3u Mathematics Marathon V 1.0 (1 Viewer)

[rama_v]

Solution to Question 29

Spoiler

(1+x)ⁿ = C₀ + C₁x + C₂x² + ...+ C_nxⁿ
multiply through by x
x(1+x)ⁿ = C₀x + C₁x² + C₂x³ + ...+ C_nxⁿ⁺¹

Differentiate with respect to x
(1)(1+x)ⁿ + (x)(n(1+x)^n-1) = C₀ + 2C₁x + 3C₂x² + ...+ (n+1)C_nxⁿ

put x = 1
LHS = 2ⁿ + n(2^n-1)
= 2^n-1(n + 2)
RHS = nCo + 2C1 + 3C2 +...+(n+1)Cn

.: 2^n-1(n+2)= C₀ + 2C₁ + 3C₂ +...+(n+2)C_n
as required

Next Question:
Using the substitution t = tan@ , or otherwise, show:

(tan2@ - tan@) / (tan2@ + cot@) = tan²@

 

[haboozin]

Next Question:
Using the substitution t = tan@ , or otherwise, show:

(tan2@ - tan@) / (tan2@ + cot@) = tan²@

Spoiler

donno why u would use that sub
so i'll go with or otherwize

[{2sin@cos^2@ - sin@(cos^2@ - sin^2@)}/{(cos^2@ - sin^2@)cos@}]/[{2sin^2@cos@ + cos@(cos^2@ - sin^2@)}/{(cos^2@ - sin^2@)sin@}]

=

{2sin^2@cos^2@ - sin^2@(cos^2@ - sin^2@)}/{2sin^2@cos^2@ + cos^2{cos^2@ - sin^2@)}

=sin^2@/cos^2@* (cos^2@ + sin^2@)/(cos^2@ + sin^2)
= sin^2@/cos^2@ * 1
= tan^2@

A 6 is scored in cricket game when the ball is hit over the boundary fence on the full. A ball is hit from 0 with velocity of 32 ms^-1 at an angle @ to the horizontal and towards 1m high boundary fence 100 meters away.

i . derive the equations of motion for the ball in flight. g = 10ms^-2.

ii, show that the ball just clears the boundary fence when 5000tan^@ + 102400tan@ + 51024 = 0
iii, in what range must @ lie for a 6 to be scored.
iv, if during the flight of the ball, its velocity is reduced by piercing an extremly thin board, show by a sketch how its path is altered. Without further calculation, discuss qualitatively the effect of air resistance on your answer in iii...

 

[Stefano]

haboozin, you'd use that sub to make things NEATER, not easier. Just that.

 

[PiGMAN]

don't you mean 50000 in pt ii haboozin

 

[EvilDude]

Q30, projectile / cricket:

I think he meant a 50000 Pig.

Spoiler

(i)

At instant of projection:
V_x = 32cos@
V_y = 32sin@
(From triangle)

a_x = 0
v_x = ∫0 dt = c
when t=0, v_x = 32cos@ therefore c = 32cos@
v_x = 32cos@
x = ∫32cos@ dt = 32tcos@ + c
but when t=0, x=0, therefore c = 0
x = 32tcos@

a_y = -10
v_y = ∫-10 dt = -10t + c
when t=0, v_y = 32sin@ therefore c = 32sin@
v_y = -10t + vsin@
y = ∫-10t + vsin@ dt = -10/2 * t^2 + 32tsin@ + c
when t=0, y = 0 therefore c=0
y = -5 * t^2 + 32tsin@

(ii)

x = 32tcos@, therefore t = x / (32cos@)
Subbing in t = x / (32cos@) into y :

Y = -5x^2 / (32cos@)^2 + 32sin@ * x / (32cos@)
= -5x^2/32^2 * sec^2@ + xtan@
= -5/32^2 * x^2 (tan^2@ + 1) + xtan@
Clears when Y = 1 and x = 100, therefore
1 = -50000 / 32^2 * (tan^2@ + 1) + 100tan@
Multiply both sides by 32^2 (32^2 = 1024)
1024 = -50000tan^2@ - 50000 + 102400tan@
50000tan^2@ - 102400tan@ + 51024 = 0 as required (by rearranging)

(iii) solve for tan@
Say x = tan@
50000x^2 – 102400x + 51024 = 0

Use quadratic formula and the two values are:
X = 0.85638 or x = 0 1.19162

Since tan@ = x, @ =
40.6° or 50°

Therefore the range is in between 40.6° -> 50°

(iv)

Not a parabola. I don’t like drawing (especially on computers!)

