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1993 Paper 1 Q6
Let \(N=10^{100}.\) The graph of \[ \mathrm{f}(x)=\frac{x^{N}}{1+x^{N}}+2 \] for \(-3\leqslant x\leqslant3\) is sketched in the following diagram. \noindent
1992 Paper 3 Q4
A set of curves \(S_{1}\) is defined by the equation \[ y=\frac{x}{x-a}, \] where \(a\) is a constant which is different for different members of \(S_{1}.\) Sketch on the same axes the curves for which \(a=-2,-1,1\) and \(2\). A second of curves \(S_{2}\) is such that at each intersection between a member of \(S_{2}\) and a member of \(S_{1}\) the tangents of the intersecting curves are perpendicular. On the same axes as the already sketched members of \(S_{1},\) sketch the member of \(S_{2}\) that passes through the point \((1,-1)\). Obtain the first order differential equation for \(y\) satisfied at all points on all members of \(S_{1}\) (i.e. an equation connecting \(x,y\) and \(\mathrm{d}y/\mathrm{d}x\) which does not involve \(a\)). State the relationship between the values of \(\mathrm{d}y/\mathrm{d}x\) on two intersecting curves, one from \(S_{1}\) and one from \(S_{2},\) at their intersection. Hence show that the differential equation for the curves of \(S_{2}\) is \[ x=y(y-1)\dfrac{\mathrm{d}y}{\mathrm{d}x}. \] Find an equation for the member of \(S_{2}\) that you have sketched.
1991 Paper 2 Q2
The equation of a hyperbola (with respect to axes which are displaced and rotated with respect to the standard axes) is \[ 3y^{2}-10xy+3x^{2}+16y-16x+15=0.\tag{\(\dagger\)} \] By differentiating \((\dagger)\), or otherwise, show that the equation of the tangent through the point \((s,t)\) on the curve is \[ y=\left(\frac{5t-3s+8}{3t-5s+8}\right)x-\left(\frac{8t-8s+15}{3t-5s+8}\right). \] Show that the equations of the asymptote (the limiting tangents as \(s\rightarrow\infty\)) are \[ y=3x-4\qquad\mbox{ and }\qquad3y=x-4. \] {[}Hint: You will need to find a relationship between \(s\) and \(t\) which is valid in the limit as \(s\rightarrow\infty.\){]} Show that the angle between one asymptote and the \(x\)-axis is the same as the angle between the other asymptote and the \(y\)-axis. Deduce the slopes of the lines that bisect the angles between the asymptotes and find the equations of the axes of the hyperbola.
Solution: \begin{align*} && 0 &= 3y^{2}-10xy+3x^{2}+16y-16x+15 \\ \Rightarrow && 0 &= 6y \frac{\d y}{\d x} - 10x \frac{\d y}{\d x} - 10y + 6x+ 16 \frac{\d y}{\d x } - 16 \\ &&&= \frac{\d y}{\d x} \left (6y - 10x +16 \right) - (10y-6x+16) \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{5y-3x+8}{3y-5x+8} \\ \Rightarrow && \frac{y-t}{x-s} &= \frac{5t-3s+8}{3t-5s+8} \\ && y &= \left(\frac{5t-3s+8}{3t-5s+8}\right)x -\left(\frac{5t-3s+8}{3t-5s+8}\right)s + t \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{(5ts-3s^2+8s)-(3t^2-5st+8t)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{(5ts-3s^2+8s)-(3t^2-5st+8t)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8s-8t-(3s^2+3t^2-10st)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8s-8t-(16s-16t-15)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8t-8s+15}{3t-5s+8} \\ \end{align*} While \(x \to \infty\) we still have \(3 \frac{y^2}{x^2} - 10 \frac{y}{x} + 3 + 16 \frac{y}{x^2} - 16\frac{1}{x} + 15 \frac{1}{x^2} = 0\), ie if \(\frac{y}{x} = k\), then \(3k^2 - 10k + 3 \to 0 \Rightarrow k \to 3, \frac13\). Therefore, as \(s \to \infty\) we can write \begin{align*} && y &= \left(\frac{5\frac{t}{s}-3+8\frac{1}{s}}{3\frac{t}{s}-5+8\frac1{s}}\right)x - \frac{8\frac{t}s-8+15\frac{1}{s}}{3\frac{t}{s}-5+8\frac{1}{s}} \\ k \to 3: &&& \to \left(\frac{15-3}{9-5}\right)x - \frac{24-8}{9-5} \\ &&&= 3x - 4 \\ k \to \frac13: && &\to \left(\frac{\frac53-3}{1-5}\right)x - \frac{\frac83-8}{1-5} \\ &&&= \frac13 x - \frac43 \end{align*} Therefore the equations are \(y = 3x-4\) and \(3y=x-4\) The lines are parallel to \(y = 3x\) and \(y = \frac13x\), so by considering the triangles formed with the origin and a point \(1\) along the \(x\) or \(y\) axis we can see the angles are identical. This means the line \(y = x\) is parallel to one axis and \(y = -x\) is parallel to the other. They must meet where our two lines meet which is \((1,-1)\), so our lines are \(y = x-2\) and \(y = -x\)
1990 Paper 3 Q7
The points \(P\,(0,a),\) \(Q\,(a,0)\) and \(R\,(a,-a)\) lie on the curve \(C\) with cartesian equation \[ xy^{2}+x^{3}+a^{2}y-a^{3}=0,\qquad\mbox{ where }a>0. \] At each of \(P,Q\) and \(R\), express \(y\) as a Taylor series in \(h\), where \(h\) is a small increment in \(x\), as far as the term in \(h^{2}.\) Hence, or otherwise, sketch the shape of \(C\) near each of these points. Show that, if \((x,y)\) lies on \(C\), then \[ 4x^{4}-4a^{3}x-a^{4}\leqslant0. \] Sketch the graph of \(y=4x^{4}-4a^{3}x-a^{4}.\) Given that the \(y\)-axis is an asymptote to \(C\), sketch the curve \(C\).
