Create your own Questions for Revision

[Saracen Rue]

This is an idea I've been thinking of for a while. While providing a question for an individual a question to complete works fine as revision, numerous studies have proven that creating a question is, in itself, a more beneficial way to revise. Instead of simply recalling a process or equation to solve the problem, you have to craft the problem itself - doing this requires a much more in-depth understanding of the overall concepts of the topic.

This is why I'm creating this thread; to prompt people out there to create their own questions for other users to answer. Not only will the person forming the question achieve a better understanding of the topic than simply revising, they will also be exposed to the thought processes of other people. There are countless ways to solve a problem; by putting a question out on the internet to be solved by others you will not only be aiding other people in revising certain areas, but you will also achieve a greater overall understanding of the topic and will be exposed to problem solving process you had never even thought of before.

I'll pose a mathematical related question which addresses multiple year 12 course areas as an example:

Question
A function, ##f(x)=2ax^3-a^2x## intersects its inverse at the origin, point ##S(-b,f(-b))## and point ##T(b,f(b))##. A probability density function, ##p(x)=f(x)-f^{-1}(x)##, can be formed over the domain ##[0, b]##. Determine, correct to 4 decimal places:
a) The value of the constant, ##a##, and the coordinates of points ##S## and ##T##.
b) The mean, variance and standard deviation of ##p(x)##
c) The probability that the contentious random variable ##X## lies within ##|a|## standard deviations either side of the mean (i.e. ##Pr(μ-|a|σ≤X≤μ+|a|σ)##)

Answers

Spoiler: Click here to see answers

a) ##a=-0.2253, S\left(-1.4515,\ 1.4515\right),\ T\left(1.4515,\ -1.4515\right)##
b) ##μ=0.6692, Var(X)=0.5673,## ##σ=0.7532##
c) ##Pr(μ-|a|σ≤X≤μ+|a|σ)=0.3147##

 

[Andy Resnick]

This is an idea I've been thinking of for a while. While providing a question for an individual a question to complete works fine as revision, numerous studies have proven that creating a question is, in itself, a more beneficial way to revise. Instead of simply recalling a process or equation to solve the problem, you have to craft the problem itself - doing this requires a much more in-depth understanding of the overall concepts of the topic.

This is an excellent idea- I sometimes suggest to my students that they study by designing test-like questions. They often don't realize how difficult that is, but they do see the value very quickly.

 

[martane]

Wow, this is a really interesting idea! I've never thought of creating my own questions for others to answer as a way to revise. I can definitely see how it would be more beneficial in terms of understanding the topic and exposing yourself to different problem-solving processes.

As for the question you posed, it's definitely a challenging one. I'm not sure if I have the skills to solve it, but I'll give it a try.

a) To find the value of ##a##, we can set the function equal to its inverse and solve for ##a##. This gives us the equation ##2ax^3-a^2x=x^3-2ax##. Simplifying, we get ##x^3-2ax=0##. This is true for all values of ##x##, so we can choose any value to solve for ##a##. Let's choose ##x=1##. Substituting, we get ##1-2a=0##, or ##a=1/2##.

To find the coordinates of points ##S## and ##T##, we can use the fact that the function and its inverse intersect at these points. Substituting ##a=1/2## into the function, we get ##f(x)=x^3-x##. Setting this equal to ##-b## and solving for ##x##, we get ##x=-b## and ##x=b##. Therefore, the coordinates of points ##S## and ##T## are ##(-b, f(-b))## and ##(b, f(b))## respectively.

b) To find the mean of ##p(x)##, we can use the formula ##\mu=\int_{0}^{b}xp(x)dx##. Substituting in the formula for ##p(x)##, we get ##\mu=\int_{0}^{b}x(f(x)-f^{-1}(x))dx##. Using the fact that ##f(x)=x^3-x## and ##f^{-1}(x)=x^3+x##, we can simplify this to ##\mu=\int_{0}^{b}x(x^3-x^3-x)dx##. Solving this integral, we get ##\mu=-b^4/4##.

To find the variance, we can use the formula ##\sigma^2=\int_{0}^{b}(