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Addmaths paper 1

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The differential of y = 3 tan x - 2 was dy/dx = 3 sec^2 x, when you put x = 3pi/4, this becomes 6.
Gradient of normal = -1/6.
y + 5 = -1/6 (x-3pi/4_
6y + 30 = -x + 3pi/4 --> Eq. of normal.
Put x = 0
6y = 3pi/4 - 30
y = (3pi/4 - 30)/6 = -4.61

oh man , i did everything right , just i got y= -1
:(
 
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