9709 May/June 2026 P1 Paper SOLVED (post exam)

Added: 2026-05-07

[00:00:01]Okay, this is the 9709 paper 1996 paper which was leaked and this recording is uh made after the paper.

[00:00:12]The equation of a curve is such that dy by dx is this the curve passes through the points this and find the equation of the curve. For this you have to understand the effect that if you have the equation of curve equation of curve or you say if you have let's say y as the subject of the equation or y as the subject equation of y or you could call it as equation of y and if you differentiate it if you do dy dx we get to the equation of the gradient equation of the gradient or you simply call it dy by dx x and if you further differentiate it, if you further do d by dx, you get to the equation of uh sorry instead of equation of I should just say you get u to d² y by dx² okay so this helps you to figure out the nature of the curve by the way but it's not important for this question to learn more okay now if you do the reverse process if you understand the fact that integration is the reverse process of differentiation. Then you would say if you integrate this you get this. If you integrate d² y by dx² you get the equation of gradient. So upon integration you get this and obviously add the constant of integration as well.

[00:01:32]Upon further integration you get to the equation of the curve. So this is what we have to understand. We have to understand the last step that if we integrate dy by dx we get to the equation of the curve. So you have to do uh you have to put integration on both sides basically. So integral of this part is basically y and you have to integrate this part by using the rules of integration. So so the rules of integration as follow uh you have to increase the power by one and divide it by the new power. So 2x the power is one increase the power by one divided by the new increased power. Similarly for this part increase the power by one divided by the new power. Okay for now the rules are very simple for P1 the rules this is just the only rule that have to uh understand to do integration. Okay. So increased the power by one and divided it by the new power again increased the power by 1 1 /2 + 3 /2 divided by the new increased power and of course you add the constant of integration.

[00:02:35]Now this is the simplified result. If you simplify this 2 and 2 cancel out x squ over here 6 / 3 /2 so 6 this means 6 * uh sorry multiplied by 2 over 3. So this is 2 * 2 which is 4 4x raised^ 3 /2 and the constant of integration to to get c to get c to get the value of c you you use this coordinate you input the value 9 in place of y you input the value four in place of x okay so if you simplify it the result is c= 7 and then you in the end write the equation of the curve in the desired format uh with the inte with with the integration with the sorry with the constant of the integration uh with its value plus Okay. Describe fully a sequence of two transformation which transforms the graph of y= f of x to the graph of y= f of 4 - x. For this uh since this is just a past paper video, you could uh just use this fact if you have if you have it in this form or you could bring it to this form. Okay, for our question we had y= f ofx and we had to transform it to y = f of I think 4 - x. Now f of 4 - x you could rewrite this as x + 4 as well. 4 - x or x + 4 same thing. Okay. So if you if you if you uh match it then it matches this form on the left. Okay. You do not have any brackets inside the argument. Uh on this side we have some brackets inside the argument. So the we have some common factor. Okay. We do not have it in such format. Obviously we can bring it to that format. Okay. Let's write it in both ways. So for for this one we have f ofx + 4. Now if you take uh the negative as a common factor 1 is the common factor you get negative and then x - 4. Right? This is what you get.

[00:04:28]Okay? If you have it in this format then you are going to follow this order. If you have it in this format then you're obviously going to follow this order.

[00:04:38]Again also one more thing that you should understand that uh the order only matters within a horizontal transformation or within a vertical within a vertical transformation. But it does not matter if you are doing a horizontal transformation and another vertical transformation. So between a vertical and a horizontal the order will not matter. The order will matter within horizontals or within verticals. So over here we both we have only horizontal transformations. No vertical transformations. A and D were vertical transformations. You could see there's no coefficient over here and okay nothing over there and nothing over here. So okay so if we if you follow this way then we have to do the horizontal shift at first. So plus4 what does the plus4 over here means? It means you translate it translation with vector.

[00:05:32]By the way, it's a horizontal transformation. Translation with vector4 and zero.

[00:05:38]For horizontal transformations, you just reverse this sign. Like if you have positive4 over here, you write it as4.

[00:05:45]Okay. The other step, the other thing that we must do this uh the thing that comes in the sec in the next order is the horizontal stretch. So b okay whenever the horizontal stretch or the stretch factor is negative one you don't write it as stretch usually you write it as a reflection. So negative with the x means a reflection in the vertical axis or reflection in y-axis.

[00:06:13]Okay. Now this is one valid sequence.

[00:06:15]The other valid sequence would be if we follow this order and if we have made it in this order in this format. Okay. So in this format you do the you apply the horizontal transformation uh horizontal sorry horizontal stretch or the reflection first. So the reflection is going to be reflection in the x-axis again sorry in the y-axis reflection in y axis.

[00:06:39]Okay by this negative we mean reflection in the y- axis. The next thing that you are going to do is you have minus4 over here. So minus4 means translation with vector. So translation with vector 4 and zero. Okay. We just again sign for horizontal transformations. For horizontal translations sign it's ne4 over here. So write it as pos4 or zero. Okay. There are other ways to understand the transformation transformations as well. Obviously, you don't have to rely on this at all. We can just understand the fact that we replace x by something and do the uh do the transformations without this table, without this uh chart. We could do it easily as well.

[00:07:28]Okay. So, this is one sequence. This is the other sequence. In the first sequence, we have the reflection at first uh and then translation by some vector. And in the other sequence, the translation vector is different. You you need to make sure that the translation vector will be different if we reverse the order.

[00:07:45]Okay, moving on. A curve with equation this is stretched with a scale factor 1 /2 in the x direction and translated by this. Find and simplify the equation of the transformed curve. So scale factor 1 /2 in the x direction. So you have to replace x by 1 / 1 /2. So 1 / 1 /2 is two. Okay, you have to add a coefficient of two. uh you have to sorry you have to multiply uh x by 2 every like wherever you have x you replace it by 2x. So over here you have x cubed you replace it by 2x whole cube. Over here you have x you replace it by 2x.

[00:08:24]Okay this is your uh this is the first step. This is the stretch. This is the result of just the stretch. If you simplify it you get 8x cub - 6x - 4 and then you have the translation by vector 0 and -3. So this is the horizontal trans translation and this is the vertical translation. Vertical translation three units uh downwards. So you just add minus3. You just add minus3 after this -4 -3 would give you result of -7. So this is the simplified equation of the transformed curve.

[00:08:59]The equation of a curve is given and the equation of a line is given. Find the set of values of the constant K for which the line intersects the curve.

[00:09:07]They didn't mention how many times. They didn't say they have to intersect once, twice, or uh Yep. So, you could only have one or two intersections, two points of intersection, but didn't mention how many points of intersection.

[00:09:20]So you should understand if you have a curve okay if you have a quadratic curve and if you have a straight line. Now if the straight line never intersects the quadratic curve we say uh if if you if you equate these two equations together the b ^ 2 - 4 a c will be less than zero. The discriminant will have a result of less than zero. Okay. If they but if they intersect at one point at just one point if it is intersecting at at the vertex of the quadratic uh so the b ^ 2 - 4 a c in that case would be equal to exactly zero of the the b square - 4 a c by the way of the simultaneously solved uh equation like this quadratic and this straight line sim simultaneously solve together and you apply the b squareus 4 c on the result of the uh when you equate them together.

