Welcome to this study IB site for students of the IB Mathematical Studies course. This short page is designed as a brief introduction to the site to help you understand what is here and how you can use it. The site is designed to be a 'virtual teacher' and a 'practice and revision center'. The virtual teacher aspect is a huge bank of slides and videos aimed at explaining the key concepts of the course. These are then backed up by revision flashcards, hundreds of practice questions with advice and feedback and exam style questions with full worked solutions. In short, everything you need to support the studying and assessment of this course.
Premium Subscription - 50 Euros only for full access to all of this. Our aim is to provide an indispensible study support grade for the whole course for the same price of 1 hours private tuition. We think it is extrodinary value for money
What is here?
- 200+ Teaching videos covering the key concepts on the syllabus.
- 200+ Slides with visual explanations and examples.
- 200+ Onscreen practice questions with feedback
- Revision Flashcards
- 50+ original exam style questions with video solutions
- Regular updates with more videos and questions going up every month.
And more.....
- You can also find a comprehensive section on how to use your Your Graphical Display Calculator
- There is a really helpful chapter on how to consider your IA & technology
- A crucial section on Exam Advice
- With a subsection on how to get Sixes & sevens
Teaching Videos
These allow you to review key syllabus concepts that you have learned in class. Slow, careful demonstrations and explanations that you can watch again and again. Designed to make you think and help you understand and improve. Some are designed to explain concepts from the beginning and others are focussed on very specific skills
Teaching Slides
Each topic page starts with a gallery of teaching slides that can be easily scanned so that you can find a slide that offers you a quick explanation, if you don't want to watch the whole video. Like flicking through the pages of a book, this section is desigend to be a quick reference guide to what you are looking for. You can click on any image to make it fill the screen and then swipe through them - this works really well on a mobile device.
On Screen Quizzes
This is a huge section of the site allowing lots and lots of practice. You can get instant feedback on your answers including advice about how to tackle the question and worked solutions. See the example below.
START QUIZ!
Question 1
Consider the function \(f(x)={ x }^{ 2 }-4x+7\)
then \(f'(x)\quad =\quad ax\quad +\quad b\)
a) What are the values of a and b?
b) What is the value of x when the gradient is zero.
c) And so if the coordinates of the vertex are (x,y), what are the values of x and y?
a) a = , b =
b) x =
c) x = , y =
a) Function differentiates to \(f'(x)\quad =\quad 2x-4\) (careful, since it is -4, but the equation says +b, then b must be negative.
b) \(when\quad gradient\quad =\quad 0,\\ f'(x)\quad =\quad 2x-4\quad =\quad 0\\ so,\quad 2x\quad =\quad 4\\ and\quad x\quad =\quad 2\)
c) \(At\quad vertex,\quad f'(x)\quad =\quad 0\quad so\quad x\quad =\quad 2\\ Substitute\quad x\quad =\quad 2\quad into\quad f(x)={ x }^{ 2 }-4x+7\\ So,\quad y\quad =\quad 3\) (You might use your table function here)
Question 2
Consider the function \(f(x)={ x }^{ 2 }+6x+3\)
then \(f'(x)\quad =\quad ax\quad +\quad b\)
a) What are the values of a and b?
b) What is the value of x when the gradient is zero.
c) And so if the coordinates of the vertex are (x,y), what are the values of x and y?
a) a = , b =
b) x =
c) x = , y =
a) Function differentiates to \(f'(x)\quad =\quad 2x+6\)
b) \(when\quad gradient\quad =\quad 0,\\ f'(x)\quad =\quad 2x+6\quad =\quad 0\\ so,\quad 2x\quad =\quad -6\\ and\quad x\quad =\quad -3\)
c) \(At\quad vertex,\quad f'(x)\quad =\quad 0\quad so\quad x\quad =\quad -3\\ Substitute\quad x\quad =\quad -3\quad into\quad f(x)={ x }^{ 2 }+6x+3\\ So,\quad y\quad =\quad -6\) (You might use your table function here)
Question 3
Consider the function \(f(x)=\frac { 1 }{ 3 } { x }^{ 3 }-\frac { 1 }{ 2 } { x }^{ 2 }-12x+4\)
Then \(f'(x)=a{ x }^{ 2 }+b{ x }+c\)
a) What are the values of a, b and c?
