**Unformatted text preview: **10 MiniHSC4 2
y=2x-8x g§=4X'8 -20=4x-8 .'.x:-3 . 24 7
cose . Sine =-2-g. 2—5 25 7
1__6_8
=625
24
i) 7x - x2 = 0
x(7-x)= '. x=0,7 ii) |3+x|=1=> lx--3|=1
i.e. the distance between x and -3
isl x=-4,-2
1 1 -4 -3 -2 iii) 4x +9x-13=0
(x - 1)(4x + 13) = -13
..x=T,1
l) f(‘i/2+1’§)=V—6a+ﬂl—4—4+2\r
-- 1 __§ 2_3
‘1) '3‘32'3‘ 3a
1) is -x—)=1-x ii) Edd-1:6; +6x l)=20x3 -6x 2=20x3-E2
x iii) a (2x - 1)2 = 2(2x - 1)1. 2 = 4(2x - 1) i) y = 40°, co-interior angles are
supplementary, AB/lDF ii) Angle DCE = 180 - x, straight angle BCD
(180 - x) + y = 100,
exterior angle theorem i.e. 180 - x + 40 = 100 ‘. x = 120° i) Substituting : 2
LHS=2(8) - 19(8)+24=0=RHS
'. x=8isaroot In ACDE : ii) Since x = 8 is a root, x - 8 must be a
factor 2x2 - 19x + 24 = 0
=> (x-8)(2x-3)=0 - x3, 8
OR Let the other root be B
Then 8. 24:3 22—4 ,using of): :1 OR We could simply solve the
quadratic equation
Required tangent must have a gradient of 2 y=8x-x2=> §§=8-2x => 2:8-2x
=> x = 3 Ifx = 3, y = 8(3) - (3)2 = 15 Required tangent passes through (3,15) with agradient of 2:
y-15=2(x-3) => y=2x+9 sin45+cos45 = é+ﬁ = % = 1’2 i) T1=6+3(1) =9
T2=6+3(2) =12
T3=6+3(3) :15 11 ii) iii) Tn > 1000 => 6 + 3n > 1000 => 11 > 331%
'. n = 332 since 11 is a positive
integer
T332 =1002 is the Mt term > 1000
Tn = 6 + 3n 4198 = 6+3n n = 1397% . Since 11 must be an integer, 4198 cannot be a member of the sequence 2 2 2
= b +c -2bccosA 2 2 2
10 = 9 + 13 - 2(9)(13) cosA
100: 81 + 169 - 234 cosA
_1_50 '. A = 50° 8' The gradient of
this line is 1 ...

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- Spring '14