IGCSE Add Maths (0606/12/F/M/26) / Feb 2026 (fm26) / Paper 12 non-calculator

Añadido: 2026-05-11

[00:00:00]Hello ladies and gentlemen, welcome to the session. Today we will go for at maths FM 2026 paper one non-calculator paper. Let's go.

[00:00:10]First question, factors of polynomial.

[00:00:13]It says a polynomial P is given by this.

[00:00:16]Show that x + 2 is a factor. So we need to use factor theorem.

[00:00:21]So we sub in x equals to minus two to the equation.

[00:00:27]Proof that it is zero, then we can confirm this is a factor. Part two, solve the equation px. This is power three, then we use long division.

[00:00:44]This thing that we confirm, they ask us to solve, then we need to continue to factorize.

[00:00:51]Because this is a squared minus b squared, then we can split it into a plus b, a minus b. Then we can confirm these three answers. Done.

[00:01:03]Question two, on the axis sketch this quadratic equation that is modulus.

[00:01:09]State the intercept with the coordinate axis. So we know this is a smiley face, so we need to find what is the x intercept and y intercept. So when x equals to zero, y equals to zero squared minus two times zero minus three. So y equals to minus three. So initially it will cut through minus three. And then next we find when y equals to zero.

[00:01:32]It will cut through minus one and three.

[00:01:35]So the graph initially looks like this.

[00:01:38]Then because it's modulus, so everything below we will remove and go up. It will become three here. After confirm all the coordinate, we can join.

[00:01:48]And remember to label here as three. So make sure the x intercept and y intercept clearly indicate.

[00:01:56]Part B, they say the function is defined by this equation for all real X. Ask you to find the expression for G inverse.

[00:02:03]So, we change the subject.

[00:02:12]Question three. In this question, time T is in second and displacement is in meter. A particle start from rest at the origin O. So, if start from rest and the velocity is confirmed start from zero and move in a straight line with constant positive acceleration. Positive acceleration means it will move up in a straight line. They say when time equals to four, the displacement is six. Means that the total distance that they travel will be six because it start from origin.

[00:02:41]So, to count the displacement here, we need to use area under graph and we know this base is four, then we can find the H by using the triangle formula, half base times height equals to total distance which is six.

[00:02:55]So, we know when time equals to four, the height would be three. So, the the velocity here is three at last.

[00:03:03]And it must be a straight line.

[00:03:05]Part B, hence find acceleration. So, acceleration is just the difference of velocity divided by the difference of time. So, it's just three over four.

[00:03:17]Question number four. In an arithmetic progression, the common difference is minus two.

[00:03:22]The sum of the first four terms is equals to the sum of the first 10 terms.

[00:03:27]Find the first term which means what is A. So, we apply the formula. Formula is given in the formula sheet.

[00:03:34]2 A plus N minus one. 4 minus one is three. 3 D equals to half times N. 2 A plus N minus one. 10 minus one is nine.

[00:03:46]Then we can find what is A.

[00:03:52]A equals to 13.

[00:03:56]Question number five, a curve of an equation of this. P and Q are the stationary point of the curve. When you see stationary point means dy/dx equals to zero.

[00:04:06]Find dy/dx and find the exact length of line PQ. So we need to find dy/dx and find the coordinate of P and Q. So dy/dx is a quotient rule.

[00:04:20]Quotient rule is this time this minus this time this divided by 3 minus x squared. And then stationary point means dy/dx equals to zero.

[00:04:32]Denominator comes here will become no more.

[00:04:46]After common P and Q, we can find the exact length.

[00:04:56]2 square root 5 is the exact length.

[00:04:59]But of course you can write your answers in square root 20 because they do not ask for the simplest exact length. So both usually they will accept.

[00:05:10]Next is the modulus question. Solve this equation. So we need to split into two part.

[00:05:18]Done. And then next solve the inequalities of this. Usually we just square both side.

[00:05:27]Divide by three.

