)\). In the case \(m=5\,\), sketch the curves \(y = \f (x)\) and \(\displaystyle y = \frac1 { \f (x )}\;\).

Show Solution

---

Solution: If \(m\) is even, clearly that expression is positive since it's the sum of two (different) squares. If \(m\) is odd, then we can expand it as a sum of powers of \(x^2\) with a leading coefficient of \(1\) so it is also positive. \begin{align*} && f (x) = \frac{ (1+x )^m - ( 1-x )^m}{ (1+x )^m + (1-x )^m} \\ && f'(x) &= \frac{(m(1+x )^{m-1} + m( 1-x )^{m-1})((1+x)^m + (1-x)^m ) - ((1+x )^m - ( 1-x )^m)(m(1+x)^{m-1} - m(1-x)^{m-1} )}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{2m(1+x)^m(1-x)^{m-1}+2m(1+x)^{m-1}(1-x)^m}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{2m(1+x)^{m-1}(1-x)^{m-1}(1+x+1-x)}{\l (1+x)^m + (1-x)^m \r^2} \\ &&&= \frac{4m(1+x)^{m-1}(1-x)^{m-1}}{\l (1+x)^m + (1-x)^m \r^2} \\ \end{align*}

+ - ↺

Find the coordinates of the turning point on the curve \(y = x^2 - 2bx + c\,\). Sketch the curve in the case that the equation \(x^2 - 2bx + c=0\) has two distinct real roots. Use your sketch to determine necessary and sufficient conditions on \(b\) and \(c\) for the equation \(x^2 - 2bx + c = 0\) to have two distinct real roots. Determine necessary and sufficient conditions on \(b\) and \(c\) for this equation to have two distinct positive roots. Find the coordinates of the turning points on the curve \(y = x^3 - 3b^2x + c\) (with \(b>0\)) and hence determine necessary and sufficient conditions on \(b\) and \(c\) for the equation \(x^3 - 3b^2x + c = 0\) to have three distinct real roots. Determine necessary and sufficient conditions on \(a\,\), \(b\) and \(c\) for the equation \(\l x - a \r^3 - 3b^2 \l x - a \r + c = 0\) to have three distinct positive roots. Show that the equation \(2x^3 - 9x^2 + 7x - 1 = 0\) has three distinct positive roots.

Show Solution

---

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.

+ - ↺

Let \(f(x) = x^m(x-1)^n\), where \(m\) and \(n\) are both integers greater than \(1\). Show that the curve \(y=f(x)\) has a stationary point with \(0 < x < 1\). By considering \(f''(x)\), show that this stationary point is a maximum if \(n\) is even and a minimum if \(n\) is odd. Sketch the graphs of \(f(x)\) in the four cases that arise according to the values of \(m\) and \(n\).

Show Solution

---

Solution: \begin{align*} && f'(x) &= mx^{m-1}(x-1)^n + nx^m(x-1)^{n-1} \\ &&&= (m(x-1)+nx)x^{m-1}(x-1)^{n-1} \\ &&&= (x(m+n) - m)x^{m-1}(x-1)^{n-1} \\ \end{align*} Therefore when \(x = \frac{m}{m+n}\) there is a stationary point with \(0 < x < 1\). \begin{align*} && f''(x) &= m(m-1)x^{m-2}(x-1)^n + 2mnx^{m-1}(x-1)^{n-1} + n(n-1)x^{m}(x-1)^{n-2} \\ &&&= (m(m-1)(x-1)^2 +2mnx(x-1)+n(n-1)x^2)x^{m-2}(1-x)^{n-2} \\ \Rightarrow && f'' \left ( \frac{m}{m+n} \right) &= \left ( m(m-1) \frac{n^2}{(m+n)^2} - 2mn\frac{mn}{(m+n)^2} + n(n-1) \frac{m^2}{(m+n)^2} \right) \frac{m^{m-2}}{(m+n)^{m-2}} \frac{(-1)^{n-2}n^{n-2}}{(m+n)^{n-2}} \\ &&&= (-1)^{n-2}\frac{m^{m-1}n^{n-1}}{(m+n)^{m+n-2}} \left ( (m-1)n-2mn+(n-1)m\right) \\ &&&= (-1)^{n-2}\frac{m^{m-1}n^{n-1}}{(m+n)^{m+n-2}} \left ( -m-n\right) \\ &&&= (-1)^{n-1} \frac{m^{m-1}n^{n-1}}{(m+n)^{m+n-3}} \end{align*} Therefore this is positive (and a minimum) when \(n\) is odd and negative (and a maximum) when \(n\) is even.

