i'm stuck on qn 1 a) and b) ii. Qn 2 i'm stuck on iii) . Pls help! gaah
i'm stuck on qn 1 a) and b) ii. Qn 2 i'm stuck on iii) . Pls help! gaah
Hi there. U taking Olvls this year?
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
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]
Wahh... sweat u beat me to it
U pple all taking the O's this year?
Anw haf to rush off for chem practical Prelims. Check back 2nite.
Haha~! What a nice response from tingchiyen!!Originally Posted by tingchiyen
Anyway, I'd had the exact same response when I saw the question too! Mathematics is beyond me...
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.
Hee... I jus happen to see this thread and try to help out if i can. Btw, I finished my O level liao.Originally Posted by Timber Wolf
If my answer is wrong corret me ah.
gosh....
i even have problems solving it. on the other hand, maybe it's just that i'm lazy to do so.
thks guys! help appreciated
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