Top 30 JUPEB Mathematics Questions & Answers | Exam Revision

Added: 2026-05-27

[00:00:00]Hi guys, welcome to today's tutorial.

[00:00:03]Welcome to all CBT. Today we'll be working on 30 likely mathematics questions.

[00:00:14][music] [music] [music] Our first question says the position of an object in motion at any time t is given by s = 3 t cq minus 5 tus 2. Obtain the velocity of the object after 2 seconds.

[00:00:50]Now in order to obtain the velocity of the object after 2 seconds we have s = 3t^ 3 - 5 t - 2. We first find the the the t ddt of this. So our dt of our equation will be 90 t^ 2 - 5. All right. Now where are t is 2 seconds. So substituting into this value we have it as 9 bracket 2^ 2 - 5. This gives us 9 * 4 - 5 which is 36 - 5 which gives us 31 m per seconds. So our position our velocity of the object after 2 seconds will be 31 m/ second.

[00:02:04]The next question says find the derivative of 2 x - 5 x^ 2 + 2. All right. So we have this as 2 x cub - 5x^2 + 2. All right. So this multiplied by this this times this will give us 6 x.

[00:02:31]Okay. For us to do this let's expand a bit. So this will be power by this. This will be 3 * 2 x^ 3 - 1 - 5 * 2 2 * 5 x^ 2 - 1. Well, since there's no x here, this becomes zero. Now this gives us 6 x^ 2 - 10 x^ 1 which is 10 x. So our answer will be 6 x^ 2 - 10 x.

[00:03:22]All right. For this it says find the derivative of y = 3 + 2x and 1 - x. So for this we use product rule okay where we have it as we find the the derivative of this keeping this constant plus the derivative of this keeping this constant. Okay, let's do d the x of 3 + 2x * 1 - x + 3 + 2x * dx of 1 - x.

[00:04:14]Okay. Now having this as this, our dx here will be 2 bracket 1 - x + 3 + 2 x bracket dx of this will be -1.

[00:04:37]Okay.

[00:04:41]All right. So from here we open up the bracket. If open this brackets 2 * 1 is 2 - 2 x + 3 * 3 * 1 will give us - 3.

[00:05:00]So this becomes -3 - * + - 2 x. Okay. Now collecting items this will be.

[00:05:09]>> Get the cheapest science materials from all schoollabs.com.

[00:05:14]2 - 3 gives us -1 - 2 - 2 will give us -4 x. So our derivative of that will be -1 - 4x.

[00:05:29]Okay. Next we have is x + y^2 = 5. Okay. For this we'll be using implicit differentiation where we have our question is x + y squared = 5. Okay. So here we will have this as two brackets x + y. All right. Now times dx of x + y. Okay. = 5. Now if we do this, this becomes 2 bracket x + y.

[00:06:18]dx of x will give us 1 1 + dy / dx which gives us 5. Okay. And since x + y = z. Okay, since x + y = 0. Okay, if you do our d x of 5 will be zero.

[00:06:44]Sorry. So since x + y = 0, this becomes 2 bracket 0 bracket 1 + d y / x = 0. So this will be zero. canceling us leaving us with 1 + dy / dx which equals z. So our dy dx will be -1. So the different differentiation of this will be minus1.

[00:07:22]Okay. Next we have evaluate the limit of this. Now before we do the um limits we have to find the different different we have to differentiate our equation. So we have for the numerator we have x^2 - 25. If we do ddx of it this becomes 2 x okay will be our ddx for this. While if we do have x - 5 and we do ddx of it, we are having it as 1. So we can write this equation to our limit x ts towards 5 to be 2x / 1. All right. Now substituting when x is 5 2 bracket 5 over 1 would give us 10. So our answer would be [clears throat] 10.

[00:08:31]Okay, for this it says if y = x - 1 exponential 1 - x find the y dx.

[00:08:44]Now for us to do this we use our product rule. Remember in our product rule we find the differentation of one keeping one constant then find the other differentation keeping the other one constant. So this we have as dx of x -1 e^ - x + x -1 dx of e^ - x.

[00:09:21]Okay. Now if we differentiate this we are having it as one bracket e^ - x and if we differentiate this this is x - 1 differentiating e^ - x will give us - e^ - x.

[00:09:51]Okay. So here we can open this up to be uh e^ - x + x - 1 - e^² - x. Now from here so we collect our like terms this becomes e^ - x bracket open 1 + x + one.