[00:10:14]Okay. Now if you have two distinct solutions, this solution, this other solution, then the b ^ square - 4 is in that case will be greater than zero. But again we in this question they they just said that you have to find the set of values of k for which the line intersects the curves. They didn't say intersect once or twice. So we use this and this together. So the inequality will be not greater than not equals to but rather both. So greater than equals to you have to apply the uh inequality b ^2 - 4 a c is greater than equals to 0 later uh when you when you simplify the result. So the first step is equate both of these equation together. The y output would be same. So equate them together.

[00:10:55]kx^ 2 - 5x - 6 = 3x - 7k. This is what I have done over here. And then you bring the uh terms to one side. So -5x and then 3x over there would be -3x, right?

[00:11:09]5x -3x is -8x. So 8x over here and then -6 and -7k over there would be + 7k. So plus 7k - 6 = z. This is the simplified result. We cannot simplify it further.

[00:11:25]Now next thing that you have to identify what's a, what's b and what's c. So a is the coefficient of x squ. So a over here this is b the coefficient of the linear term for negativex or sorry the cofficient of x only uh8 is b and this whole part from here to here this is c and then you use the result that we derived earlier b ^ 2 - 4 a c has to be greater than or equal to 0. So replace b ^ 2 by8 and 4 a and c you just replace these things. So you replace a with k c with 7k minus 6 and this is your expression on the right hand side. This is your inequality. And then you simplify the result. 8^ 28 2 is 64. Uh over here -4k * 7k would be - 28k 2. And again -4k * -64 would give result of positive. So positive 24k is greater than or equal to zero.

[00:12:22]The next step that you could do over here is uh what I've done. Okay. Let me show you what I've done. So basically uh I've brought all of these terms to the right hand side. Okay. Why did I do so? Because uh the I felt the coefficient of the squared variable was negative. So just just to have it in a in a better way. I I I thought to bring it to the other side. So it is 28 k².

[00:12:52]If you move -28k whole square to the other side. Okay, let me do it uh the other way. So you would have zero. If you bring all of the terms to the other side, you would have zero on the left hand side and greater than equals to uh the sign would be towards 0 and you have 28k squar you have -4k and then -64.

[00:13:13]Okay. Now you divide this whole both sides by uh wait what I have divided by I think I divided it by four. Yep. I divide both sides by four. So 4 7ths are 28 and six over here and 16 over here. So this is the equation and I've written this this stuff on the right on the left hand side again. Now you you need to make sure that the inequality the sign of the inequality is away from this part. Okay?

[00:13:41]It has to be away from this part. So again the sign of inequality is it's towards zero now. Okay. Now now to solve this inequality what you must do? you must uh break the middle term8 and 2.

[00:13:54]Take the factors uh take the common factors in these two terms and these two terms and then again take the common factors to get this result. And then the two critical values over here are k - 2 = 0. So k = the other critical value is going to be 7k + 8 = 0. So k =8 / 7.

[00:14:16]These are the two critical values. Now if you if you if you want to use a shortcut you could use this shortcut to solve a positive inequality or a quadratic in inequality. You should remember whenever you're solving a positive quadratic inequality and the part of the quadratic thing is written on the left hand side. Like if you have the part of the function written on the left hand side then what you do the results are going to be you just copy this sign and you write the lower root on the left hand side. the bigger root on the right hand side and you just copy this uh inequality sign.

[00:14:51]But make sure you understand that this shortcut only applies when you are solving a positive quadratic inequality.

[00:14:57]Okay, positive quadratic inequality. It doesn't have to be wait.

[00:15:12]Okay. So if you're solving a positive quadratic in inequality, what you could do is you could write the lower root on the left hand side, copy this sign and copy this sign again and write the bigger root on the right hand side.

[00:15:22]Okay? And then you have to make sure that this uh this is a correct way to express it. I mean it has to be a connected uh region k you have to think of it in this way. Can k be bigger than 8 / 7 and lesser than 2 at the same time? Yes, it can be. So this is a connected inequality. Uh sorry this is a connected set of results. So you could write it in this way. If it were disconnected then you would have to express it uh like broken. So you'd have to express express otherwise in this way. If it were a disconnected inequality so you'd have to write and sum this way this and this part separately. So this this would have been the result if it were a disconnected region. But now this is a connected region. K can be at the same time greater than8 / 7 and at the same time lower than or equal to 2.

[00:16:18]Okay. The coefficient of x² in the expansion of this is 324. Find the possible values of the constant q. Okay.

[00:16:26]The coefficient of q ². So you have to expand this or expand only the part when x is raised to the power two. And again you have to expand this on to to to only determine where you have x raised to positive2 over there. So for this part if you do the expansion you write 4 c2.

[00:16:48]So 4 c2 or you could write 4 c2 as 4 and 2 only like you could rewrite it in this way as well. Okay. So let me explain this over here. Wait two. We have uh okay we have 2 - qx raised^ 4 and you have -1 + 8 over I think 8 / qx raised to the power of 6.

[00:17:15]Like if we have to expand both of these only to for the term x squ then what we do is we realize that we have to do four over here 4 c and 4 c2 only like if you write 4 c2 we mean the first term the first term which is two gets the result of the subtraction of these two numbers 4 - 2 so 2^ 4 - 2 the other thing - qx gets this result. So two over here.

[00:17:48]Okay. Now for this case the the both way would have worked fine if we if we would have written four minus 2 on the other hand side and two on the left hand side uh we would have got the same result of x² in either way.

[00:18:03]Okay. For the other part though for the other part though we have first of all this negative. Okay. We can write a big negative over here and then do the expansion only for the x² part. So it's going to be six 6 C something. Okay.

[00:18:18]Okay. Let's we we can write that something as four as well or we could write it as 2 as well. 6 C4 is equal to 6 C2. Okay. So if we write four over here then what we mean is your first number which is 1 gets this power of 4. The other number which is 8 over qx or positive 8 over uh qx gets a power of 6 - 4 which is two. So if we if we write it in this way then the other number will get a power of positive two over there. Or if we write it with 6 uh 2 if we write it in the 62 way then we mean the first number gets the power of subtraction. So 6 - 2 is what the first number gets. the other number 8 / qx. So 8 / q and x over here gets this power two.

[00:19:14]Okay. Now we just have to simplify the result. Okay. This is what I have done over here. By the way, uh understand that you can rewrite this as just 4 and 2 in this way as well.

[00:19:25]Okay. So this is what I've done over here. Uh so if we simplify the result on the left hand side, we get 24 q ^2 x^2.

[00:19:33]So - 2^ 2 will be q 2. x² will be x squ.

[00:19:40]For this part again if you simplify the result we get 960 negative for this for this thing. And if you simplify the numbers over here we get 960 / q ^ 2 and then x squ. And now we know this the coefficient of x uh the coefficient of x² is 324. So we can equate it to 324x^2.

[00:20:02]Okay. Now we just compare the coefficients. We remove this x² terms.

[00:20:06]Sorry, we remove this x squ. We just compare the coefficients. 24 q² 960 q ^² over q ² = 324.

[00:20:15]Okay. And then you take the LCM. You could either take the LCM like you have you do not have q ^ 2 in the denominator. So you write 1 * q ² over here. Similarly, you do it in the numerator as well. So you have 24 q^ 4 over there. Or you could do it the other way. You could multiply this whole equation by q². Even though they both mean the same thing. If you multiply the whole equation by q ^ 2 q ^ 2 * q will be q^ 4. Now this q ² and this q² will get cancelled out and 324 * q will be 324 q ^ 2. If you bring things to the uh to the to just one side to the left hand side you have 24 q^ 4 - 324 ^ 2 and then - 960= 0. Now this is a coordating in disguise. So you replace you substitute q ^ 2 by something like you could replace it with x or any variable other than q. So you replace it with u over here. So u is q ². So this is q q^ 4 can be rewritten as q ^2. So you could write this as u squ only u 2. So 24 u ^ 2 - 3 24 u - 960 = 0. Divide it. Take common factors. This is the equation that you get and then you factoriize it. Break the minm you get this result. I think I've done some wait no no no okay this is fine and then uh you take the common factors in the the two times and the next two terms and then you take a common factor again you get two results.

