Binomial theorem, proof & complex numbers – IB Maths exam questions with mark scheme

1CoreNon-calculator3 marks

Expand $(x+2)^4$, giving your answer in descending powers of $x$.

Uses binomial coefficients $1,4,6,4,1$ with powers of 2M1
At least three correct termsA1
$x^4+8x^3+24x^2+32x+16$A1
Answer: $x^4+8x^3+24x^2+32x+16$
Examiner tip: Remember $2^2=4$ and $2^3=8$ are multiplied by the coefficient, not replaced by it.

2StandardNon-calculator4 marks

Find the constant term in the expansion of $\left(x^2+\dfrac{2}{x}\right)^9$.

General term $\binom9r(x^2)^{9-r}\left(\dfrac2x\right)^r=\binom9r2^rx^{18-3r}$M1
Sets $18-3r=0$M1
$r=6$A1
$\binom96\times2^6=84\times64=5376$A1
Answer: $5376$
Examiner tip: Combine the powers of $x$ first, then set the exponent to zero.

3StandardNon-calculatorHL only5 marks

Use proof by contradiction to show that $\sqrt3$ is irrational.

Assumes $\sqrt3=\dfrac pq$ where $p,q\in\mathbb{Z}$, $q\neq0$, with no common factor (allow $\gcd(p,q)=1$ / fraction in lowest terms)R1
Squares: $p^2=3q^2$M1
So $p^2$ is a multiple of 3, hence $p$ is a multiple of 3 (since 3 is prime; accept 'if $p$ is not a multiple of 3 then neither is $p^2$')R1
Sets $p=3m$: $9m^2=3q^2$, so $q^2=3m^2$ and $q$ is a multiple of 3M1
Both multiples of 3 contradicts no common factor, so $\sqrt3$ is irrationalR1
Answer: Proof (contradiction)
Examiner tip: Say why $p^2$ divisible by 3 implies $p$ divisible by 3, and state the contradiction explicitly.

4StandardNon-calculatorHL only5 marks

Let $z=3+4i$ and $w=1-2i$.

(a) Find $zw$. [2]

(b) Find $\dfrac zw$ in the form $a+bi$. [3]

Expands $(3+4i)(1-2i)=3-6i+4i-8i^2$M1
$11-2i$A1
Multiplies numerator and denominator by $1+2i$M1
$\dfrac{-5+10i}{5}$A1
$-1+2i$A1
Answer: (a) $11-2i$ (b) $-1+2i$
Examiner tip: Replace $i^2$ by $-1$, and multiply by the conjugate of the denominator.

5StandardNon-calculatorHL only6 marks

Prove by mathematical induction that, for all $n\in\mathbb{Z}^+$,

$$\sum_{r=1}^{n}r(r+1)=\frac{n(n+1)(n+2)}{3}.$$

$n=1$: LHS $=1\times2=2$, RHS $=\dfrac{1\times2\times3}{3}=2$, so true for $n=1$A1
Assumes true for $n=k$: $\sum_{r=1}^k r(r+1)=\dfrac{k(k+1)(k+2)}{3}$ (must be stated as an assumption; 'let $n=k$' alone does not earn this mark)M1
Considers $n=k+1$ and adds the next term $(k+1)(k+2)$M1
Correct intermediate step, e.g. $\dfrac{k(k+1)(k+2)+3(k+1)(k+2)}{3}$ or $(k+1)(k+2)\left(\dfrac k3+1\right)$ (accept expanded $\dfrac{k^3+6k^2+11k+6}{3}$)A1
$=\dfrac{(k+1)(k+2)(k+3)}{3}=\dfrac{(k+1)\big((k+1)+1\big)\big((k+1)+2\big)}{3}$, i.e. the formula with $n=k+1$A1
True for $n=1$, and true for $n=k\Rightarrow$ true for $n=k+1$, so true for all $n\in\mathbb{Z}^+$ by induction (award only if both M1 marks have been earned)R1
Answer: Proof by induction
Examiner tip: The final conclusion needs all three ideas: true for $n=1$, true for $k$ implies $k+1$, hence all $n$.

6HardNon-calculatorHL only7 marks

Let $z=1+\sqrt3\,i$.

(a) Find the modulus and argument of $z$. [2]

(b) Use de Moivre's theorem to find $z^5$ in the form $a+bi$. [3]

$|z|=2$A1
$\arg z=\dfrac\pi3$A1
$z^5=2^5\left(\cos\dfrac{5\pi}3+i\sin\dfrac{5\pi}3\right)$M1
$32\left(\tfrac12-\tfrac{\sqrt3}2i\right)$A1
$16-16\sqrt3\,i$A1
Requires $\dfrac{n\pi}3=m\pi$ (argument a multiple of $\pi$)M1
$n=3$A1
Answer: (a) $2$, $\pi/3$ (b) $16-16\sqrt3\,i$ (c) $n=3$
Examiner tip: Raise the modulus to the power and multiply the argument by the power.

7StretchNon-calculatorHL only7 marks

Let $z=\cos\theta+i\sin\theta$.

(a) Use de Moivre's theorem to show that $z^n+z^{-n}=2\cos n\theta$. [3]

(b) By considering $\left(z+\dfrac1z\right)^4$, show that $\cos^4\theta=\dfrac{\cos4\theta+4\cos2\theta+3}{8}$. [4]

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