IB Maths AA SL · Paper 1

1. Work out the lower fence using $Q_1=66$ and the IQR of $24$.
2. Only values strictly below the fence are outliers, so the fence value itself still counts as a normal value.

1. $AC$ is opposite the known angle at $B$, with the two given sides either side of it, so use the cosine rule.
2. Substitute $5$, $6$ and $\cos \hat{B}=\frac{1}{5}$, then take the positive square root.

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

1. The general term is $\left(\genfrac{}{}{0ex}{}{8}{r}\right)(mx{)}^r$, so the $x^2$ coefficient is $\left(\genfrac{}{}{0ex}{}{8}{2}\right)m^2$ and the $x^3$ coefficient is $\left(\genfrac{}{}{0ex}{}{8}{3}\right)m^3$.
2. Find $\left(\genfrac{}{}{0ex}{}{8}{2}\right)=28$ and use $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=56$ from part (a).
3. Set the $x^3$ coefficient equal to eight times the $x^2$ coefficient and solve. Since $m>0$, you can divide by $m^2$.

1. Write the number $3$ as a base-3 logarithm: $3={\log }_{3}27$.
2. Combine the two logs on the right with the product law, then compare the insides.

Another way. Move the logs together to get ${\log }_3\frac{p}{r}=3$, then convert to exponential form: $\frac{p}{r}=3^3$.

1. Turn the second equation into exponential form: ${\log }_{r}p=4$ means $p=r^4$.
2. Replace $p$ with $27r$ from part (a) and divide by $r$ (allowed, since $r>1$).
3. Solve for $r$, then find $p$.

1. Spot that the bracket $x^3-\cos 4x$ is exactly a quarter of the derivative from part (a).
2. Let $u=x^4-\sin 4x$, so $\mathrm{d}u=4(x^3-\cos 4x)\mathrm{d}x$ and the second bracket becomes $\frac{1}{4}\mathrm{d}u$.
3. Integrate $\frac{1}{4}{u}^7$ and put $x$ back in.

1. Write $y=f(x)$, swap $x$ and $y$, then make $y$ the subject.
2. Isolate the fraction first, flip both sides, then undo the $3y+2$.

1. Apply $f^{-1}$ to both sides of the given condition, which leaves $g$ of $\frac{x}{3}$ equal to $f^{-1}(x)$.
2. To get $g(x)$ itself, replace every $x$ with $3x$.
3. Substitute $3x$ into your answer to part (a).

Another way. Let $a=\frac{x}{3}$, so $x=3a$; then $f(g(a))=3a$, giving $g(a)=f^{-1}(3a)$ directly.

Another way. Expand: $-(x+2)(x-4)=-x^2+2x+8=-(x-1{)}^2+9$.

1. At $x=-2$ the curve is on the $x$-axis, so the line passes through $(-2,0)$.
2. Substitute that point into $y=mx+4$ and solve for $m$.

Another way. The line meets the $y$-axis at $(0,4)$, so the gradient from $(-2,0)$ to $(0,4)$ is $\frac{4}{2}=2$. Check with $x=2$: $f(2)=8$ and $2(2)+4=8$.

1. Notice $-x^2+2x+8=f(x)$, so the equation is $f(x)=p$: count where the horizontal line $y=p$ meets the parabola.
2. (i) The parabola has a maximum of $9$ at the vertex, so any line below $9$ cuts it twice.
3. (ii) For both crossings to have $x>0$, the line must also sit above the $y$-intercept $f(0)=8$; at $p=8$ one root is $0$, which is not positive.

Another way. For (i), use the discriminant of $-x^2+2x+(8-p)$: $4+4(8-p)>0$ gives $p<9$.

1. Rewrite both sides in terms of $x$: $\sin 2x=2\sin x\cos x$ and $1-\cos 2x=2{\sin }^{2}x$.
2. Bring everything to one side and factorise out $2\sin x$. Do not divide by $\sin x$, or you lose solutions.
3. Solve $\sin x=0$ and $\tan x=1$ in the interval $0\le x\le \pi$.

1. $y=\sin 2x$: one full sine wave on $[0,\pi ]$, starting at $(0,0)$, peaking at $\left(\frac{\pi }{4},1\right)$, back to $0$ at $\frac{\pi }{2}$, lowest at $\left(\frac{3\pi }{4},-1\right)$, and ending at $(\pi ,0)$.
2. $y=1-\cos 2x$: an upside-down cosine lifted up by $1$, so it never goes below $0$. It starts at $(0,0)$, rises to a maximum of $2$ at $\frac{\pi }{2}$, and falls back to $(\pi ,0)$.
3. The curves meet at the three solutions from (a): label $(0,0)$, $\left(\frac{\pi }{4},1\right)$ and $(\pi ,0)$. Between $\frac{\pi }{4}$ and $\pi$ the second curve lies above the first.

1. Here $a=\frac{\pi }{4}$ and $b=\pi$. On this interval $g$ is on top, so integrate $g(x)-f(x)$.
2. Integrate term by term: $\cos 2x$ gives $\frac{1}{2}\sin 2x$ and $\sin 2x$ gives $-\frac{1}{2}\cos 2x$.
3. Substitute the limits and subtract, using $\sin 2\pi =0$, $\cos 2\pi =1$, $\sin \frac{\pi }{2}=1$ and $\cos \frac{\pi }{2}=0$.

1. Use the quotient rule with $u={\mathrm{e}}^{2x}+x$ and $v=x$, so $u^{\prime }=2{\mathrm{e}}^{2x}+1$ and $v^{\prime }=1$.
2. Expand the numerator. The $+x$ and $-x$ cancel, leaving two exponential terms.
3. Factorise out ${\mathrm{e}}^{2x}$.

Another way. Split first: $h(x)=\frac{{\mathrm{e}}^{2x}}{x}+1$, so only one quotient is needed and the constant differentiates to $0$.

1. $h$ is increasing where $h^{\prime }(x)>0$.
2. For $x>0$, both ${\mathrm{e}}^{2x}$ and $x^2$ are positive, so the sign depends only on $2x-1$.

1. Find the tangent gradient by substituting $x=1$ into $h^{\prime }(x)$.
2. The normal gradient is the negative reciprocal.
3. Use $y-y_1=m(x-x_1)$ through the given point.

1. A point of inflexion needs $h^{\prime \prime }(x)=0$ (with a change of concavity), so try to solve $h^{\prime \prime }(x)=0$.
2. The factor $2{\mathrm{e}}^{2x}$ is never zero and $x^3\ne 0$ for $x>0$, so only the quadratic $2x^2-2x+1$ could be zero.
3. Its discriminant is negative, so it has no real roots. Therefore $h^{\prime \prime }(x)$ is never zero and there is no point of inflexion.

Another way. Complete the square: $2x^2-2x+1=2{\left(x-\frac{1}{2}\right)}^2+\frac{1}{2}>0$, so $h^{\prime \prime }(x)>0$ throughout and the curve is always concave up.