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Q1
A verified core-concept review across the full AA HL syllabus
Each question opens with the concept and the formula you need to memorise, then a quick worked example, before the real exam-style question. Every answer below has been verified twice — algebraically and numerically — before being included.
The Binomial Theorem expands $(a+b)^n$ without multiplying it out term by term. The general term is $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$.
Quick example: Find the coefficient of $x^2$ in $(x+2)^4$.
$4-k=2\Rightarrow k=2$, so coefficient $=\binom{4}{2}\cdot2^2=6\cdot4$.
The term in $x^2$ needs $5-k=2$, so $k=3$.
$T_4=\binom{5}{3}(2x)^2(-3)^3=10\cdot4x^2\cdot(-27)=-1080x^2$.
Coefficient $=-1080$. Choosing $+1080$ drops the negative sign on $(-3)^3$; choosing $720$ accidentally uses the $x^3$ term ($k=2$) instead; choosing $-270$ forgets to include $\binom{5}{3}$ or the factor $2^2$.
De Moivre's Theorem lets you raise a complex number to a power using its modulus-argument (polar) form: $z=r(\cos\theta+i\sin\theta) \Rightarrow z^n=r^n(\cos n\theta+i\sin n\theta)$.
Quick example: $|6-8i|=\sqrt{36+64}=\sqrt{100}$.
Answer: 10$z=1+i$ has $r=\sqrt{2}$ and $\theta=\dfrac{\pi}{4}$.
$z^8=(\sqrt2)^8\left(\cos 8\cdot\tfrac{\pi}{4}+i\sin 8\cdot\tfrac{\pi}{4}\right)=16(\cos2\pi+i\sin2\pi)=16(1+0i)=16$.
$-16$ comes from reducing the angle to $\pi$ instead of $2\pi$; $256$ comes from computing $2^8$ instead of $(\sqrt2)^8=2^4$; $16i$ forgets to evaluate $\sin2\pi=0$.
A geometric series converges to a finite sum only when $|r|<1$. The sum to infinity is $S_\infty=\dfrac{a}{1-r}$, where $a$ is the first term and $r$ the common ratio.
Quick example: $a=8,\ r=\tfrac12 \Rightarrow S_\infty=\dfrac{8}{1-\frac12}=\dfrac{8}{\frac12}$.
Answer: 16$S_\infty=\dfrac{12}{1-\frac23}=\dfrac{12}{\frac13}=36$.
$18$ comes from mistakenly using $r=\tfrac13$; $4$ comes from computing $a(1-r)$ instead of dividing; $20$ is a simple arithmetic slip.
Solving a logarithmic equation usually means combining logs with the product rule $\log_a m+\log_a n=\log_a(mn)$, converting to exponential form, then — critically — checking that every solution satisfies the original domain ($\text{argument}>0$).
Quick example: $\log_3 x+\log_3 2=2\Rightarrow\log_3(2x)=2\Rightarrow 2x=9\Rightarrow x=4.5$.
Answer: 4.5$\log_2[x(x-2)]=3\Rightarrow x(x-2)=2^3=8\Rightarrow x^2-2x-8=0\Rightarrow(x-4)(x+2)=0$.
So $x=4$ or $x=-2$ — but the domain requires $x>2$ (so that $x-2>0$). Reject $x=-2$.
Answer: $x=4$. Picking both roots is the classic error of forgetting the domain check.
A composite function $(f\circ g)(x)=f(g(x))$ is evaluated from the inside out: compute the inner function first, then substitute that result into the outer function.
Quick example: $f(x)=x+4,\ g(x)=2x$. $g(1)=2$, so $f(g(1))=f(2)=6$.
Answer: 6$g(2)=2^2-3=1$. Then $f(1)=2(1)+1=3$.
$9$ comes from forgetting the $-3$ in $g(2)$ (using $g(2)=4$); $22$ is $(g\circ f)(2)$ — the composition done in the wrong order; $6$ comes from adding $f(2)+g(2)$ instead of composing.
To find an inverse function, write $y=f(x)$, swap $x$ and $y$, then solve algebraically for $y$. The result is $f^{-1}(x)$.
Quick example: $f(x)=2x-3 \Rightarrow f^{-1}(x)=\dfrac{x+3}{2}$, so $f^{-1}(7)=\dfrac{10}{2}$.
Answer: 5Let $y=\dfrac{3x+1}{x-2}$. Then $y(x-2)=3x+1 \Rightarrow yx-2y=3x+1 \Rightarrow x(y-3)=2y+1 \Rightarrow x=\dfrac{2y+1}{y-3}$.
So $f^{-1}(y)=\dfrac{2y+1}{y-3}$, and $f^{-1}(7)=\dfrac{15}{4}$.
The other options come from sign slips and arithmetic errors while isolating $x$.
For a rational function, a vertical asymptote occurs where the denominator equals zero (and the numerator doesn't also vanish there).
Quick example: $g(x)=\dfrac{x+5}{x-1}$ has vertical asymptote at $x-1=0$.
Answer: x = 1The denominator is zero when $x+3=0$, i.e. $x=-3$.
