IB Maths AA HL · Paper 1

1. Work out the lower fence $Q_1-1.5\times \text{IQR}$.
2. A value equal to the fence is not below it, so it is not an outlier. The minimum of $28$ sits below the fence, so the fence value itself is the smallest allowance that survives.

Another way. For (i), draw a right-angled triangle with adjacent side $1$ and hypotenuse $5$: the opposite side is $\sqrt{24}$ by Pythagoras.

1. Move the logs to one side and combine them with the quotient law.
2. Then switch to exponential form.

Another way. Write $3={\log }_{3}27$, so the right-hand side is ${\log }_{3}27+{\log }_{3}r={\log }_{3}27r$.

1. Turn the second equation into exponential form: $p=r^4$.
2. Substitute $p=27r$ from part (a) and divide by $r$ (allowed, as $r>1$).
3. Take the cube root, then find $p$.

1. Write $75^{\circ }=45^{\circ }+30^{\circ }$ and expand $\cos (A+B)$ with exact values.
2. Secant is the reciprocal of cosine, so flip the fraction.
3. Rationalise the denominator by multiplying top and bottom by $\sqrt{6}+\sqrt{2}$.

Another way. Use the half-angle route: ${\cos }^{2}75^{\circ }=\frac{1+\cos 150^{\circ }}{2}=\frac{2-\sqrt{3}}{4}$, so ${\sec }^{2}75^{\circ }=\frac{4}{2-\sqrt{3}}=8+4\sqrt{3}=(\sqrt{6}+\sqrt{2}{)}^2$.

1. Spot that $x^3-\cos 4x$ is a quarter of the derivative from part (a).
2. Substitute $u=x^4-\sin 4x$, so $du=4(x^3-\cos 4x)dx$.
3. Integrate $u^7$ and put $x$ back in.

1. Write $y=f(x)$, swap $x$ and $y$, then make $y$ the subject.
2. Isolate the fraction first, then flip both sides.

1. Apply $f^{-1}$ to both sides: $g\left(\frac{x}{3}\right)=f^{-1}(x)$.
2. Let $t=\frac{x}{3}$, so $x=3t$ and $g(t)=f^{-1}(3t)$.
3. Substitute $3t$ into your part (a) answer and rename $t$ as $x$.

1. Write the given relation with the formulas $S_k=\frac{u_1(1-r^k)}{1-r}$ and $S_{\infty }=\frac{u_1}{1-r}$.
2. Cancel $\frac{u_1}{1-r}$ from both sides to get $r^k$.
3. The term after the $k$th is $u_{k+1}=u_1{r}^k$. Substitute $r^k=\frac{1}{2}$.

1. The real part is positive and $k<0$, so $z$ is in the fourth quadrant: $\arg z=\arctan \frac{k}{3}$ lies between $-\frac{\pi }{2}$ and $0$.
2. By De Moivre, $\arg (z^5)=5\arg z$, which lies between $-\frac{5\pi }{2}$ and $0$.
3. $z^5$ is real when its argument is a multiple of $\pi$. In that interval that means $-\pi$ or $-2\pi$.
4. Divide by $5$, then use $k=3\tan (\arg z)$.

1. Base case $n=2$: the sum is $\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$, which satisfies the inequality.
2. Assume the result holds for $n=k$: the sum $S_k$ from $\frac{1}{k+1}$ to $\frac{1}{2k}$ is at least $\frac{7}{12}$.
3. For $n=k+1$, the sum runs from $\frac{1}{k+2}$ to $\frac{1}{2k+2}$. Compared with $S_k$ it loses $\frac{1}{k+1}$ and gains $\frac{1}{2k+1}$ and $\frac{1}{2k+2}$.
4. Combine the extra terms: they come to a positive amount, so $S_{k+1}>S_k\ge \frac{7}{12}$.
5. Conclude: true for $n=2$, and true for $k$ implies true for $k+1$, so it holds for every integer $n\ge 2$.

1. Use $\sin 2x=2\sin x\cos x$ and $1-\cos 2x=2{\sin }^{2}x$.
2. Bring everything to one side and factorise. Do not divide by $\sin x$, or you lose solutions.
3. Solve each factor on $0\le x\le \pi$.

