Need help on this maths question. Pls help!
- Thread starter wind
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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:
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:
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]
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 :what:
U pple all taking the O's this year?
Anw haf to rush off for chem practical Prelims. Check back 2nite.
U pple all taking the O's this year?
Anw haf to rush off for chem practical Prelims. Check back 2nite.
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.
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.
Timber Wolf said:
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.
gosh.... :bigeyes:
i even have problems solving it. on the other hand, maybe it's just that i'm lazy to do so.
i even have problems solving it. on the other hand, maybe it's just that i'm lazy to do so.
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