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2u Mathematics Marathon v1.0 (3 Viewers)

switchblade87

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Question 12:

d/dx 3^3x+1 dx
= (3^3x+2) * ln 3
= 3^3x+2.ln3
It seems hard to 'show' because it just is, is there any other steps to make it look like I was 'showing'?

Question 13: Find the area bounded by the curve y = x^4 + 1, the y-axis and the lines y = 1 and y = 3 in the first quadrant, correct to 2 significant figures.
 

aaaman

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Riviet said:
Here's the solution to Q3:
a=0.5, r=0.2
Sum (infinity)=a/(1-r)
=0.5/(1-0.2)
=0.5/0.8
=0.625m
.: The frog jumps a total distance of 0.625m

Q4: If the 11th term of an AP is 120 and the 4th term is 71, find the first term and common difference. Hence find the sum of the first 100 terms.

no hes not,the series starts when the frog jumps that .1
 

haboozin

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Q13

x^4 = y - 1
x = (y - 1)^1/4

I(from 1 to 3) (y - 1)^1/4 dy

= [4/5(y-1)^5/4] from 1 to 3

=4/5*2^5/4

1.9027....
= 1.9 .... 2 sig fig.


Q14.

By using the substitution u = lnx or otherwize,

Integrate : 1/(xlnx) dx
 

klaw

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switchblade87 said:
It seems hard to 'show' because it just is, is there any other steps to make it look like I was 'showing'?
Let y=33x+1
lny=(3x+1)ln3
lny/ln3=3x+1
lny/(3ln3)-1/3=x
dx/dy=1/y(3ln3)
dx/dy=1/33x+1(3ln3)
.:dy/dx=33x+2*ln3
Q14:
let u=ln x
du/dx=1/x
.: dx=x du
int [1/(xlnx)]dx=int [x/(xu) du]
=int 1/u du
=ln u+c
=ln (ln x)+c
Q15:
Differentiate sin3(2x-3)
 

Riviet

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Solution to Q15:
d/dx[sin(2x-3)]3]=3[sin(2x-3)]2.dy/dx[sin(2x-3)]
=3sin2(2x-3).2cos(2x-3)
=6sin2(2x-3)cos(2x-3)

Q16 Find, to the nearest degree, the acute angle between the curves y=4-x2 and y=4x-x2 at their point of intersection.

P.S I hope this question is 2u, or at least hsc 2u. :rolleyes:
 

klaw

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Yes I think it's 3U work
Q16:
At the intersection pt, equ 1=equ 2
.:4-x2=4x-x2
4x-4=0
x=1

y1=4-x2
y1'=-2x
y2=4x-x2
y2'=4-2x

When x=1, y1'=-2, y2'=2
tan @=|[-2-2]/[1*-4]|
@=45 degrees

Q17:\A 5m fence stands 4m from the wall of a house. A farmer wishes to reach a point A on the wall by the use of a ladder L that can reach from the ground outside the fence to the wall. Find the length of the shortest ladder that can reach from the ground outside the fence to the wall
 
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word.

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Q17 Answer
When the ladder touches the fence and wall, the length of the ladder between the wall and the fence is given by Sqrt{(A - 5)2 + 16} (pythagoras's theorem)

The triangle with the whole ladder as a side and the triangle above the fence to the wall is similar (AA)
By similar triangles,
(A - 5)/A = Sqrt{(A - 5)2 + 16}/L

L = A*Sqrt{(A - 5)2 + 16}/(A - 5)
L = A*Sqrt{A2 - 10A + 41}/(A - 5)

The length of the shortest ladder that can reach from the ground to the wall on point A touches the fence and hence is given by the eqn.
L = A*Sqrt{A2 - 10A + 41}/(A - 5)

Question 18
Find the focus of the parabola given by 4x = 8y - y2.
 

Kd14

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I'll try... yeh?
4x=8y-y^2
x=2y-y^2/4
x=(2-y/4)y

focus --->
4a=2-y/4
a=1/2 - y/16 [1]

when y=0 , a=1/2

therefore (3 triangular dots) focus (0,1/2)

o_O not sure.. not thinking well yet.. area of study side effects.
 
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Kd14

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Forgot to post a the question, so here goes.

Q19

Differentiate ln(x^3 + 1/x^2). =D
 

klaw

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Q19:
d/dx ln(x^3 + 1/x^2)=(3x^2-2x^-3)/(x^3+1/x^2)
Q20:
express cos2@ in terms of tan@
 
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Riviet

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Solution to Q20:
cos2@=cos2@-sin2@
=1/sec2@ - 1/cosec2@
=(cosec2@-sec2@)/(sec2@cosec2@)
=(1/tan2@-tan2@)/(2+1/tan2@+tan@)
=(1-tan4@)/tan2@ divided by (2tan2@+1+tan3@)/tan2@
=(1-tan4@)/(tan3@+2tan2@+1)

Q21 Show that tan(45o+x/2)=secx+tanx
 

DeanM

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word. said:
Question 18
Find the focus of the parabola given by 4x = 8y - y2.
its been awhile but i think:-
Answer:
k, heres what i think
4x = 8y - y^2
y^2 - 8y = 4x
therefore complete the square so:
y^2 - 8y + 16 = 4x + 16
(y-h)^2 = 4a(x-k) ; where h, and k = vertex
(y+4)^2 = 4(x-4)
so 4a=4
therefore focal length = 1, vertex = (-4,4)
so focus = (-5,4) OR (-3,4)
 
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DeanM

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Question 22.

find a number which,when added to each of 2,6, and 13 gives 3 numbers in geometric sequence
 

Slidey

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to prove: tan(pi/4+x/2)=secx+tanx

Using t results and the addition theorems (both 3u):
LHS=(1+tanx/2)/(1-tanx/2)=(1+tanx/2)^2/(1-tan^2(x/2))=(1+t)^2/(1-t^2)
RHS=(1+sinx)/cos=(1+2t+t^2)/(1+t^2) * (1+t^2)/(1-t^2)=(1+t)^2/(1-t^2)=LHS

Be interested in an elegant 2u way of doing it.
c=cos(x/2)
s=sin(x/2)
LHS=(1+tanx/2)/(1-tanx/2)=(cosx/2+sinx/2)/cos(x/2)*cos(x/2)/(cosx/2-sinx/2)=(c+s)/(c-s)=(c+s)^2/(c^2-s^2)=(1+2sc)/(c^2-s^2)=(1+sinx)/(cos)=1/cosx+sinx/cosx=secx+tanx

Still uses the double angle identities (3u): sin2x=2sinxcosx and cos2x=cos^2x-sin^2x
 

MarsBarz

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Solution to Q22:
x+2,x+6,x+13.....
Geo-serie so Tn+1/Tn = Tn+2/Tn+1
Hence
(x+6)/(x+2) = (x+13)/(x+6)
(x+6)(x+6) = (x+13)(x+2)
x2+12x+36 = x2+15x+26
Simplifiying,
.:. x = 10/3

Q23:
Prove that e^(ln x) = x
 

word.

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close DeanM,
but remember a parabola can't have two focusses
4x = 8y - y2
you missed the negative sign.
so y - 8y = -4x
(y - 4)2 = -4x + 16
(y - 4)2 = -4(x - 4)
focal length 1; concave left
hence the focus is (3,4)

Anyway:
Soln. Q23
by definition, if y = ax, then x = logay
consider y = lnx = logex
then x = ey = elnx

Question 24
The sum of the first n terms of an arithmetic sequence is given by the formula: Sn= 2n2 - n.
Find the formula for the nth term.
 

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