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Reply 1
Help, Cant solve these questions :mad: . This was my mock exam in school.

Q1)
For the curve C with equation y=f(x)
f '(x)=(3-√x)²
Given that C passes thriugh the point (4,7), Find f(x)

Q2)
For the curve C with equation y=2x³-2x+4/x, x>0
(a) find dy/dx and d²y/dx²
(b) Verify that the gradient of C at x=1 is zero
(c) Find, in the form y=mx+c, the equation of the tangent to C at the point where x=2

Hope you could help, please put some explanations next to it if you could. Thank you very much. :smile:
Reply 2
ahmadz5
Help, Cant solve these questions :mad: . This was my mock exam in school.

Q1)
For the curve C with equation y=f(x)
f '(x)=(3-√x)²
Given that C passes thriugh the point (4,7), Find f(x)

f'(x)=(3-√x)²
f(x)=∫(3-x1/2 dx
f(x)=∫ x-6x1/2+9 dx
f(x)=½x²-4x3/2+9x+c

At (4,7)
7=8-32+36+c
c=-5

f(x)=½x²-4x3/2+9x-5
Reply 3
Thanx alot Malik. Could any one help me with question 2.
Reply 4
Hey guys, this is a tough S1 question that we had in our mock exam.

Q1) Records show that the length, in mm, of a particular type of building block can be modelled by L where L~N(450,3²).
Find the probability that a randomly chosen block will be:
(a) longer than 455mm
(b) betweeen 444 and 452mm long
(c) Find the probability that of three randomly chosen blocks at least one will be longer than 455mm
(d) Determine the value of k such that 95% of the block will have their lenght in the range (450±k)mm

Well hope you could solve this one for me. Thanks very much and could you please put explanaisons next to the answer please. Could some one also try and sole question 2 of the C1. Thanks :smile:

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By the way Malik, the answer you gave me makes no absolute scence man! Where did u get the 9 and the x and 6x^1/2???????????
Reply 5
ahmadz5
By the way Malik, the answer you gave me makes no absolute scence man! Where did u get the 9 and the x and 6x^1/2???????????


he expanded the brackets
Reply 6
ahmadz5

By the way Malik, the answer you gave me makes no absolute scence man! Where did u get the 9 and the x and 6x^1/2???????????


(3-√x)²
= (3-x1/2
= (3-x1/2)(3-x1/2)
=9-3x1/2-3x1/2+x
=9-6x1/2+x

It looks more complicated written down but only takes a few seconds to do in your head.
Q2)
For the curve C with equation y=2x³-2x+4/x, x>0
(a) find dy/dx and d²y/dx²
(b) Verify that the gradient of C at x=1 is zero
(c) Find, in the form y=mx+c, the equation of the tangent to C at the point where x=2


a) y= 2x3 - 2x+ 4x-1
dy/dx= 6x2 -2 -4x-2
d2y/d2x= 12x + 8x-3

b) sub x=1 into dy/dx= 6x2 -2 -4x-2
6(1)2 - 2 - 4(1)-2 = 0

c) At x=2 gradient of curve C =
6(2)2 - 2 - 4(2)-2 = 21

So gradient of tangent at point C is -1/21.
At x=2, the value of y (on curve C) is-
y= 2(2)3 - 2(2)+ 4(2)-1= 14.

So the tangent with the gradient -1/21 meets the curve at point (2,14)

y=mx+c
14= -1/21 (2) + c
c= 14 2/21

So y= -1/21x + 14 2/21
y= -x + 296

.........
Can someone plz confirm this...
Reply 8
"scence" - duuude...
Reply 9
Thanx guys for the replys but could some1 help me with the statistics questions. Thank you
Reply 10
Q1) Records show that the length, in mm, of a particular type of building block can be modelled by L where L~N(450,3²).
Find the probability that a randomly chosen block will be:
(a) longer than 455mm
(b) betweeen 444 and 452mm long
(c) Find the probability that of three randomly chosen blocks at least one will be longer than 455mm
(d) Determine the value of k such that 95% of the block will have their lenght in the range (450±k)mm


a) you need to normalise the data by applying the transformation Z = (X - mu) / sd. So for a) this would be Z = 5/3. The question is asking for probability above the point 5/3. It's easy to make a little sketch at this point highlighting the area you want to find. Use your tables to find p when z = 5/3 and subtract that number from 1 to get your answer.

Its quite hard to explain through without the diagrams so if you wish ill try and write a model answer in paint or something. I've got S1 in January too and have just been revising the normal distribution (2nd weakest topic after probability :biggrin:). Good luck in January!
Reply 11
ahmadz5
Q1) Records show that the length, in mm, of a particular type of building block can be modelled by L where L~N(450,3²).
Find the probability that a randomly chosen block will be:
(a) longer than 455mm
(b) betweeen 444 and 452mm long


Well this is the normal distribution and S2 for me:

L ~ N(450, 3²)
(a) P(L > 455) = (445 - 450)/3 = -5/3 (1.67)
.............. => 1- Φ(1.67) = 1 - 0.9525 = 0.0475 (Because curve is symmetrical we can use the table and this is the smaller part of the area)

(b) P(444 < L < 452) = [(444 - 450)/3] - [1-(452 - 450)/3]
............................ = (-2) - [1-0.67]
............................ => &#934;(2) - [1-&#934;(0.67)]
............................ => 0.9772 - [1-0.7486] = 0.7258
Reply 12
thanx every 1 for your posts. If any one would like to add some questions they are welcome.
Reply 13
I've attached my walkthrough to part a). I hope it's correct as im a little prone to the odd mistake. If you find it helpful, i'd gladly do it for the others, it is christmas :wink:

Bob
Reply 14
but could some 1 answer part c and d.
Reply 15
ahmadz5
Q1) Records show that the length, in mm, of a particular type of building block can be modelled by L where L~N(450,3²).
Find the probability that a randomly chosen block will be:
(d) Determine the value of k such that 95% of the block will have their lenght in the range (450±k)mm


(d) 95% = 2 standard deviations.

... => (x - 450)/ 3 = 2
... => x = 456
... => (y - 450)/ 3 = -2
... => y = 444

so (450 ± 6)mm.

^_^

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ahmadz5
(c) Find the probability that of three randomly chosen blocks at least one will be longer than 455mm


P(L > 455) = 0.0475

...=> 3 blocks = 3 x 0.0475 = 0.1425
Reply 16
Thanx alot for you great posts. Thanx robfinlay for the great diagrams and explanasions. I could gladly help you with probablity cuz its my best topic. Anywasy thanx alot, but need part c and d cuz they so hard.

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Thanx V.P.Keys, i sent the replay after you sent the reply by a bit, before i saw it. Thanx alot.
Reply 17
Ok, i did this question and i got it wrong. Its the specimen paper for C1. I did it and saw the answer but got something totally different. So here's the question.

Q) The points A and B have coordinates (3,4) and (7,-6) respectively. The straight line l passes through A and is perpendicular to AB.

Find an equation for l, giving your answer in the forum ax+by+c=0, where a,b and c are intergers.

Thanx, and hope you can help. :smile:
Reply 18
ahmadz5
Q) The points A and B have coordinates (3,4) and (7,-6) respectively. The straight line l passes through A and is perpendicular to AB.

Find an equation for l, giving your answer in the forum ax+by+c=0, where a,b and c are integers.


Equation for line AB => passes through (3,4) and (7,-6)
gradient of AB = (-6-4)/(7-3) = -10/4 => -5/2
gradient of l => 2/5

l => (y-4) = 2/5(x-3)
............=> y = 2/5x + 14/5
............=> 5y = 2x + 14
............=> 0 = 2x - 5y + 14
Reply 19
line l is perpendicular to AB, gradients of two perp lines multiply to -1.
First find gradient of AB, which is -10/4 = -5/2
The gradient of l = -1 / (gradient of AB) = 2/5
You know l passes through point A, and now know it's gradient.
So use the formula to find equation of line from gradient and one point:-
y - 4 = 2/5(x - 3)

5y - 20 = 2x - 6

rearrange into form ax+by+c=0:-

2x - 5y + 14 = 0

Ah! I have once again been defeated! You've beaten me to it.