IB Maths AA HL · Paper 3

1. Both $x^4$ and $x^2$ are never negative, so the smallest value of $y$ comes at $x=0$, where $y=1$.
2. Only even powers appear, so the curve is symmetrical in the $y$-axis. It looks like a flattened U with its minimum at $(0,1)$.
3. Label the $y$-intercept $1$. There are no $x$-intercepts to label.

1. Spot the perfect square: $x^4-2x^2+1=(x^2-1{)}^2$, so $y\ge 0$ and it equals zero at $x=\pm 1$.
2. The curve is a W shape, symmetrical in the $y$-axis, with minima touching the $x$-axis at $(-1,0)$ and $(1,0)$ and a local maximum at $(0,1)$.
3. Label the $x$-intercepts $\pm 1$ and the $y$-intercept $1$.

1. A function $g$ is odd when $g(-x)=-g(x)$ for every $x$.
2. Replace $x$ with $-x$ in $f^{\prime }(x)$ and simplify.
3. Take out a factor of $-1$ and recognise $f^{\prime }(x)$.

1. At a double root both $f(\alpha )=0$ and $f^{\prime }(\alpha )=0$, so the curve meets the $x$-axis at $x=\alpha$ with zero gradient.
2. On the graph, that means the curve touches the $x$-axis at $x=\alpha$ without crossing it: it has a turning point sitting on the axis there.

1. You need two facts about $-\alpha$: that $f(-\alpha )=0$ and that $f^{\prime }(-\alpha )=0$.
2. $f$ has only even powers, so it is an even function: $f(-\alpha )=f(\alpha )=0$.
3. From part (c)(ii), $f^{\prime }$ is odd, so $f^{\prime }(-\alpha )=-f^{\prime }(\alpha )=0$.
4. Both conditions hold, so $-\alpha$ is a double root too.

Another way. The quartic has no $x^3$ term, so its four roots add to zero; with $\alpha$ counted twice, the other double root must be $-\alpha$.

1. Set $f^{\prime }(\alpha )=0$ and factorise out $2\alpha$.
2. $\alpha$ cannot be $0$, because $f(0)=d\ne 0$, so $0$ is not even a root.
3. That leaves the bracket equal to zero; rearrange for ${\alpha }^2$.

Another way. Complete the square, $f(x)={\left(x^2+\frac{b}{2}\right)}^2+d-\frac{b^2}{4}$; for a repeated root the constant must vanish and ${\alpha }^2+\frac{b}{2}=0$.

1. Now use $f(\alpha )=0$. Since ${\alpha }^4=({\alpha }^2{)}^2$, substitute ${\alpha }^2=-\frac{b}{2}$ throughout.
2. Simplify and make $d$ the subject.

Another way. The roots are $\alpha ,\alpha ,-\alpha ,-\alpha$, and their product equals $d$, so $d={\alpha }^4=\frac{b^2}{4}$.

1. Differentiate $f^{\prime }(x)$ from part (c)(i) again.
2. Set $f^{\prime \prime }(x)=0$ and solve for $x^2$.
3. Because $b<0$, the value $-\frac{b}{6}$ is positive, so you can take square roots.

1. Let $p=\sqrt{-\frac{b}{6}}$, so $p^2=-\frac{b}{6}$ and $p^4=\frac{b^2}{36}$. Find the $y$-coordinate $f(p)$.
2. The given gradient uses $(-b{)}^{3/2}$, and $p=\frac{(-b{)}^{1/2}}{\sqrt{6}}$, so the gradient times $p$ simplifies to a multiple of $b^2$.
3. The point $(p,f(p))$ lies on the tangent, so substitute it and solve for $c$.
4. Collect the $b^2$ terms over $36$.

Another way. Use $c=f(p)-f^{\prime }(p)p$ with $f^{\prime }(p)=p(4p^2+2b)=\frac{4b}{3}p$, so $f^{\prime }(p)p=\frac{4b}{3}\left(-\frac{b}{6}\right)=-\frac{2b^2}{9}$.

1. The particle is at $O$ when $x=0$, so $\sin \left(2t+\frac{\pi }{3}\right)=0$.
2. At $t=0$ the angle is already $\frac{\pi }{3}$, so the first positive zero of sine after that is at angle $\pi$.
3. Solve for $t$.

Another way. Graph the displacement on your GDC for $0\le t\le \pi$ and read off the first $t$-axis crossing.

1. Differentiate with the chain rule: the inner function $nt+b$ has derivative $n$.
2. Differentiate again, then spot $A\sin (nt+b)=s$ and finish the sentence.

1. Separate the variables: $v$ with $\mathrm{d}v$, $s$ with $\mathrm{d}s$, and integrate both sides.
2. Multiply through by $2$ and rename the constant $D=2C$.
3. Find $D$ from a known point of the motion. At an end point, $s=\pm A$ and the particle is momentarily at rest, so $v=0$.
4. Substitute $D$ back and factorise.

Another way. Use $t=0$ instead: $s=A\sin b$ and $v=nA\cos b$, so $D=n^2{A}^2({\cos }^{2}b+{\sin }^{2}b)=n^2{A}^2$.

1. From $v^2=n^2(A^2-s^2)$, speed is largest when $s^2$ is smallest, that is at $s=0$ (passing through $O$).
2. Take the positive square root for speed.

Another way. From part (b), $v=nA\cos (nt+b)$, and $|\cos |\le 1$ gives the same maximum speed $nA$.

1. Integrate the density between the two given limits.
2. Use the standard integral $\int \frac{1}{\sqrt{A^2-s^2}}\mathrm{d}s=\arcsin \frac{s}{A}$ from the formula booklet.
3. Substitute the limits: both reduce to $\arcsin$ of $\pm \frac{1}{2}$.
4. Use $\arcsin \frac{1}{2}=\frac{\pi }{6}$ and that $\arcsin$ is odd.

Another way. The integrand is even, so you can double the integral from $0$ to $\frac{A}{2}$, which gives $\frac{2}{\pi }\cdot \frac{\pi }{6}$.

1. From part (c)(ii), the speed at displacement $s$ is $n\sqrt{A^2-s^2}$.
2. Rewrite the density with that expression in the denominator by multiplying top and bottom by $n$.
3. Read off $k$ as the numerator.

1. $f(s)$ is an even function, so $sf(s)$ is odd.
2. The integral of an odd function over the symmetric interval from $-A$ to $A$ is zero, since the areas either side of $O$ cancel.