Unformatted text preview: 10 MiniHSC4 2
y=2x8x g§=4X'8 20=4x8 .'.x:3 . 24 7
cose . Sine =2g. 2—5 25 7
1__6_8
=625
24
i) 7x  x2 = 0
x(7x)= '. x=0,7 ii) 3+x=1=> lx3=1
i.e. the distance between x and 3
isl x=4,2
1 1 4 3 2 iii) 4x +9x13=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—)=1x ii) Edd1:6; +6x l)=20x3 6x 2=20x3E2
x iii) a (2x  1)2 = 2(2x  1)1. 2 = 4(2x  1) i) y = 40°, cointerior 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
=> (x8)(2x3)=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=8xx2=> §§=82x => 2:82x
=> x = 3 Ifx = 3, y = 8(3)  (3)2 = 15 Required tangent passes through (3,15) with agradient of 2:
y15=2(x3) => 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

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