Problems

Two curves have equations \(\; x^4+y^4=u\;\) and \(\; xy = v\;\), where \(u\) and \(v\) are positive constants. State the equations of the lines of symmetry of each curve. The curves intersect at the distinct points \(A\), \(B\), \(C\) and \(D\) (taken anticlockwise from \(A\)). The coordinates of \(A\) are \((\alpha,\beta)\), where \(\alpha > \beta > 0\). Write down, in terms of \(\alpha\) and \(\beta\), the coordinates of \(B\), \(C\) and \(D\). Show that the quadrilateral \(ABCD\) is a rectangle and find its area in terms of \(u\) and \(v\) only. Verify that, for the case \(u=81\) and \(v=4\), the area is \(14\).

Show Solution

---

Solution: The curve \(x^4 + y^4 = u\) has lines of symmetry:

• \(y = 0\)
• \(x = 0\)
• \(y = x\)
• \(y = -x\)

The curve \(xy = v\) has lines of symmetry:

• \(y = x\)
• \(y = -x\)

+ - ↺

The points are \(A = (\alpha, \beta), B = (\beta, \alpha), C = (-\alpha, -\beta), D = (-\beta, -\alpha)\) \(AD\) has gradient \(\frac{\beta+\alpha}{\alpha+\beta} = 1\), \(BC\) has the same gradient. \(AB\) has gradient \(\frac{\alpha-\beta}{\beta-\alpha} = -1\), as does \(CD\). Therefore it has two sets of perpendicular and parallel sides, hence a rectangle. The area is \(|AD||AB| = \sqrt{2(\alpha+\beta)^2}\sqrt{2(\alpha-\beta)^2} = 2(\alpha^2-\beta^2)\) The squared area is \(4(\alpha^4+\beta^4 - 2 \alpha^2\beta^2) = 4(u - 2v^2)\) ie the area is \(2\sqrt{u-2v^2}\) When \(u = 81, v = 4\) we have the area is \(2 \sqrt{81 - 2 \cdot 16} = 14\) as required.

The curve \(C\) has equation \[ y= a^{\sin (\pi \e^ x)}\,, \] where \(a>1\).

1. Find the coordinates of the stationary points on \(C\).
2. Use the approximations \(\e^t \approx 1+t\) and \(\sin t \approx t\) (both valid for small values of \(t\)) to show that \[ y\approx 1-\pi x \ln a \; \] for small values of \(x\).
3. Sketch \(C\).
4. By approximating \(C\) by means of straight lines joining consecutive stationary points, show that the area between \(C\) and the \(x\)-axis between the \(k\)th and \((k+1)\)th maxima is approximately \[ \Big( \frac {a^2+1}{2a} \Big) \ln \Big ( 1+ \big( k-\tfrac34)^{-1} \Big)\,. \]

Prove that \[ \tan \left ( \tfrac14 \pi -\tfrac12 x \right)\equiv \sec x -\tan x\,. \tag{\(*\)} \]

1. Use \((*)\) to find the value of \(\tan\frac18\pi\,\). Hence show that \[ \tan \tfrac{11}{24} \pi = \frac{\sqrt3 + \sqrt2 -1}{\sqrt3 -\sqrt6+1}\;. \]
2. Show that \[ \frac{\sqrt3 + \sqrt2 -1}{\sqrt3 -\sqrt6+1}= 2+\sqrt2+\sqrt3+\sqrt6\,. \]
3. Use \((*)\) to show that \[ \tan \tfrac1{48}\pi = \sqrt{16+10\sqrt2+8\sqrt3 +6\sqrt6 \ }-2-\sqrt2-\sqrt3-\sqrt6\,. \]

The polynomial \(\p(x)\) is of degree 9 and \(\p(x)-1\) is exactly divisible by \((x-1)^5\).

1. Find the value of \(\p(1)\).
2. Show that \(\p'(x)\) is exactly divisible by \((x-1)^4\).
3. Given also that \(\p(x)+1\) is exactly divisible by \((x+1)^5\), find \(\p(x)\).

Expand and simplify \((\sqrt{x-1}+1)^2\,\).

