IB Maths AI HL · Paper 2

The bottom is 10 m in from the edge on each side, over a depth of 6 m.

(i) By similar triangles, or using $\tan \theta =0.6$ with the half-width 40:

(ii) The reservoir is the large pyramid with apex V minus the small pyramid below it, whose height is $24-6=18$.

The tower is 90 m high and the blade is 40 m long.

(i) 12 rotations a minute means one every 5 seconds.

(ii) $360^{\circ }$ in 5 seconds.

In $h(t)=90-40\cos (72t^{\circ })$:

Start at the minimum, $(0,50)$, since $\cos 0=1$ gives $h=50$. The graph rises to the maximum $(2.5,130)$ and falls back to a minimum at $(5,50)$. It is a smooth wave, like an upside-down cosine.

(i) At $t=2$:

(ii) Solve $90-40\cos (72t)=100$, which is $\cos (72t)=-0.25$, so $72t=104.5^{\circ }$ and $255.5^{\circ }$ in the first rotation.

Point C is above 110 m when $90-40\cos \varphi >110$, that is $\cos \varphi <-0.5$, so for $\varphi$ between $120^{\circ }$ and $240^{\circ }$. That is $120^{\circ }$ of the rotation, which now takes 1 second.

With $X\sim \mathrm{N}(10,3^2)$:

The throws are independent, so multiply.

The chance of a point on a turn is $0.8143^3$. A turn without a point is the complement.

(i) A turn scores with probability $0.8143^3\approx 0.5399$. Over ten turns the score is $Y\sim \mathrm{B}(10,0.5399)$.

(ii) At least 5 points:

(iii) At least 5 and less than 8, which is 5, 6 or 7 points:

(iv) A conditional probability.

(i) Anticlockwise rotation by $\frac{\pi }{6}$ ($A$), the reflection in the line $y=\frac{x}{\sqrt{3}}$ through angle $\frac{\pi }{6}$ ($F$), and the clockwise rotation by $\frac{\pi }{3}$ ($C$):

(ii) The overall effect applies $A$ first, then $F$, then $C$.

(iii) Square it.

(iv) Doing the overall movement twice returns the drone to where it started.

(i) Plot $1-\mathrm{i}$ at the point $(1,-1)$ on the Argand diagram.

(ii) Its modulus is $\sqrt{2}$ and its argument is $-\frac{\pi }{4}$.

(i) Since ${\mathrm{e}}^{-\mathrm{i}\pi /2}=-\mathrm{i}$:

(ii) Use part (a): $1-\mathrm{i}=\sqrt{2}{\mathrm{e}}^{-\mathrm{i}\pi /4}$.

Since $y=\dot{x}$, we have $\dot{y}=\ddot{x}$. Substitute into $\ddot{x}+3\dot{x}+1.25x=0$.

Write $\dot{x}=y$ and $\dot{y}=-1.25x-3y$ as a matrix equation.

(i) Solve $\det (A-\lambda I)=0$.

(ii) For $\lambda =-2.5$: the first row gives $2.5a+b=0$, so an eigenvector is $\left(\begin{array}{c}2\\ -5\end{array}\right)$. For $\lambda =-0.5$: the first row gives $0.5a+b=0$, so an eigenvector is $\left(\begin{array}{c}2\\ -1\end{array}\right)$.

(i) Let $m$ be the mean number of patients arriving per minute.

(ii) Over 3 hours (180 minutes) the mean under $H_0$ is 270, so the number of arrivals is $Y\sim \mathrm{Po}(270)$. The sample had 320 patients.

Since $0.00166<0.05$, reject $H_0$. There is sufficient evidence that the arrival rate exceeds 1.5 per minute, so Loreto should employ extra staff.

(i) Let $p$ be the proportion of patients who see a nurse in under 20 minutes.

(ii) Under $H_0$ the number who wait more than 20 minutes is $W\sim \mathrm{B}(150,0.05)$. Eleven waited that long.

Since $0.132>0.10$, we fail to reject $H_0$. There is insufficient evidence that the health centre is missing its target of 95% seen in under 20 minutes.