+ - ↺

+ - ↺

+ - ↺

+ - ↺

Give a sketch of the curve \( \;\displaystyle y= \frac1 {1+x^2}\;\), for \(x\ge0\). Find the equation of the line that intersects the curve at \(x=0\) and is tangent to the curve at some point with \(x>0\,\). Prove that there are no further intersections between the line and the curve. Draw the line on your sketch. By considering the area under the curve for \(0\le x\le1\), show that \(\pi>3\,\). Show also, by considering the volume formed by rotating the curve about the \(y\) axis, that \(\ln 2 >2/3\,\). [Note: \(\displaystyle \int_0^ 1 \frac1 {1+x^2}\, \d x = \frac\pi 4\,.\;\)]

Show Solution

---

Solution:

+ - ↺

\begin{align*} && y &= (1+ x^2)^{-1} \\ \Rightarrow && y' &= -2x(1+x^2)^{-2} \\ \text{eqn of tangent}:&& \frac{y - (1+t^2)^{-1}}{x-t} &= -2t(1+t^2)^{-2} \\ \text{passes thru }(0,1): && \frac{1-(1+t^2)^{-1}}{-t} &= -2t(1+t^2)^{-2} \\ \Rightarrow && (1+t^2)^2-(1+t^2) &= 2t^2 \\ \Rightarrow && t^4-t^2 &= 0 \\ \Rightarrow && t &= 0, \pm 1 \\ \Rightarrow && \frac{y - \frac12}{x - 1} &= -\frac12 \\ && y &=1 -\tfrac12 x \end{align*} There can be no further intersections since the equation is equivalent to the cubic \((1-\frac12 x)(1+x^2) = 1\) and we have already found \(3\) roots. \begin{align*} && A &= \int_0^1 \frac{1}{1 + x^2} = \frac{\pi}{4} \\ && A &> \frac12 \cdot 1 \cdot (1 + \tfrac12) = \frac34 \\ \Rightarrow && \pi &> 3 \\ \\ && V &=\pi \int_{\frac12}^1 x^2 \d y \\ &&&= \pi \int_{\frac12}^1 \left ( \frac{1}{y}-1 \right) \d y \\ &&&= \pi \left [\ln y \right]_{1/2}^1-\frac12 \\ &&&= \pi \ln 2 - \frac{\pi}{2} \\ && V &> \frac13 \pi 1^2 \frac{1}{2} \\ &&&= \frac{\pi}{6} \\ \Rightarrow && \ln 2 &> \frac{2}{3} \end{align*}

Find \(y\) in terms of \(x\), given that: \begin{eqnarray*} \mbox{for \(x < 0\,\)}, && \frac{\d y}{\d x} = -y \mbox{ \ \ and \ \ } y = a \mbox{ when } x = -1\;; \\ \mbox{for \(x > 0\,\)}, && \frac{\d y}{\d x} = y \mbox{ \ \ \ \ and \ \ } y = b \ \mbox{ when } x = 1\;. \end{eqnarray*} Sketch a solution curve. Determine the condition on \(a\) and \(b\) for the solution curve to be continuous (that is, for there to be no `jump' in the value of \(y\)) at \(x = 0\). Solve the differential equation \[ \frac{\d y}{\d x} = \left\vert \e^x-1\right\vert y \] given that \(y=\e^{\e}\) when \(x=1\) and that \(y\) is continuous at \(x=0\,\). Write down the following limits: \ \[ \text{(i)} \ \ \lim_ {x \to +\infty} y\exp(-\e^x)\;; \ \ \ \ \ \ \ \ \ \text{(ii)} \ \ \lim_{x \to -\infty}y \e^{-x}\,. \]

Let \[\f(x) = a \sqrt{x} - \sqrt{x - b}\;, \] where \(x\ge b >0\) and \(a>1\,\). Sketch the graph of \(\f(x)\,\). Hence show that the equation \(\f(x) = c\), where \(c>0\), has no solution when \(c^2 < b \l a^2 - 1 \r\,\). Find conditions on \(c^2\) in terms of \(a\) and \(b\) for the equation to have exactly one or exactly two solutions. Solve the equations

1. \(3 \sqrt{x} - \sqrt{x - 2} = 4\, ,\)
2. \(3 \sqrt{x} - \sqrt{x - 3} = 5\;\).

