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Mathematics: Post your doubts here!

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luniaz226

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Assalamo'alaikum, I need help with this question:

Solve the equation
ln(x + 5) = 1 + lnx,
giving your answer in terms of e.

i'm home studying and no one around can help me right now.
 
yousef

yousef

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parthrocks said:
Its quite simple for the question 7 part 2 just do it like
3/(1 + 2x)^2 < 1/3 and then solve for x in the next step you will get like this

(1 + 2x)^2 (>)9
then
once when you solve it using the (a+b)^2 formulae you will get it like 4x^2 + 4x +1>9
you will get like
4x^2 +4x -8 and then solve it quadratically......ok get it like x > 1, x < –2 ok gt it?
Click to expand...


This question was repeated many times > and i didnt now it ....
so any help plz ? <>> http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_41.pdf

q7 >>> last part ... (iii) Find the time from the instant the string becomes slack until it becomes taut again
 
syed1995

syed1995

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Mayedah said:
Question 7 (ii)
Click to expand...


There is only one way to solve this.

You need to remember that for a line to be tangent it needs to be like

tangent.gif


Meaning that the line y=mx+c needs to intersect the curve at exactly one which .. (that point being the minimum/maximum point on the curve) Meaning both the roots need to be same.

that is only true in b^2-4ac=0 .. in b^2-4ac=0 .. there is only 1 root.

so to solve this question

y = m(x – 2)
y = mx - 2m ===> 1
y = x^2 − 4x + 5 ===> 2

Substitute 1 in 2..

mx -2m = x^2 -4x +5 // shifting it to one side gives us..

x^2-4x + 5 - mx + 2m=0 // arranging them in order of powers gives us

x^2 -(4+m)x +5 + 2m=0

a= 1
b=4+m
c=5+2m

using b^2-4ac =0 we get..

(4+m)^2-4(1)(5+2m)=0 // now opening the brackets
16+8m+ m^2 -20-8m=0 // +8 , -8 get cancelled
16+m^2-20=0
m^2-4=0
m^2=4
m=√4
m= 2 or m = -2

now we need to substitute m into the equation to get the co-ordinates of x.

x^2 -(4+m)x +5 + 2m=0
x^2 - (4+2)x +5 +2(2)=0 ------- or ----- x^2 -(4-2)x +5 +2(-2)=0
x^2-6x+5+4=0 ------- or ---------- x^2 -2x +5 -4
x^2-6x+9=0 ------------ or --------- x^2 -2x +1 =0
x^2-3x-3x+9 ------------ or --------- x^2-x-x+1
x(x-3)-3(x-3) ------------ or --------- x(x-1)-1(x-1)
(x-3)(x-3) ------------ or --------- (x-1)(x-1)

x=3 or 1 ..

Now finally equate the values of x into y=x^2 − 4x + 5 to get the y-coordinates of the co-ordinates..


I know it seems a little too lengthy but usually they stop at values of m and don't ask the co-ordinates itself.. it's the first time i have seen them ask the co-ordinate as well.

As far as derivate goes.. that's not possible and will yield wrong answer since we can only get 1 value of m via that method and you won't get full marks.. the only correct way to solve this question is via b^2-4ac=0
 
syed1995

syed1995

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Alice123 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf
In qs 5i , the qs says 94% of the letters are within 12g of mean, so isnt it supposed to be (20+12<x<20-12)?? my answer's not matching with the mark scheme......:cry:
pleaseee help Dug syed1995 Ashique
also tell me how to find the p is 0.97
Click to expand...

That is one question which I still haven't understood.. I myself need help in that question :p

Tkp , Dug , PhyZac you guys got any idea how this one would be solved?
 
P

PhyZac

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syed1995 said:
That is one question which I still haven't understood.. I myself need help in that question :p

Tkp , Dug , PhyZac you guys got any idea how this one would be solved?
Click to expand...
This is very tricky question. Took me time to understand, anyway,


Alice123
you are right about (20+12<x<20-12)

now you have to imagine the graph in your head, and you have to come into a conclusion that it is symmetrical.