b) What are the values of x when the gradient is zero? (Give your answers in numerical order)
c) What are the coordinates of the local maximum and minimum for this function
a) a = , b = , c =
b) x = , x =
c) Local maximum x = , y = (give y to 3sf)
Local Minimum, x = , y = (give y to 3sf)
a) \(f'(x)={ x }^{ 2 }-{ x }-12\)(Pay attention to the negative signs)
b) \(When\quad the\quad gradient\quad is\quad zero\\ f'(x)={ x }^{ 2 }-{ x }-12\quad =\quad 0\\ f'(x)=(x+3)(x-4)\quad =\quad 0\\ x\quad =\quad -3\quad or\quad 4\) You can use the polynomial solver on your GDC to solve this equation
c) Read the gradient function from your GDC for x = -3 and x = 4
Question 4
Consider the function \(f(x)={ 2x }^{ 2 }+\frac { 32 }{ x } \)
a) This can be written as, \(f(x)={ 2x }^{ 2 }+m{ x }^{ n }\), what are the values of m and n
b) The derivative of the function is given by \(f'(x)\quad =\quad ax+b{ x }^{ c }\)what are the values of a, b and c?
c) At what value of x is the gradient = 0?
d) What are the coordiantes of this local minimum?
a) m = , n =
b) a = , b = , c =
c) x =
d) x = , y =
a) BEWARE you are NOT differentiating here, you are just expressing it as a negative index. Remember...
\(\frac { 1 }{ { x }^{ n } } ={ x }^{ -n }\quad and\quad \frac { 3 }{ { x }^{ n } } ={ 3x }^{ -n }\quad and\quad \frac { 2 }{ { 5x }^{ n } } =\frac { 2 }{ 5 } { x }^{ -n }\) so \(f(x)={ 2x }^{ 2 }+\frac { 32 }{ x } \quad =\quad f(x)={ 2x }^{ 2 }+32{ x }^{ -1 }\)
...
b) \(if,\quad f(x)={ 2x }^{ 2 }+32{ x }^{ -1 },\quad then,\quad f'(x)\quad =\quad 4x-32{ x }^{ -2 }\)
...
c) \(when\quad gradient\quad =\quad 0,\\ f'(x)\quad =\quad 4x-32{ x }^{ -2 }\quad =\quad 0\\ f'(x)\quad =\quad 4x-\frac { 32 }{ { x }^{ 2 } } =0\\ so\quad 4x\quad =\quad \frac { 32 }{ { x }^{ 2 } } \\ and,\quad 4{ x }^{ 3 }\quad =\quad 32\\ and\quad { x }^{ 3 }\quad =\quad 8,\quad \\ so\quad x\quad =\quad 2\) Again, your equation solver on the GDC can do this for you.
...
d) \(At\quad vertex,\quad f'(x)\quad =\quad 0\quad so\quad x\quad =\quad 2\\ Substitute\quad x\quad =\quad 2\quad into\quad f(x)={ 2x }^{ 2 }+\frac { 32 }{ x } \\ So,\quad y\quad =\quad 24\) Again, the table function will help you to avoid mistakes
Question 5
The Volume of water (V) in a container varies with some given conditions against time (t), The Volume, \(V{ cm }^{ 3 }\) , is given by \(V=400+3t-{ t }^{ 2 }\)
a) If \(\frac { dV }{ dt } =at+b\) then what are the values of a and b?
b) At what value of t will the volume be at a maximum?
c) What will the maximum volume be?