$x=3$ is a sign error; $x=\tfrac12$ and $x=-\tfrac12$ both incorrectly use the numerator $2x-1=0$ instead of the denominator.
When the numerator and denominator of a rational function have the same degree, the horizontal asymptote is the ratio of their leading coefficients (found by taking the limit as $x\to\infty$).
Quick example: $g(x)=\dfrac{x+5}{x-1}\to$ ratio of leading coefficients $\dfrac{1}{1}$.
Answer: y = 1As $x\to\infty$, $h(x)\to\dfrac{5x}{2x}=\dfrac{5}{2}$.
$\tfrac25$ inverts the ratio; $y=0$ wrongly assumes the numerator's degree is smaller; $-\tfrac27$ mistakenly uses the constant terms instead of the leading coefficients.
Some trig equations are quadratic in disguise. Let $s=\sin x$ (or $\cos x$), factorise the quadratic, then solve each resulting simple trig equation across the given domain using the unit circle / CAST rule.
Quick example: $2\cos x-1=0$ for $0\le x\le2\pi \Rightarrow \cos x=\tfrac12 \Rightarrow x=\tfrac{\pi}{3},\tfrac{5\pi}{3}$.
Answer: π/3, 5π/3Factorise: $(2\sin x+1)(\sin x-1)=0 \Rightarrow \sin x=-\tfrac12$ or $\sin x=1$.
$\sin x=1\Rightarrow x=\dfrac{\pi}{2}$. $\sin x=-\tfrac12$ has reference angle $\dfrac{\pi}{6}$, true in quadrants III and IV: $x=\pi+\tfrac{\pi}{6}=\dfrac{7\pi}{6}$ and $x=2\pi-\tfrac{\pi}{6}=\dfrac{11\pi}{6}$.
Full set: $\dfrac{\pi}{2},\dfrac{7\pi}{6},\dfrac{11\pi}{6}$. Using quadrants I/II instead of III/IV is the classic CAST-rule mix-up; missing a root or stopping after one solution are the other common errors.
For a circle sector, both formulas require the angle in radians: arc length $l=r\theta$ and sector area $A=\tfrac12 r^2\theta$.
Quick example: $r=5,\ \theta=\dfrac{\pi}{3}\Rightarrow l=\dfrac{5\pi}{3},\ A=\dfrac12(25)\dfrac{\pi}{3}=\dfrac{25\pi}{6}$.
Answer: l = 5π/3, A = 25π/6$l=r\theta=8\cdot\dfrac{5\pi}{6}=\dfrac{40\pi}{6}=\dfrac{20\pi}{3}\text{ cm}$.
$A=\dfrac12 r^2\theta=\dfrac12(64)\dfrac{5\pi}{6}=\dfrac{320\pi}{12}=\dfrac{80\pi}{3}\text{ cm}^2$.
Doubling $l$ comes from using the diameter instead of the radius; halving $A$ to $\tfrac{40\pi}{3}$ comes from forgetting the $\tfrac12$ factor; doubling $A$ to $\tfrac{160\pi}{3}$ comes from using $A=2r^2\theta$.
The angle between two vectors comes from rearranging the dot product formula: $\cos\theta=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$. A negative dot product means the angle is obtuse.
Quick example: $\mathbf{a}=(1,0,0),\ \mathbf{b}=(0,1,0)\Rightarrow\mathbf{a}\cdot\mathbf{b}=0\Rightarrow\theta=90°$.
Answer: 90°$\mathbf{a}\cdot\mathbf{b}=2(1)+(-1)(4)+3(-2)=2-4-6=-8$.
$|\mathbf{a}|=\sqrt{4+1+9}=\sqrt{14}$, $|\mathbf{b}|=\sqrt{1+16+4}=\sqrt{21}$.
$\cos\theta=\dfrac{-8}{\sqrt{14}\sqrt{21}}\approx-0.4663 \Rightarrow \theta\approx117.8°$.
$62.2°$ is the supplementary (reference) angle obtained by ignoring the negative sign — the single most common error on this question type.
The cosine rule finds a missing side when you know two sides and the included angle (SAS): $a^2=b^2+c^2-2bc\cos A$.
Quick example: $b=5,\ c=6,\ A=60° \Rightarrow a^2=25+36-60(0.5)=31$.
Answer: a ≈ 5.57$a^2=7^2+9^2-2(7)(9)\cos65°=49+81-126(0.4226)\approx76.75$.
$a=\sqrt{76.75}\approx8.76$.
$13.54$ comes from adding instead of subtracting the cosine term; $10.17$ comes from forgetting the factor of $2$; $11.40$ comes from using Pythagoras and ignoring the angle entirely.
For a normal distribution $X\sim N(\mu,\sigma^2)$, standardise with $z=\dfrac{x-\mu}{\sigma}$, then use $P(X>k)=1-\Phi(z)$.
Quick example: $X\sim N(50,5^2)$, $P(X>55)$: $z=1\Rightarrow P\approx1-0.8413$.
Answer: ≈ 0.1587$z=\dfrac{82-70}{8}=1.5$.
$P(X>82)=1-\Phi(1.5)=1-0.9332=0.0668$.