Another way. Rearrange to $\sin 2x+\cos 2x=1$, square to get $1+\sin 4x=1$, solve $\sin 4x=0$ and reject the extra roots that squaring brings in.

1. $y=\sin 2x$: one full sine wave over $[0,\pi ]$. It starts at the origin, peaks at $\left(\frac{\pi }{4},1\right)$, crosses the axis at $x=\frac{\pi }{2}$, dips to $\left(\frac{3\pi }{4},-1\right)$ and ends at $(\pi ,0)$.
2. $y=1-\cos 2x$: a single hump between $0$ and $2$. It starts at the origin, rises to a maximum of $\left(\frac{\pi }{2},2\right)$ and falls back to $(\pi ,0)$. It never goes below the $x$-axis.
3. Draw both as smooth curves on one set of axes and label the meeting points $(0,0)$, $\left(\frac{\pi }{4},1\right)$ and $(\pi ,0)$.

1. Between $a=\frac{\pi }{4}$ and $b=\pi$, the graph of $g$ lies above $f$ (test $x=\frac{\pi }{2}$: $g=2$, $f=0$).
2. Integrate $g(x)-f(x)$ over that interval.
3. Substitute the limits and subtract.

1. Write $L_2$ in parametric form using A and its direction.
2. Substitute into $2x+y-2z=2$ and solve for the parameter.
3. Put the parameter back into the line.

1. Find the angle $\theta$ between the direction of $L_2$ and the normal $\mathbf{n}=\left(\begin{array}{c}2\\ 1\\ -2\end{array}\right)$.
2. The angle with the plane is the complement: $\alpha =90^{\circ }-\theta$, so $\sin \alpha =\cos \theta$.

1. $L_1$ is parallel to the plane, so its distance from the plane equals the distance from A.
2. A, N and the foot of the perpendicular from A form a right-angled triangle with hypotenuse AN and angle $\alpha$ at N, so the distance is $\text{AN}\sin \alpha$.

Another way. Use the point-to-plane distance: $\frac{|2(7)+(-1)-2(1)-2|}{3}=\frac{9}{3}=3$.

1. The reflected line is parallel to $L_1$, so it keeps direction $\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)$. You just need one point on it: the image of A.
2. Run a line from A along the normal and find where it meets the plane (the foot F).
3. Go the same distance again past F to get the image A${}^{\prime }$.
4. Write the new line through A${}^{\prime }$.

1. Differentiate with the chain rule: the outside is $\tan u$, which gives ${\sec }^{2}u$, and the inside $1-e^{3x}$ gives $-3e^{3x}$.
2. Replace ${\sec }^{2}u$ with $1+{\tan }^{2}u$, and $\tan (1-e^{3x})$ is just $y$.

Another way. Write $\arctan y=1-e^{3x}$ and differentiate implicitly: $\frac{1}{1+y^2}\frac{dy}{dx}=-3e^{3x}$.

1. Differentiate the part (a) result with the product rule. The factor $1+y^2$ needs implicit differentiation: its derivative is $2y\frac{dy}{dx}$.
2. At $x=0$: $y=\tan 0=0$, and then $\frac{dy}{dx}=-3(1)(1+0)=-3$.
3. Substitute these into the second derivative.

1. Replace the numerator by its series from part (c) and divide through by $x$.
2. Let $x\to 0$: every remaining term with an $x$ vanishes.

Another way. The limit is $\frac{0}{0}$, so L'Hôpital's rule works too: it gives $f^{\prime }(0)=-3$.

1. Factor out $a$ and expand $(1+\frac{b}{a}x{)}^{-1}$ with the extended binomial theorem.
2. Match the coefficient of $x$, then of $x^2$, with part (c).

Another way. Treat $-3x-\frac{9}{2}{x}^2$ as the start of a geometric series with first term $-3x$ and ratio $\frac{3}{2}x$; its sum $\frac{-3x}{1-\frac{3}{2}x}$ rearranges to $\frac{x}{-\frac{1}{3}+\frac{1}{2}x}$.