a) a = , b =
b) t = (give the exact answer as a decimal)
c) Maximum Volume =
a) \(if\quad V=400+3t-{ t }^{ 2 }\quad then,\frac { dV }{ dt } =3-2t\)
b) \(at\quad maximum,\quad \frac { dV }{ dt } =3-2t=0\\ so,\quad 3=2t\\ and\quad t=\frac { 3 }{ 2 } =1.5\)
c) \(Maximum\quad occurs\quad at\quad t\quad =\quad 1.5\\ when\quad t=1.5\\ V=400+3(1.5)-{ (1.5) }^{ 2 }\\ V=400+4.5-2.25\\ V=402.25,\quad V=402{ cm }^{ 3 }\quad (3sf)\)
Using the GDC, if you enter the Volume function and have the derivative switched on then you can see that at t = 1.5, Volume is 402.25 and the derivative is zero.
Question 6
A company's profit, P, in 1000s of euros can be modelled by the the function, \(P(x)=19x-0.07{ x }^{ 2 }-10\)where \(x\), is a measure of units of product for the company.
a) if \(\frac { dP}{ dx } =ax+b,\)what are the values of a and b?
b) At what value of x while the profit be at a maximum?
c) Use the rounded answer to part b) to calculate the value of the maximum profit.
a) a = , b =
b) x = (give your answer to 3sf)
c) Maximum profit, P = (give your answer as the number of 1000s to 3sf)
a) \(if\quad P(x)=19x-0.07{ x }^{ 2 }-10,\quad then,\frac { dP }{ dx } =19-0.14x\)
b) \(At\quad maximum,\quad \frac { dP }{ dx } =19-0.14x=0\\ 19=0.14x\\ x=\frac { 19 }{ 0.14 } =135.71428.....\\ x=\quad 136\quad (3sf)\)
c) \(maximum\quad profit\quad occurs\quad at\quad x\quad =\quad 136\\ Substituting\quad in\quad to\quad P(x)=19x-0.07{ x }^{ 2 }-10\\ gives\quad V=1279.2\\ V=1280\quad (3sf)\)
Question 7
Consider the formula A = bh which is subject to the constraint that b - h=10.
a) Which of the expressions is the correct expression for A in terms of b?
b) If \(\frac { dA }{ db } =mb+n\), what are the values of m and n?
c) For what value of b is A a minimum?
d) What is the minimum value of A?
b) m = , n =
c) b =
d) Minimum value of A =
a) \(A=bh\\ b-h=10\quad so\\ b=10+h\quad and\\ b-10=h\\ Substitute\quad h\quad =\quad b-10\\ A=b(b-10)\\ A={ b }^{ 2 }-10b\)
b) \(if\quad A={ b }^{ 2 }-10b\\ then\quad \frac { dA }{ db } =2b-10\)
c) \(at\quad minimum\quad value,\quad \frac { dA }{ db } =2b-10=0\\ so\quad 2b=10\\ and\quad b=5\)
d) \(Substitute\quad b=5\quad into\quad A={ b }^{ 2 }-10b\\ so\quad A=-25\)
Question 8
Consider the formula \(A={ x }^{ 3 }+2xh\), subject to the condition that \(h=\frac { 3 }{ { x }^{ 3 } } \)
a) A can be expressed in terms of x only like this, \(A={ x }^{ 3 }+p{ x }^{ q }\), what are the values of p and q?
b) \(\frac { dA }{ dx } =m{ x }^{ 2 }-n{ x }^{ -3 }\), what are the values of m and n?
c) For what value of x is A at a local minimum?