Solution: \begin{align*} && 0 &= xy^{2}+x^{3}+a^{2}y-a^{3} \\ \frac{\d }{\d x} : && 0 &= y^2+2xyy' + 3x^2+a^2y' \\ \Rightarrow && y' &= -\frac{y^2+3x^2}{a^2+2xy} \\ \\ \frac{\d^2 }{\d x^2}: && 0 &= 2yy'+2yy'+2x(y')^2+2xyy''+6x+a^2y'' \\ \Rightarrow && y'' &= -\frac{4yy'+2x(y')^2+6x}{a^2+2xy} \\ \\ P: && y &= a \\ && y' &= -\frac{a^2}{a^2} = -1 \\ && y'' &= -\frac{-4a}{a^2} = \frac{4}{a} \\ \Rightarrow && y &\approx a - h+\frac{2}{a}h^2 \\ \\ Q: && y &= 0 \\ && y' &= -\frac{3a^2}{a^2} = -3 \\ && y'' &= -\frac{18a+6a}{a^2} = -\frac{24}{a} \\ \Rightarrow && y &\approx 0-3h-\frac{12}{a}h \\ \\ R: && y &= -a \\ && y' &= -\frac{a^2+3a^2}{a^2-2a^2} = 4 \\ && y'' &= -\frac{-16a+32a+6a}{a^2-2a^2} = \frac{22}{a} \\ \Rightarrow && y &\approx -a+4h + \frac{11}{a}h^2 \end{align*} Alternatively: \begin{align*} && 0 &= xy^{2}+x^{3}+a^{2}y-a^{3} \\ P(0,a): && y &\approx a + c_1h + c_2h^2 \\ && 0 &= h(a+c_1h)^2 + a^2(a + c_1h + c_2h^2)-a^3 \\ &&&= a^3-a^3 + (a^2+a^2c_1)h+(2ac_1+a^2c_2)h^2 \\ \Rightarrow && c_1 &=-1, c_2 =\frac{2}{a} \\ \Rightarrow && y &\approx a - h + \frac{2}{a}h^2 \\ \\ Q(a,0): && y &\approx c_1h + c_2h^2 \\ && 0 &= (a+h)(c_1h)^2+(a+h)^3+a^2(c_1h + c_2h^2 )-a^3 \\ &&&= a^3-a^3+(3a^2+a^2c_1)h + (ac_1^2+3a+a^2c_2)h^2 + \cdots \\ \Rightarrow && c_1 &=-3, c_2 = -\frac{12}{a} \\ \Rightarrow && y &\approx -3h -\frac{12}{a}h \\ \\ R(a,-a): && y &\approx -a + c_1h + c_2h^2 \\ && 0 &= (a+h)(-a + c_1h+c_2h^2)^2+(a+h)^3+a^2(-a + c_1h + c_2h^2)-a^3 \\ &&&= (a^2-2a^2c_1+3a^2+a^2c_1)h+(-2ac_1+c_1^2+\cdots)h^2 \\ \Rightarrow && c_1 &=4, c_2 = \frac{11}{a} \\ \Rightarrow && y &\approx -a + 4h + \frac{11}{a} \end{align*}
1989 Paper 2 Q4
The function \(\mathrm{f}\) is defined by \[ \mathrm{f}(x)=\frac{\left(x-a\right)\left(x-b\right)}{\left(x-c\right)\left(x-d\right)}\qquad\left(x\neq c,\ x\neq d\right), \] where \(a,b,c\) and \(d\) are real and distinct, and \(a+b\neq c+d\). Show that \[ \frac{x\mathrm{f}'(x)}{\mathrm{f}(x)}=\left(1-\frac{a}{x}\right)^{-1}+\left(1-\frac{b}{x}\right)^{-1}-\left(1-\frac{c}{x}\right)^{-1}-\left(1-\frac{d}{x}\right)^{-1}, \] \((x\neq0,x\neq a,x\neq b)\) and deduce that when \(\left|x\right|\) is much larger than each of \(\left|a\right|,\left|b\right|,\left|c\right|\) and \(\left|d\right|,\) the gradient of \(\mathrm{f}(x)\) has the same sign as \((a+b-c-d).\) It is given that there is a real value of real value of \(x\) for which \(\mathrm{f}(x)\) takes the real value \(z\) if and only if \[ [\left(c-d\right)^{2}z+\left(a-c\right)\left(b-d\right)+\left(a-d\right)\left(b-c\right)]^{2}\geqslant4\left(a-c\right)\left(b-d\right)\left(a-d\right)\left(b-c\right). \] Describe briefly a method by which this result could be proved, but do not attempt to prove it. Given that \(a < b\) and \(a < c < d\), make sketches of the graph of \(\mathrm{f}\) in the four distinct cases which arise, indicating the cases for which the range of \(\mathrm{f}\) is not the whole of \(\mathbb{R}.\)
