4U Revising Game (1 Viewer)

[joker252]

well yes they're right but would u mind posting up a question, its no fun posting them up and not being able to solve other peoples.

shut up.

 

[duy.le]

huh?

 

[QuLiT]

oh sorry i keep forgetting, i'll find one. try this one:

integrate between pi/2 -> 0 sqrt(1+sin2x) dx

 

[vds700]

oh sorry i keep forgetting, i'll find one. try this one:

integrate between pi/2 -> 0 sqrt(1+sin2x) dx

=I[pi/2->0] sqrt(sin^2 x + cos^2 x + 2 sinxcos)dx
=I[pi/2->0]sqrt[(sinx + cosx)^2]dx
=I[pi/2->0](sinx + cosx)dx
=[-cosx + sinx]{pi/2->0}
=1 -(-1)
=2

New Q: A die is biased so that on any single roll, the probablilty of getting an even score is p, where p =/= 0.5. In 12 rolls of this die, the probablilty of getting exactly 4 even scores is 3 times the probablilty of getting exactly 3 even scores. Find the value of p.

 

[3.14159potato26]

New Q: A die is biased so that on any single roll, the probablilty of getting an even score is p, where p =/= 0.5. In 12 rolls of this die, the probablilty of getting exactly 4 even scores is 3 times the probablilty of getting exactly 3 even scores. Find the value of p.

P(X=4) = 12C4((1-p)^8)(p^4)
P(X=3) = 12C3((1-p)^9)(p^3)
Since P(X=4) = 3 * P(X=3),
495*((1-p)^8)(p^4) = 3*220*((1-p)^9)(p^3)
495p = 660(1-p)
495p+660p = 660
p = 660/1155
p = 4/7

New Question:
Prove that (a^2-b^2)(c^2-d^2) <= (ac-bd)^2

 

[duy.le]

new question:
If a>1, b>1 and c>1 such that a / b >= c / a,
show that log a /log b >= log c /log a

note its log base e

ive attached the answer to previous.

 

[3.14159potato26]

If a>1, b>1 and c>1 such that a / b >= c / a,
show that log a /log b >= log c /log a.

If a/b >= c/a,
a^2/bc >= 1 (true because a>1 and c>1)
log(a^2/bc) >= log 1 (base e)
log(a^2/bc) >= 0
Since a^2/bc = (a/b)*(a/c),
log((a/b)*(a/c)) >= 0
log(a/b) + log(a/c) >= 0
log(a/b) >= -log(a/c)
log(a/b) >= log(c/a)
a/b >= c/a.

New Question:
Given that a+b+c = 1 and a+b+c >= 3(abc)^1/3,
prove that 1/a + 1/b + 1/c >= 9.

 

[duy.le]

If a/b >= c/a,
a^2/bc >= 1 (true because a>1 and c>1)
log(a^2/bc) >= log 1 (base e)
log(a^2/bc) >= 0
Since a^2/bc = (a/b)*(a/c),
log((a/b)*(a/c)) >= 0
log(a/b) + log(a/c) >= 0
log(a/b) >= -log(a/c)
log(a/b) >= log(c/a)
a/b >= c/a.

New Question:
Given that a+b+c = 1 and a+b+c >= 3(abc)^1/3,
prove that 1/a + 1/b + 1/c >= 9.

how? say if a = 2, b=10000, c=1000000. clearly not true

 

[u-borat]

yeah that proof doens't make any sense.

 

[3.14159potato26]

First off, the proof is crap and doesn't make sense, so ignore it.

 

[3.14159potato26]

if a = 2, b=10000, c=1000000. clearly not true

if so, then the first condition a/b >= c/a doesn't fit in the first place, so that line is correct. Beyond that, its crap

 

[u-borat]

New Question:
Given that a+b+c = 1 and a+b+c >= 3(abc)^1/3,
prove that 1/a + 1/b + 1/c >= 9.

a+b+c >= 3(abc)^1/3,

substitute a+b+c=1 into it.

therefore

(abc)^1/3 <= 1/3
if you inverse this

(1/abc)^1/3 >= 3

now

1/a+1/b+1/c>= 3 (1/abc) ^1/3 (substitute a=1/a, b=1/b etc into a+b+c >= 3(abc)^1/3)

now sub in (1/abc)^1/3 >= 3

and 1/a+1/b+1/c >= 9

question:

show that for x>0, x- x^3/3 < inverse tan x < x- x^3/3 - x^5/3

btw duyle, do you have the answer to your log question?

 

[3.14159potato26]

show that for x>0, x- x^3/3 < inverse tan x < x- x^3/3 - x^5/3

If so, then x - x^3/3 < x - x^3/3 - x^5/3.
0 < -x^5/3, which cannot be true since x > 0.
Is there something wrong?