[00:10:35]Okay.

[00:10:37]So from here we have it as e^ - x bracket x.

[00:10:49]Okay, sorry. This will give us - x.

[00:11:06]Okay, let me explain this. From here we have this we have as e^ - x plus this multiplying this will give us - e - e^ - x x + 1. Okay. So plus sorry that will not be + one plus e^ - x. Okay. So if we collect like terms here we have this as e^ - x + - e + e^ - x bracket - x + 1. Okay. Now from here we have it as e connecting these two e^ - x plus okay e^ - x bracket open since we have a common term bracket open 1 + 1 - x + 1 Okay.

[00:12:33]Okay. Okay. So opening this we have it as e^ - x bracket 1 - x + 1 which gives us which now gives us e^ - x bracket 2 - x.

[00:12:58]Okay.

[00:13:01]So that's how we arrived at this 2 - x. Okay. So from here our answer will be e^ e^ - x 2 - x.

[00:13:21]Okay. Our next question says find the modulus of 2 I + 3 J - 4 K.

[00:13:35]Okay. Now here our modulus will be the root of I 2 + J 2 + K 2. So this would be 2 + 3 2 + okay 4 squ. So this we having as 4 + 9 + 16 4 + 16 is 20. So this gives us root 29.

[00:14:17]So the modus will be what? 29. Okay. Our question says find the scalar product of a 2 i + 3 j b i + 4 j.

[00:14:38]Okay. So our a b will be 2 * 1 + 3 * 4. Now 2 * 1 will give us 2 + 3 * 3 * 1.

[00:15:14]Okay, this is - one.

[00:15:18]This is -1 sorry. So that will be - 2. 3 * 4 will give us 12. Now this will give us 12 - 2 will give us 10.

[00:15:31]Okay. So our answer would be 10.

[00:15:40]Okay. Our next question says find the value for which n Find the value for of n for which the vector s i + n j - 3 k and n i >> run quick tests and analysis. Visit analysis.affrica Africa >> minus J + 5 K R perpendicular. So for our perpendicular vectors the dot product is always equal to zero. So bringing this we have s i + n j - 3 k n i - j + 5 k = z. Okay. So from here we expand. All right. Now uh S Sn and N will be equals to what?

[00:17:03]This is collecting this Okay. So from here we have common terms.

[00:17:58]So this becomes s n bracket sus one = 15.

[00:18:08]So this and this forms s n. This and this forms our n and this and this forms our minus5.

[00:18:19]Okay. So from here we have n = 15 all over s -1.

[00:18:29]So there we have it as 15 / s -1 as our answer.

[00:18:54]Find the center and radius of the circle.

[00:19:02]8 x^2 + 8 y^ 2 - 24 x - 14 y + 18 = 0.

[00:19:15]Okay. So for this we'll first of all move our constant over to the other side. So this comes 8 x^2 + 8 y^ 2 - 24x - 40 y = - 18.

[00:19:37]Okay. Now having it as -8, we will now do what? Divide through by 8. Or we can factoriize this as 8 bracket x^2 + y^2 - 3x - 5 y = - 18. So by 8 we have 2 x^2 + y^ 2 - 3x - 5 y = -8 / >> get the cheapest science materials from all schoollabs.com.

[00:20:15]Okay. So if we break this down, we have it as x^2 + y^ 2 - 3x - 5 y = 9 - 9 / 4.

[00:20:30]Okay, now let's look at our factors for x.

[00:20:38]for x. Okay, we look at it as x - 3x all squared equals what?

[00:20:53]R x - 3 / 2 all 2. Now - 9 / 4. Okay. Now for for y we have it as y^ 2 - 3 a - 5x sorry y^2 - 5 y = y - 5 / 2 all^ 2 - 25 / 4. Now this is application of completing the square method. That's the formula we are using here. Okay. So now we can solve this out. Okay. Here we have we are substituting this into this equation. Okay. So here now for x - 3x.

[00:22:07]So for x - 3x we have it as x - 3 / 2^ 2 - 9 / 4 = sorry + y - 5 / 2 - 25 4 = - 9 / 4. Okay. So if we collect collect like we have x - 3 / 2 sorry we have x - 3 / 2 all 2 + y - 5 / 2 2 - 9 9 + 25 would give us 34 over 4 = - 9 / 4.