Solution: Notice that \(\ln f(x) = \ln (x - a) + \ln (x-b) - \ln (x-c) - \ln (x-d)\) therefore: \begin{align*} \frac{\d}{\d x}: && \frac{f'(x)}{f(x)} &= (x-a)^{-1}+(x-b)^{-1}-(x-c)^{-1} - (x-d)^{-1} \\ &&&= \frac{1}{x} \left ( (1-\frac{a}{x})^{-1}+(1-\frac{b}{x})^{-1}-(1-\frac{c}{x})^{-1} - (1-\frac{d}{x})^{-1}\right) \end{align*} Multiplying by \(x\) gives the desired result. When \(|x|\) is very large then: \begin{align*} \frac{x f'(x)}{f(x)} &\approx 1 + \frac{a}{x} + o(\frac{1}{x^2})+ 1 + \frac{b}{x} + o(\frac{1}{x^2})-(1 + \frac{c}{x} + o(\frac{1}{x^2}))-(1 + \frac{d}{x} + o(\frac{1}{x^2})) \\ &= \frac{a+b-c-d}{x} + o(x^{-2}) \end{align*} Dividing by \(x\) we obtain \(\frac{f'(x)}{f(x)} \approx \frac{a+b-c-d}{x^2} + o(x^{-3})\) if \(|x|\) is sufficiently large this will be dominated by the \(\frac{a+b-c-d}{x^2}\) term which will have the same sign as \((a+b-c-d)\). When \(|x|\) is very large all of the brackets will have the same sign, and therefore \(f(x)\) will be positive, and so \(f'(x)\) must have the same sign as \(a+b-c-d\). To prove this result, we could set \(f(x) = k\) and rearrange to form a quadratic in \(x\). We could then check the discriminant is non-zero. Case 1: \(a < c < d < b\) and \(a+b > c+d \Rightarrow\) not all values reached and approx asymtope from below on the right and above on the left.
1989 Paper 3 Q4
Sketch the curve whose cartesian equation is \[ y=\frac{2x(x^{2}-5)}{x^{2}-4}, \] and give the equations of the asymptotes and of the tangent to the curve at the origin. Hence, or otherwise, determine (giving reasons) the number of real roots of the following equations:
- \(4x^{2}(x^{2}-5)=(5x-2)(x^{2}-4)\);
- \(4x^{2}(x^{2}-5)^{2}=(x^{2}-4)^{2}(x^{2}+1)\);
- \(4z^{2}(z-5)^{2}=(z-4)^{2}(z+1)\).
Solution:
- \begin{align*} && 4x^{2}(x^{2}-5)&=(5x-2)(x^{2}-4) \\ && \frac{2x(x^2-5)}{x^2-4} &= \frac{5x-2}{2x} = \frac52 -\frac1x \end{align*} Therefore it will have two roots as the hyperbola will intersect our graph in two places. (In the upper left and lower right quadrants).
- \begin{align*}
&& 4x^{2}(x^{2}-5)^{2}&=(x^{2}-4)^{2}(x^{2}+1) \\
&& \frac{2x(x^2-5)}{x^2-4} &= \frac{(x^2-4)(x^2+1)}{2x}
\end{align*}
No solutions \begin{align*} && 4z^{2}(z-5)^{2}&=(z-4)^{2}(z+1) \\ && \frac{2z(z^2-5)}{z^2-4} &= \frac{(z+5)(z-4)(z+1)}{(z-5)(z+4)} \end{align*}
5 solutions
1988 Paper 3 Q1
Sketch the graph of \[ y=\frac{x^{2}\mathrm{e}^{-x}}{1+x}, \] for \(-\infty< x< \infty.\) Show that the value of \[ \int_{0}^{\infty}\frac{x^{2}\mathrm{e}^{-x}}{1+x}\,\mathrm{d}x \] lies between \(0\) and \(1\).
Solution:
