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Thread: Need help on this maths question. Pls help!

  1. #1

    Default Need help on this maths question. Pls help!

    i'm stuck on qn 1 a) and b) ii. Qn 2 i'm stuck on iii) . Pls help! gaah


  2. #2
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    Ummm...

  3. #3

    Default Answer to 1a

    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

  4. #4

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    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]

  5. #5

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    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.

  6. #6
    Senior Member The_Cheat's Avatar
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    Quote Originally Posted by tingchiyen
    Ummm...
    Haha~! What a nice response from tingchiyen!!

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

  7. #7

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    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.

  8. #8

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    Quote Originally Posted by Timber Wolf
    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.
    Hee... I jus happen to see this thread and try to help out if i can. Btw, I finished my O level liao.
    If my answer is wrong corret me ah.

  9. #9

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    gosh....

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

  10. #10

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    thks guys! help appreciated

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