a) p = , q =
b) m = , n =
c) x = (Answer to 3sf)
a) Substitute the value for h in to the expression for A,
\(if\quad A={ x }^{ 3 }+2xh\quad and\quad h=\frac { 3 }{ { x }^{ 3 } } \\ Substituting\quad gives\\ A={ x }^{ 3 }+\frac { 2x\times 3 }{ { x }^{ 3 } } \\ A={ x }^{ 3 }+\frac { 6x }{ { x }^{ 3 } } \\ A={ x }^{ 3 }+\frac { 6 }{ { x }^{ 2 } } \\ A={ x }^{ 3 }+6{ x }^{ -2 }\)
b) Differentiate A with respect to x
\(if\quad A={ x }^{ 3 }+6{ x }^{ -2 }\\ \\ then,\frac { dA }{ dx } =3{ x }^{ 2 }-12{ x }^{ -3 }\)
c) Solve for gradient = 0
\(At\quad minimum,\quad \frac { dA }{ dx } =3{ x }^{ 2 }-12{ x }^{ -3 }=0\\ 3{ x }^{ 2 }-12{ x }^{ -3 }=0\quad (use\quad GDC\quad solver)\\ x=1.3195....\\ x=1.32\quad (3sf)\)
Question 9
A closed cuboid box has Volume, \(V={ x }^{ 2 }h\). Its surface area, A, is limted to \(600{ cm }^{ 2 }\)
Dimensions shown are in cm
a) A is given by \(A=m{ x }^{ 2 }+nxh\), what are the values of m and n?
b) Since A = 600, we can say that \(\frac { a-{ x }^{ 2 } }{ bx } =h\), what are the values of a and b?
c) Use this expression for h to express the volume in terms of h only, where \(V=px-\frac { 1 }{ 2 } { x }^{ q }\), what are the values of p and q?
d) For what value of x is the Volume at a maximum?
e) What is the maximum volume?
f) What will the value of h be at this volume?
a) m = , n =
b) a = , b =
c) p = , q =
d) x = cm
e) Maximum Volume = \({ cm }^{ 3 }\)
f) h = cm
a) \(Area\quad is\quad the\quad 6\quad faces\quad added\quad together\\ A\quad =\quad 2{ x }^{ 2 }+2xh+2xh\\ A=2{ x }^{ 2 }+4xh\)
b) This part is tricky but is all about making h the subject of the formula
\(A=2{ x }^{ 2 }+4xh\\ 600=2{ x }^{ 2 }+4xh\\ 600-2{ x }^{ 2 }=4xh\\ \frac { 600-2{ x }^{ 2 } }{ 4x } =h\\ \frac { 300-{ x }^{ 2 } }{ 2x } =h\)
c) This stage involves substitutuing the expression for h in to the expression for volume.
\(Since\quad V{ =x }^{ 2 }h\quad and\quad h=\frac { 300-{ x }^{ 2 } }{ 2x } \\ V=\frac { { x }^{ 2 }(300-{ x }^{ 2 }) }{ 2x } \\ V=\frac { 300{ x }^{ 2 }-{ x }^{ 4 } }{ 2x } \\ (Divide\quad through\quad by\quad 2x)\\ V=150x-\frac { 1 }{ 2 } { x }^{ 3 }\)
d) Differentiate the function and solve for gradient = 0
\(if\quad V=150x-\frac { 1 }{ 2 } { x }^{ 3 }\quad then,\frac { dV }{ dx } =150-\frac { 3 }{ 2 } { x }^{ 2 }\\ Volume\quad is\quad a\quad maximum\quad when\quad \frac { dV }{ dx } =150-\frac { 3 }{ 2 } { x }^{ 2 }=0\\ 150=\frac { 3 }{ 2 } { x }^{ 2 }\\ 300=3{ x }^{ 2 }\\ 100={ x }^{ 2 }\\ x=\quad 10\quad (-10\quad is\quad not\quad in\quad the\quad valid\quad range)\)
e) This involves substituting the value of x that gives a minimum into the expression for volume.