[00:05:29]Paper one we can't use calculator so simplified would be easier for us to do.

[00:05:35]So this more than zero. More than zero we need to choose the upper part. So it's this region and this region. So it's this and this.

[00:05:46]Question seven, the variable x and Y are related by this. Use calculus to find the approximate change in Y when X is increases from 4 to 4 + H where H is small.

[00:05:59]Small changes we use this formula. So, we need to know what is dy dx first since we already know small change of X is H.

[00:06:09]Power to front, power minus one.

[00:06:12]Power to front, power minus one.

[00:06:15]Then because they already tell us the X is 4, then we must sub in in the application.

[00:06:27]11 over 16.

[00:06:29]So, small change of Y over small change of X is H. dy dx is this. Small change of Y is this.

[00:06:37]Done.

[00:06:40]Question eight, a curve have this equation. Ask you to find dy dx. This is a product rule, so U and V.

[00:06:48]Differentiate maybe is two times two.

[00:06:50]Then my dy dx is this time this plus this time this.

[00:06:56]Part B, the line L is the tangent to the curve at point X equals to 1 over 2.

[00:07:01]Whenever you see tangent means that you need to form the straight line equation.

[00:07:06]So, you need three info, X equals to 1 over 2.

[00:07:09]Y you need to find.

[00:07:13]And then we need to have the gradient.

[00:07:15]Gradient of tangent is equals to dy dx.

[00:07:20]Then sub in X equals to half.

[00:07:25]Then we can find the equation of tangent.

[00:07:28]After we get equation of tangent, they say let's find the exact coordinate of the two points when L meet the axis.

[00:07:36]Meet the axis means either X equals to 0 or Y equals to 0. So, when x equals to zero, e squared e squared can cancel off, and then 1 over 2 divided by 2 equals to x.

[00:07:56]x equals to 1 over 4.

[00:07:58]So, we can confirm the two exact coordinate include x equals to zero, y is this, and when x equals to 1 over 4, y equals to zero. So, these are the two exact coordinate.

[00:08:15]Part C, use your answer to 8A to find this. 8A inverse this.

[00:08:22]Then we can check if differentiation this will become this. So, integrate of this will go back to this, right?

[00:08:31]Then we just write this part.

[00:08:33]After we write this, the question ask us to find this, which means we need this part only.

[00:08:40]So, this part obviously we can throw to the right, and then these two also extra. Then we can write in this way first.

[00:08:50]And then these two, how to remove? We can everything times half for the whole equation. So, left hand side no more two, and then here the whole thing times two.

[00:09:02]And then remember at last the answer plus C for our hands integrate situation. So, this is my final answer.

[00:09:12]Question nine. In this question, time is second and distance are in meter. At time t equals to zero, a particle starts from the origin and move with a velocity of this. At the same time, B start from the point with position vector of this and velocity of this. Write down the position vector of B at time t.

[00:09:31]Position vector at time t means initial, so the initial position for B is this.

[00:09:38]Plus V times time.

[00:09:41]So this for B. And then determine whether or not the particle collide. So we need to know the position vector of A after time T first. It's not from origin. Initial plus V is 2 1 T.

[00:09:54]So if they collide, they will equate to each other. So we equate them.

[00:10:01]Then we compare I to I.

[00:10:05]T is 3 over 2. This is for the part I.

[00:10:08]And then for the part J, 0 plus 1 T means T equals to this.

[00:10:14]Since both of them do not have the same time, so we can make a conclusion do not collide.

[00:10:22]Part B. Q have this position vector. So we can confirm OQ is this. R is this.

[00:10:30]S is this.

[00:10:32]Then they say is S lies on the straight line of QR.

[00:10:37]So you can tell me the gradient of QS equals to the gradient of QR.

[00:10:42]QS first. We find QS.

[00:10:48]And QR. [snorts] So the gradient of QS is Y over X. Gradient of QR is Y over X also.

[00:11:03]Then we can solve that.