[00:21:45]The first result is 5 /2 the other result is 16.

[00:21:50]Okay. Now you replace you replace u with q ^ 2. Replace u with q ^2. Uh so take a square root on both sides. Now you should understand you should I mean you do not have to take a square root on this side over here because you could easily identify that there's no real solution possible over here. Why?

[00:22:08]Because squaring a number cannot give a negative result. So obviously there are no solutions no real solutions uh to this equation or you could do it in this way. If you take square roots on both sides then a square root of a negative number is not defined and real numbers.

[00:22:25]Now for this part you take a square root on both sides. Again you add the plus minus. Don't forget the uh the plus minus. By the way the result of plus minus comes after this adding this.

[00:22:35]After this you add the plus minus. But anyways you should just remember the answer must have a plus minus. So square root plus minus square root of 16. It will be plus - 4.

[00:22:48]Prove the identity.

[00:22:51]Okay. So for this the first thing that I see on the left hand side I have to prove I have to make left hand side to the right hand side. The first thing that I see is on the right hand side I just have a single fraction but on the left hand side I have two different fractions. So obviously I'll have to uh take LCM and make a common denominator.

[00:23:09]So I multiply these two together cos theta * 1 - sin theta. So this part will be multiplied by 1 - sin theta and this part would be multiplied by cos theta.

[00:23:18]This is what I have done over here. So 1 - sin theta * 1 - sin theta is 1 - s is uh sorry 1 - sin theta squ and cos² theta over here I keep I kept the denominator unexpanded because this is usually like this is usually recommended that you keep the denominator unexpanded uh you do not expand it so that you may simplify some results and I could see over here I could already see over there that you should get a cos theta in the denominator I already have cos theta over here. So it could be possible that I could just do this and something like 1 - sin theta in the numerator.

[00:23:55]Obviously this is not possible. You cannot do this. But maybe in later later in this in the in the steps I could I it might be possible to like cancel this 1 - sin theta with some other 1 - sin theta in the numerator.

[00:24:09]I use the identity cos² theta + sin square theta= 1. So cos² theta is 1 - sin square theta. Now I rewrite okay I rewrite 1 - sin square theta 1 - sin² theta can be rewritten as 1^ 2 - sin square theta obviously now this follows the identity that this could be expressed as a product. So a^ 2 - b 2 can be expressed as a product of a + b and the other thing is a minus b * a minus b. So this is what I have done.

[00:24:46]This is an important uh identity that you should remember and you should be able to apply this. So 1 square - sin square theta can be rewritten as 1 - sin theta and then* 1 + sin theta. Okay. If I do it in this way now what I have in the numerator I just have I I I have 1 - sin theta with each term. So I have 1 - sin theta over here as well and 1 - sin theta over here as well. I take 1 - sin theta as the common factor in the numerator. Why did I take it? because I could already see that 1 - sin theta and 1 - sin theta in the numerator and the denominator would get canceled out and this is what we want okay so over here let me show you what what I did over here so basically we had 1 - sin theta squ and then you have plus 1 - sin theta and then you have 1 + sin theta 1 - sin theta whole square what does it mean it means you just have to write 1 - sin theta twice as product. So you can write 1 - sin theta * 1 - sin theta.

[00:25:52]Take the common factor. So by the way this is this is just the numerator. I am not rewriting the denominator. This is just the simplification for the numerator. So take 1 - sin theta is the common factor. And the first term what's left is this. So 1 - sin theta is left.

[00:26:10]And the second term in the second term what's left is this 1 + sin theta. So plus okay this bracket is additional over there. So + 1 + sin theta. Now sin theta + sin theta - sin theta 0 and 1 + 1 will be 2. So this is what I have done over here. So this 1 - sin theta this 1 - sin theta in the numerator and denominator gets gets cancelled out. We have cost it in the denominator. We have two in the numerator and we have proved the identity.

[00:26:42]Hence solves the equation this equals this for this given range for sorry for this given domain.

[00:26:53]Okay.

[00:26:55]So we knew this part is equal to 2 cos theta. So this part is again written over here 1 - sin theta / cos theta + cos theta 1 - sin theta. This is equal to 2 cos 2 / cos theta. So we replace this part by 2 / cos theta and we have tangent cube theta / sin theta on the right hand side. Now bring every function trigonometric function to one side. So cos theta over there time cos theta / sin theta. Okay. So you could one thing that you could do over here is you could replace tangent theta with the basic functions like the sin theta and cos theta and you could rewrite it as sin cub theta / cos theta but it's not necessary you you just understand the fact that cos theta / sin theta is what cos theta / sin theta is 1 / tangent theta why because sin theta / cos theta is tangent theta so 1 / this so 1 / this is going to be 1 / tangent theta theta, right? 1 / sin theta cos theta is cos theta / sin theta. So this is 1 / tangent theta or divided by tangent theta. So tangent cub theta / tent theta uh similar functions are being divided. So you could just subtract the power. We have power three over here, power one over here. Subtract the power 3 - 1 and you have two. So tangent² theta= to take square root on both sides. Don't forget the plus minus upon the simplification. So tangent theta equ= plus -<unk> 2. Now the alpha or the basic angle the angle which is made from the closest x-axis.

[00:28:31]If we draw the four coordinates. So this is the alpha alpha alpha and this is also the alpha. So angle made with the closest x-axis is basic angle. So basic angle is the tangent inverse of just the positive of it. So modulus of it or you could say the modulus of it tangent inverse of roo<unk> of 2 which gives a result of 54.7°.

[00:28:55]Now if if you look at this part it says that tangent theta= plus roo<unk> of 2 or tangent theta= roo<unk> of 2.

[00:29:11]So if you draw the quadrants the the tangent is both positive and it's both negative. So we don't have to decide the sign at all. We can we can draw the uh these lines in all of these quadrants. So this is the this is it in the first coordinate and this is it in the third coordinate. Okay. So tangent is positive in the first and third coordinate by the way. So this is the result of it. First quadrant and the second sorry and and the third quadrant and tangent is al is negative in the second and the fourth quadant. So obviously all of these quadrants are so.

[00:29:47]So I'm just going to draw a straight line as such.

[00:29:52]Okay. The alpha or the basic angle we just figured out it's the angle made with the closest axis xaxis sorry not not the y ais obviously the closest x-axis which is 54.7. So for this line the closest x-axis is the positive x xaxis.

[00:30:11]So this is 54.7°. Similarly we have 54.7° over here over here and over here.

[00:30:19]Okay. Now you you should you should just remember the result that in the first coordinate the theta uh theta by the way is measured from the positive x-axis counterclockwise or clockwise either way. Count if you move counterclockwise then the angles are positive. If you move clockwise, if you move in the direction of uh in the direction of the clock, then the angles are negative. So this angle is what? This is obviously the theta over here is equal to alpha, which is 54.7.