Question 31

Find all real x such that

|4x - 1| > 2 sqroot[x( 1-x)]

 

[100percent]

Question 31

Find all real x such that

|4x - 1| > 2 sqroot[x( 1-x)]

Spoiler

|4x - 1| > 2 sqroot[x( 1-x)]
(4x-1)²>4(x-x²)
16x²-8x+1>4x-4x²
20x²-12x+1>0
(10x-1)(2x-1)>0
x<1/10, x>1/2

question 32, find
lim_x->0 (1-cos2x)/x²

 

[Stefano]

<b> Question 31 </b>

Spoiler

(4x-1)² > 4[x-x²]

Thus, 20x²-12x+1>0
(10x-1)(2x-1)>0
Hence; 0<=x<1/10 OR x>1/2 (where x=>0, x=>1 are conditions)
*Edit: Thanks Rama. I realised that before I started working on the problem, but it takes so long to type up I forgot! lol...

EDIT: 100percent beat me to it.

<b> Question 32 </b>

Spoiler

2

<b> Question 33 </b>
Prove: cotan(@) + cotan ($) = 1-cotan(@).cotan($)

 

[100percent]

you forgot the square root

 

[rama_v]

Sorry guys, both ur answers are incorrect

You forgot the condition that x(1-x)=> 0
Then it becomes right

The final solution is 0<=x<1/10 or 1/2 < x =<1

 

[Stefano]

you forgot the square root

What square root? I can't find a mistake anywhere...

 

[Stefano]

The standing question is:

<b>Question 33</b>
Prove: cotan(@) + cotan ($) = 1-cotan(@).cotan($)

 

[EvilDude]

Is the question a solve for @ and $?

I tried to prove LHS = RHS, but it seemed wrong to me.
I tried to substitue @ = pi/4 and $ = pi/4 and LHS = 2, RHS = 0. So is it a solve for @ and $?

 

[100percent]

hey, i tried and i end up with sin(@+$)= -cos(@+$) and i don't think LHS=RHS

 

[jake2.0]

ohwell the thread goes on

Q34
using the graph y=cosx 0=< x =<2pi, or otherwise, find those values of x satisfying 0=< x =<2pi for which the geometrical series

1 + 2cosx + 4cos²x + 8cos³x + ...

has a limiting sum

 

[Stefano]

Is the question a solve for @ and $?

I tried to prove LHS = RHS, but it seemed wrong to me.
I tried to substitue @ = pi/4 and $ = pi/4 and LHS = 2, RHS = 0. So is it a solve for @ and $?

sorry guys, it was a mistake. I read the question incorrectly.

Continue with Q34. My apologies.

 

[Estel]

-1<2cosx<1
-1/2 < cosx < 1/2
pi/3 < x < 2pi/3 4pi/3 < x < 5pi/3

Q35
Asina = xy
Acosa = xz^2

Prove that y^2cos^2a + z^4sin^2a = yz^2sin2a
i) by considering the value of A
ii) by considering the values of sina and cosa.

 

[Slidey]

Spoiler

y=Asin(a)/x -> y^2cos^2(a)=A^2sin^2(a)cos^2(a)/x^2
z^2=Acos(a)/x -> z^4sin^2(a)=A^2cos^2(a)sin^2(a)/x^2

Adding these together:

LHS = 2A^2sin^2(a)cos^2(a)/x^2

y=Asin(a)/x -> ycos(a)=Asin(a)cos(a)/x
z^2=Acos(a)/x -> z^2sin(a)=Asin(a)cos(a)/x

Multiply the above together:

RHS/2 = yz^2sin(2a)/2 = A^2sin^2(a)cos^2(a)/x^2 = LHS/2

.'. LHS=RHS.

I don't quite get what you mean by "considering the value of", sorry, Estel.

Next question:

Find the derivative of xe^[x^2] and hence show that:
Integral 2x^2.e^[x^2] dx = xe^[x^2] - Integral e^[x^2] dx

Hence find: Integral (2x^2+1)e^[x^2] dx

 

[who_loves_maths]

mmm... just to break the stagnancy of this thread...

Spoiler

Question 36 (Slide_Rule's)

d[x.e^[x^2]]/dx = e^[x^2] + 2x^2.e^[x^2] = (2x^2 + 1).e^[x^2]

ie. Int[(2x^2 + 1).e^[x^2] dx] = x.e^[x^2] + C

also, Int[(2x^2 + 1).e^[x^2] dx] = Int[2x^2.e^[x^2] dx] + Int[e^[x^2] dx]

ie. Int[2x^2.e^[x^2] dx] = x.e^[x^2] - Int[e^[x^2] dx]

note: constant "C" is 'absorbed' into the integrals in the second result.

here's new question below...