[00:11:07]So this is our first equation from part one. And part two they say is length of RS is square root 5. Then we find RS also.

[00:11:20]The distance of RS is the X squared plus Y squared. And this thing is square root 5.

[00:11:29]We can subs one into two.

[00:11:32]This part we can cancel off the square root.

[00:11:37]This is eight.

[00:11:47]Everything divided by five.

[00:11:55]Possible B is five and three. Possible A is minus three and one. So, the working is a bit long.

[00:12:02]You just make sure you write correctly and if you do not have sufficient paper, you can request for extra paper and write on it.

[00:12:13]10A is a log question. So, we see we have log base five, log base five, and log base X.

[00:12:20]We make everything to log base five first. We change base here. Then, the five put up, the small numbers there put down.

[00:12:27]This and this is just two chicken minus one chicken will left one more chicken.

[00:12:34]So, we can let the complicated thing equals to something that is simpler.

[00:12:40]Everything we time this log base five X will become this thing square minus this.

[00:12:46]This thing also need to times the log base five X, so it's this.

[00:12:52]And remember, every time you let, at last you need to replace it back.

[00:12:58]125 This is the two value of X.

[00:13:07]Part B, solve the equation. So, this question is indices. We make eight and two both become two power something.

[00:13:17]Since here both also two power something, then we can combine the power. Is times relationship, the power will become plus.

[00:13:29]The challenging part is there is a three extra here, so we cannot make everything into base two. Then we have no choice to make it a quadratic situation. Quadratic situation can only happen if this power and this power is a times two relationship. We know 3x * 2 is already 6x, but times two become four here must make it into two, then how can we make 3x + 4 into 3x + 2?

[00:13:59]The extra plus two we separate it out.

[00:14:03]Become like this.

[00:14:04]Here is 4 * 3 is 12.

[00:14:08]If like this, we can let the power smaller part become A.

[00:14:13]Then here will become A squared minus 12 A minus 64 equals to zero.

[00:14:24]This thing is not possible to change because if you want to solve a power unknown with if the LG both side, LG minus four is max error. So, this part we don't have answer. We only have answer as X equals to 2 over 3.

[00:14:41]Question 11, a curve have this equation where A is a positive constant. The area of the region enclosed by the curve and X axis is this. So, we know this part they say this is 9 over 16. So, we need to know these two coordinate.

[00:14:58]So, we need to factorize this part.

[00:15:01]We make everything to positive first.

[00:15:04]We need to find this thing and this thing. So, this is obviously X times X. This is obviously 2A times A.

[00:15:12]We don't have any other option. And then at last, we need to form this. So, this time this, this time this. Since at last is a negative, so I make the relationship as minus. This time this is minus 2 ax. This time, this is minus ax.

[00:15:26]So, it's just nice. Then, we can factorize it in this way.

[00:15:31]Then, we can confirm here is A, here is 2 A.

[00:15:34]So, to find the area under curve is integration from A to 2 A of this equation.

[00:15:41]Power plus one divided by new power.

[00:15:44]Power plus one divided by new power.

[00:15:47]at X.

[00:15:49]We know this thing he says area is 9 over 16.

[00:15:54]Then, we can substitute 2 A and A.

[00:16:23]So, we can confirm A equals to 3 over 2.

[00:16:26]It's again a bit long.

[00:16:31]Question 12. Find the coordinate of point intersection of circles with equation of this and this. So, we solve simultaneous first.

[00:16:40]Equation one, equation two.

[00:16:42]We can use equation one minus equation two.

[00:16:50]So, after we form this equation, we can substitute into either of the equation to continue solve it. Let's say I substitute this into this.

[00:17:10]Everything divided by 50.

[00:17:21]>> So this are the two intersection point of the circle.

[00:17:27]So this is the end of FM 2016 paper one non-calculator paper. Hope you learn something from here and all the best in your exam. See you in the next class.

[00:17:36]Bye-bye.