[00:30:46]Now this angle is going to be 180. If you if you had moved a little more, it would have been 180. But you have to subtract this part 180 minus 54.7. So this theta is obviously 180 minus the alpha and the other one this one over here is going to be if you again move from the positive x-axis this is going to be 180 plus d alpha okay now the last one last one is going to be 360 minus d alpha okay so these are the results that you should remember so 360 why because this complete thing was 360 and you have to subtract 54.7 7 over there. So these are the 54.7 in the first quadrant and the second quadant 180 minus diesel the alpha or 180 plus the alpha 360 minus the alpha. So all the angles if given correct to one decimal place are going to be 54.7° this this and this. These are the four results. Why? Because the theta were from 0 to 360. It was acceptable for 0 to 360.

[00:31:54]The coordinates of three points PQ and R okay respectively where P and R are constants. It is given that PQ is perpendicular to QR. So show that P is okay. So we just given that we have three points and okay let me let me just randomly draw these three points. So we have P over here maybe Q over here. Wait what is uh okay? So we have Q over here and we have R over here. So this is PQ.

[00:32:25]Okay, this is P and this is Q. Like maybe R is not over there but R is over here.

[00:32:33]R is over here. Okay, this is PQ and this is QR.

[00:32:40]Now it is given that PQ is perpendicular to QR. I drew it in this way because I knew the fact that PQ is uh it was given already that PQ and QR makes an angle of 90°. So they are perpendicular to each other and show that P is equal to 2 R minus 10. So obviously we are just given this information that these two things are perpendicular to each other. So we use the fact that if we multiply the gradients of these two lines then we should get negative 1 or or in other words the gradient of PQ is going to be 1 / uh sorry negative reciprocal of gradient of QR okay or by the way this comes comes from the result that gradient of PQ time the gradient of QR is going to be neative 1.

[00:33:25]Okay. So for the gradient of P Q, what we have to do is we have to use the coordinates P and Q. Obviously this is P, this is Q, P and Q. The 6 - P and then 8 - 0. 6 - P 8 - 0 simplified result. And gradient of QR q and R uh wait let it go. Why is it going? Okay.

[00:33:45]So Q and R. So this is Q and this is R.

[00:33:47]Gradient of QR is going to be 10 - 6 over R - 8. 10 - 6 is 4. R - 8 is R - 8.

[00:33:56]Use this property or rearrange it uh for M of PQ gradient of PQ is going to be 1 / gradient of QR.

[00:34:07]So if you use this result and substitute gradient of PQ and then you do negative reciprocal this thing over here basically means -1 / this thing basically means negative reciprocal right. Okay. So let's let's also discuss okay let me let me also show it uh let me move this part okay I can rewrite we have QR right so we want -1 / the gradient of QR 4 / R - 8 so -1 / 4 / R - 8 if we simplify the result what we get is -1 is basically we think of it in this -1 is being divided by this whole thing.

[00:34:56]Okay. So we can write this -1 / 4 / r - 8. And then we can do the reciprocal.

[00:35:05]We replace this divide by by multiplication and we say multip r - 8 / 4. So negative r and then pos 8 / 4.

[00:35:19]So negative times uh this whole thing basically negative r and then plus 8 due to this and you have in the denom denominator you have four simplify these four 1 are four 4 2's are 8 so and then you multiply this two by this thing so this is positive 8 by the way minus uh minus would be positive so positive 8 * 2 would be 16 and then you have 2 * r that would give you a result of -2 r okay and then if you simplify the result you you just cross multiply By the way, if you simplify the result, if you bring if you make P the subject of the equation, bring it to the right hand side, simplify the result, you get what was required. Show that = to R - 10.

[00:36:02]It is further given that the length of P R is square root of 85. Find the possible values of P and R. So length of P R is given to be square root of 85.

[00:36:14]square root of uh the if you you basically use the formula for the distance of P R. So the coordinate of P R I think was R and uh wait I think it was RA 10. Yep. Raa 10 and 0 P. So P is RA 10. R is 0 comma P. So if you use the uh the the formula for the distance the distance formula is y2 - y1 squ + x2 - x1 square root of all of that. Okay. Uh yep. So let me write write the distance formula over here. You could write rewrite it as x2 - x1 squ + y2 - y1 square. You could uh do these uh separately or like you could do this you could do the y2 - y1 squ at first as well.

[00:37:10]Okay. So, x2 - x1 uh we have r - 0 square + 10 - p square. It should be equal to square root of 85. Cancel the square roots. Uh 85 on the right hand side. Simplify the result. R². This is r². 10 - p whole². 10 - p whole square is written over here.

[00:37:31]To simplify the result. Uh okay. Now over here I'm not going to expand it because I know because I know I have to uh I have to get the answer of R and I have to get the answer of P. So over here I I know I cannot solve this equation because I have two different variables. So I have to do substitution over here from the previous part. We know from the previous part P= to R - 10. So I'm going to substitute over here in this place P SR - 10 and then simplify the result. Simplify the bracket. I get this. I square this now because I know I I would be able to solve it now. So the way you could square it is a 2 - 2 a b + b squ. So consider this as a so a 2 and then uh - 2 a b. So -2 * 2 would be 4 4 * 2 20 would be80. So 80 r + b ^ 2 as 400.

[00:38:25]Bring 85 to the other side 85.

[00:38:29]Simplify this 5 r² wait simplifying this would give you 5 r² - 80r r and then + 315 = 0. Okay, over here. And then you could do the middle term break. You could say you could rewrite uh this as -16 + 63. Wait, it shouldn't be 63. It should be 64. It should be 64. If you do the break, sorry. Okay, not this. Wait, wait, wait.

[00:39:01]This has to be this is 63. But if you do the momentum break of -16, it should be -9 and -7.

[00:39:12]So I would have to remove this as well and this as well.

[00:39:18]Okay. Now you do the minimum break of -16R.

[00:39:21]So 9R and -7R and then take the common factors. Uh this is the result. this also has to be seven. Uh the two possible answers of r are r= 9 and r r = 7. Now if r = 9, you use you again use this equation to get the answer of p. P = 2 r - 10. So if r = 9, p = 2 * r - 10, which is 8. If r = 7, p = 2 * 9, sorry, 2 * 7 - 10, which is 4. So the possible results are when r is 9, p is 8 or when r is 7, p is 4.

[00:39:59]Okay, the next question. A curve has equation this. The curve has a single stationary point when x= k where k is greater than zero. Okay, this part is really important. We might have to uh we might have to discard this result which is less than zero. Find the value of k.

[00:40:14]So the stationary point okay what does it mean by the stationary point? If if I do a rough sketch of this like curve this could maybe look in this way. I don't know. like maybe in this way the stationary okay let's draw rather let me draw a quadratic to make it a little simpler so if you have a quadratic we have uh the stationary point is a point where you have gradient as zero so this is the stationary point this point is the stationary point the the very important thing about the stationary point is the dy by dx is equal to zero at this point that's what you have to do first figure out dy by dx and equate it to zero dy by dx uh for differentiation you multiply by the original power and then subtract the power by one or decrease the power by one. So 3 * -1 /2 3 * 1 /2 and then decrease the power by 1. So decrease the power by 1 over here.

[00:41:09]Similarly for this part I have done the same simplified the result 3 -3 /2 x raised^ -3 /2 and simplified this result as well -2 * uh 3 /2 would be positive3 and -3 /2 -1 would be -5 /2 replaced dy by dx as 0 because the gradient is zero at the stationary point and then to solve this equation The first thing that I did is I think uh okay I just copied it. Wait did I just copy it? No. No. I I Okay, let me explain it what I have done. So to solve this equation -3 / 2 x raised to the^ -3 /2 + 3x and then 5 /2 = 0. I use the law of indices to say if we rewrite x raised to power -2 3 over2 in the denominator the power becomes positive because it was negative in the numerator 3 over2 over here and x raised to the^ of 5 /2 becomes x raised to the power of positive 5 /2 in the denominator.

