For Question 11 it seems much simpler to just do: (PS)² + (PQ)² = (SQ)² , it simplifies to x²=yz very quickly without having to do any circle theorem stuff.
@@sez_srdn7749 I mean: (PS)² = x² + z², (PQ)²= x² + y², (SQ)² = (z+y)², so sub into: (PS)² + (PQ)² = (SQ)², we get: x² + z² + x² + y² = (z+y)², => 2x² + z² + y² = z² + 2zy + y² => 2x² = 2zy => x² = yz, which is necessary and sufficient to show angle SPQ is right angled. Answer E doesn't relate to the side lengths of triangle SPQ.
A small comment about the comparison of (1/2)^(1/2) versus (sqrt(3)/2)^(sqrt(3)/2) in Q18. While it is true that the former is the smaller quantity, the reasoning of 'bigger base to bigger power' doesn't quite work because for a base between 0 and 1 (as both 1/2 and sqrt(3)/2 are), increasing the exponent will actually make the result *smaller*. So, when we grow both the base and the power (as from 1/2 to sqrt(3)/2) it isn't clear whether the result ought to get bigger or smaller overall. By differentiating the relevant function, x^x, implicitly, one could in fact show that it is decreasing until 1/e and growing after that point (which means in this case the logic you used does work out, because 1/2 and sqrt(3)/2 both exceed 1/e), but this isn't obviously the case without that calculation which requires knowledge beyond the TMUA specification. Instead, note that R is quite flat near x = 0 while S is quite steep there. Since cosine is flat near x = 0, one can see without needing any differentiation etc. that cos(x)^(cos(x)) should also be pretty flat near x=0, because both the exponent and the base are only changing slowly. This leads us to the same conclusion: that cos(x)^cos(x) is R.
Useful piece of knowledge that can save some time with question 2 (0:32) is that the sum of coefficients in a binomial expansion is the same as evaluating the expression at x = 1, here you get 2^5 = 32. We can use this result as the right side doesn't hold any coefficients, meaning there is one possible combination for each element from the two brackets that multiplies to a fifth order term.
An algebraic way of doing q8: a = 1, d = 3, two distinct terms in S = a and b Sa = 1 + (a-1)3 Sb = 1 + (b-1)3 1 + 3a -3 + (1 + 3b - 3) = 74 2 - 6 + 3a + 3b = 74 3a + 3b = 78 3(a + b) = 78 a + b = 26 Because we are looking for the smallest value we need to minimise both a and b We can try 1 + 25, 2 + 24 ... until we get to 12 + 14 where 12 + 14 is the smallest combination of a and b we can get Note 13+13 is not a valid solution as we are looking for two "distinct" values Therefore we should have 14 terms in our sequence and they will add up to 74
on question 13) you can simply algebraically collect terms to get an inequality in for x, then its pretty easy to see x cannot be 1 (taking into account swapping of inequality signs)
For Q16, how do we know that the sequence of numbers can be literally anything. Is this something which can just be assumed if it isn't specified in the question?
Hello sir big fan of your videos, Could you pls make a video on how you would recommend doing multiple choice tests - for example in what situations would you choose to eliminate options vs doing the question "properly" Many thanks
In question 13,by "there exists a real y"do they mean that there is only a single value of y for all x or for a particular value of x there is a y which satisfies the above relation (means there can be multiple values of y
The bound 0 < k < 1 is only for the value of k, it doesn’t affect x, I think you may be thinking that this implies 0 < x < 1 but that is not the case. E.g. if k = 0.5, X < 0.5 => x could be -1 -1 < 0.5 is true but then 1 < 0.5 is false.
I dont understand question 7, how is step IV wrong, if you cant factorise it it shows its prime, because it has no factors and cannot be divided, therefore prime as prime numbers have no other factors other than 1 and itself
@@rtwodrew2 Thank u drew. One more question, if I have an identity, lets call it A, in terms of x. And another equation in terms of x, in the form x = B(x). I can substitute x = B(x) into the identity in as many instances of x as I want to form different identities which work for all x such that x = B(x). However, if I take each one of these identities to be an equation which I can solve for x, I get the solutions such that x = B(x), and, more commonly than not, I get some extra solutions. Is there a true reason as to why this happens? Or is it just something that cant be avoided and best we can say is that these equations work (for certain) for all x such that x = B(x), and potentially some others. If this makes no sense I can elaborate further with an example/
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
For Question 11 it seems much simpler to just do: (PS)² + (PQ)² = (SQ)² , it simplifies to x²=yz very quickly without having to do any circle theorem stuff.
But it could also imply option e aswell, even more to be honest as it is exactly Pythagoras
@@sez_srdn7749 I mean: (PS)² = x² + z², (PQ)²= x² + y², (SQ)² = (z+y)², so sub into: (PS)² + (PQ)² = (SQ)², we get: x² + z² + x² + y² = (z+y)², => 2x² + z² + y² = z² + 2zy + y² => 2x² = 2zy => x² = yz, which is necessary and sufficient to show angle SPQ is right angled. Answer E doesn't relate to the side lengths of triangle SPQ.
A small comment about the comparison of (1/2)^(1/2) versus (sqrt(3)/2)^(sqrt(3)/2) in Q18.
