11 June FP1 Model
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Transcript of 11 June FP1 Model
8/12/2019 11 June FP1 Model
http://slidepdf.com/reader/full/11-june-fp1-model 1/9
otl. f@)=3 'a3 t - 7
(a) Show that the equation f(-r) = 6 1lu' a root d between x =I and x =2'
(2)
(b) Startingwith the interval [1,2], use ntervalbisectionwiceto find an intervalof
width0.25whichcontains . (3)
8/12/2019 11 June FP1 Model
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.- - - ' , J. l
Thesolutionso thequadratic quation
z ' ) l0z +28= 0
ate z, and zr.
(c) Find z, and, ., giving your answersn the form p + i q' wherep andq are ntegers'
(d) Show,on an Argand diagram, he points representing our complexnumbers z" zt
andzr.e)
a)_ zrl= lz+lrL = rl Zt+\a '{5
(a) Find hemodulus f z'.
(b) Find, n radians,heargument f z',
(1)
givingyour answero 2 decimal laces(2)
--k;"(-|) ' -O''as3bt bu"E--"bO.B. ) 0rE*-? +TT ar5Ca)= 2'68"
) =>Z=l,rriG,_
:. Zz --5-r ,Cs Z3 " 5:iG
zr(-1t)
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(a) Given hat
A=f J' l
[ {2- t
)
(i) find A'?,
(ii) describeully thegeometricalransformationepresentedy A2
O) Given hat( 0 -1 )
"=[.-' o l
describeul1y hegeometricalransformationepresentedy B'
(c) Given hat
/k+ l 12.c=l r o I
\ "' )
where is a constant,ind thevalueofft for which he matrixC is singular'
(4)
(2)
(3)
.t 3 CcntrQ-ryrSrn
=-X-
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f ( : r ) - ;6 '?+] - : t -1 ' x+o2x
(a) Usedifferentiationo find f'(.;r).(2)
Therootaof heequation("r)=6 1;., n the nterval[0.2,O.e].
(b) Taking0.8asa first approximationo a, apply heNewton-Raphsonrocess nce of(:r) to obtaina second pproximationo a. Giveyour answero 3 decimalplaces.
(4)
trtr; .:-:---*r-.-sT-,;O'8 T-a= O:8
/2-z O. s61 crap)
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I - A , ' \
A=l ; ; l . whereaandare ons tan ls '\ b -z )
Given that the matrix A maps he point with coordinatesa, 6)
coordinates2, -8),
(a) find thevalueofa and he valueof6.
A quadrilateral hasarea30 square nits.
It ii transformednto another uadrilateralSby thematrixA'
Usingyour valuesof a andb,
(b) find theareaof quadrilateral .
onto the point with
-t$+ s =̂2 ') a;3jb -tz ---8ab=\
E-3=5
(4)
(4)
-c t 4
b-Ll(+
fivo-=513,0 =lS0 sq.r^,e u-..1-
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hatz = x + iy, find the valueofx and he valueofy such hat
z+3 i z . = -1+ l3 i
wherez. is the complex onjugate fz.(7)
,t*U)l3i(r.-.ig) = rgrisx {i'J
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, F - ,+ ,(af USe ne esultsor
Lrand
Zr ' to Snow nat
\ i l) (2r - l l ' = - n(2n+l\(2n - l )
J
for all positive ntegers .
(b) Hence how hat
. s l ^ , / ,
\ ' ( 2 " - f t t = n (an '+b \/ - J
- '1 ' \
- l
where a and b are ntegers o be found.
(6)
+1-6n -6+3
2^-\) #
:trltzJ\ tD(t-.)f zll* I
' - l) - t+n2 \
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TheparabolaC hasequation' = 48x.
Thepoint P(12f , 24t) is a general oint on C.
(a) Find theequation fthe directrixofC
(b) Showhat heequatlonfthe tangento C at P(12f,24t) is
x - t y+12 t2=0(4)
The angento C at thepoint (3, 12) meetshedirectrixof C at thepointX
(c) Find hecoordinatesfX(4)
4 '5'cra8?q ;- La:[- 2 q--\z drrectcu< f*-\z:o
'5$eD =)agfr=+e=)9g t
(2)
a.F* =L-\Z* =) ;g--t$t\Ztz- =OtL
Il t{,rzl + 3 -\z+- b. . 2trrtc.:\zb'+
*Ln a }'3 :-19 * a ,, { $= 2lc+6
gzrr)f,aluedrx A-r:tL i,3:4-*th=\8 -X
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ProvebY induction, that fot neZ',
/3 o\ ' ( t " o)ra)lo rJ
=[lr:, rr rJ
(b) f (n)=7'?"+5 is divisible Y 12'
) n:l o'i
36
_- J,,, lneP4or *\
tA',llt;ftI,..t)=tru ?t ;':. i-.""-a-. a-^r \gV: bslltf{ttl1
b\ tr,) =?FL 5 = z=
lzrl -" tru+ eqn=\
fi"*D-.Trnro.lsl. +"11is t f-.-;a
a(+ltr.-r) f
q,-.[ir)_r4r PtD-t tPsv6=\ , tfrrc- {of n=LL+\ tkJC99- {oc n=G*