IB Maths AA HL · Paper 2

1. Half of $52$ is $26$, so draw across from a cumulative frequency of $26$ to the curve.
2. Drop down to the height axis and read off the value.

Another way. Type the midpoints and frequencies into two GDC lists and run one-variable statistics to get the mean directly.

1. A function is decreasing where its derivative is negative, so solve $x^2-x-1<0$.
2. Find the roots of $x^2-x-1=0$ with the quadratic formula (or the GDC).
3. The parabola opens upwards, so it is negative between its roots.

1. The height oscillates about the centre of the wheel, so $a$ is the centre height, $23$.
2. The amplitude is the radius, so $b=20$. Starting at the bottom matches $a-b\cos (ct)$ at $t=0$.
3. The angle is in degrees and one turn takes $4$ minutes, so $c\times 4=360$.

1. The sine version must give the same heights, so substitute a known point, such as $h=3$ at $t=0$.
2. Solve for the smallest positive $d$.

Another way. A cosine curve is a sine curve shifted by a quarter of a period, and a quarter of $4$ minutes is $1$ minute.

1. Graph $y=x^3-3x+2\sin x$ on the GDC and use the maximum and minimum tools.
2. The function is odd, so the two points are reflections of each other through the origin.

1. The line through the origin and $B$ has gradient $-0.46856\dots /0.70010\dots$.
2. Between $x=0$ and $x=0.700$ the line sits above the curve, so integrate line minus curve.
3. Let the GDC do the integral, using the stored $x$-coordinate of $B$.

Another way. Find the area between the curve and the $x$-axis from $0$ to $0.700$ (about $0.2047$), then subtract the triangle under the line, $\frac{1}{2}(0.700)(0.469)\approx 0.1640$.

1. From $P(H>171)=0.4$, standardise and use inverse normal on $0.6$.
2. The second fact is conditional: $P(H>175\mid H>171)=0.35$. Since $H>175$ already implies $H>171$, the top of the fraction is just $P(H>175)$.
3. So $P(H>175)=0.35\times 0.4=0.14$; use inverse normal on $0.86$.
4. Solve the two linear equations in $\mu$ and $\sigma$ together.

1. Write $P=135V^{-1}$ and differentiate with respect to $V$.
2. Link the rates with the chain rule: $\frac{dP}{dt}=\frac{dP}{dV}\times \frac{dV}{dt}$.
3. The volume is decreasing, so $\frac{dV}{dt}=-8$. Substitute $V=20$.

Another way. Differentiate $PV=135$ implicitly with the product rule: $V\frac{dP}{dt}+P\frac{dV}{dt}=0$, with $P=6.75$ when $V=20$.

1. The total area under a probability density function is $1$.
2. Here the graph is a triangle with base $5$ and height $b$.

1. Use $E(X)=\int xf(x)dx$ over both pieces, with $b=0.4$.
2. Integrate each piece and substitute the limits, leaving everything in terms of $a$.
3. Set the total equal to $3.2$ and solve for $a$ (GDC solver is fine).

Another way. For a triangular density on $[0,5]$ with peak at $a$, the mean is the average of the three corners, $\frac{0+a+5}{3}=3.2$, which gives $a=4.6$ at once.

1. Cube the given root. In polar form it has modulus $4$ and argument $\frac{2\pi }{3}$.
2. Cubing multiplies the argument by $3$ and cubes the modulus.

1. The coefficients of $z^3-64$ are real, so the complex conjugate of the given root is also a root.
2. The third root is the real cube root of $64$.

1. Each value of $iz$ must be one of the three roots from parts (a) and (b).
2. So divide each root by $i$, which is the same as multiplying by $-i$.

1. The three points are equally spaced on a circle of radius $4$, so the triangle is equilateral.
2. Split it into three triangles from the origin, each with two sides of $4$ and an angle of $\frac{2\pi }{3}$.

