Mathematics: Post your doubts here!

[manya]

plz help me with part 2

 

[qwertypoiu]

pk this is a really dumb question
ive done this before
but i completely forgot how to solve this
can anyone explain me how to find the inverse of a cubic function?
ex: y = 5x^3 - 9x^2 - 2

Well cubic equations are not one-one functions in general. So it seems to me that your function cannot possibly have an inverse function unless you restrict the domain. If the domain IS restricted, there's probably a way but I feel that this requires solving a cubic equation, which I think isn't in our syllabus. Do you have a past paper question like this?

 

[RadzMau]

Can someone please help me with this M1 question?

The top of a cliff is 40 metres above the level of the sea. A man in a boat, close to the bottom of the cliff, is in difficulty and fires a distress signal vertically upwards from sea level. Find (i) the speed of projection of the signal given that it reaches a height of 5 m above the top of the cliff, [2]
(ii) the length of time for which the signal is above the level of the top of the cliff. [2]

The man fires another distress signal vertically upwards from sea level. This signal is above the level of the top of the cliff for 17 s.
(iii) Find the speed of projection of the second signal.

 

[Sarosh Jameel]

ca

Can someone please help me with this M1 question?

The top of a cliff is 40 metres above the level of the sea. A man in a boat, close to the bottom of the cliff, is in difficulty and fires a distress signal vertically upwards from sea level. Find (i) the speed of projection of the signal given that it reaches a height of 5 m above the top of the cliff, [2]
(ii) the length of time for which the signal is above the level of the top of the cliff. [2]

The man fires another distress signal vertically upwards from sea level. This signal is above the level of the top of the cliff for 17 s.
(iii) Find the speed of projection of the second signal.

can u tell me the paper year ??

 

[qwertypoiu]

Can someone please help me with this M1 question?

The top of a cliff is 40 metres above the level of the sea. A man in a boat, close to the bottom of the cliff, is in difficulty and fires a distress signal vertically upwards from sea level. Find (i) the speed of projection of the signal given that it reaches a height of 5 m above the top of the cliff, [2]
(ii) the length of time for which the signal is above the level of the top of the cliff. [2]

The man fires another distress signal vertically upwards from sea level. This signal is above the level of the top of the cliff for 17 s.
(iii) Find the speed of projection of the second signal.

I) s=45, v=0, a= -10
v^2 = u^2 + 2as
0 = u^2 - 900
u = 30

ii) s = ut + 1/2*a*t^2
s = 30t - 5t^2
For ball to be above cliff:
s > 40
30t - 5t^2 > 40
-5t^2 + 30t - 40 > 0
t^2 - 6t + 8 < 0
(t-2)(t-4) < 0
2<t<4
So between 2nd and 4th second, the ball was above the cliff.
Length of time = 4-2 = 2s

iii) s = ut - 5t^2
ut - 5t^2 > 40
ut - 5t^2 - 40 > 0
5t^2 - ut + 40 < 0
Let's assume the two roots of this quadratic expression are α and β. Then we have:
α<t<β
β-α = 17
So how do we determine the difference of roots of a quadratic expression?
A general quadratic equation is:
ax^2 + bx + c = 0
With the roots:
x1 = [ -b + (b^2 - 4ac)^(1/2) ] / 2a
x2 = [ -b - (b^2 - 4ac)^(1/2) ] / 2a
x1 - x2 = (b^2 - 4ac)^(1/2) / a
So in our case,
β-α = [u^2 - 4(5)(40)]^(1/2) / 5
= (u^2 - 800)^(1/2) / 5 = 17
(u^2 - 800)^(1/2) = 85
u^2 - 800 = 7225
u^2 = 8025
u = 89.6

This last method may not be the best method, but I thought it was neat and no one else was answering so hope it makes sense

 

[RadzMau]

I) s=45, v=0, a= -10
v^2 = u^2 + 2as
0 = u^2 - 900
u = 30

ii) s = ut + 1/2*a*t^2
s = 30t - 5t^2
For ball to be above cliff:
s > 40
30t - 5t^2 > 40
-5t^2 + 30t - 40 > 0
t^2 - 6t + 8 < 0
(t-2)(t-4) < 0
2<t<4
So between 2nd and 4th second, the ball was above the cliff.
Length of time = 4-2 = 2s

iii) s = ut - 5t^2
ut - 5t^2 > 40
ut - 5t^2 - 40 > 0
5t^2 - ut + 40 < 0
Let's assume the two roots of this quadratic expression are α and β. Then we have:
α<t<β
β-α = 17
So how do we determine the difference of roots of a quadratic expression?
A general quadratic equation is:
ax^2 + bx + c = 0
With the roots:
x1 = [ -b + (b^2 - 4ac)^(1/2) ] / 2a
x2 = [ -b - (b^2 - 4ac)^(1/2) ] / 2a
x1 - x2 = (b^2 - 4ac)^(1/2) / a
So in our case,
β-α = [u^2 - 4(5)(40)]^(1/2) / 5
= (u^2 - 800)^(1/2) / 5 = 17
(u^2 - 800)^(1/2) = 85
u^2 - 800 = 7225
u^2 = 8025
u = 89.6

This last method may not be the best method, but I thought it was neat and no one else was answering so hope it makes sense

Thank you so much

 

[Shujaat Khan]

plz help me with part 2View attachment 57072

You got your drawing of a circle and a line with angle π/4, do you not get a right angle triangle ? If you do just use cos/sin π/4=x/|z|, where x is either the adjacent or opposite depending on the sketch, sorry I'm on mobile can't draw it for you maybe tommorow I do it

 

[Shujaat Khan]

pk this is a really dumb question
ive done this before
but i completely forgot how to solve this
can anyone explain me how to find the inverse of a cubic function?
ex: y = 5x^3 - 9x^2 - 2

Pretty sure no inverse for cubic because it is not one-one, but if you want to find try this.
Let y=F(x)
Find a factor of x where remainder is 0
Use long division
You got 3 factors just solve inverse like you usually do.
This is where the prob starts, theres 3factors and you cant just choose one and ignore the rest, thus inverse not possible imo

 

[RadzMau]

Can someone please help me with this M1 question??