[00:23:24]Okay. So from now from here we collect like terms again moving a negative over this becomes x^2 - 3 / 2 + y^ 2 - sorry this shouldn't be x² this should be x okay + Y + y + 5 / 2 all^ 2 = 34 / 4 - 9 / 4. So here we have x - 3 / 2 all^ 2 + y + 5 / 2 all² = 25 / 4.

[00:24:31]Okay. So from here we can take our square roots. Okay. So if we take remember for us our formula is x - h all² + y + k all 2 = r².

[00:24:57]Okay. So from here our center will be 3 / 2 - 3 / 2 and 5 / 2. This forms our center and our radius will be radius squar is 25 / 4. So therefore our radius is 25 / 4 which is 5 / 2. So our radius is 5 / 2 and our center is - 3 / 2 and 5 / 2.

[00:25:42]Okay. Our question says find the equation of the tangent to the circle.

[00:25:47]we have 2 x^2 + 2 y^ 2 = 30. Okay. At point -3 and six. All right. So from here we can reduce this equation by breaking it into two as x2 + y 2 = 15.

[00:26:12]All right. So if we have it as 15, we know that for our gradient, all right, our gradient, our gradient is y 2 - 1 y1 all over x2 - x1. So here this becomes 6 - 0 all over - 3 - 0 which is 6 - 3 which gives us -2 okay and for a for a tangent okay is perpendicular to our point. So our radius for tangent will be the inverse of this which will be 1 / 2. Okay. So if we have our slope our end slope our slope will be y - y1 = m bracket x - x1. All right. Now in this case this is y - 6 = 1 / 2 bracket x - - 3. Okay - gives us plus which gives us y - 6 = x + 3. Okay. Now having it as x + 3 we expand this we expand this as y - y - 6 * 2 = x + 3. So this becomes 2 y - 12 = x + 3. Okay. So moving all over to one side and equating to zero. This becomes x + 3 - 2 y + 12 = 0. So this will be x - 2 y + 15.

[00:28:58]= 0. So this forms tangent. So that would be x - 2 y + 15= 0. So that forms the equation of the tangent. Okay. It says find the gradient of the curve y = xq - 6 x^2 + 11 x - 6. So for us to find the gradient at point 1 and 0 we do our dx.

[00:29:32]This gives us 3 x - 12 x 3x^ 2 - 12 x + 11. Okay. I write it as 3x 2 - 12 x + 11. So at.1 okay this becomes 3 bracket 1^ 2 - 12 bracket 1 + 11. So this we have as 3 - 12 + 11.

[00:30:05]Okay. So 3 - 12 3 + 11 gives us 14 - 12 which gives us 2. So our answer would be 2.

[00:30:30]Given set A as a B 1 3 and set B as a 2 4 A union B. Now when we talk about union of a set it's the combination of the elements in both sets. So our A union B would be since A is common we'll write it as one A. Then B 1 2 3 and 4.

[00:31:13]So this will be our A union B which is A 1 2 3 and 4.

[00:31:23]Let P be the set of prime factors of 42.

[00:31:27]So when we look at let's find the prime factors of 42 starting with two this is 21 starting with three this is 7 and 7 1 so prime factors of 42 are 2 3 and 7 all right and so this is let's call this as set P all right and prime factors of 45 are set Q let's look at prime factors of 45 that is 5 that gives us 9 3 3 1 so 45 we have it prime factors as three and five all right that's set Q now it says find P intersection Q now talk about intersection we are just looking for what is common among the two sets so P intersection Q would be since three is the only thing in common here would be three.

[00:32:34]So 3 is our P intersection Q. Okay. Our next question says for what values of K will the equation Y^ 2 - K - 2 bracket Y + 1 = - 2K. Okay. Let's equate this to Z. So this becomes y^ 2 - k - 2 y + 1 + 2 k = z.

[00:33:09]Okay. So now this will be y^ 2 - k - 2 y + 2 k + 1 = z. Okay. So here our a is 1, b will be minus k - k - 2. Okay, that to be our b and c will be 2k + 1.

[00:33:47]Okay, that forms our c. Now for us to find our determinant we have it as b - 4 a c. Okay our b - 4 a c will give us our determinant sorry b ^ 2 our determinant will be b ^ 2 - 4 a c. All right this was out of the quadratic formula. So now at this pres run quick tests and analysis visit analysis.affrica >> - 2^ 2 - 4 * 1 * 2 k + 1. Okay.

