2u Mathematics Marathon v1.0 (1 Viewer)

[switchblade87]

Question 12:

Spoiler

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]

Here's the solution to Q3:

Spoiler

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

 

[switchblade87]

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

Just to clear it up, the answer is:

Spoiler

0.625m

 

[haboozin]

Q13

Spoiler

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]

It seems hard to 'show' because it just is, is there any other steps to make it look like I was 'showing'?

Spoiler

Let y=3^3x+1
lny=(3x+1)ln3
lny/ln3=3x+1
lny/(3ln3)-1/3=x
dx/dy=1/y(3ln3)
dx/dy=1/3^3x+1(3ln3)
.:dy/dx=3^3x+2*ln3

Q14:

Spoiler

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]

Solution to Q15:

Spoiler

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.

 

[klaw]

Yes I think it's 3U work
Q16:

Spoiler

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

 

[Riviet]

Q16:

Spoiler

tan @=|[-2-2]/[1*-4]|
@=45 degrees

Check the denominator.

 

[word.]

Q17 Answer

Spoiler

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]

I'll try... yeh?

Spoiler

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)

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

 

[Riviet]

Answer to 16:

Spoiler

53 degrees

 

[Kd14]

Forgot to post a the question, so here goes.

Q19

Differentiate ln(x^3 + 1/x^2). =D

 

[klaw]

Q19:

Spoiler

d/dx ln(x^3 + 1/x^2)=(3x^2-2x^-3)/(x^3+1/x^2)

Q20:
express cos2@ in terms of tan@

 

[Riviet]

Solution to Q20:

Spoiler

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(45^o+x/2)=secx+tanx

 

[DeanM]

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

its been awhile but i think:-
Answer:

Spoiler

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)

 

[DeanM]

Question 22.

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

 

[Riviet]

Q21 still needs someone to answer n_n

 

[Slidey]

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]

Solution to Q22:

Spoiler

x+2,x+6,x+13.....
Geo-serie so T_n+1/T_n = T_n+2/T_n+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.]

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

Spoiler

by definition, if y = a^x, then x = log_ay
consider y = lnx = log_ex
then x = e^y = e^lnx

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