 

[tommykins]

should be a +x^5/3

 

[duy.le]

(a+b+c)/3 >=(abc)^1/3..........(1)
(1/a+1/b+1/c)/3 {} >= (1/abc)^1/3.........(2)

(1)x(2):

[(a+b+c)(1/a+1/b+1/c)]/9 >= 1

hence;

1/a+1/b+1/c >= 9 {hehe but my one is shorter than u-borat's}

get a load of this bad boy; note that it comes from one of the texts books but dont try to find it

The point P(cp,c/p), where p=/=1, is a point on the hyperbola xy=c^2, and the normal to the hyperbola at P intersects the second branch at Q. The line through P and the origin o intersects the second branch at R.

I) Show that the equation for the normal at P is;
py-c=p^3(x-cp)
II) Show that the x-coordinates of P and Q satisfy the equation x^2-c(p-[1/p^3])x-c^2/p^2=0
III)Find the coordinates of Q and deduce that angle QRP is a right angle

have fun.

 

[duy.le]

if so, then the first condition a/b >= c/a doesn't fit in the first place, so that line is correct. Beyond that, its crap

oh sorry forgot about that condition, i just did it another way and didnt recognize it. my bad. well done. oh and my way is something along those line just less confusing, use am>gm for 2 when u get logb+logc

dam and u-borat beat me in answering it while i was typing it, should of just scanned it.

 

[u-borat]

duyle your solution is missing a few lines

[(a+b+c)(1/a+1/b+1/c)]/9 >= 1

that should be divided by 3 instead of 9.

then you use the a/b+b/a >=2 formula three times and you get the answer.

 

[duy.le]

If a/b >= c/a,
a^2/bc >= 1 (true because a>1 and c>1)
log(a^2/bc) >= log 1 (base e)
log(a^2/bc) >= 0
Since a^2/bc = (a/b)*(a/c),
log((a/b)*(a/c)) >= 0
log(a/b) + log(a/c) >= 0
log(a/b) >= -log(a/c)
log(a/b) >= log(c/a)
a/b >= c/a.

New Question:
Given that a+b+c = 1 and a+b+c >= 3(abc)^1/3,
prove that 1/a + 1/b + 1/c >= 9.

lol just realised u proved the wrong thing.

heres how u do it;

Spoiler

method:
1.get in form a^2>bc
2.take logs
3.use log laws
4.use am>gm on RHS
5.sub back in
6.simplify

 

[tommykins]

The point P(cp,c/p), where p=/=1, is a point on the hyperbola xy=c^2, and the normal to the hyperbola at P intersects the second branch at Q. The line through P and the origin o intersects the second branch at R.

I) Show that the equation for the normal at P is;
py-c=p^3(x-cp)
II) Show that the x-coordinates of P and Q satisfy the equation x^2-c(p-[1/p^3])x-c^2/p^2=0
{III)Find the coordinates of Q and deduce that angle QR(P)? is a right angle}

sorry about the third part, i just realised i haven't done it, and i fail at reading my own hand writing...

so just the first two parts. yet me do it again and ill edit it later. have fun.

py-c=p³(x-cp) (1)
xy = c² (2)

using (1), multiply by x.

pxy - cx = p³3x(x-cp)
but xy = c²
pc² - cx = p³x(x-cp)
pc² -cx = p³x² - cxp^4 divide through by p³
c²/p² - cx/p³ = x² - cpx
x² - c²/p² + cx/p³ - cpx = 0
.'. x² - cx(p-1/p³) - c²/p²

 

[3.14159potato26]

show that for x>0, x- x^3/3 < inverse tan x < x- x^3/3 - x^5/3

Actual question: show that for x>0, x- x^3/3 < inverse tan x < x- x^3/3 + x^5/3.
Let f(x) = arctan(x) - x + (x^3)/3
f'(x) = 1/(1+x^2) - 1 + x^2
f'(x) = (1+(x^4-1))/(1+x^2)
f'(x) = (x^4)/(1+x^2)
Therefore, f'(x) > 0 for x > 0.
f(0) = arctan(0) - 0 + 0
f(0) = 0
Since f'(x) > 0 for x > 0,
f(x) > 0 for x > 0.
Hence,
arctan(x) - x + (x^3)/3 > 0
arctan(x) > x - (x^3)/3 --(1)
Let g(x) = x - (x^3)/3 + (x^5)/3 - arctan(x)
g'(x) = 1 - x^2 + (5/3)x^4 - 1/(1+x^2)
3g'(x) = 3 - 3x^2 + 5x^4 - 3/(1+x^2)
3g'(x) = ((3 - 3x^2 + 5x^4)(1+x^2)-3)/(1+x^2)
g'(x) = (3 - 3x^2 + 5x^4 + 3x^2 - 3x^4 + 5x^6 - 3)/3(1+x^2)
g'(x) = (5x^6+2x^4)/3(1+x^2)
Therefore, g'(x) > 0 for x > 0.
g(0) = 0 - 0 + 0 - arctan(0)
g(0) = 0
Since g'(x) > 0 for x > 0,
g(x) > 0 for x > 0.
Hence,
x - (x^3)/3 + (x^5)/3 - arctan(x) > 0
x - (x^3)/3 + (x^5)/3 > arctan(x) --(2)
By comparing (1) and (2),
x- x^3/3 < arctan(x) < x- x^3/3 + x^5/3 for x > 0