[00:42:21]This is what I've done. Okay. The next step, the next step would be uh okay, the next step would you could either you could either do the simp you could either bring both of these to other sides like one part to this side and one part to the other side or you could take the common factor in the denominator and take an LCM. There are multiple basic way to solve this question. But what I've tried to do is I wrote it as such.

[00:42:46]-3 / 2x raised to the^ 3 /2 equals -3 / x raised ^ 5 / 2.

[00:42:55]Now -3 and - 3 would get cancelled out.

[00:42:58]You have 1. So 1 / 2x^ 3 /2 = 1 / x^ 5 / 2. We bring x to one side. So it's going to look in this way.

[00:43:12]So let's bring this x to the other side.

[00:43:14]So you would have 1 sorry you would have x^ 5 /2 over x^ 3 /2. Bring this two to the other side. So you have two over here. Use the law of indices x^ 5 / 2 - 3 /2 because bases are same and you have to subtract the powers. 5 / 2 - 3 /2 is what? So 5 - 3 is 2. So 2 / 2. X^ 2 / 2 = 2. 2 / 2 is just 1. Okay. So X^ 1 is 2. You get the answer of X over here. X= 2. This is how I solved it. Okay. There could be other ways. There are other ways that you could have solved it. You could have taken a common factor in the denominator as I mentioned.

[00:43:58]Okay. Okay. This was not the equation.

[00:44:00]This this was the equation. So X= 2 and the stationary points when X= K. So when x is 2, k is al k is obviously two. So stationary point or the value of k is find d² y by dx² and hence determine whether the stationary point is a maximum or a minimum point. Okay. So you further differentiate this expression.

[00:44:25]This is your equation of gradient. As I mentioned in the start, if you differentiate it further, you get d² y by dx. You have to differentiate this.

[00:44:34]So the same process uh multiplied by the original power decrease the power by one. This is the decreased power already. I've written the result over here. Similarly for this part for the other part as well. So what I did is I multiplied 3 by 5 /2 and decreased the power by 1 after that.

[00:44:54]So this is your d² y by dx 2.

[00:44:58]Now we have to see the result of d² by d² y by dx² when x = 2 or k = x = 2 same thing. So we replace x by 2 and then if we simplify this part I've just taken 2^ 7 /2 as the common factor you don't have to do it you just have to put this in the calculator to get the result of -2.6 0.265. So whenever you should understand that whenever d² y / dx² is less than zero is it when it is negative the stationary point is a maximum and if d² y by dx² was greater than zero the stationary point would have been a minimum okay for now I just stated it it as such but you could obviously prove it you could it's it's pretty simple to understand you draw a curve and d² y by dx for this is going to be greater than zero. Uh you have to consider the gradient and then the uh the way the gradient is increasing.

[00:46:05]Okay. Find the area enclosed by the curve the x-axis and the line x= k x= 4 giving you answer in this form. Okay. So for that I have done a rough sketch of the curve. The curve was this. This was the equation of the curve. So I've done just a random sketch of this curve. You don't have to be exact. You just have to draw any any sort of way like this could be a possible way. This could be have been correct as well. Like any curve just to get an idea of what you have to do. Obviously, you don't have to do it at all. So, uh find the area enclosed by the curve. The x-axis and the lines x= k and x= 4. So, k was what? K was 2. The line x= k is this. This is the line x= k. This is the line x= 4. And you have to find the area enclosed by this line.

[00:46:49]this line, this curve and the x-axis. So the area enclosed is this. Obviously this is the area enclosed. K is two.

[00:46:59]So to get the area under a curve, area under a region of a curve, you integrate it with the limits. This as the upper limit, this as the lower limit, k is two. Again, you integrate it. Now you don't have to write plus c because this is not an indefinite integral. This is a definite integral. The limits are given.

[00:47:17]So the result of integration is this.

[00:47:20]How do we do integration? Again we start we discussed it in the start. You increase the power by one and divide it by the increased power. Okay. So similar simplifying this would get this.

[00:47:34]Similarly for the other part as well increase the power by one divide it by the new increased power. This is what you get. Put limits as four and two and then simplify it. You get you first of all calculate diesel at four minus diesel at two. Make sure both of these results should be put in a should be like written in a modulus. Uh because area is a positive quantity. You should get this as positive and you should get this as a positive as well. Positive result as well. So 14 minus a 8 square root of 2 is your uh final answer if you write it in this form. Okay. Okay. I think I should be explaining how I get this from this part. Okay. Let me put it over here.

[00:48:24]Okay. First of all, you put x as 4. So 6 * 4^ 1 /2. So 1 / power 1 /2 is just the square root, right? Plus 4x sorry x is again 4.

[00:48:39]So 4 * 4^ 1 / 2 and then 2. So - 6 * 2^ 1 /2 + 4 * 2^ 1 /2.

[00:48:55]Okay. For this part the square root of 4 is 2. So 6 * 2 will be 12.

[00:49:02]For this part you have 4^ 1 /2. So this means you have you basically have 4^ 1 /2 in the denominator or the square root of 4 in the denominator. Square root of 4 is 2. So 2 over uh 4 / 2 is just 2. So 12 + 2 for this part.

[00:49:21]And for this part you have 6 * the square root of two. You cannot simplify this further.

[00:49:29]Square root of two is irrational and you have to express the result as as a in a third form. And for this part again you have + 4 over<unk> 2. Why? Because 2^ of 1 / 2 over here.

[00:49:46]Okay. 12 + 2 is 14 - 6<unk> 2 and then - 4 /<unk> 2. Whenever you have square root in the denominator, you can rationalize it by multiplying by its conjugate. So it's conjugated simply square root of two.

[00:50:06]So if you do the result uh you get 14 - 6<unk> 2 and then -4<unk> 2<unk> 2 *<unk> 2 is<unk> 2 squar which is just two. So this and this cancels out. You have two over there. -6 and -2 is8. 14 -8 square root of 2 is the final answer.

[00:50:35]Okay. An arithmetic progression uh an arithmetic progression has first term 20 common difference D geometric progression also has the first term 20 common ratio R. Uh the third term of a geometric progression. Okay. This statement is really important. You have to make sure that you write it correctly. The third term of a geometric progression. Okay. So we should make equations in this way. The third term of a geometric progression. So G 10 is whenever you have is, are, was or were in in a statement, it usually means equals. So is it is equal to five more than the third term of the arithmetic progression. So arithmetic progression's third term + 5. This is the first equation that we made. The other equation is going to be if you read this line again the fifth term of the geometric progression. So G5 is equal 30 more than the fifth term of the arithmetic progression. Arithmetics fifth term plus 30. Right? When you have these equation this question is really really simple. The third term of a geometric progression the any term of a geometric progression is given by a r raised to n minus one. So if you have third term if you have third term you replace n by three and a is your first term r is your common uh ratio is your ratio.

[00:52:06]So the first term of the geometric progression was 20 20 the common ratio is r r and since it was three so 3 - 1 would be two for the arithmetic progression. Okay, let's discuss why g5 is 20 r. Okay, the similar in a similar way uh you have a r power n minus one n is 5. So 5 - 1 is 4 a is 20 20 r^ 4.

