IB Maths AI HL · Paper 1

1. Let $X$ be the number of heads. Each of the 12 coins is independent with the same chance of heads, so $X\sim B(12,0.73)$.
2. Use the binomial probability function on your GDC for $X=8$.

1. More heads than tails out of 12 means at least 7 heads.
2. Use the cumulative binomial function: $P(X\ge 7)=1-P(X\le 6)$.

Another way. Count tails instead: with $T\sim B(12,0.27)$ you need $P(T\le 5)$, which gives the same value.

1. Treat the garden as the full $6x$ by $(2x+10)$ rectangle with the house corner removed.
2. The missing corner is $6x-2x=4x$ wide and $(2x+10)-(x+4)=x+6$ tall.
3. Subtract and simplify to reach the given expression.

Another way. Split the L into a $6x$ by $(x+4)$ strip and a $2x$ by $(x+6)$ strip, then add: $6x^2+24x+2x^2+12x$.

1. Set the area equal to 72 and solve the quadratic (factorise or use the GDC).
2. A length must be positive, so reject the negative root.

1. The fence runs along the four outer edges only: $6x$, $x+4$, $2x+10$ and $2x$.
2. Add them and substitute $x=1.5$.

1. Integrate $P^{\prime }(t)$ term by term and include a constant $c$.
2. Use $P(5)=254$ to find $c$.

1. The population is eliminated when $P(t)=0$.
2. Solve the cubic with the GDC and keep only the positive root.

1. Fill in the given values: 2 heavy and lost, 24 heavy in total, 5 lost in total, 80 overall.
2. Complete the rest by subtraction so every row and column adds up.

1. Let $W\sim N(16,3^2)$ be the weight. Heavy means $W>18$, found with the normal CDF.
2. Multiply by the chance a heavy item is lost, $\frac{1}{12}$ from (b)(ii).

Another way. For (ii), form $M+M^2+M^3$ and note it has no zero entries off the diagonal.

1. Each bisector passes through the midpoint of the segment and is perpendicular to it.
2. $[CD]$ is horizontal with midpoint $(3.5,1.5)$, so its bisector is vertical.
3. $[AD]$ has gradient $-1$ and midpoint $(2.5,2)$, so its bisector has gradient $1$.
4. $[BD]$ has gradient $1$ and midpoint $(3.5,2)$, so its bisector has gradient $-1$.

1. D's cell is the region closer to D than to A, B and C: below $y=x-0.5$, below $y=-x+5.5$ and left of $x=3.5$.
2. Its top vertex is $(3,2.5)$, equidistant from A, B and D, and its right corner is $(3.5,2)$, equidistant from B, C and D.
3. Keep the old A-B edge only above $(3,2.5)$ and the old B-C edge only to the right of $(3.5,2)$. The old A-C edge disappears inside D's cell.

1. Find the scalar product $\mathbf{r}\cdot \mathbf{F}$ and both magnitudes.
2. Use $\cos \theta =\frac{\mathbf{r}\cdot \mathbf{F}}{|\mathbf{r}||\mathbf{F}|}$.

Another way. Use $|\mathbf{T}|=|\mathbf{r}||\mathbf{F}|\sin \theta$ with the angle from (a): $\sqrt{252}\sin 67.79^{\circ }\approx 14.7$.

1. The cost is $C=12Y+5$ with $Y\sim Po(2.1)$, so $E(Y)=Var(Y)=2.1$.
2. (i) $E(aY+b)=aE(Y)+b$.
3. (ii) $Var(aY+b)=a^{2}Var(Y)$: the fixed 5 dollars has no effect.

Another way. Or note that C can only take values like 5, 17, 29, …, not every non-negative integer.

1. Write the system as $\dot{\mathbf{x}}=A\mathbf{x}$ with $A=\left(\begin{array}{cc}1& 2\\ 5& 4\end{array}\right)$.
2. (i) Solve $\det (A-\lambda I)=0$; one root is the given $-1$.
3. (ii) Substitute the new eigenvalue into $(A-\lambda I)\mathbf{v}=0$ and read off a direction.

Another way. The trace is 5, so the eigenvalues add to 5; with one equal to $-1$, the other is 6.

1. The eigenvalues have opposite signs, so the origin is a saddle point.
2. Draw the line $y=-x$ with arrows pointing in towards the origin ($\lambda =-1$).
3. Draw the line $y=2.5x$ with arrows pointing away from the origin ($\lambda =6$).
4. In each of the four regions, draw a curve that comes in roughly parallel to $y=-x$, bends near the origin, and heads out alongside $y=2.5x$ without crossing either line.

1. When the card is rolled up, each arc becomes a circle, so each arc length equals a circumference.
2. The inner arc has radius $x$ and the outer arc radius $x+10$, both with angle $\theta$.
3. Subtract the equations to find $\theta$, then substitute back for $x$.

Another way. The shape unrolls to a trapezium-like strip: average arc length times slant height, $\frac{18+24}{2}\times 10=210$.

1. Write $y=4-x^2$, swap $x$ and $y$, and make $y$ the subject.
2. Take the positive square root, since the inverse's outputs must be the original domain $0\le x\le 2$.
3. The inverse's domain is the range of $f$: from $f(2)=0$ to $f(0)=4$.

Another way. Use the inverse from (a): $\pi {\int }_0^4{\left(f^{-1}(x)\right)}^{2}dx$ is the same integral.

1. Moving parallel to the $y$-axis means the horizontal velocity is zero: $\dot{x}=0$.
2. Solve $\sin (3t)=\frac{1}{2}$ in radians and take the smallest positive $t$.

1. Integrate each component separately, with its own constant.
2. Use the starting position $(0,0)$ at $t=0$ to find both constants.

1. Start at $(-0.5,3)$ and repeat $y_{n+1}=y_n+0.1\left(3x_n^2+2y_n-6.75\right)$, $x_{n+1}=x_n+0.1$.
2. From $x=-0.5$ to $x=1.6$ is 21 steps; run them in a GDC table or spreadsheet.

1. At $(0,4)$ the gradient is $0+8-6.75=1.25>0$.
2. Following the slope marks both ways from $(0,4)$, the gradient stays positive across the whole window, so this solution increases throughout, with no maximum or minimum, climbing very steeply for $x>0$.

1. Look at the given second derivative: the numerator 40 and $(4+x^2{)}^{3/2}$ are both positive for every $x$.
2. So $\frac{d^{2}T}{d{x}^2}>0$ at $x=0.408$, which makes it a minimum.

Another way. Check the sign of the first derivative either side of 0.408: it changes from negative to positive.

1. Triangle ABC has sides $AB=|w-z|=6.5$, $BC=|w|=7.0646\dots$ and $AC=|z|=3$.
2. Use the cosine rule for the angle at B, opposite side AC.

Another way. Find angle ACB as $2-0.833=1.167$ rad, then use the sine rule with $\frac{\sin B}{3}=\frac{\sin C}{6.5}$.