 

[Shujaat Khan]

Can someone please help me with this M1 question??

Fairly easy im just giving the logic you work the rest
A. It says velocity same, time also is same. Equate with v=u+at and solve, i got t=1s.
Then replace t in equation s=ut +0.5at^2
S=9m

B. Distance same time same use s=ut +0.5t^2
T=0 or t=2s
Replace in v=u+at
V=4m/s

Make sure you understand the logic that Time is always same for these cases, thats how you proceed with the question.
Its a mistake to think distance is same for part A and also a mistake to think velocity is same for part B.

 

[RadzMau]

Fairly easy im just giving the logic you work the rest
A. It says velocity same, time also is same. Equate with v=u+at and solve, i got t=1s.
Then replace t in equation s=ut +0.5at^2
S=9m

B. Distance same time same use s=ut +0.5t^2
T=0 or t=2s
Replace in v=u+at
V=4m/s

Make sure you understand the logic that Time is always same for these cases, thats how you proceed with the question.
Its a mistake to think distance is same for part A and also a mistake to think velocity is same for part B.

I was a bit confused. Now I understood. Thank you very much

 

[RadzMau]

Help please? :/

 

[Shujaat Khan]

Help please? :/

When you do these type of M1 questions, always relax, take your pencil and draw the scene that's happening in the question. It's very easy when you can visualise it your brain processes much better.( You also make examiners understand your work better)

1. P.E=mgh
= 18*15000*10
= 2.7*10^6 J

2. Velocity constant.
Dx( Drivingforce*distance)= Rx(resistance*distance) - P.E
Dx= (4.8-2.7)*10^6 J
Dx= 2.1*10^6 J

3. I would advise you remember and use this formula everytime( it works even for part 2 if you can't think like physics students)
KE(initial) + PE(initial) + Dx= KE(final) + PE(final)+ Rx
so we already know ke initial and final, i'm just roughly going through the calculations.
Dx= 0.5(15000)(16^2-14^2) + (1600*2500)
Dx= 4.45*10^6 J
Remember: Common mistake people do is they find D, which is driving FORCE, NOT WORK DONE!
You might confirm answers with other people, I might have missed a piece of information

 

[musiclover gurl]

Hi, can someone can help with this question please? Actually its part (iii) which i can't do. I know its easy but i can't find the logic in it.
(June 2015 9709/32)

 

[Shujaat Khan]

Hi, can someone can help with this question please? Actually its part (iii) which i can't do. I know its easy but i can't find the logic in it.
(June 2015 9709/32)View attachment 57088

You have the equation already use t= infinity and you will see e^-t becomes 0, therefore value of x reaches a constant which i got 47.8<48
If you can't do it in exams just press e^-100000 on your calculator.

 

[musiclover gurl]

You have the equation already use t= infinity and you will see e^-t becomes 0, therefore value of x reaches a constant which i got 47.8<48
If you can't do it in exams just press e^-100000 on your calculator.

So anything to the power of infinity becomes zero, even though there is a minus before t?? Am a bit confused!

 

[qwertypoiu]

So anything to the power of infinity becomes zero, even though there is a minus before t?? Am a bit confused!

Please try thinking about it:
5^4 = 625
5^3 = 125
5^2 = 25
5^1 = 5
5^0 = 1
5^-1 = 1/5 = 0.2
5^-2 = 1/25 = 0.04
5^-3 = 1/125 = 0.008
5^-4 = 1/625 = 0.0016

We can conclude a few things from the pattern above:
1) if the power increases the final number also increases.
2) if the power decreases, the final number also decreases.
3) even if the power is negative, the final answer is positive.
4) switching the sign of the power is same as taking reciprocal

Anyhow e^inf is infinitely large. But for e^-inf, the power is infinitely negative, so it's gonna be zero. Or you could think of it as 1/e^inf, which is 1/inf , or zero

 

[Shujaat Khan]

So anything to the power of infinity becomes zero, even though there is a minus before t?? Am a bit confused!

It's not EVEN THOUGH, it's BECAUSE there is a Negative sign before the power that it it tends to zero. Qwertypoiu gave a detailed explanation about it hope it helps

 

[musiclover gurl]

Please try thinking about it:
5^4 = 625
5^3 = 125
5^2 = 25
5^1 = 5
5^0 = 1
5^-1 = 1/5 = 0.2
5^-2 = 1/25 = 0.04
5^-3 = 1/125 = 0.008
5^-4 = 1/625 = 0.0016

We can conclude a few things from the pattern above:
1) if the power increases the final number also increases.
2) if the power decreases, the final number also decreases.
3) even if the power is negative, the final answer is positive.
4) switching the sign of the power is same as taking reciprocal

Anyhow e^inf is infinitely large. But for e^-inf, the power is infinitely small, so it's gonna be zero. Or you could think of it as 1/e^inf, which is 1/inf , or zero

Thank you! Got it!

 

[musiclover gurl]

Tha

It's not EVEN THOUGH, it's BECAUSE there is a Negative sign before the power that it it tends to zero. Qwertypoiu gave a detailed explanation about it hope it helps

Thankss!