2u Mathematics Marathon v1.0 (1 Viewer)

[Riviet]

That's correct switchblade87 ^_^

 

[switchblade87]

That's correct switchblade87 ^_^

Woo!
You should've answered the next one!

 

[adgala]

Spoiler

36a) dy/dx = - 6 sin(3x - pi/2)
36b) 5(lnx^2) = 5.lnx^2 .: dy/dx = 0 + (2x/x^2) * 5 = 10x/x^2

Q 37, Somebody borrows $30000 with 15% interest Per Anuum. He has to pay it all off after 20 years in equal monthly repayments. He also gets the first 6 months interest free. Find
a) the monthly fee
b) how much he pays altogether

 

[凍鴛鴦]

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

Answer:

Spoiler

(a)

y = 2 cos (3x - pi/2)

let u = 3x - pi/2
du = 3 dx

therefore, y = 2 cos(u)
dy = -2 sin(u) du
= -2 sin(3x - pi/2) * 3 dx

therefore,
dy/dx = -6 sin(3x - pi/2)

(b)

y = 5 (ln[x])^2

let u = ln[x]
du = dx / x

y = 5 u^2
dy = 10 u du
= 10(ln[x]) / x dx

therefore,

dy/dx = 10(ln[x]) / x

Q37(b): (harder 2U, or 3U question)

you can apply your knowledge of 2U exp. decay in this question:

dP / dt = kP(H-P)

Show that the equation P = (AH) / (A + e^-kHt) is a solution of the equation (A, k and H are constants)

 

[凍鴛鴦]

Oops looks like I was too slow. And adgala, there's a mistake in your second answer. Anyway, make my last question on Exponential Decay 37(b) =D

Answer to 37(a):
Q 37, Somebody borrows $30000 with 15% interest Per Anuum. He has to pay it all off after 20 years in equal monthly repayments. He also gets the first 6 months interest free. Find
a) the monthly fee
b) how much he pays altogether

Spoiler

P = 30,000
r = 15% per annum
= (15/12)% per month
n = 12 yrs = 240 months

Let A(n) be the amount owing after n months
Let M be the equal monthly repayment
Let R = 1 + 15/1200

The first 6 months are interest free, therefore:

A(1) = 30000 - M
A(2) = A(1) - M = 30000 - 2M
...
A(6) = 30000-6M

interest starts being charged at the 7th month:

A(7) = A(6) * R - M
= (30000-6M)R - M
A(8) = A(7) * R - M
= (30000-6M)R² - MR - M
= (30000-6M)R² - M (R + 1)
...
A(n) = (30000-6M)R^{n - 6} - M (1+ R + R² + ... + R^n-7)

Let Q = 1+ R + R² + ... + R^n-7

Now, Q is a geometric series,

a = 1
r = R
number of terms = n-7 + 1 = n-6

Sum(Q) = a (rⁿ - 1) / (r-1)
= (R^n-6 - 1) / (R - 1)

Therefore,

A(N) = (30000 - 6M) * R^n-6 - M( {R^n-6 - 1} / {R - 1} )

Now, since it is to be paid off in 240 months, A(240) = 0

Therefore,

A(240) = (30000-6M) * R²³⁴ - M ( {R²³⁴ - 1} / {R-1} ) = 0


30000 * R²³⁴ - 6M * R²³⁴ - M( {R²³⁴ - 1} / {R-1} ) = 0

therefore,

30000 * R²³⁴ = M { 6 * R²³⁴ + ( {R²³⁴ - 1} / {R-1} ) }

M = {30000 * R²³⁴} / { 6 * R²³⁴ + ( {R²³⁴ - 1} / {R-1} ) }

Substitute R = 1 + 15/1200, we have:

M = 367.5196965
= $367.52 (2dp)

(a) Monthly repayment = $367.52 (2dp)
(b) Total amount paid = M * 240 = $367.52 * 240 = $88,204.73 (2dp)

Please check for mistakes, I'm bound to have made some since it was so long =D

Now Question 38:

Prove that (cot[x] + cosec[x])² = (1 + cos[x]) / (1- cos[x])

Hence or otherwise, solve the equation:

(cot[2x] + cosec[2x])² = sec[2x], where 0 <= x <= 2pi

 

[Riviet]

I didn't see a question 38?
Ah well, answer to Q38:

Spoiler

LHS=(cosx/sinx+1/sinx)²
=[(cosx+1)/sinx]²
=(cosx+1)²/sin²x
=(1+cosx)(1+cosx)/(1+cosx)(1-cosx)
=(1+cosx)/(1-cosx)
=RHS
Hence,
(1+cosx)/(1-cosx)=1/cosx
cosx+cos²x=1-cosx
cos²x+2cosx-1=0
using quadratic formula
cos x= -1+sqrt2 or cosx=-1-sqrt2 for 0<2x<4pi
2x=65^o32', 294^o28', 425^o32', 654^o28'
.: x=32^o46', 147^o14', 212^o46', 327^o14'

Q39: Given that the 2 roots of a quadratic are α and ß, show that the sum of the roots of the quadratic equation ax²+bx+c=0 is given by -b/a and the product of the roots is given by c/a.

