Q4 always factorise rather than multiply out to save yourself the cubic factoring at the end. Not meaning to nitpick, I always direct my students to your videos. Thanks a billion for your efforts.
Another way to do Q6 is if a cubic has two real roots you can see by the graph you showed that the y-values of the the stationary points multiplied together are zero, so differentiate, find x-values then the y values and solve. Similarly if it has one root product is >0 and it has three real roots if
For questions similar to question 5, is it worth considering when x0 and just flipping the inequality sign accordingly as opposed to multiplying out by (x+p)^2?
For Q8, I used the following method and it was really quick, please correct if I'm mistaken. Since sinx has lowest value of -1, would it not be easier to use sinx = -1, then it's done nice and quick. Again, if I'm mistaken I would appreciate some correction 😄
this could be wrong as with different co-efficents, that might not be the largest. e.g if the quadratic produced has a minimum at x = 2, then sin x = 1 and so your answer would be wrong.
Hello Sir, I wanted to point out for Q8 when you factored out the - from the expression then you expanded it back on to the expression giving us -(2^sinx -1)^2 + 13/3 would you not get a "negative quadratic"? So I prefer doing substitution if we were to sub a variable and follow your method we would something like -(u - 1)^2 + 13/3 but if you were you expand that you would get a negative quadratic therefore, we would be looking for a maximum and not a minimum. I might be horribly wrong though If I am I apologise. The method I did was I multiplied the entire expression from -1 then you do your usual stuff for finding the minimum but the I would get -13/3 which is none of the options so I am really confused by this question.
I think the confusion arises from the difference between an expression and an equation. The maximum value of the expression -(u-1)^2 + 13/3 will be 13/3. However, if I multiply the expression by minus one, I will now be working with a different expression, which has a MINIMUM value of -13/3. This is because the graph of y=expression has been reflected in the x axis. So with your approach you need to change the sign at the end. The min value of -y is the max value of y. But note that there is a subtlety to this question because 'u' can't take all real values. Only values between 0.5 and 2, because u=2^sinx. Hence why at the end, we need to think about the shape of the graph for the quadratic expression we're minimising, so we can see that it will be smallest at the extremes. test both u=0.5 and u=2 and you'll see its smallest when u=2. Hope that helps.
Im curious as well, are the papers the exact same on both days? Because if not thecheating can take place. Or do you get random questions from a large database of questions. So its more ir less different for everyone. Or is there a different structure?
@@guywithglasses-g1q It's standardised so I think everyone gets different questions of equal difficulty, most likely from a database as you have guessed.
Q4 always factorise rather than multiply out to save yourself the cubic factoring at the end. Not meaning to nitpick, I always direct my students to your videos. Thanks a billion for your efforts.
Ohh I see what you mean you can take out x(x+3) near the start which just makes it easier
by factoring out will you not lose a solution?
never mind im stupid
Another way to do Q6 is if a cubic has two real roots you can see by the graph you showed that the y-values of the the stationary points multiplied together are zero, so differentiate, find x-values then the y values and solve. Similarly if it has one root product is >0 and it has three real roots if
For questions similar to question 5, is it worth considering when x0 and just flipping the inequality sign accordingly as opposed to multiplying out by (x+p)^2?
Do you have a link to the rest of the papers?
Question 6 becomes much easier if we consider:
at a turning point k = -3x^2 (after differentiating)
and at a turning point y = 0
at 2:45 , wouldnt you be able to minus the right hand side and then have both expressions on one side and solve from there, or wouldnt that work?
It would work!, infact im pretty sure, when you get to there everything just factorises and you get everysolution he got but much faster
Another way of doing q7 is using roots of polynomials law. Helpful for any further maths students!
For Q8, I used the following method and it was really quick, please correct if I'm mistaken. Since sinx has lowest value of -1, would it not be easier to use sinx = -1, then it's done nice and quick. Again, if I'm mistaken I would appreciate some correction 😄
same, I plugged in sinx = -1, and just to check also sinx= 0 and sinx=1
this could be wrong as with different co-efficents, that might not be the largest. e.g if the quadratic produced has a minimum at x = 2, then sin x = 1 and so your answer would be wrong.
do you think for question 1 it would make sense to try and find counter examples for values of p and q which would/wouldn't work
Are there more questions like this for specific topics
Hello Sir, I wanted to point out for Q8 when you factored out the - from the expression then you expanded it back on to the expression giving us -(2^sinx -1)^2 + 13/3 would you not get a "negative quadratic"? So I prefer doing substitution if we were to sub a variable and follow your method we would something like -(u - 1)^2 + 13/3 but if you were you expand that you would get a negative quadratic therefore, we would be looking for a maximum and not a minimum. I might be horribly wrong though If I am I apologise. The method I did was I multiplied the entire expression from -1 then you do your usual stuff for finding the minimum but the I would get -13/3 which is none of the options so I am really confused by this question.
I think the confusion arises from the difference between an expression and an equation. The maximum value of the expression -(u-1)^2 + 13/3 will be 13/3. However, if I multiply the expression by minus one, I will now be working with a different expression, which has a MINIMUM value of -13/3. This is because the graph of y=expression has been reflected in the x axis. So with your approach you need to change the sign at the end. The min value of -y is the max value of y.
But note that there is a subtlety to this question because 'u' can't take all real values. Only values between 0.5 and 2, because u=2^sinx. Hence why at the end, we need to think about the shape of the graph for the quadratic expression we're minimising, so we can see that it will be smallest at the extremes. test both u=0.5 and u=2 and you'll see its smallest when u=2.
Hope that helps.
for q5, if we sub in x = -p we get the inequality 0< -1?
Yes, well x=-p is not a valid solution and is not in the range of
Do u only book one day for tmua this year. As in are both papers on 16th October and both papers on 17th as well
So I guess no-one does the TMUA
Yes, book only one day.
Im curious as well, are the papers the exact same on both days? Because if not thecheating can take place. Or do you get random questions from a large database of questions. So its more ir less different for everyone. Or is there a different structure?
@@srinivasmedida107 thank you
@@guywithglasses-g1q It's standardised so I think everyone gets different questions of equal difficulty, most likely from a database as you have guessed.
These questions are trivial, surely the real TMUA is much harder than this?
ive been flicking through and yeah i agree