Edexcel Pure Year 2

Treat $\sin x$ as the variable while factorising, then find every angle in the stated interval.

$2\sin^2x+\sin x-1=(2\sin x-1)(\sin x+1)$. Hence $\sin x=\tfrac12$ or $\sin x=-1$.

In the given interval, $\sin x=\tfrac12$ gives $x=\tfrac\pi6$ and $x=\tfrac{5\pi}6$. The equation $\sin x=-1$ gives $x=\tfrac{3\pi}2$.

Answer: $x=\dfrac\pi6,\ \dfrac{5\pi}6,\ \dfrac{3\pi}2$.

Use $\mathrm{d}y/\mathrm{d}x=(\mathrm{d}y/\mathrm{d}t)/(\mathrm{d}x/\mathrm{d}t)$, and find the point as well as the gradient.

At $t=2$, $x=5$ and $y=2$, so the required point is $(5,2)$.

$\mathrm{d}x/\mathrm{d}t=2t$ and $\mathrm{d}y/\mathrm{d}t=3t^2-3$. Thus $\mathrm{d}y/\mathrm{d}x=(3t^2-3)/(2t)$, which is $9/4$ at $t=2$.

The tangent through $(5,2)$ with gradient $9/4$ is $y-2=\tfrac94(x-5)$.

Answer: $y-2=\dfrac94(x-5)$.

In integration by parts, differentiate $x$ and integrate $e^x$.

Take $u=x$ and $\mathrm{d}v=e^x\,\mathrm{d}x$, so $\mathrm{d}u=\mathrm{d}x$ and $v=e^x$.

$$\begin{aligned}\int_0^1 xe^x\,\mathrm{d}x&=[xe^x]_0^1-\int_0^1e^x\,\mathrm{d}x\\&=e-(e-1)=1.\end{aligned}$$

Answer: $1$.

The integral is correctly calculated as zero: the contribution below the axis is $-1/2$ and the contribution above the axis is $1/2$.

Total area adds their magnitudes. Equivalently, $\int_{-1}^1|x|\,\mathrm{d}x=\tfrac12+\tfrac12=1$.