1. Evaluate \[ \int_{5}^{10} \frac{ \sqrt{x+2\sqrt{x-1} \;} + \sqrt{x-2\sqrt{x-1} \;} } {\sqrt{x-1}} \,\d x\;. \]
2. Find the total area between the curve \[ y= \frac{\sqrt{x-2\sqrt{x-1}\;}}{\sqrt{x-1}\;} \] and the \(x\)-axis between the points \(x=\frac54\) and \(x=10\).
3. Evaluate \[ \int_{\frac54}^{10} \frac{ \sqrt{x+2\sqrt{x-1}\;} + \sqrt{x-2\sqrt{x+1}+2 \;} } {\sqrt{x^2-1} } \;\d x\;. \]

The Fibonacci sequence \(F_1\), \(F_2\), \(F_3\), \(\ldots\) is defined by \(F_1=1\), \(F_2= 1\) and \[ F_{n+1} = F_n+F_{n-1} \qquad\qquad (n\ge 2). \] Write down the values of \(F_3\), \(F_4\), \(\ldots\), \(F_{10}\). Let \(\displaystyle S=\sum_{i=1}^\infty \dfrac1 {F_i}\,\). \begin{questionparts} \item Show that \(\displaystyle \frac 1{F_i} > \frac1{2F_{i-1}}\,\) for \(i\ge4\) and deduce that \(S > 3\,\). Show also that \(S < 3\frac23\,\). \item Show further that \(3.2 < S < 3.5\,\). \end{questionpart}

Let \(y= (x-a)^n \e^{bx} \sqrt{1+x^2}\,\), where \(n\) and \(a\) are constants and \(b\) is a non-zero constant. Show that \[ \frac{\d y}{\d x} = \frac{(x-a)^{n-1} \e^{bx} \q(x)}{\sqrt{1+x^2}}\,, \] where \(\q(x)\) is a cubic polynomial. Using this result, determine:

1. $\displaystyle \int \frac {(x-4)^{14} \e^{4x}(4x^3-1)} {\sqrt{1+x^2\;}} \, \d x\,;\(
2. \)\displaystyle \int \frac{(x-1)^{21}\e^{12x}(12x^4-x^2-11)} {\sqrt{1+x^2\;}}\,\d x\,;\(
3. \)\displaystyle \int \frac{(x-2)^{6}\e^{4x}(4x^4+x^3-2)} {\sqrt{1+x^2\;}}\,\d x\,.$

The non-collinear points \(A\), \(B\) and \(C\) have position vectors \(\bf a\), \(\bf b\) and \(\bf c\), respectively. The points \(P\) and \(Q\) have position vectors \(\bf p\) and \(\bf q\), respectively, given by \[ {\bf p}= \lambda {\bf a} +(1-\lambda){\bf b} \text{ \ \ \ and \ \ \ } {\bf q}= \mu {\bf a} +(1-\mu){\bf c} \] where \(0<\lambda<1\) and \(\mu>1\). Draw a diagram showing \(A\), \(B\), \(C\), \(P\) and \(Q\). Given that \(CQ\times BP = AB\times AC\), find \(\mu\) in terms of \(\lambda\), and show that, for all values of \(\lambda\), the the line \(PQ\) passes through the fixed point \(D\), with position vector \({\bf d}\) given by \({\bf d= -a +b +c}\,\). What can be said about the quadrilateral \(ABDC\)?

1. A uniform lamina \(OXYZ\) is in the shape of the trapezium shown in the diagram. It is right-angled at \(O\) and \(Z\), and \(OX\) is parallel to \(YZ\). The lengths of the sides are given by \(OX=9\,\)cm, \(XY=41\,\)cm, \(YZ=18\,\)cm and \(ZO=40\,\)cm. Show that its centre of mass is a distance \(7\,\)cm from the edge \(OZ\).
   

   + - ↺
2. The diagram shows a tank with no lid made of thin sheet metal. The base \(OXUT\), the back \(OTWZ\) and the front \(XUVY\) are rectangular, and each end is a trapezium as in part (i). The width of the tank is \(d\,\)cm.
   

   + - ↺
   Show that the centre of mass of the tank, when empty, is a distance \[ \frac {3(140+11d)}{5(12+d)}\,\text{cm} \] from the back of the tank. The tank is then filled with a liquid. The mass per unit volume of this liquid is \(k\) times the mass per unit area of the sheet metal. In the case \(d=20\), find an expression for the distance of the centre of mass of the filled tank from the back of the tank.

\(\,\)