Show Solution

---

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*}

Find all the solution curves of the differential equation \[ y^4 \l {\mathrm{d}y \over \mathrm{d}x }\r^{\! \! 4} = \l y^2 - 1 \r^2 \] that pass through either of the points

1. \(\l 0, \, \frac{1}{2}\sqrt3 \r\),
2. \(\l 0, \, \frac{1}{2}\sqrt5 \r\).

Show Solution

---

Solution: \begin{align*} && y^4 \left (\frac{\d y}{\d x} \right)^4 &= (y^2 - 1)^2 \\ \Rightarrow && y^2 \left (\frac{\d y}{\d x} \right)^2 &= |y^2 - 1| \\ && y \left (\frac{\d y}{\d x} \right) &= \pm \sqrt{|y^2-1|} \\ \Rightarrow &&\int \frac{y}{\sqrt{|y^2-1|}} \d y &= \int \pm 1 \d x \\ \Rightarrow && \pm \sqrt{|y^2-1|} &= \pm x + C \\ \end{align*}

1. Since \(y^2 < 1\), our solution curve should be of the from \(-\sqrt{1-y^2} = \pm x + C\) Plugging in \((0, \tfrac12 \sqrt{3})\), we obtain \(-\tfrac12 = C\), therefore our solution curves are \(\pm x = \frac12 - \sqrt{1-y^2}\)
2. Since \(y^2 > 1\), our solution curve should be of the from \(\sqrt{y^2-1} = \pm x + C\) Plugging in \((0, \tfrac12 \sqrt{5})\), we obtain \(\tfrac12 = C\), therefore our solution curves are \(\pm x = \sqrt{y^2-1}-\frac12\)

Clearly if \(y = \pm 1\), \(y'=0\) and the equation is satisfied.

+ - ↺

Sketch, without calculating the stationary points, the graph of the function \(\f(x)\) given by \[ \f(x) = (x-p)(x-q)(x-r)\;, \] where \(p < q < r\). By considering the quadratic equation \(\f'(x)=0\), or otherwise, show that \[ (p+q+r)^2 > 3(qr+rp+pq)\;. \] By considering \((x^2+gx+h)(x-k)\), or otherwise, show that \(g^2>4h\,\) is a sufficient condition but not a necessary condition for the inequality \[ (g-k)^2>3(h-gk) \] to hold.

Show Solution

---

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.

In a cosmological model, the radius \(\rm R\) of the universe is a function of the age \(t\) of the universe. The function \(\rm R\) satisfies the three conditions: $$ \mbox{\({\rm R}(0)=0\)}, \ \ \ \ \ \ \ \ \ \mbox{\({\rm R'}(t)>0\) for \(t>0\)}, \ \ \ \ \ \ \ \ \ \ \mbox{\({\rm R''}(t)<0\) for \(t>0\)}, \tag{*} $$ where \({\rm R''}\) denotes the second derivative of \(\rm R\). The function \({\rm H}\) is defined by \[ {\rm H} (t)= \frac{{\rm R}'(t)}{{\rm R}( t)}\;. \]

1. Sketch a graph of \({\rm R} (t)\). By considering a tangent to the graph, show that \(t<1/{\rm H}(t)\).
2. Observations reveal that \({\rm H}(t) = a/t\), where \(a\) is constant. Derive an expression for \({\rm R}(t)\). What range of values of \(a\) is consistent with the three conditions \((*)\)?
3. Suppose, instead, that observations reveal that \({\rm H}(t)= b t^{-2}\), where \(b\) is constant. Show that this is not consistent with conditions \((*)\) for any value of \(b\).

Sketch the graph of the function \([x/N]\), for \(0 < x < 2N\), where the notation \([y]\) means the integer part of \(y\). (Thus \([2.9] = 2\), \ \([4]=4\).)

1. Prove that \[ \sum_{k=1}^{2N} (-1)^{[k/N]} k = 2N-N^2. \]
2. Let \[ S_N = \sum_{k=1}^{2N} (-1)^{[k/N]} 2^{-k}. \] Find \(S_N\) in terms of \(N\) and determine the limit of \(S_N\) as \(N\to\infty\).