Question 37:

Given that when applying the iterative method of Newton's to find a zero of a function f(x), the output values of successive iterations follow a linear pattern, irrespective of the starting input value, given by:
y = mx + b ; where 'y' is the output value, and 'x' is the input value. and 'm' & 'b' are any reals.

Find:

(a) ONE such function f(x) where m = 1 and b is non-zero.

(b) ONE such function f(x) where b = 0 and m is non-zero.

 

[richz]

mmm... just to break the stagnancy of this thread...

Spoiler

Question 36 (Slide_Rule's)

d[x.e^[x^2]]/dx = e^[x^2] + 2x^2.e^[x^2] = (2x^2 + 1).e^[x^2]

ie. Int[(2x^2 + 1).e^[x^2] dx] = x.e^[x^2] + C

also, Int[(2x^2 + 1).e^[x^2] dx] = Int[2x^2.e^[x^2] dx] + Int[e^[x^2] dx]

ie. Int[2x^2.e^[x^2] dx] = x.e^[x^2] - Int[e^[x^2] dx]

note: constant "C" is 'absorbed' into the integrals in the second result.

here's new question below...

Question 37:

Given that when applying the iterative method of Newton's to find a zero of a function f(x), the output values of successive iterations follow a linear pattern, irrespective of the starting input value, given by:
y = mx + b ; where 'y' is the output value, and 'x' is the input value. and 'm' & 'b' are any reals.

Find:

(a) ONE such function f(x) where m = 1 and b is non-zero.

(b) ONE such function f(x) where b = 0 and m is non-zero.

Spoiler

lol, tricky, i'll have a go, i have no idea if this is rite.
let f(x)=mx+b
a) f'(x) = m
since m = 1 .`. f'(x) = 1
f(x)=x+b
so x₁=-b
f(x)=-b

b) f'(x)=m
f(x)=mx
so, x₂= x-(mx/m)
= 0
so f(x)=0

Q 38:

a) Find the Volume if the solid formed if the curve y=cos^-1x is rotated about the x-axis between x=0 and x=1 (use simpson's rule with 3 function values).

b) The Volume of an expanding balloon is increasing by a constant rate of 15cm³s^-1. Find the rate of increase in its surface area when the baloon's is 5cm

 

[who_loves_maths]

Answer to Question 38:

Spoiler

Posed originally by xrtzx

Part (a):

limits of integration at x = 0, x = 1, middle ordinate is: x = 0.5

ie. Area = pi*DInt[(ArcCos(x))^2 dx] = pi*[(1/6)((ArcCos1)^2 + 4(ArcCos0.5)^2 + (ArcCos0)^2)]
= (pi/6)[(0)^2 + 4(pi/3)^2 + (pi/2)^2] = (pi^3/6)(4/9 + 1/4) = (pi^3/6)(25/36)
= (25pi^3)/216 units^3

Part (b):

assuming the balloon is spherical (which was not mentioned in question); let its radius = r

ie. Volume of balloon = (4/3)pi*r^3 ; dV/dr = 4pi.r^2 = surface area of balloon.
ie. d(SA)/dr = 8pi.r
we know dV/dt = 15 when r = 5; but, dV/dt = dV/dr * dr/dt
---> 15 = 4pi.(5)^2 * dr/dt ---> dr/dt = 3/(20pi)

and, d(SA)/dt = d(SA)/dr * dr/dt = 8pi.(5) * 3/(20pi) = 6
therefore, surface area of balloon is increasing at 6 cm^2/s

Next question... will be Question 37.

xrtzx's solutions to both parts of the question were, unfortunately, incorrect... xrtzx, do not let the function f(x) = mx + b .
the formula y = mx + b is a function that describes the relationship between successive output values of Newton's method on f(x). ie. 'y' is 'x(2)' and is output value, while 'x' denotes 'x(1)' and is input value.

Here's the question again for ppl:

Question 37:

Given that when applying the iterative method of Newton's to find a zero of a function f(x), the output values of successive iterations follow a linear pattern, irrespective of the starting input value, given by:
y = mx + b ; where 'y' is the output value, and 'x' is the input value. and 'm' & 'b' are any reals.

Find:

(a) ONE such function f(x) where m = 1 and b is non-zero.

(b) ONE such function f(x) where b = 0 and m is non-zero.

[HINT: f(x) need not be a linear function.]