2u Mathematics Marathon v1.0 (2 Viewers)

[rama_v]

Solution to question 24.

Spoiler

T_n = S_n - S_n-1
= 2n² - n - (2(n-1)² - (n-1))
= 2n² - n - (2(n² - 2n + 1) - n + 1)
= 2n² - n - (2n² - 4n + 2 - n + 1)
= 2n² - n - (2n² - 5n + 3)
= 2n² - n - 2n² + 5n - 3
= 4n - 3

Question 25.
Find the derivative of y = ln [(x² + 1)/(x² + 3)]

 

[mjhilton]

Solution to Question 26

Spoiler

y = ln[(x^2 + 1)/(x^2 + 3)
= ln(x^2 + 1) - ln(x^2 + 3)
dy/dx = 2x/(x^2 + 1) - 2x/(x^2 + 3)
= [2x(x^2 + 3) - 2x(x^2 + 1)]/[(x^2 + 3)(x^2 + 1)]
= 4x/(x^2 + 3)(x^2 + 1)

Question 27:
Solve tanx = 2 for 0 < x < 2pi. Express in radian measure to 2dp

 

[word.]

Solution to Q27-

Spoiler

tan^-12 = 1.107...,
consider tanx = 2,
x = 1.107, pi + 1.107 = 1.12, 4.25 radians (to 2 decimal places)

Question 28
Find a primitive for tan{2^-1x}.

 

[rama_v]

Solution to Q 28

Spoiler

tan (x/2)

= sin (x/2) / cos (x/2)
= - ln(cos(x/2)) + C

 

[switchblade87]

Question rama_v?

 

[Riviet]

I'll post one:
Q29 Differentiate y=sin²x from first principles.

 

[MarsBarz]

Q29 Answer:

Spoiler

y' = 2sinxcosx

Ok so I cheated haha. Why would you do it in first principle seriously...

 

[word.]

note: first principles is strictly extension-1

Spoiler

f(x) = sin²(x)
f'(x) = lim[h-->0] { [f(x + h) - f(x)]/h }
= lim[h-->0] { [sin²(x + h) - sin²x]/h }
= lim[h-->0] { [(sinxcosh + cosxsinh)² - sin²x]/h }
= lim[h-->0] { [sin²xcos²h + 2sinxcosxsinhcosh + cosx²sin²h) - sin²x]/h }
= lim[h-->0] { [sin²x(cos²h - 1) + (1/2)sin(2x)sin(2h) + cosx²sin²h]/h }
= lim[h-->0] { [(-sin²xsin²h] + cosx²sin²h + (1/2)sin(2x)sin(2h)]/h }
= lim[h-->0] { [(sin²h(cos²x - sin²x) + (1/2)sin(2x)sin(2h)]/h }
= lim[h-->0] { (sinhcos2x * sinh/h) + (sin(2x) * sin(2h)/(2h)) }
= (0 * cos2x * 1) + (sin2x * 1)
= sin2x

Question 30
If f'(x) = cot(x) + x and f(pi/2) = 0, find an expression for f(x).

 

[MarsBarz]

Q30 Answer:

Spoiler

f'(x) = cot(x) + x
f'(x) = 1/tanx + x
f'(x) = cosx/sinx + x
f(x) = ln(sinx) + (x²)/2 + C
when x = pi/2, f(x) = 0
0 = ln[sin(pi/2)] + [(pi/2)²]/2 + C
0 = ln1 + [(pi/2)²]/2 + C
C = - [(pi/2)²]/2
C = - pi²/8
.:. f(x) = ln(sinx) + (x²)/2 - pi²/8

Question 31:
Find the derivative of 2^x
Hint: This question involves logarithms.

 

[switchblade87]

note: first principles is strictly extension-1

No, its in the 2U textbook, just not used much.

 

[Riviet]

Lol, that firist principles question is a hard 2u one or 3u question.
Q31 answer:

Spoiler

d/dx(2^x)=2^x.ln2

Q32: Consider the function f(x)=10x⁵-24x³. Find any stationary points or points of inflexion and determine their nature.