The curve \(C_1\) passes through the origin in the \(x\)--\(y\) plane and its gradient is given by $$ \frac{\d y}{\d x} =x(1-x^2)\e^{-x^2}. $$ Show that \(C_1\) has a minimum point at the origin and a maximum point at \(\left(1,{\frac12\, \e^{-1}} \right)\). Find the coordinates of the other stationary point. Give a rough sketch of \(C_1\). The curve \(C_2\) passes through the origin and its gradient is given by $$ \frac{\d y}{\d x}= x(1-x^2)\e^{-x^3}. $$ Show that \(C_2\) has a minimum point at the origin and a maximum point at \((1,k)\), where \phantom{} \(k > \frac12 \,\e^{-1}.\) (You need not find \(k\).)

Consider the equation \[ x^2 - b x + c = 0 \;, \] where \(b\) and \(c\) are real numbers.

1. Show that the roots of the equation are real and positive if and only if \(b>0\) and \phantom{} \(b^2\ge4c>0\), and sketch the region of the \(b\,\)-\(c\) plane in which these conditions hold.
2. Sketch the region of the \(b\,\)-\(c\) plane in which the roots of the equation are real and less than \(1\) in magnitude.

In this question, the function \(\sin^{-1}\) is defined to have domain \( -1\le x \le 1\) and range \linebreak \( - \frac{1}{2}\pi \le x \le \frac{1}{2}\pi\) and the function \(\tan^{-1}\) is defined to have the real numbers as its domain and range \( - \frac{1}{2}\pi < x < \frac{1}{2}\pi\).

1. Let $$ \g(x) = \displaystyle {2x \over 1 + x^2}\;, \ \ \ \ \ \ \ \ \ \ -\infty
2. Let \[ \displaystyle \f \l x \r = \sin^{-1} \l {2x \over 1 + x^2} \r \;,\ \ \ \ \ \ \ \ \ -\infty < x < \infty\;. \] Show that $ \f(x ) = 2 \tan^{-1} x\( for \) -1 \le x \le 1\,\( and \)\f(x) = \pi - 2 \tan^{-1} x \( for \)x\ge1\,$. Sketch the graph of \(\f(x)\).

Show that the equation \(x^3 + px + q=0\) has exactly one real solution if \(p \ge 0\,\). A parabola \(C\) is given parametrically by \[ x = at^2, \: \ \ y = 2at \: \: \: \ \ \ \ \ \ \l a > 0 \r \;. \] Find an equation which must be satisfied by \(t\) at points on \(C\) at which the normal passes through the point \(\l h , \; k \r\,\). Hence show that, if \(h \le 2a \,\), exactly one normal to \(C\) will pass through \(\l h , \; k \r \, \). Find, in Cartesian form, the equation of the locus of the points from which exactly two normals can be drawn to \(C\,\). Sketch the locus.

Show Solution

---

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*}

+ - ↺

Sketch the graph of the function \(\ln x - {1 \over 2} x^2\). Show that the differential equation \[ {\mathrm{d} y \over \mathrm{d} x} = {2xy \over x^2 - 1} \] describes a family of parabolas each of which passes through the points \((1,0)\) and \((-1,0)\) and has its vertex on the \(y\)-axis. Hence find the equation of the curve that passes through the point \((1,1)\) and intersects each of the above parabolas orthogonally. Sketch this curve. [Two curves intersect orthogonally if their tangents at the point of intersection are perpendicular.]

Show Solution

---

Solution:

+ - ↺

\begin{align*} && y' &= \frac{2xy}{x^2-1} \\ \Rightarrow && \int \frac{1}{y} \d y &= \int \frac{2x}{x^2-1} \d x \\ \Rightarrow && \ln |y| &= \ln |x^2-1| + C \\ \Rightarrow && y &= A(x^2-1) \end{align*} which is a family of parabolas each passing through \((\pm1, 0)\) and with a vertex on the \(y\)-axis. The curve we seek must satisfy \begin{align*} && y' &= \frac{1-x^2}{2xy} \\ \Rightarrow && \int2 y \d y &= \int \left ( \frac{1}{x} - x \right) \d x \\ \Rightarrow && y^2 &= \ln x - \tfrac12 x^2 + C \\ (1,1): && 1 &= -\tfrac12+C \\ \Rightarrow && C &= \frac32 \\ \Rightarrow && y^2 &= \tfrac32 + \ln x - \tfrac12 x^2 \end{align*}

+ - ↺