Now imagine the z of 20+12 that is 32 is x
So the z of 20-12 that is 8 will be a negative number -x

Both of them are x but one is negative, because they are the same distance from the mean

Let the probability of x is p
so the probability of -x is 1-p

we know that difference of the two probability is 0.94

so p - (1-p) = 0.94
p - 1 + p = 0.94
2p = 0.95 + 1 = 1.94
p = 0.97

Now is p is 0.97 z is from the table, 1.882 (or 1.881)

now continue normally

1.882 = 32-20 / sd
sd = 12/1.882
=6.38
 
syed1995

syed1995

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PhyZac said:
This is very tricky question. Took me time to understand, anyway,


Alice123
you are right about (20+12<x<20-12)

now you have to imagine the graph in your head, and you have to come into a conclusion that it is symmetrical.

Now imagine the z of 20+12 that is 32 is x
So the z of 20-12 that is 8 will be a negative number -x

Both of them are x but one is negative, because they are the same distance from the mean

Let the probability of x is p
so the probability of -x is 1-p

we know that difference of the two probability is 0.94

so p - (1-p) = 0.94
p - 1 + p = 0.94
2p = 0.95 + 1 = 1.94
p = 0.97

Now is p is 0.97 z is from the table, 1.882 (or 1.881)

now continue normally

1.882 = 32-20 / sd
sd = 12/1.882
=6.38
Click to expand...

Damn! that's real tricky.. I would never have reached that conclusion ever!
 
Dudu

Dudu

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Help on this one please?

On a certain road 20% of the vehicles are trucks, 16% are buses and the remainder are cars.
(i) A random sample of 11 vehicles is taken. Find the probability that fewer than 3 are buses. [3]
(ii) A random sample of 125 vehicles is now taken. Using a suitable approximation, find the
probability that more than 73 are cars.
 
Iffat

Iffat

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Kumkum said:
oct/nov 08
2. 1+sinx/cosx = cosx/1+sinx take common denominator
= [(1+sinx)^2 + (cos^2)x]/cosx(1+sinx)
= (1 + 2sinx + (sin^2)x + (cos^2)x)/cosx(1+sinx) here an identity needs to be used : (cos^2)x +(sin^2)x = 1, so (sin^2)x = 1 - (cos^2)x
= (1 + 2sinx + 1 - (cos^2)x + (cos^2)x )/ cosx(1+sinx) , the '(cos^2)x' cancels out
= 2 + 2sinx/cosx(1+sinx)
= [2(1 + sinx)]/cosx(1+sinx) , the '1+sinx' cancels out
= 2/cosx hence equal to right hand side
hope you understood :)
Click to expand...
thanx(y)...wat bout the oda 3qs?
 
syed1995

syed1995

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Dudu said:
Help on this one please?

On a certain road 20% of the vehicles are trucks, 16% are buses and the remainder are cars.
(i) A random sample of 11 vehicles is taken. Find the probability that fewer than 3 are buses. [3]
(ii) A random sample of 125 vehicles is now taken. Using a suitable approximation, find the
probability that more than 73 are cars.
Click to expand...


Probability of buses = 0.16
Probability of trucks = 0.2
Probability of cars = 0.64

for i) we will use binomial distribution..

n=11
buses = 0.16 = p
not buses= 0.84 = q

Fewer than 3 are buses means .. there can be either 0,1 or 2 buses.

0.84^11 + 11C1*(0.16)*(0.84)^10 + 11C2*(16)^2*(0.84)^9

= 0.7479
= 0.748 (3sf)

for part ii) we use normal approximation to binomial..

n=125
Cars = 0.64 = p
Not Cars = 0.36 = q

first we check if both np >5 and nq > 5 or not..

np = 125*0.64 = 80 and nq = 125*0.36 = 45 so normal can be used...

np = 80 = mean or µ
npq = 28.8 = variance (SD^2)

=>P(X>73)
= P(z>73.5-80/√28.8) // REMEMBER since it's Normal approximation to binomial .. we need to do continuity correction.. i always forget that.. (z>73 means + 0.5 z<73 means - 0.5)


=P(z>-6.5/√28.8)
=P(z>-1.211)
=P(z<1.211)
=Phi(1.211)
=0.8869 + 2*10^-4 (from the table)
=0.8871
=0.887 (3sf) Answer
 
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