Resources - TL Maths

Free IB Mathematics questions and practice materials from TL Maths.

Question of the Day

Question of the day 7.09

AA HL · Integration

Find the indefinite integral:

$\int \sqrt{2x-x^2}\,dx$

Hint

First complete the square:

$2x-x^2=1-(x-1)^2.$

Then use the substitution

$x-1=\sin\theta.$

Solution

First complete the square:

$2x-x^2=1-(x-1)^2.$

Therefore

$\int \sqrt{2x-x^2}\,dx = \int \sqrt{1-(x-1)^2}\,dx.$

Let

$x-1=\sin\theta.$

Then

$dx=\cos\theta\,d\theta.$

Also,

$\sqrt{1-(x-1)^2} = \sqrt{1-\sin^2\theta} = \cos\theta.$

Hence

$\int \sqrt{1-(x-1)^2}\,dx = \int \cos^2\theta\,d\theta.$

Using the identity

$\cos^2\theta=\frac{1+\cos 2\theta}{2},$

we obtain

$\int \cos^2\theta\,d\theta = \frac{\theta}{2} + \frac{\sin 2\theta}{4} +C.$

Since

$\theta=\sin^{-1}(x-1),$

and

$\frac{\sin 2\theta}{4} = \frac{1}{2}\sin\theta\cos\theta = \frac{1}{2}(x-1)\sqrt{2x-x^2},$

the final answer is

$\boxed{ \int \sqrt{2x-x^2}\,dx = \frac12(x-1)\sqrt{2x-x^2} + \frac12\sin^{-1}(x-1) +C }$

Question of the day 4.09

AA HL · Calculus · Related Rates

Water is evaporating from a cup in the shape of an inverted cone. The rate of evaporation is proportional to the area of the surface of the water.

Show that the depth of the water decreases at a constant rate that does not depend on the dimensions of the cup.

Hint

Let $h$ be the depth of the water and $r$ the radius of its surface. Use similar triangles to express $r$ in terms of $h$ .

The evaporation rate is proportional to the exposed surface area:

$\frac{dV}{dt}=-k\pi r^2.$

Solution

Let $h$ be the depth of the water and $r$ the radius of its surface. Since the water forms a cone similar to the cup,

$\frac{r}{h}=c,$

where $c$ is a constant. Hence

$r=ch.$

The volume of the water is

$V=\frac13\pi r^2h.$

Substituting $r=ch$ ,

$V=\frac13\pi c^2h^3.$

Differentiating with respect to time,

$\frac{dV}{dt} = \pi c^2h^2\frac{dh}{dt}.$

Since $r=ch$ ,

$c^2h^2=r^2,$

$\frac{dV}{dt} = \pi r^2\frac{dh}{dt}.$

The rate of evaporation is proportional to the area of the surface of the water, so for some constant $k>0$ ,

$\frac{dV}{dt}=-k\pi r^2.$

Therefore,

$\pi r^2\frac{dh}{dt} = -k\pi r^2.$

Hence

$\boxed{\frac{dh}{dt}=-k}.$

Thus the depth of the water decreases at a constant rate, independent of the dimensions of the cup.

Question of the day 3.09

AA SL · Sequences · Logarithms

An arithmetic sequence has first term $\ln a$ and common difference $\ln 3$ . The 13th term of the sequence is $8\ln 9$ .

Find the value of $a$ .

Hint

Use the formula for the $n$ th term of an arithmetic sequence:

$u_n=u_1+(n-1)d.$

You will also need to simplify $\ln 9$ .

Solution

Using

$u_n=u_1+(n-1)d,$

we have

$\ln a+12\ln 3=8\ln 9.$

Since

$\ln 9=\ln(3^2)=2\ln 3,$

it follows that

$\ln a+12\ln 3=16\ln 3.$

Hence

$\ln a=4\ln 3=\ln(3^4).$

Therefore,

$\boxed{a=81}.$

Question of the day 2.09

AI HL · Probability · Poisson Distribution

The number of marking errors made by a teacher on a piece of homework follows a Poisson distribution with mean 1.6. The teacher marks one piece of homework from each of 13 pupils.