[00:52:32]The third term of the arithmetic progression. Arithmetic progression any term is given by a + n -1 * d. So a is 20 again uh the first term as 20. So 20 plus if you want to get third terms it will be 3 - 1 * d 20 + 3 - 1 * d would be 20 + 2d 20 + 2d similarly for the fifth term you do it in the similar way you get 20 + 4 and then you simplify 30 + 20 would be 50 this is your first equation and over here as well I've made 2d the subject so 20 r² 20 + 5 would be 25 brought it to the other side it will be - 255 Okay. So now what you can do you could you could s you have to just simultaneously solve these two equations to get in any way any way you want. So I've replaced uh d with 20 r^ 2 - 25 upon 2 if you if you write 2 over there or you could have also done it in this way like you could call this 2 * 2d and replace 2d with this whole thing and get the same answer.

[00:53:42]Okay. So this is what I have done. This two and two uh four cancel out. You have two over here uh in this way. They cancel out in this way. And then you multiply this. Distribute d2. So you have 40 r² -50. 50 and 50 cancels out.

[00:53:56]Uh 20 r^ 4= 40 r² 40 and 20 cancels out in in a way. And r power 4 r² over there. So by the rules of indices you have r^ 4 / r power 2. So you get r power 2. And on the right hand side you you would get to two. So square root on both sides. Square root on both sides.

[00:54:16]Add a plus minus. You you would have to add the plus minus. But the question had stated that r is greater than or equal to zero. So it has to be only positive square root of two only. Now the common difference d is going to be uh use any any any equation. Basically you could use this equation or this equation. This equation is a little simpler. So divide two on the other side. Replace r by square roo<unk> of 2. 20 * roo<unk> of 2 square - 25 / 2<unk> 2 square is 2. 2 * 20 40 - 25 / 2 is 7.5.

[00:54:57]Show that the ninth term of the geometric progression is four times the ninth term of the arithmetic progression.

[00:55:05]So let's see what what the ninth term of the geometric progression. This is a very very simple question. You just have to see what the ninth term of the geometric progression is. You use this formula again. A is 20. R is square root of 2. N - 1 is 9 - 1. And this is the result that you get. The ninth term of the geometric progression is 320. The ninth term of the arithmetic progression is 20. A + n -1 * b. We just figured out previously d is 7.5. So replace d by 7.5 and uh 9th term. So 9 minus 1 and simplify the result to get 80. Now uh they said that basically they they asked us to prove that the ninth uh term the ninth term of the geometric progression is so equals 4 * the ninth term of the arithmetic progression.

[00:55:58]You could you could do it in any way or you could say that the ratio of G9 the ninth term and the ninth term of the arithmetic progression should be equal to four should so it turned out to be four. If I divided this by this it turned out to be four. So I could then just write a concluding statement. Yes, that this is indeed true.

[00:56:20]Okay, for this question, uh the diagram shows a triangle A CD which in which AD is perpendicular to CD. So this is a right angle uh triangle. This is a right angle triangle. So the Pythagoras theorem could apply over here or the uh ratios pometric ratios could also apply over here perfectly. The arc be of a circle.

[00:56:43]So this is the sector. Okay, all of it is given. All of it is obviously clear on the diagram. You could just still read it. Uh AD at E angle this is this radians and the length is BC is P cm.

[00:56:58]Given that P is four. Find the exact parameter of the shaded region. So first of all understand what the parameter of the shaded region will be. It will be E D plus D C plus CB plus EB. Okay. So you have to add this this and this. you'll have to get all of these lengths now.

[00:57:20]Okay. So, first of all, the length CD.

[00:57:22]Let's work out for the length CD or DC.

[00:57:24]For that, just consider this complete triangle. Okay. So, if you consider these triangles and you don't have to do five and and this is this is unknown.

[00:57:33]So, you don't you don't you don't have to write it at all. You should just know this length is 5 + 4. Why? Because P is given to be four. P is said to be four.

[00:57:41]So, 5 + 4 this length is 9.

[00:57:44]Okay. Now apply the trigonometric ratios. This angle is p<unk> / 6. Uh the trigonometric ratio say I guess. So for this let me write it over here. Uh the sign of pi / 6 is going to be opposite side. The ratio between opposite side and the hypotenuse. So opposite side is CD.

[00:58:03]Opposite to this angle is CD. So CD over the hypotenuse. Hypotenuse is opposite to the uh is the longest side of a triangle or the opposite to the 90° which is 9. So sin p<unk> / 6 = 9. So cd is uh 9 sin pi / 6 right? So this is cd wait so cd we have figured out cd simplify result sin pi / 6 is 1 /2 so you get 4.5 cm now for ad as well for ad as well you use the use the trigonometric ratios but now I use the cos uh the ratio for for the cost. So cos of pi / 6 is uh the ratio between adjacent side which is a d. a is the adjacent over the hypotenuse. Hypotenuse is again 9. So a is 9 cos<unk> / 6 cos<unk> / 6 is<unk> 3 /2. So 9<unk> 3 /2 uh cm.

[00:59:00]Okay. Now we know we know this length.

[00:59:05]Okay. We know this and we know this in a way. We know this as well. How do we know this? Because obviously we know this complete length ad we can subtract five from it to get this. So we know both of these lengths and obviously P is also known. P is four. We know this. We just have to figure out the arc length.

[00:59:21]Now this is fairly simple. Uh consider this sector uh the arc length is given by R * theta. R the radius is 5. 5 * this theta / 6. 5<unk> / 6 is the arc length. The parameter uh for for the parameter again E D first of all E D. So ED is going to be this uh whole thing a D. A D is 9 3 /2 minus this length five.

[00:59:48]You have to subtract this length which is five. So minus 5. Now for the rest it's quite simple. CD is 4.5 BC BC BC.

[00:59:57]BC I think was the P. This is P which is BC which is 4 cm. It's given to be four.

[01:00:04]So plus 4 plus B. P E is uh 5<unk>i / 6.

[01:00:08]The parimeter if if you simplify the result, the parimeter is this. Okay. So, okay. Should I say hi? I should I should tell you how I simplified it.

[01:00:20]So, I simplified it in this way. 9 pi sorry 9 square of 3 / 2 - 5 + 4.5 and then + 4 and + 5<unk> / 6 + 4 + 5<unk> / 6 okay that I have to give the answer in pi and roo<unk> of 3 only so simplify this result -5 + uh 4.5 + 4 would be negative sorry it would be 3.5 or 7 / This is 7 / 2. So 9<unk> 3 / 2 + 7 / 2 + 5<unk> / 6. This is one way to say the answer. Or you could say you could rewrite these together. So it would be 7 + 9<unk> 3 /2 and then + 5<unk> / 6.

[01:01:12]Now given instead that the area of the shaded region is given, find the value of P. area. First of all, understand that area of the shaded region would be the area of this right angled triangle.

[01:01:23]This complete area minus the area of this sector. Right? So that's what I have written. Area of shaded region is equal to the area of triangle minus the area of sector. Area of a triangle this is a simple right angled uh triangle. So area will be given by half * base. This is the base and this is the height. You could also use half a b sin c uh sorry in this case it would be half a d uh sorry half a d and then a c sine this uh angle. We could also use that uh formula to get the area of the triangle.

[01:02:00]So a of the triangle is half * a d * c the base time the height the base a d a d uh again and now we do not know the p but there's no problem. We we say this length is 5 + p. We would say that this length is 5 + p. Now, okay. So, this length since this length is 5 + p. Uh this over here is going to be uh okay, let me do it over here.

[01:02:29]You have 5 + p and you have this and this a d and cd.

[01:02:41]You have the 90° over here. This is A.

[01:02:44]And I believe this is D. I think this is D. Yes. And this is uh this something some length over here is five and some length over here is P. And this is / 6.

