Edexcel Core Pure 2

Write $8=8e^{2k\pi i}$, then take cube roots with three distinct arguments.

The roots have modulus $8^{1/3}=2$ and arguments $2k\pi/3$, for $k=0,1,2$.

Using $z=2(\cos(2k\pi/3)+i\sin(2k\pi/3))$ gives $2$, $-1+i\sqrt3$ and $-1-i\sqrt3$. These are distinct and account for all roots of the cubic.

Answer: $2,\ -1+i\sqrt3,\ -1-i\sqrt3$.

Use the exponential definition of $\sinh x$, and set $u=e^x>0$.

$(e^x-e^{-x})/2=3/4$. With $u=e^x>0$, this becomes $u-1/u=3/2$.

Multiplying by $2u$ gives $2u^2-3u-2=0$, or $(2u+1)(u-2)=0$.

The root $u=-1/2$ is impossible because $e^x>0$. Thus $u=2$, so $x=\ln2$.

Answer: $x=\ln2$.

Find the two auxiliary roots, then use both initial conditions to determine the constants.

The auxiliary equation is $m^2-3m+2=(m-1)(m-2)=0$, giving distinct roots $1$ and $2$.

The general solution is $y=Ae^x+Be^{2x}$, with $y^{\prime}=Ae^x+2Be^{2x}$.

The initial conditions give $A+B=1$ and $A+2B=0$. Subtraction gives $B=-1$, then $A=2$. Hence $y=2e^x-e^{2x}$.

Answer: $y=2e^x-e^{2x}$.

The student has integrated $r$; the correct area formula is $\tfrac12\int r^2\,\mathrm{d}\theta$.

Here the area is $$\begin{aligned}A&=\frac12\int_{-\pi/2}^{\pi/2}4\cos^2\theta\,\mathrm{d}\theta\\&=\int_{-\pi/2}^{\pi/2}(1+\cos2\theta)\,\mathrm{d}\theta\\&=\left[\theta+\frac12\sin2\theta\right]_{-\pi/2}^{\pi/2}\\&=\pi.\end{aligned}$$

As a check, multiplying the polar equation by $r$ gives $x^2+y^2=2x$, or $(x-1)^2+y^2=1$: a circle of radius $1$, whose area is $\pi$.