While it is true that the former is the smaller quantity, the reasoning of 'bigger base to bigger power' doesn't quite work because for a base between 0 and 1 (as both 1/2 and sqrt(3)/2 are), increasing the exponent will actually make the result *smaller*. So, when we grow both the base and the power (as from 1/2 to sqrt(3)/2) it isn't clear whether the result ought to get bigger or smaller overall. By differentiating the relevant function, x^x, implicitly, one could in fact show that it is decreasing until 1/e and growing after that point (which means in this case the logic you used does work out, because 1/2 and sqrt(3)/2 both exceed 1/e), but this isn't obviously the case without that calculation which requires knowledge beyond the TMUA specification.
Instead, note that R is quite flat near x = 0 while S is quite steep there. Since cosine is flat near x = 0, one can see without needing any differentiation etc. that cos(x)^(cos(x)) should also be pretty flat near x=0, because both the exponent and the base are only changing slowly. This leads us to the same conclusion: that cos(x)^cos(x) is R.
thanks
Useful piece of knowledge that can save some time with question 2 (0:32) is that the sum of coefficients in a binomial expansion is the same as evaluating the expression at x = 1, here you get 2^5 = 32. We can use this result as the right side doesn't hold any coefficients, meaning there is one possible combination for each element from the two brackets that multiplies to a fifth order term.
Thanks for the video too, it's nice to be able to check solutions with someone explaining so well.
An algebraic way of doing q8:
a = 1, d = 3, two distinct terms in S = a and b
Sa = 1 + (a-1)3
Sb = 1 + (b-1)3
1 + 3a -3 + (1 + 3b - 3) = 74
2 - 6 + 3a + 3b = 74
3a + 3b = 78
3(a + b) = 78
a + b = 26
Because we are looking for the smallest value we need to minimise both a and b
We can try 1 + 25, 2 + 24 ... until we get to 12 + 14 where 12 + 14 is the smallest combination of a and b we can get
Note 13+13 is not a valid solution as we are looking for two "distinct" values
Therefore we should have 14 terms in our sequence and they will add up to 74
on question 13) you can simply algebraically collect terms to get an inequality in for x, then its pretty easy to see x cannot be 1 (taking into account swapping of inequality signs)
For Q16, how do we know that the sequence of numbers can be literally anything. Is this something which can just be assumed if it isn't specified in the question?
Maybe go over AQA A Level Further Maths Papers if you're looking for content to make!
2022 was horrible so might a good couple of videos
Hello sir big fan of your videos,
Could you pls make a video on how you would recommend doing multiple choice tests - for example in what situations would you choose to eliminate options vs doing the question "properly"
Many thanks
For Question 20, for the M2, why can’t you do pi/2 to get the value equals to 1. Why does it have to be less than 1?
Cos x which is inside the brackets cannot be equal to pi/2 because that’s larger than 1, and cos x is restricted between -1 and 1.
Why isn’t Q13 D as y = -x seems to be a valid answer. x - xy + y = x + x^2 - x = x^2 which is never negative. Correct me if I wrong
In question 13,by "there exists a real y"do they mean that there is only a single value of y for all x or for a particular value of x there is a y which satisfies the above relation (means there can be multiple values of y
You should go over some bmo papers
Hello, I have a doubt in Question 9, shouldn't this be true if k is between 0 and 1
Yes same
It states, for all numbers x, in that conditional statement it means all x values
The bound 0 < k < 1 is only for the value of k, it doesn’t affect x,
I think you may be thinking that this implies 0 < x < 1 but that is not the case.
E.g. if k = 0.5,
X < 0.5 => x could be -1
-1 < 0.5 is true but then
1 < 0.5 is false.
for q20, you say that sin maxes out at pi, but we cant evaluate greater than 1, do you mean pi/2 is the max
9:30 proof by obviousness lmao that made me laugh
I dont understand question 7, how is step IV wrong, if you cant factorise it it shows its prime, because it has no factors and cannot be divided, therefore prime as prime numbers have no other factors other than 1 and itself
ruclips.net/video/1h4Nj0XVTFA/видео.html
@@rtwodrew2 thank you
Can you explain question 7 further please? Perhaps an extra video about deducing when polynomials output prime numbers etc?
Thank u so much
ruclips.net/video/1h4Nj0XVTFA/видео.html
Further reading:
mathworld.wolfram.com/Prime-GeneratingPolynomial.html
@@rtwodrew2 Thank u drew. One more question, if I have an identity, lets call it A, in terms of x. And another equation in terms of x, in the form x = B(x). I can substitute x = B(x) into the identity in as many instances of x as I want to form different identities which work for all x such that x = B(x). However, if I take each one of these identities to be an equation which I can solve for x, I get the solutions such that x = B(x), and, more commonly than not, I get some extra solutions. Is there a true reason as to why this happens? Or is it just something that cant be avoided and best we can say is that these equations work (for certain) for all x such that x = B(x), and potentially some others. If this makes no sense I can elaborate further with an example/
right angles that intersect at 90 degrees
thank you so much, amazing video
what was that paper??
thank you
Q12, Q13, Q14, Q18, Q19
hello
Saviour
thanks your a goat