[00:34:46]So from here if we expand this.

[00:34:52]So from here if we expand this we have this as >> [snorts] >> Okay. So if we expand, let's expand this. We have k² k - 2. Okay.

[00:36:45]All squ. So this will give us - k + 2 - k + 2. So this will give us k^ 2 - 2 k - 2 k + 4 - 2 k + 2 k sorry that - 2 - ++.

[00:37:32]So this gives us k² - 4 k + 4. Okay.

[00:37:38]Replacing this for this.

[00:37:45]Here we have it as k^ 2 - 4 k + 4 - 12 sorry what is - 8k - 8k - 4 = So + 4 will cancel - 4 leaving us with k^ 2 - 12 k = 0. So factorizing we have k bracket k - 12 = 0. So this is k = 0 or k - 12 = 0. So our k will be 0 or 12.

[00:38:43][snorts] So the value of our k will be 0 or 12.

[00:39:05]Okay, the next question says which of the following is correct? We have radian 180° equals to radian pi radian.

[00:39:16]All right. So for 270° that will be 270 / 118 which gives us 3 / 2 radian 3 2 pi. Okay. So 270 is in 2 pi. So this is wrong.

[00:39:36]Now you have 360° which will give us 360 over 180 which gives us 2 pi. So this is three. So this is wrong.

[00:39:52]we have 90° which is 90 / 180 which is 1 / 2 pi. All right. So this is also wrong. So let's check if this is correct. 720° will give us 720 over 180.

[00:40:13]So this will give us 4 pi. So this is the correct answer.

[00:40:21]720° = to 4 pi radian.

[00:40:28]The data presentation in which the bars are joined together is called okay. So what we have here the bars in this are joined together in this form is referred to as histogram. Okay. Back charts are where the columns are separated.

[00:40:50]The columns of the bar chart are separated while that of the hog is joined together.

[00:41:00]If f(x) is 3x^2 - 5x + find f_sub_2. So our f_sub_2 would be 3 brackets 2^ 2 - 5 bracket 2 + 2. So this gives us 3 * 4 - 10 + 2. 3 * 4 is 12 - 10 + 2. So 2 + 2 gives us 4.

[00:41:32]So our f_sub_2 will be 4.

[00:41:39]It says solve for x if 2^ x + 1 = 16.

[00:41:47]Okay. So for this we have to bring this down to the base of 2 which gives us 2^ x + 1 = 2^ 4. All right. 2 * 2 4 * 2 8 3 * 2 16. So equating our basis here we have x + 1 = 4. So x = 4 - 1 which gives us three. So the value of x here will be three.

[00:42:21]Find the gradient of the line joining points this and this. So our gradient we have it as y2 - y1 all over x2 - x1 where this is x1 y1 y 2 x2. So here we have it as 11 - 3 all over 6 - 2. 11 - 3 gives us 8 / 6 - 2 4. So our gradient will be 2.

[00:42:59]If sin theta is 3, sin theta is 3 over 5. And we know that s is opposite over hypotenus.

[00:43:11]Okay. It says we should find cos theta.

[00:43:14]All right. So let's look at this in a right angle triangle. If our hypotenus is five and the opposite is three. Okay.

[00:43:23]Using Pythagoras theorem we have a hypotenus = opposite 2 + square. All right. Where our hypotenus is 5^ 2 = 3 2 + x². All right. So this is 25 - 9 = x². So this gives us 16 = x². So our x will be <unk>6 which is 4. That means our adjacent is 4. And we know that cos theta is adjacent over hypotenus which gives us 4 / 5. So our cos theta would be 4 / 5.

[00:44:16]Evaluate log 1000.

[00:44:19]Okay. We have log base 10,000.

[00:44:23]Now we change this to the powers of 10.

[00:44:26]This becomes log base 10 10^ 3. All right. Dropping this down here. This becomes 3 log 10 10. Now we know that mathematically log 10 10 = 1. So this becomes 3 * 1 which is 3.

[00:44:47]So our answer will be three.

[00:44:52]The sum of the first 10 numbers first first 10 terms of an arithmetic progression is 155. Find the first five 10 if the first 10 is five and we should find the common difference. It says the sum of the first 10 terms of an arithmetic progression is 155. If the first term is five, find the common difference. So for sum of terms in arithmetic progression, we have as n / 2 bracket 2 a + n -1 d. Now in this case we are looking for the common difference where we are told s 10 10 / 2 bracket 2 * 5 + 10 - 1 d.