$0.9332$ is $P(X<82)$ — the complement was never taken; $0.1336$ doubles the tail incorrectly; $0.5$ assumes $z=0$, i.e. forgets to standardise at all.
For a binomial distribution $X\sim B(n,p)$ with fixed, independent trials: $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$.
Quick example: $X\sim B(5,0.5)$, $P(X=2)=\binom{5}{2}(0.5)^2(0.5)^3=10(0.03125)$.
Answer: 0.3125$P(X=4)=\binom{10}{4}(0.3)^4(0.7)^6=210(0.0081)(0.117649)\approx0.2001$.
$0.2668$ is actually $P(X=3)$, picked by miscounting which term was needed; $0.0010$ comes from forgetting the binomial coefficient $\binom{10}{4}=210$ entirely; $0.3$ confuses the probability of success $p$ with $P(X=4)$.
Conditional probability: $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$ — direction matters. $P(A|B)$ is not the same as $P(B|A)$.
Quick example: $P(A)=0.6,P(B)=0.5,P(A\cap B)=0.3 \Rightarrow P(A|B)=\dfrac{0.3}{0.5}$.
Answer: 0.6$P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.2}{0.5}=0.4$.
$0.5$ is actually $P(B|A)=\dfrac{0.2}{0.4}$ — the conditioning was flipped; $0.2$ just restates $P(A\cap B)$ without dividing; $0.7$ confuses the question with $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
The expected value of a discrete random variable is a probability-weighted average: $E(X)=\sum xP(X=x)$ — not a simple average, and not the mode.
Quick example: $X:0,1$ with $P=0.3,0.7 \Rightarrow E(X)=0(0.3)+1(0.7)$.
Answer: 0.7$E(X)=1(0.1)+2(0.3)+3(0.4)+4(0.2)=0.1+0.6+1.2+0.8=2.7$.
$2.5$ is the simple, unweighted average of $1,2,3,4$; $3.0$ is the mode (the value with the highest probability); $10.0$ just sums the $x$-values and forgets the probability weights altogether.
The chain rule differentiates a composite function: $\dfrac{d}{dx}\big[f(g(x))\big]=f'(g(x))\cdot g'(x)$ — "outer derivative times inner derivative."
Quick example: $y=(2x+1)^3 \Rightarrow \dfrac{dy}{dx}=3(2x+1)^2\cdot2$.
Answer: 6(2x+1)²$\dfrac{dy}{dx}=4(3x^2-1)^3\cdot6x=24x(3x^2-1)^3$.
At $x=1$: $24(1)(2)^3=24(8)=192$.
$32$ forgets to multiply by the inner derivative $6x$; $48$ forgets the outer coefficient $4$; $12$ misapplies the power rule directly to the bracket's contents.
The Fundamental Theorem of Calculus evaluates a definite integral as $\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)$, where $F$ is an antiderivative of $f$.
Quick example: $\displaystyle\int_0^1 3x^2\,dx=\big[x^3\big]_0^1=1-0$.
Answer: 1$F(x)=x^4-3x^2+2x$.
$F(2)-F(0)=(16-12+4)-0=8$.
$56$ comes from using $4x^4$ as the antiderivative term instead of dividing by the new power; $32$ comes from a sign slip on $-6x \to +3x^2$; $4$ drops the $+2x$ term during evaluation.
To optimise a function, differentiate, set $f'(x)=0$ to find critical points, then confirm a minimum or maximum with the second derivative test ($f''>0\Rightarrow$ min).
Quick example: $S(x)=x+\dfrac9x,\ x>0$: $S'(x)=1-\dfrac{9}{x^2}=0\Rightarrow x=3$, and $S(3)=6$ (minimum, since $S''(3)>0$).
Answer: minimum 6 at x = 3$S'(x)=2x-\dfrac{16}{x^2}=0 \Rightarrow 2x^3=16 \Rightarrow x^3=8 \Rightarrow x=2$.
$S''(x)=2+\dfrac{32}{x^3}>0$ at $x=2$, confirming a minimum. $S(2)=4+8=12$.
$2$ is the $x$-value, not the minimum value — a common mix-up between "where" and "what"; $8$ only evaluates $\tfrac{16}{x}$ and forgets the $x^2$ term; $20$ comes from an arithmetic slip while substituting.
In kinematics, if $s(t)$ is displacement, then $v(t)=s'(t)$ is velocity and $a(t)=v'(t)$ is acceleration. "Instantaneously at rest" means $v(t)=0$.
Quick example: $s(t)=t^2-4t$: $v(t)=2t-4=0\Rightarrow t=2$, $a(t)=2$.
Answer: a = 2$v(t)=3t^2-12t+9=0 \Rightarrow t^2-4t+3=0 \Rightarrow (t-1)(t-3)=0 \Rightarrow t=1,3$.
The second time at rest is $t=3$. $a(t)=6t-12$, so $a(3)=18-12=6\text{ m/s}^2$.
$-6$ comes from using the first rest time $t=1$ instead of the second; $0$ confuses acceleration with the (zero) velocity at that instant; $18$ forgets to subtract the constant term $12$.