\(Maximum\quad occurs\quad at\quad x=\quad 10\\ substitute\quad x=10\quad into\quad V=150x-\frac { 1 }{ 2 } { x }^{ 3 }\\ V=1000{ cm }^{ 3 }\)
f) Now substitute the value of x in to the expression for h.
\(Since\quad \frac { 300-{ x }^{ 2 } }{ 2x } =h\quad and\quad x=10\\ then,\quad \frac { 300-100 }{ 20 } =h\\ 10=h\)
Question 10
Consider a closed cylinder, radius r, height, h, whose volume must be 400\({ cm }^{ 3 }\). The aim is to minimise the Surface area, A, required to make the cylinder.
a) if the height h, can be expressed as \(h=\frac { a }{ \pi { r }^{ b } } \), what are the values of a and b?
b) The surface area of the cylinder can be expressed as \(A=2\pi { r }^{ 2 }+p{ r }^{ q }\), what are the values of p and q?
c) What value of r will give the minimum surface area?
d) Use the rounded answer to the part c) to work out,
i) The value of h when A is a minimum
ii) The minimum value of A
a) a = , b =
b) p = , q =
c) r = (Answer correct to 3sf)
d) i) h = cm , ii) A = cm2 (answers to 3sf)
a) Use the formula for volume of a cylinder and rearrange.
\(V=\pi { r }^{ 2 }h\quad and\quad is\quad given\quad as\quad 400{ cm }^{ 3 }\\ then\\ 400=\pi { r }^{ 2 }h\\ and\quad so,\quad h=\frac { 400 }{ \pi { r }^{ 2 } } \)
b) Use the formula for surface area of a cylinder and substitute the expression for h, then simplify.
\(A=2\pi { r }^{ 2 }+2\pi { r }h\quad and\quad h=\frac { 400 }{ \pi { r }^{ 2 } } \\ Substituting\quad for\quad h,\\ A=2\pi { r }^{ 2 }+\frac { 2\pi { r }\times 400 }{ \pi { r }^{ 2 } } \\ Cancelling\quad common\quad factors\\ A=2\pi { r }^{ 2 }+\frac { 800 }{ { r } } \\ A=2\pi { r }^{ 2 }+800{ r }^{ -1 }\)
c) Solve for gradient = 0. Perfectly acceptable to use the solve function to do this.
\(If\quad A=2\pi { r }^{ 2 }+800{ r }^{ -1 },\quad then\quad \frac { dA }{ dr } =4\pi { r }-800{ r }^{ -2 }\\ At\quad minimum,\quad \frac { dA }{ dr } =4\pi { r }-800{ r }^{ -2 }=0\\ 4\pi { r }=800{ r }^{ -2 }\\ 4\pi { r }=\frac { 800 }{ { r }^{ 2 } } \\ 4\pi { r }^{ 3 }=800\\ { r }^{ 3 }=\frac { 800 }{ 4\pi } \\ r=\sqrt [ 3 ]{ \frac { 800 }{ 4\pi } } \\ r=3.9929454...\\ r=3.99\quad cm\quad (3sf)\)
d) Substitutions can be done using the table function on your GDC or otherwise.
i) \(Substitute\quad r=3.99\quad in\quad to\quad h=\frac { 400 }{ \pi { r }^{ 2 } } \\ h=7.997685...\\ h=8.00cm\quad (3sf)\)
ii) \(Substitute\quad r=3.99\quad in\quad to\quad A=2\pi { r }^{ 2 }+800{ r }^{ -1 }\\ A=300.5301915...\\ A=\quad 301{ cm }^{ 2 }\quad (3sf)\)
Revision Cards
We reconginse that when preparing for assessments, that you will want shorter and shorter references. For that reason, each topic comes with a short set of revision cards for really quick reference.
Exam Style Questions
The end of each chapter has a selection of questions like the ones you can expect in your assessments. You can download and print a version of these from the top of the topic page and then watch video solutions! See this example from the probability topic.
...
Twitter 
Facebook 
LinkedIn