 

[凍鴛鴦]

Oops sorry, changed to Q38. BTW, theres a Q37(b)! Try it =D

 

[icycloud]

dP / dt = kP(H-P)

Show that the equation P = (AH) / (A + e^-kHt) is a solution of the equation (A, k and H are constants)

Answer to 37(b):

Spoiler

P = AH / (A + e^-kHt)
= AH / (A + 1/e^kHt)
= AH / ([Ae^kHt + 1]/e^kHt)
= AHe^kHt / (Ae^kHt + 1)

Let Q = Ae^kHt
dQ/dt = Q' = AkHe^kHt
= kHQ

So, P = QH / (Q+1)

Let u = QH, u' = Q'H
v = Q+1, v' = Q'

Using the quotient rule,
P' = (Q'H(Q+1) - QHQ') / (Q+1)^2
= (Q'H(Q+1-Q))/(Q+1)^2
= Q'H / (Q+1)^2
= kQH^2 / (Q+1)^2 ------ (1)

But, P = QH / (Q+1)
So, PQ + P = QH
P = QH - PQ
= Q(H-P)

We have,

P/Q = H-P ------ (2)
H = P(Q+1)/Q ------ (3)

Now, substituting (3) into (1),

dP/dt = P' = kQH^2 / (Q+1)^2
= kQ(P^2*(Q+1)^2/Q^2) / (Q+1)^2
= {kQ * P^2 * (Q+1)^2} / {Q^2 * (Q+1)^2}
= kP^2 / Q
= kP (P/Q) ------ substitute (2)
= kP (H-P)
#

Q39: Given that the 2 roots of a quadratic are α and ß, show that the sum of the roots of the quadratic equation ax2+bx+c=0 is given by -b/a and the product of the roots is given by c/a.

Answer to Question 39:

Spoiler

ax^2 + bx + c = 0

Divide through by a,

x^2 + (b/a)x + (c/a) = 0 ----- (1)

Since α and ß are roots,
(x-α)(x-ß)=0
x^2 - αx - ßx + αß = 0
x^2 + (-α-ß)x +αß = 0 ----- (2)

Equations (1) and (2) are congruent, therefore equating coefficients we have:

Coefficient of x:
-α-ß = b/a
-(α+ß) = b/a
α+ß = -b/a

Constant term:
αß = c/a
#

Question 40:

Given a^2 + b^2 = 7ab, express ([a+b]/3)^2 in terms of a and b.
Hence show that ln([a+b]/3) = 1/2 (ln[a] + ln).

 

[klaw]

Q40:

Spoiler

a^2 + b^2 = 7ab
(a+b)^2-2ab=7ab
.: (a+b)^2=9ab

([a+b]/3)^2=(a+b)^2/9
=ab

ln ([a+b]/3)=(1/2)ln ([a+b]/3)^2
=1/2 ln (ab)
=1/2(ln[a]+ln)

Q41:
Jeremy goes to the casino and plays a new game involving two unbiased dice in which the sum of numbers on the uppermost faces is recorded after each throw. If the sum is 7 or 11 after the first throw, he wins at once. If the sum is 2, 3, or 12 on the first throw he loses at once. If the sum is any other number (apart from 2, 3, 7, 11, or 12) on the first throw, this number is noted and becomes his "score". Jeremy then wins by throwing his "score" again or loses by throwing a sum of 7. Show that in the long run the odds for winning the game are in the casino's favour.

 

[icycloud]

Q41:
Jeremy goes to the casino and plays a new game involving two unbiased dice in which the sum of numbers on the uppermost faces is recorded after each throw. If the sum is 7 or 11 after the first throw, he wins at once. If the sum is 2, 3, or 12 on the first throw he loses at once. If the sum is any other number (apart from 2, 3, 7, 11, or 12) on the first throw, this number is noted and becomes his "score". Jeremy then wins by throwing his "score" again or loses by throwing a sum of 7. Show that in the long run the odds for winning the game are in the casino's favour.

Answer: (please check for correctness)

Spoiler

P(sum of 7) = 6/36 = 1/6
P(sum of 11) = 2/36 = 1/18

Thus, P(win on 1st throw) = 1/6 + 1/18 = 4/18 = 2/9

P(sum of 2) = 1/36
P(sum of 3) = 2/36 = 1/18
P(sum of 12) = 1/36

Thus, P(lose on 1st throw) = 1/36 + 1/18 + 1/36 = 4/36 = 1/9

1st throw
So the probability of winning on the 1st throw is 2/9, while the probability of losing is 1/9

2nd throw
P(lose on 2nd throw) = P(sum of 7) = 1/6
NB: Does not depend on the previous event (i.e. 1 * 1/6)

P(sum of 4) * P(sum of 4) = P(sum of 10) * P(sum of 10) = (3/36)^2 = (1/12)^2 or
P(sum of 5) * P(sum of 5) = P(sum of 9) * P(sum of 9) = (4/36)^2= (1/9)^2 or
P(sum of 6) * P(sum of 6) = P(sum of 8) * P(sum of 8) = (5/36)^2

Thus, P(win on 2nd throw):
2 * (1/12)^2 = 1/72 or
2 * (1/9)^2 = 2/81 or
2 * (5/36)^2 = 25/648

P(win on 2nd throw) = 1/72 + 2/81 + 25/648 = 50/648 = 0.077

You are nearly twice as likely to lose than to win on the second throw.

