IB Maths AI HL · Paper 1

One assumption: the distances per litre for both fuel types come from normal populations with equal variances (and the days are independent).

(i) The null hypothesis is that there is no difference in the mean distance per litre.

(ii) The alternative hypothesis, since Jacob is testing for further travel using premium.

(i) Run a pooled two-sample $t$-test on the calculator.

(ii) The $p$-value is less than the significance level, $0.0264<0.05$, so reject $H_0$. There is sufficient evidence at the 5% level that Jacob's car travels further per litre on premium fuel than on standard fuel.

Each box removes 6 tarts.

After the first box has been sold, $400-6=394$ tarts remain.

The number of tarts remaining cannot be negative.

$T\sim N(5,1^2)$.

(i) Shade the right tail of the curve, beyond the point to the right of 5 that cuts off 15% of the area.

(ii) Use the inverse normal with an area of $0.85$ to the left.

Use the cosine rule for the angle at $A$, opposite side $BC=8$.

Expand, using $i^2=-1$.

Inverse variation means $V=\frac{k}{t}$. Use $V=300$ at $t=5$.

The total volume is 300, shared equally when each has 150.

The graph of $y=P(t)$ starts at $(5,0)$ and increases, passing through about $(10,150)$. It is a smooth curve that is concave down and approaches 300 for large $t$.

The pond holds whatever has left the reservoir.

(i) Write $B(x)=20+32x^{-1}+2x$ and differentiate.

(ii) Set the derivative to zero.

This is in the domain $x\ge 1$ and there is no local maximum, so it gives the minimum.

Double every $y$-value and subtract 1. The maximum $(-1,3)$ goes to $(-1,2(3)-1)$, the left endpoint $(-4,1)$ goes to $(-4,1)$ and the right endpoint $(4,0)$ goes to $(4,-1)$.

Write $f(2x-2)=f(2(x-1))$. The $x$-coordinates are halved and then 1 is added; the $y$-coordinates stay the same.

In 30 minutes the number of calls $N\sim \text{Po}(17)$. Some calls are not answered if more than 22 arrive.

On any day the chance that all calls are answered is $1-0.09527=0.90473$. The number of such days among 10 is $D\sim B(10,0.90473)$.

Write $f(x)=6{\left(2x^2+5x+4\right)}^{-1}$ and use the chain rule.

(i) Set $\frac{dy}{dx}=\frac{x}{2}+\frac{y}{3}=0$.

(ii) On the slope field, draw the straight line through the origin with gradient $-1.5$, for example from $(-4,6)$ to $(4,-6)$. All the little segments on this line are horizontal.

(i) $G$ is not simple, because there are two edges between $A$ and $B$.

(ii) $G$ is not connected, because there is no path from $A$ to $F$ (the vertices $E$ and $F$ form a separate piece).

The unknown entries are $A$–$B$, $B$–$A$, $B$–$C$, $C$–$B$. Two edges join $A$ and $B$, and one edge joins $B$ and $C$. The matrix is symmetric.

The direction vector $(7,-24,0)$ has length $\sqrt{49+576}=25$. The velocity with speed 800 is $\frac{800}{25}(7,-24,0)=32(7,-24,0)$.

At $x=5$, $y=2$ use $\frac{dx}{dt}=2.2x-2.6y$ and $\frac{dy}{dt}=3.4x-2.2y$.

From taxi the probabilities of going to bus, bicycle or taxi are $0.3$, $q$ and $0.4$, and these add to 1.

From bus on day 1, the probability of taxi on day 2.

From the diagram, bus goes to bus with probability 0.2, bicycle to bicycle 0.1 and taxi to taxi 0.4. On day 2 he uses bus, bicycle or taxi with probabilities 0.2, 0.1 and 0.7. Day 3 matches day 2 in three ways.

Write each column as the probabilities of going from that mode (Bus, Bicycle, Taxi) to bus, bicycle and taxi.

The height varies by 15 above and below the centre height 16.

(i) Integrate the speed from 0 to 4.5. Use the calculator, with the absolute value, because the wheel turns the same way throughout.

(ii) Differentiate $h$ with respect to $\theta$.

Evaluate at $\theta =1.1239$ and also the rate of turning at $t=4.5$.

Multiply, using the chain rule.