Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section A

The factor theorem gives a factor $x-1$. Polynomial division and then factorization giveConsequently all three roots of a polynomial are

The chain rule for the cosine gives

The product rule and the derivative of the exponential function give

With the change of variables $u=3+2x$, $du=2dx$, an antiderivative isHere $C$ is the constant of integration.

Integration by parts with $u=\ln x$ and $dv=xdx$ givesThe remaining integral is $\frac{1}{2}\int xdx$.

Completing the square puts the equation of the circle in the formHence its centre is $(0,-3)$ and its radius is $2$.

This arithmetic series has $101$ terms, from $1$ to $201$. Thus

The geometric series has first term $1$ and common ratio $1/3$, whose absolute value is less than one. Therefore

On this parametric curve, $x=1$ permits $t=0$ or $t=-2$, but only $t=0$ also gives $y=0$. The chain rule gives the slopeAt the required point, $t=0$, so

The double-angle identity for the sine factors the equation as $2\cos \theta (\sin \theta -\cos \theta )=0$. The first factor gives $\theta =\pi /2$ in the specified interval; the second gives $\tan \theta =1$ and hence $\theta =\pi /4$. ThusKeeping the first factor avoids losing the solution where division by $\cos \theta$ would be invalid.

Call the centre of the circle $O$, the left endpoint on the diameter $A$, the point of tangency $T$, and the external intersection $E$. Since $OA=OT$ are radii, the isosceles triangle $AOT$ has equal base angles: $\angle OAT=\angle ATO=30^{\circ }$. Therefore $\angle AOT=120^{\circ }$, and, because $OA$ and $OE$ point in opposite directions, $\angle TOE=60^{\circ }$. The tangent to a circle $TE$ is perpendicular to the radius $OT$, so in the right triangle $OTE$,

Rationalizing a denominator by multiplying by $\sqrt{5}-1$ gives $2/(\sqrt{5}+1)=(\sqrt{5}-1)/2$. Consequently the sum inside the parentheses is $\sqrt{5}$, and the entire expression is

The function is nonnegative and satisfies ${\cos }^2(\pi -x)={\cos }^{2}x$. Its derivative is $-\sin (2x)$, so it decreases from $(0,1)$ to its minimum $(\pi /2,0)$ and then increases to $(\pi ,1)$. These three points all have horizontal tangent lines.

The denominator vanishes only at $x=0$, giving a vertical asymptote. Since $1-e^x=-x+O(x^2)$, the function tends to $+\infty$ from the left and $-\infty$ from the right. Its derivative isso each branch is increasing. The horizontal asymptotes are $y=1$ as $x\to -\infty$ and $y=0$ as $x\to +\infty$. The left branch lies above $1$; the right branch lies below $0$, as in the preceding figure.

If tangency occurs at $x_{\ast }$, the slope must satisfy both $k=2e^{2x_{\ast }}$ and $k{x}_{\ast }=e^{2x_{\ast }}$. Eliminating $k$ gives $2x_{\ast }=1$, so the tangent through the origin to an exponential curve has

The preceding tangent line calculation gives $x_{\ast }=1/2$ and $y_{\ast }=e^{2x_{\ast }}=e$. The point of tangency is