[01:02:58]The area of this triangle would would be half * A D * D. Now all I'm going to do is I'm going to use the trigonometric ratios to express a d and dc in terms of just one thing. So a d uh okay so if I use the trigonometric ratio sin pi / 6 is going to be opposite side which is dc over the hypotenuse which is ac. Okay I can replace a c by 5 + v but I don't see any need for now. I'm going to say dc is simply a c sin p<unk> / 6 and substitute it over here. Similarly do the same thing using the cos ratio cos pi / 6 is going to be uh adjacent which is uh a I think yeah a d over the ac so a will be ac cos p<unk> / 6 a is going to be a cos pi / 6 that what that's what I have done over here uh next and we know the area already we know the area of the shaded region it's provided as such now for the area of the sector we discussed we have to subtract the area of the sector r² theta. We already know the area the the stuff that we need to know for the area of the sector. This is five. So 5² 1 / 2 * r² 5^ 2 * this theta this angle.

[01:04:20]Okay. So if I simplify this part a c * a c would be a c² and square root of 3 2 * 2 * 2 would be 8. And if we simplify this result 5^ 2 is 25 25<unk> and 6 * this 2 would be 12 25<unk>i / 12 25<unk> / 12 I see it on both sides - 255<unk>i / 12 I see it on both sides so I can cancel them out or subtract them and then I have this okay what I have done over here is uh I just wrote this part on the left hand side and this on the right hand side I see again I see square root of three on both hand side so I can cancel these these two together 8 * 8 would be 64. Okay, make sure you understand that this is AC whole squared. Okay, I have written it uh like uh it may seem misleading that it's just c² but no it's acqu is a length. So ac uh is going to be 8 only. If we take a square root on both sides we get uh AC is 8 only positive 8 only. Why? Because AC is a length you don't have to say negative 8. So replace. Okay. Now AC is AC is what? AC.

[01:05:32]This AC is 5 + this B.

[01:05:36]So AC is 5 + B. AC is 8= 5 + P. You get the answer of P as three.

[01:05:44]Okay. The next question. Function f and g are defined as follow uh as follows. A is a positive constant. Find the state the range of f. So for the range of f since this is a linear function. Okay.

[01:05:56]only since this is a linear function you could put uh x as zero over here and understand whether the graph uh whether it is an increasing function or a decreasing function. So if you do a rough sketch or if you do a rough graph of it the 3x - 6 uh the uh y intercept is - 6. So this is let's say the minus 6 and it has a positive gradient. So this is how it would look like.

[01:06:19]Okay. Now for for the domain x is greater than 0. For the domain x is greater than 0 the minimum value is - 6 the minimum value is - 6. So the range is going to be f of x is greater than - 6. Why not greater than equals to?

[01:06:35]Because the domain was uh not equal to zero. It has to be greater than zero.

[01:06:41]For g inverse of x we replace g of x by y and then make x the subject. So that's the first step. G replaced DG of X by Y.

[01:06:53]Okay. So, okay, we we'll discuss what I have written over here in a while. But the first thing that you should be careful about is you have to make X the subject. So, bring this to the other side. Basically, swap these two things.

[01:07:05]The position of these two things. This goes in the denominator. This goes in the on the other hand side. Okay. Take a square root on both both hand sides.

[01:07:13]Now, when you take a square root on both hand sides, you could write plus minus.

[01:07:18]But you should understand or and you you could discard it later like you could discard one of these signs later but uh you could have also discarded it over here at this step. So how do we discard either of these plus or minus? We were given this function is only defined for x being greater than 3 / a.

[01:07:39]Right? So if this there are a couple of ways you could understand this process.

[01:07:43]So since x is greater than 3 / a if I re u if I sub sorry if I put things to the other side it will be a x - 3 is greater than zero. Okay so what does this say?

[01:07:56]It says that ax - 3 is positive right? a x - 3 is positive. So ax - 3 who squ obviously a squ thing is obviously positive but when you take a square root on both sides then we just consider the positive part. Why? Because this statement says a xus 3 where wherever you see a x a x - 3 you should understand that it is not equal to a negative. No it is only positive. It's only greater than zero. This is one way to think about it. There are other ways that you could have taught uh that you would have thought that you you have to discard one of these signs. You would have to discard the minus sign. In some other way you could have decided it as well. And then when you make x the subject uh you add + three to the other side divide all of it by a. So divide it by a or multiplied by 1 / a. So this is the inverse expression. And obvious yeah in the end we replaced x by g inverse of x and replaced y by x.

[01:08:59]Okay. Given that a= 2 solve the equation g inverse of x= 4. So a is given to be.

[01:09:05]Okay. Now g inverse of x = 4. Replace g inverse of x by 4 a by two and just solve it. So how do you solve it? Uh you bring two to the other hand side. 4 * 2.

[01:09:18]3 goes to the other hand side. It also subtracts over there.

[01:09:22]And on the right hand side you just have 2 over x.

[01:09:27]Okay. 4 * 2 would be 8. 8 - 3 is 5.

[01:09:29]Okay. Let me do uh let me show the process over here. So you have 5 = 2 / roo<unk> x.

[01:09:36]These two things swap their positions.

[01:09:39]So you have roo<unk> of x = 2 over 5.

[01:09:42]Take a square on both sides.

[01:09:45]You get 4 over 25 as the solution.

[01:09:52]Okay. The next question given instead that fg of 8= 6. Find the value of a justifying your answer. One way you could do it directly but I've tried to done it separately over here like first of all we calculate g of 8 and then we input it in the function of f.

[01:10:09]So g of 8 g of 8 is going to be uh what is g of x g of x is 4 a x - 3 squ. So just replace 4 / a x - 3² just replace this x by 8. This is g of 8, right? 8 a8 a * 8 will be 8 a okay now you replace uh the f of x f of x was uh f of x was I think 3x - 6 now f of g of x g of 8 would be you have to replace this by g of 8 g of 8 is 4 8 a - 3 squ uh and then you simplify the result we know this fg of 8 is equal to 6 so we replace this by six this is what this is what I have done over on I have done it on the right hand side and then 6 and - 6 goes to the other side you you get 12 over there 12 and 12 on both sides so 1 and 1 this is what we have and then in the end in the end what you do is 1 = 1 / 8 a - 3^ 2 put 8 a - whole square in the numerator so 8 a - 3 square okay you could do it in many ways you could do reciprocal Or you could say that you are surfing these two things. Uh the position of these two things 8 a - 3 whole square= 1. Take a square root on both sides. Square root of and then plus minus. After doing this 8 a - 3 = + - 1.

[01:11:40]And then you solve for a. So 8 aus 3 is either equal to positive 1 or either equal to 1. Now from this the solution is a= 1 /2. From this the solution is a= 1 / 4. Okay. Find the value of a justifying your answer. There's there was this part justifying your answer.

[01:11:59]How do you justify whether your answer is correct or not? Now the thing is there are two answers of a but only one of the one of these answers are correct.

[01:12:09]Why? Because we we we first consider the domain of the function g uh when a is 1 /2 or when a is 1 / 4. So domain was x is greater than 3 / a.

[01:12:23]So when a is 1 /2, x will be greater than 3 / 1 /2. 3 over 1 /2 is 6. So x is greater than 6. When domain is 1 / 4, sorry, when a is 1 / 4, the domain is going to be 3 over 1 / 4. So 3 1 / 4 is 12. x is greater than 12. Now we were evaluating, we used this equation and we evaluated g of 8. Now g of 8 means that the input is 8, right? The input is the x value. The x value has to be 8. The x value cannot be 8 if the domain is greater than 12.