Find the probability that fewer than half of the pieces of homework contain at least one marking error.

Hint

First find the probability that a single piece of homework contains at least one marking error:

$P(X\geq 1)=1-P(X=0).$

Then consider the distribution of the number of pieces of homework, out of 13, that contain at least one error.

Solution

Let

$X\sim \mathrm{Po}(1.6).$

The probability that a single piece of homework contains at least one marking error is

$P(X\geq 1)=1-P(X=0)=1-e^{-1.6}.$

Let $Y$ be the number of pieces of homework, out of 13, that contain at least one marking error. Then

$Y\sim \mathrm{B}\left(13,1-e^{-1.6}\right).$

Fewer than half of 13 means at most 6, so

$P(Y\leq 6).$

Using a calculator,

$P(Y\leq 6)\approx 0.00739.$

Therefore,

$\boxed{0.00739}.$

Question of the day 1.09

AA HL · Complex Numbers · Trigonometry

(a) Use de Moivre’s theorem to prove that

$\cos 5\theta = \cos\theta \left( 16\cos^4\theta-20\cos^2\theta+5 \right).$

(b) By considering the equation $\cos 5\theta=0$ , show that

$\cos^2\left(\frac{\pi}{10}\right) = \frac{5+\sqrt5}{8}.$

Hint

For part (a), consider the real part of

$(\cos\theta+i\sin\theta)^5.$

For part (b), notice that

$16\cos^4\theta-20\cos^2\theta+5$

is a quadratic in $u=\cos^2\theta$ .

Solution

(a)

By de Moivre’s theorem,

$\cos 5\theta+i\sin 5\theta = (\cos\theta+i\sin\theta)^5.$

Taking real parts gives

$\begin{aligned} \cos 5\theta &= \operatorname{Re}(\cos\theta+i\sin\theta)^5\\ &= \cos^5\theta -10\cos^3\theta\sin^2\theta +5\cos\theta\sin^4\theta. \end{aligned}$

Using $\sin^2\theta=1-\cos^2\theta$ ,

$\begin{aligned} \cos 5\theta &= \cos^5\theta -10\cos^3\theta(1-\cos^2\theta) +5\cos\theta(1-\cos^2\theta)^2\\ &= \cos^5\theta -10\cos^3\theta +10\cos^5\theta\\ &\qquad +5\cos\theta -10\cos^3\theta +5\cos^5\theta\\ &= 16\cos^5\theta -20\cos^3\theta +5\cos\theta\\ &= \cos\theta \left( 16\cos^4\theta -20\cos^2\theta +5 \right). \end{aligned}$

Hence

$\boxed{ \cos 5\theta = \cos\theta \left( 16\cos^4\theta-20\cos^2\theta+5 \right) }.$

(b)

Consider

$\cos 5\theta=0.$

Five solutions in the interval $0<\theta<\pi$ are

$\theta= \frac{\pi}{10}, \frac{3\pi}{10}, \frac{5\pi}{10}, \frac{7\pi}{10}, \frac{9\pi}{10}.$

Using the identity from part (a),

$\cos\theta \left( 16\cos^4\theta-20\cos^2\theta+5 \right)=0.$

Therefore either

$\cos\theta=0$

$16\cos^4\theta-20\cos^2\theta+5=0.$

The first equation gives

$\theta=\frac{\pi}{2} = \frac{5\pi}{10}.$

For the remaining four solutions, let

$u=\cos^2\theta.$

Then

$16u^2-20u+5=0.$

Using the quadratic formula,

$\begin{aligned} u &= \frac{20\pm\sqrt{400-320}}{32}\\ &= \frac{20\pm\sqrt{80}}{32}\\ &= \frac{20\pm4\sqrt5}{32}\\ &= \frac{5\pm\sqrt5}{8}. \end{aligned}$