[00:45:54]Okay. So this will give us 155 = 5 bracket 10 + 10 - 1 9 d. All right. So this will be 155 = 50 + 45 D. Now collecting like this becomes 55 - 50 = 45 D. So this should be 105 = 45 g. So our d = 105 over 45. So five here will give us 25.

[00:46:52]Sorry 21.

[00:46:58]Five here will give us 21 all over 9. So three here will give us 7 over 3 which will give us two whole number 1 / 3 as a common difference.

[00:47:19]Our next question says differentiate y = 4x - 7 x + 6 with respect to x. So with respect to x we have to do this as 4 dx will be 4 * 3.

[00:47:40]Okay, let's have it as 3 * 4 x^ 3 - 1 - 7 * x Okay, note this is 1 * 7 1 * 7 x^ 1 - 1 okay + 6 since it has no x we are not differentiating it this becomes 12 x^ 2 - 7 * x^ 0 and any^ 0 is 1. So this is like 7 * 1 to give us 12 x^2 - 7. So we have it our answer as 12 x 2 - 7.

[00:48:36]Okay. Okay, for this it says find the value of K if the points 1 K 3 7 and 5 11 lie on the straight line. All right, so we have 1 K 3 7. Okay, let's group them. We have 3 7 and 5 11. All right, we're going to find the gradient of these two groups and equating them together since the colinear since our colinear points of gradients are equal. So this becomes for our gradient here we have it as 7 - k all over 3 - 1 which gives us 7 - k / 2. And for this we have 11 - 7 all over 5 - 3 which gives us 4 / 2 which gives us 2. So equating 7 - k / 2 to be = 2. Okay this gives us 7 - k = 4.

[00:49:50]So our k will be 7 - 4 which gives us 3.

[00:49:57]So our answer will be 3. Our next question says integrate 6 x^2 dx. All right. So this will be our integration will be 6 x 2 + 1 all over 2 + 1 + C. So this gives us 6 x^ 3 all over 3 + c. So this cancels out leaving us with 2x cub plus our constant c. So our answer will be 2x + constant c.

[00:50:43]8 x - 4. We have x - 4.

[00:50:50]>> Run quick tests and analysis. Visit analysis.affrica.

[00:50:55]>> Find the possible values of x. So this is x - 4 = 6 and x - 4 = -6. So for here we have it as x = 6 + 4.

[00:51:10]And for here we have it as x = -6 + 4. So this gives us 10 and this gives us -2. So our possible values will be -2 and 10.

[00:51:29]The roots of the equation are x^2 - 5x + 6 = 0. So if we factoriize this, this gives us 6 x². Now two roots that can give us five are 3 x and 2 x. Now let's consider the signs. Since it's - 5, this will be - 3 - 2 will give us - 5.

[00:51:56]3 * 2 will give us + 6. So here we have it as x^2 - 3 x - 2 x + 6. All right. So grouping these two together we have it as x is common x - 3 and this is - 2 is common x - 3. So we have it as x - 3 = 0. So x = 3 and x - 2 = 0 and x = 2. So our answer will be 2 and 3.

[00:52:38]Find the value of tan 45 + sin 30. Now tan 45 = 1 and tan 30 = 1. Sorry sin 30 sin 30 = 1 / 2. So this will be 1 + 1 / 2 which will give us 3 / 2.

[00:53:09]Okay. Now it says if a mat if a matrix A is 2 1 3 4 that we should find the determinant. So for us to find the determinant we multiply this and multiply this. So this becomes our a determinant will be 2 * 4 - 3 * 1. Now 2 * 4 gives us 8.

[00:53:39]8 - 3 gives us five. So our determinant for this matrix would be five.

[00:53:51]All right guys, with this we've come to the end of today's tutorials. If there be any questions, don't forget to put it down at the comment section below. Don't forget to like, subscribe, and click on your notification buttons for more updates. See you in our next class.

[00:54:09]Thank you.

[00:54:10]>> Hi, I'm sure you enjoyed the video you just watched. To see more content as that preparing you for your upcoming exam, make sure to smash the subscribe button and turn on notification. Also remember that you can use the comment section to send us your questions or your inquiries and we are happy to take them. I'll see you in the next video.

[00:54:28]Bye.