 

[MarsBarz]

Q32 Answer:

When you said 'or', you actually mean 'and' don't you?
Anyway:

Spoiler

[1] Finding stat.pts:
f(x)=10x⁵-24x³
f'(x)=50x⁴-71x²=0 for stat.pts
2x²(25x²-36)=0
2x²(5x-6)(5x+6)=0
x = 0, 6/5, -6/5
f''(x)=200x³-144x

When x=0, f'(0)=0, f''(0)=0 [Possible pt of horizontal inflexion, test concavity]
{x}____ -0.1__0__0.1
{f''(x)}_14.2__0__-14.2
Concavity changes,
(0,0) is a pt of horizontal inflexion

When x=-6/5, f'(-6/5)=0, f''(-6/5)=-864/5 < 0
Concave downwards,
(-6/5,10368/625) is a maximum stationary pt

When x=6/5, f'(6/5)=0, f''(6/5)=864/5 > 0
Concave upwards,
(6/5,-10368/625) is a minimum stationary pt

[2] Finding other points of inflexion:
f''(x)=200x³-144x=0 for inflexion pts
8x(25x²-18)=0
8x(5x-√18)(5x+√18)=0
x=0, (3√2)/5, (-3√2)/5

When x=0 has already been found to be a pt of horizontal inflexion

When x=(3√2)/5, f''((3√2)/5)=0, testing concavity
{x}____[(3√2)/5]-0.1__[(3√2)/5]__[(3√2)/5]+0.1
{f''(x)}___-23.90...________0________34.09...
Concavity changes,
((3√2)/5,-10.26....) is a pt of inflexion

When x=(-3√2)/5, f''((-3√2)/5)=0, testing concavity
{x}____[(-3√2)/5]-0.1__[(3√2)/5]__[(-3√2)/5]+0.1
{f''(x)}___-34.09...________0________23.90...
Concavity changes,
((-3√2)/5,10.26....) is a pt of inflexion

Fuck finally. What a homo question seriously! That question was longer than my dick ffs.

Question 33:
Show that the equation 2lnx=ln(5+4x)
Note: This is q23 from HSC(1993)

 

[MarsBarz]

God that took me forever, don't post long questions like that again lol seriously!

 

[word.]

show that the equation what?

 

[Trev]

I reckon you have to find values of x for which the equality is true.

 

[Riviet]

When you said 'or', you actually mean 'and' don't you?

Yep, i meant and/or i guess, but one word can make a big difference in a question.

God that took me forever, don't post long questions like that again lol seriously!

Sorry mate, I forgot it was a quintic, probably a cubic would be better then.

 

[Stefano]

note: first principles is strictly extension-1

Spoiler

f(x) = sin²(x)
f'(x) = lim[h-->0] { [f(x + h) - f(x)]/h }
= lim[h-->0] { [sin²(x + h) - sin²x]/h }
= lim[h-->0] { [(sinxcosh + cosxsinh)² - sin²x]/h }
= lim[h-->0] { [sin²xcos²h + 2sinxcosxsinhcosh + cosx²sin²h) - sin²x]/h }
= lim[h-->0] { [sin²x(cos²h - 1) + (1/2)sin(2x)sin(2h) + cosx²sin²h]/h }
= lim[h-->0] { [(-sin²xsin²h] + cosx²sin²h + (1/2)sin(2x)sin(2h)]/h }
= lim[h-->0] { [(sin²h(cos²x - sin²x) + (1/2)sin(2x)sin(2h)]/h }
= lim[h-->0] { (sinhcos2x * sinh/h) + (sin(2x) * sin(2h)/(2h)) }
= (0 * cos2x * 1) + (sin2x * 1)
= sin2x

Question 30
If f'(x) = cot(x) + x and f(pi/2) = 0, find an expression for f(x).

Someone has a lot of spare time! Good work.

 

[word.]

Soln Q33. Solving 2lnx=ln(5+4x)

Spoiler

x² = 5 + 4x
x² - 4x - 5 = 0
(x - 5)(x + 1) = 0
x = 5 (note: x > 0)

Question 34
Express 0.3494949... (recurring 49) as a fraction in simplest form.

 

[Riviet]

Answer Q34:

Spoiler

0.34949... = 0.3 + 0.049 + 0.00049...
Using sum of an infinite series
=> Sum= 0.3+0.049/0.99
=(0.297+0.049)/0.99
=0.346/0.99
=346/990
=173/495

Q35: Solve the inequality 10x²+27x+5<0

 

[switchblade87]

Spoiler

10x² + 27x + 5 < 0
(Using quadratic formula, x = -2.5, -0.2.)
.: -2.5 < x < -0.2 (I think thats wrong, help!)

Q 36: Differentiate:
(a) 2 cos (3x - pi/2)
(b) 5(loge x)²