Need help on this maths question. Pls help!


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wind

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Apr 30, 2002
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#1
i'm stuck on qn 1 a) and b) ii. Qn 2 i'm stuck on iii) . Pls help! gaah :sweat:

 

Oct 16, 2003
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Hi there. U taking Olvls this year? ;p

Haven't had time to figure out the rest yet, but will try to help with 1a)
For this case i'll refer to sin^2x as sin2x and cos^2x as cos2x for convenience.

4sin2x + 2sinxcosx = 3

Using identity of sin2x + cos2x =1,

4sin2x + 2sinxcosx = 3(sin2x + cos2x)

Bring equation to left side,

sin2x + 2sinxcosx - 3cos2x = 0

Factorise the equation

(sinx + 3cosx)(sinx - cosx) = 0

so either sinx + 3cosx = 0 OR sinx - cosx = 0

sinx + 3cosx = 0 can be written as tanx + 3 = 0 (divide eqn by cosx)

sinx - cosx = 0 => tanx -1 = 0 (divide by cosx)

So u end up with tanx = 1 OR -3
Find the basic angles, B.A = 45deg or 71.565 deg

When tanx = 1, tanx is positive so it's in the 1st or 3rd quadrant so x = 45deg or (180+45) = 225deg

When tanx = -3, tanx is negative so it's in the 2nd or 4th quadrant so x = (180-71.565) = 108.434 deg or (360- 71.565) = 288.435 deg

Put ur answers in ascending order. x= 45.0deg, 108.4deg, 225.0deg, 288.4deg

Whew :sweatsm:
 

sweat100

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#4
For question2,

Using completing the square
i)f(x)=2(x^2-4x)+15
=2[x^2-2x+(2^2)-(2^2)]+15
=2[(x-2)^2 -4]+15
=2(x-2)^2+7


ii) 1/f(x)=1/[2(x-2)^2+7]

Now observe that when the denominator is bigger, mean that 1/f(x) is smaller, so the smallest the denomintor can go is when x=2, then u have 1/7.

iii) let y=2(x-2)^2+7

making x the subject of the formula, u have

x=2-sqroot[(y-7)/2] or x=2+sqroot[(y-7)/2]

Since x<=k, then x<=2, there for u reject x=2+sqroot[(y-7)/2]

therefore answer is x=2-sqroot[(y-7)/2]
===>y=2-sqroot[(x-7)/2]
 

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Wahh... sweat u beat me to it :what:
U pple all taking the O's this year?

Anw haf to rush off for chem practical Prelims. Check back 2nite.
 

The_Cheat

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#6
tingchiyen said:
Ummm... :faint:
Haha~! What a nice response from tingchiyen!! :bsmilie:

Anyway, I'd had the exact same response when I saw the question too! :faint: Mathematics is beyond me... :confused:
 

sweat100

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#7
as for question1 part 2,

Remeber the double angle formula for cos(2x)=1-2sin^2(x)=2cos^2(x)-1

Now, cos(400)=2/k^2 - 1
and, cos (400)=1-2sin^2(200)
making sin(200) the subject of the formula, u have sin(200)=+-sqroot[(k^2-1)/k^2)]

since at sin (200) is negative, u reject the positive and u get sin(200)= -sqroot[(k^2-1)/k^2)]


then for tan(110)=tan(200-90)
=sin(200-90)/cos(200-90)

uing addition formula, and the values of sin(200) and cos(200), u should be able to get the answer. :)
 

sweat100

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#8
Timber Wolf said:
Wahh... sweat u beat me to it :what:
U pple all taking the O's this year?

Anw haf to rush off for chem practical Prelims. Check back 2nite.
Hee... I jus happen to see this thread and try to help out if i can. Btw, I finished my O level liao. :embrass:
If my answer is wrong corret me ah.
 

mervlam

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#9
gosh.... :bigeyes:

i even have problems solving it. on the other hand, maybe it's just that i'm lazy to do so. :p
 

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