Edexcel Pure Year 1

Factorise, then decide where the upward-opening quadratic is on or below the horizontal axis.

Factorising gives $x^2-5x+6=(x-2)(x-3)$, so the roots are $2$ and $3$.

The coefficient of $x^2$ is positive, so the graph is below the axis between the roots. Equality is allowed, so both endpoints are included.

Answer: $2\leq x\leq 3$.

Solve $f^{\prime}(x)=0$, substitute back into $f$, then use the sign of $f^{\prime\prime}$.

$f^{\prime}(x)=3x^2-12x+9=3(x-1)(x-3)$, so the stationary points occur at $x=1$ and $x=3$.

$f(1)=5$ and $f(3)=1$, giving the points $(1,5)$ and $(3,1)$.

$f^{\prime\prime}(x)=6x-12$. Since $f^{\prime\prime}(1)=-6<0$, $(1,5)$ is a local maximum. Since $f^{\prime\prime}(3)=6>0$, $(3,1)$ is a local minimum.

Answer: Local maximum $(1,5)$; local minimum $(3,1)$.

Find an antiderivative and subtract its value at the lower limit from its value at the upper limit.

An antiderivative of $2x+1$ is $x^2+x$.

$$\int_1^3(2x+1)\,\mathrm{d}x=[x^2+x]_1^3=(9+3)-(1+1)=10.$$

Answer: $10$.

At $x=0$, the left side is $\sqrt{4}=2$, whereas the claimed right side is $-2$. Squaring removes a sign, so taking the non-negative square root does not always recover the original expression.

The correct identity is $\sqrt{(x-2)^2}=|x-2|$. This equals $x-2$ when $x\geq 2$, and $2-x$ when $x<2$.