The 3rd throw and beyond
P(lose) = 1/6 = 0.167
P(win) = 2 * ((1/36)^2 + (2/36)^2 + (3/36)^2 + (4/36)^2 + (5/46)^2) = 0.085

Thus, you are nearly twice as likely to lose than to win in the long run for each subsequent throw after the first throw. Thus the odds are favourable for the casino.

 

[word.]

uh yeah looks right
except that 2 * (1/12)^2 = 1/72
just post a new question, the questionmaker will correct you if you're wrong

 

[icycloud]

uh yeah looks right
except that 2 * (1/12)^2 = 1/72
just post a new question, the questionmaker will correct you if you're wrong

Oops, hehe thanks for the correction.

Alright, next question:

Question 42
Integrate and evaluate in exact form using any method:

Sqrt(12-x) * Sqrt(12+x)
limits, 0 ---> 12

 

[MarsBarz]

Q41 Answer:

Spoiler

I'm not to sure how to format this answer, I'll do my best.

Let I stand for the integration from 0 to 12,
A = I (12-x)^^1/2.(12+x)^1/2
A = I [(12-x)(12+x)]^1/2
A = I (144-x²)^1/2
F(A) = {(144-x²)^3/2/-3x}
Sub integral limits,
and you get a division by 0 and I cry in shame not being
able to see where my mistake lies.

 

[icycloud]

Question 43
Find the values of k such that k(k + 3) + (k + 3)x - x² is a negative definite.

Answer to Question 43:

Spoiler

A quadratic equation is negative definite when a < 0 and Δ < 0.

Given equation is:

-x^2 + (k+3)x + k(k+3)

Thus, a = -1 < 0
So, it remains that Δ must be < 0

Δ = b^2 - 4ac
= (k+3)^2 - 4(-1)(k^2 + 3k)
= k^2 + 6k + 9 - [-4k^2 -12k]
= k^2 + 6k + 9 + 4k^2 + 12k
= 5k^2 + 18k + 9

Since Δ < 0, 5k^2 + 18k + 9 < 0
Thus,

(5k+15)(5k+3) / 5 < 0
(k + 3) (5k + 3) < 0

{-3 < k < -3/5}

Question 44
Prove that the sum S_n of an arithmetic series is 1/2 n (a+l) where n is the number of terms in the series, a is the first term and l is the last term.

 

[word.]

Spoiler

Sqrt{12-x} * Sqrt{12+x}
= Sqrt{(12-x)(12+x)}
= Sqrt{144 - x^2}

evaluating this integral is equivalent to the area of the first quadrant of a circle radius 12 units
so, Sqrt(12-x) * Sqrt(12+x) from 0 to 12 = 144pi/4 = 36pi

Question 43
Find the values of k such that k(k + 3) + (k + 3)x - x² is a negative definite.

 

[icycloud]

Sub integral limits,
and you get a division by 0 and I cry in shame not being
able to see where my mistake lies.

The standard form I (ax+b)^n = (ax+b)^(n+1) / a(n+1) + C is only for linear equations of x (i.e. ax + b). In the question, a quadratic is being raised to a power, so you cannot use that standard form. Please see "word."'s solution to the problem below, which is the only way to do the integral using 2U level mathematics. Note that in 3U maths, the integral can be solved using trigonometric substitution of x = 12sin(u).

 

[sumkid]

MATHS IS GAAAAAAY
I mean sure there woz a time i wantd 2 be a algebra manager & perhaps evn a top class bracket expander
BUT COMEOn PPL do we really need %90 of this shit they teach us i got all that shit loaded into my brain then wen it cums 2 da algebra in physics frgrt

 

[MarsBarz]

Ok I understand my mistake in Q41. Gosh I feel stupid now. I had forgetten that it had to be linear . Thanks for pointing it out.

 

[DeanM]

note : remember to keep it to 2u level, there has been afew 3u slipped in there

 

[MarsBarz]

Answer to question 44:
Prove that the sum S_n of an arithmetic series is 1/2 n (a+l) where n is the number of terms in the series, a is the first term and l is the last term.

Spoiler

S_n = a+(a+d)+(a+2d)+...+l
S_n = l+(l-d)+(l-2d)+...+a
2S_n = (a+l)+(a+l)+(a+l)+...+(a+l)
2S_n = n(a+l)
.:. S_n = (1/2).n(a+l)

Question 45:
Differentiate y=log₂x