Another way. Use base and height: the points $\pm 2\sqrt{3}+2i$ are $4\sqrt{3}$ apart, and $-4i$ is $6$ below them, giving $\frac{1}{2}(4\sqrt{3})(6)$.

1. Make $x$ the subject: $x=d\pm \sqrt{4-(y-2{)}^2}$. The plus sign gives the outer edge, the minus sign the inner edge.
2. Rotating about the $y$-axis gives washers: the volume is $\pi {\int }_0^4(x_{\text{outer}}^2-x_{\text{inner}}^2)dy$.
3. Simplify the difference of squares: $(u+v{)}^2-(u-v{)}^2=4uv$.

1. The remaining integral is the area of a semicircle of radius $2$, which is $2\pi$ (or evaluate it on the GDC: $6.28318\dots$).
2. Set the volume equal to $50{\pi }^2$ and solve for $d$.

1. Quarterly compounding at $4.8\%$ a year means $1.2\%$ per quarter.
2. From 1 January 2026 to 1 January 2031 is $5$ years, which is $20$ quarters.
3. Round to the nearest dollar, as asked.

Another way. Use the GDC finance solver with $N=20$, $I\%=1.2$, $PV=-15000$, $PMT=0$, one payment and one compounding per period.

1. Set the balance equal to $28000$ and solve for the number of quarters $n$.
2. Convert to years and count on from the start of 2026.

Another way. Tabulate the balance: after $52$ quarters (start of 2039) it is $27891.88$ dollars, and after $53$ quarters it is $28226.59$ dollars, so it first passes $28000$ during 2039.

1. After $x$ years the boat costs $28000(0.9365{)}^x$.
2. After $x$ years the balance is $15000(1.012{)}^{4x}$, using years rather than quarters so both sides match.
3. Solve the equation with the GDC (intersection of the two graphs), then turn $x$ into a calendar year.

Another way. Make a GDC table of both expressions in steps of a quarter of a year: at $x=5.5$ the balance ($19501.25$) is still just below the price ($19518.69$), and at $x=5.75$ it is above.

1. The line from the vertex to the centre of the base circle is perpendicular to the plane of the circle.
2. So its direction is the normal of the plane, read from the coefficients: $(1,2,-1)$.
3. Start the line at $V$.

1. The centre is where this line meets the plane.
2. Substitute the general point of the line into $x+2y-z=5$ and solve for $\lambda$.

1. Form the vectors from $V$ to $P$ and from $V$ to $P^{\prime }$.
2. Use the scalar product formula for the angle between them.

Another way. Find $P{P}^{\prime }=\sqrt{70}$ and use the cosine rule in triangle $PV{P}^{\prime }$, whose other two sides are both $\sqrt{230}$.

1. A normal to the plane is the cross product of two vectors in it, such as $\overrightarrow{VP}$ and $\overrightarrow{V{P}^{\prime }}$.
2. Simplify the normal by dividing out the common factor.
3. Find $d$ by substituting any of the three points.

Another way. For (ii), l'Hopital's rule gives ${\lim }_{x\to -1}\frac{2x+a}{1}=a-2$.

1. Divide the numerator by $x+1$ to split $f$ into a linear part plus a remainder over $x+1$.
2. The linear part is the oblique asymptote. Match it with $y=x-4$.

1. With $a=-3$, write $f(x)=x-4+\frac{b+4}{x+1}$ and differentiate.
2. A point of zero gradient needs $(x+1{)}^2=b+4$ to have a solution, so $b+4\ge 0$.
3. If $b=-4$, the numerator becomes $(x+1)(x-4)$, which cancels: the graph is the line $y=x-4$ with a hole, and there is no vertical asymptote. So $b\ne -4$.
4. Hence $b>-4$, and since $b$ is an integer, $b\ge -3$.

Another way. Use the quotient rule: the numerator of $f^{\prime }(x)$ is $x^2+2x-(b+3)$, and its discriminant $4b+16\ge 0$ gives the same condition.