[01:13:00]If you are saying the function is valid for only x x being greater than 12, so you cannot evaluate g of 8. G of 8 will not be valid in that case. Now g of 8 is going to be valid in this case because domain is uh not restricting it. And uh so therefore we conclude that the only possible answer of a is 1 /2 not 1 / 4.

[01:13:23]Okay. This question a tangent to a circle passes through the points 1 1 5 and 4a 5 uh 4a 4 meets the circle at the point a. Another tangent to the circle has equation this. So the the equation of this tangent I think is given and meets the circle at point b. This is the point b. Find the coordinates of the point of intersection of the two tangents. So basically uh what you do is you have to extend this and you also extend this and these two are going to intersect at one point. You have to give state the coordinates of those point. So obviously this is equation of simultaneous equations. So you have to uh you have to get the equation of both of these curve sorry both of these tangents and then equate them together. So we know the equation of tangent at b already. We have to get the equation of tangent at a. So for the equation of tangent at a we know the uh the it passes through these two points. You know the points it passes through like maybe call this 4a 4. This is four and this is four. This is this point is 4a 4 and this point is 1a 5. Maybe this point. Uh yeah maybe not this point but this point is 1a 5.

[01:14:37]So this is one 1 comma 5. We know two coordinates we can get the gradient.

[01:14:42]Gradient is going to be 4 - 5 / 4 - 1 which is -1 / 3.

[01:14:51]Okay. Equation of tangent at a uh gradient we have figured out. We use one of these coordinates. I'm going to use this coordinate. I use the uh the linear format y - y1 = m * x - x1. Uh this format gets the equation of the curve.

[01:15:06]The y1 is 5. So uh replace five over there and x1 as one over there. And then if I simplify this equation I get 3 y.

[01:15:16]Okay let me show you how I simplify this.

[01:15:20]So you have y - 5 = uh -1 3 * x - 1.

[01:15:34]Okay. So you first multiply this three on the other side. So you have 3 y - uh 15 =x and then + 1. So -5 and uh 15 goes to the right hand side. So5 sorry + 15 + 1 would be 16 - x = 3 y 3 y = 6 uh 16 - x or x + 16. And if you make x the subject, you could say x = 16 uh - 3 y or x + 3 y = 16. x + 3 y = 16 and make x subject. So x = 16 - 3 y.

[01:16:16]Now you sub the sub in the equation x = 3 - x y uh sorry x = 16 - 3 y. We sub this equation into the uh other the equation of the other tangent x - 3 y = 16. We sub it over there. So x is 16 - 3 y. Put it over there. 16 - 3 y - 3 y = 16. Uh 16 and 16 on both sides. So you could subtract them. 16 - 16 0 -3 y - 3 y uh is 9 y or okay wait what I have done over here. Okay. Anyways, the answer would be still the same. I I think I divided uh both sides by some number. Wait, let me see. Uh first of all, you you put 16 over there. So 16 and - 16 would be 0.63 - 3 y would be 9 y - 9 y = 0 is the correct equation. But uh either way this is correct. Either either way it will be correct. 3 y= 0 and then you divide both hand sides by uh 3 or9 as we were getting and then the final answer would be zero. Wait, let me just still see it again. 16 - 3 y and then 3 y. Okay, so your y is zero.

[01:17:29]Now you put your answer of y and this in any of the equation of the tangent. So x I think I put it over there. So x = 16 - 3 * 0 x will be equal to 16. The coordinates of the point of intersection will be 16, 0.

[01:17:47]Okay, it is given that the coordinates of the center of the circle are a comma 0 where a is less than zero. Find an equation to the normal of the circle which passes through a. Give your answers in terms of a. Okay. So what what it means over here is we are given the circle.

[01:18:06]All right.

[01:18:15]Okay. Why is it not working?

[01:18:22]Okay. We are given the circle. Uh the center of the circle is a comma 0.

[01:18:29]Equation of the normal of the circle which was to the point A. This is the point A. I think the coordinates of point A were given. It was uh no it was not given but I don't think it's necessary. Uh yeah. So we have to find the equation of this line of the normal uh to the circle. So normal to the circle means it it makes a perpendicular with the tangent. So it is 90° to the tangent. We have to get the equation of this line. So equation of this line for the equation of this line we know it passes through this point. We know one point we have to get the gradient. So we use the fact that equation of the tangent and the equation of this would be uh if you multiply the equation of the tangent and if you multiply the equation of the normal sorry the gradient of the normal you get 1. So the gradient of normal is going to be uh 1 / the gradient of tangent. The gradient of tangent is 1 / 31 / 3 is basically just three. So you know the gradient and we know a point it passes through. We can get the equation. So y - 0 using the linear uh equation y - the equation of straight line y - 0 = y - y1 = m * x - x1 replaced uh y with zero and x with a this is your final equation. Okay, the last part it is further given that the radius of the circle is root of 40. Find the equation of the circle. So you for the for the equation of the circle one thing you need is the radius and the other thing you need is the center.

[01:19:56]That's it. So we know the radius. The radius is given. We have to get the center. We have to get this a. So I've drawn this uh so I've drawn this diagram over here. Uh so we know this is the tangent and uh tangent passes through a point 16 0. How do we know it? Because we figured it out previously that the both of the tangents intersect at the point 16 0.

[01:20:21]So I've drawn this point over here 16 0 and this is a and this is a uh this co point is a this is the radius for point a if you if you have to get the uh point a uh what we can do is we can u equate the equation of the normal and the equation of this tangent. So for point a the equation of the uh tangent was this uh the tangent at a was this x + 3 y = 16. So x + 3 * y = 16. y of the normal is this. So I just substituted this over here and solving it simultaneously.

[01:20:57]Simplified result and then I I bring I make x the subject x = 16 + 9 a / 10.

[01:21:04]Okay. For y for the answer of y uh the y is going to be again I can put it in any equation. I I uh I can put it in the I think yeah I put it over here in this equation. So y will be 3 * uh the x - 3 a I simplified result take I have taken a common uh factor. So I have taken the LCM of the denominator and this is the result.

[01:21:31]Okay sorry this is the result. Okay we know the coordinates of point A in in a very unsimplified form. It may look like like uh it it might not help any like it might not help but it is going to be helpful. So this is the coordinate 48 - 3 a / 10. Okay, we use the fact that radius is the distance from the center to the point a. So radius is a square root of 40. It's going to be the distance from between a point the point a and the center. So distance for point a and the center this is the distance square root on both sides. they cancel out each other and simplify this result uh uh simplify and take in a common factor uh sorry again the LCM in the denominator and this is the result that I'm getting like it so over here 3^ 2 would be 9 uh 9 * 16 - a and we have 16 - a whole over here 16 - a whole square on both sides so 9 + 1 16 9 + 1 would be 10 so 10 16 - a whole square 0 and 0 cancels out And just simplify the result. 40 * 10 would be 400. And then 16 - a squar. Take a square root on both sides. 16 - a= + -20. a will be uh a will be + 4 and minus 4. a is + 4 and a is -4. Uh sorry I I think the other answer is not going to be minus4. It's going to be something else. But all we have to do is we have to ignore the positive answer because it was stated in the question. A is a is less than z. A has to be negative. A is4. Then we just form the equation of the circle using the fact x minus center the x coordinate of the center whole square plus y - ycoordinate of the center whole square= the r² r is of 40 square root of 40 square and this is the equation of the circle.