1.
The curve C has equation
y = (2x-3) 5
· The point P lies on C and has coordinates (w, - 32).
Find
(a) the value ofw,
(2)
(b) the equation of the tangent to Cat the point P in the form y = mx + c, where m and
c are constants.
(5)
w z;..' .
.J...
2:: ..•• -
~~;:.,.,..
boA-....
..
. ..
\
=s (-2)A( X 2. = \bD
b lan~
g(x) = e x -l + x-6
2.
(a) Show that the equation g(x) = 0 can be written as
X
= ln(6 -
X)
+ 1,
X
<6
(2)
The root of g(x) = 0 is a.
The iterative formula
XII+ I
= ln( 6 -
X II )
+ 1'
Xo
=2
is used to find an approximate value for a.
(b) Calculate the values of x1 , x 2 and x3 to 4 decimal places.
(3)
(c) By choosing a suitable interval, show that a= 2.307 correct to 3 decimal places.
(3)
_ _;_) _ X-\ ~ \~(b--x.- _~
"'LX:: \~-x.):±\ ~
__ L \(\
_b)
(A
~~~0
b- JC.. __LO._:..~
) X-_5_b _] _ _
'X.e> ~ 2. ----~)-f.(_2_:_3-QbS ~ - O·DOC>18
_____:t_, "; L_2:>_B_6~_ - - ~f1-(2-=-301-j)_~Q·Q_CQ.41
2~ = ~·i,<6~'-----------___'X.~-:; 1· 3 \1.5 ___~--"-'
-·. b~_SJ ')~k,...~e__ ·
-----~\l-L.e_, oc'\- h~
---',-Ax..~~ 1·30-bSS~b-..
'l· 30~ a-'cJ-'2 ·303:$
--------=----.
----
-- : -.. cJ.. ~b~OT-{-~p')
--------------------- --~~~~~-
3.
y
y = f(x)
0
X
Figure 1
Figure 1 shows part of the curve with equation y
= f(x), x E
IRL
The curve passes through the points Q(O, 2) and P(-3, 0) as shown.
(a) Find the value of ff(-3) .
(2)
On separate diagrams, sketch the curve with equation
(2)
(c) y=f(lxl) - 2,
(2)
(d) y
= 2f (~ X).
(3)
Indicate clearly on each sketch the coordinates of the points at which the curve crosses or
meets the axes.
-
_ c.\)_£_f._(-3) = fJ_o_,J- __,.,2.,____ _ _ _ _ __
c)
f{\~\)- 2-
d.)
4.
(a) Express 6 cos()+ 8 sin() in the form R cos(() - a.), where R > 0 and 0 < a.
< ~.
Give the value of a. to 3 decimal places.
(4)
(b)
p
4
(()) = 12 + 6 cos()
+ 8 sin() '
Calculate
(i) the maximum value of p(()),
(ii) the value of() at which the maximum occurs.
(4)
=-(_C~(9-~D
(__CJ& Coso<. +~S~QS-s, d...
~b~ :LiS~ --
RS ~~~ ~-s7~ArJ....:--i
b)
~-COs~ ~.b
__ _
R --:--~ ~-\:£~
~ {2._-= l0
e-(&) ~
4-
0~--
9 t:A :;
_ __
__
I D Cos
e -0 ·~2~-)
-~~e~) =--4 -
~---'2.+ (JOLai&-0·~)) _
-----
_.,,)
'
-- -
\2.-{\0)
:.~ ~ 2. - '2..
~--
~~--~~~---·:-~~~n
--=
-~l---~--,tr~---'(r---0 ·')~..~~ - r----- - -
:. e '; +·o:r
(i) Differentiate with respect to x
5.
(b) y=(x+sin2xl
(6)
= cot y,
Given that x
(ii) show that dxdy
~
_b)
l+ x
----- ---
~ =- JC~ \[:;.
--r
--
=~
(5)
\C\ 'L:x__
-:)~ = 2~~
:-v-z.\1\l.'X.+ ~'!.
~~-- f -4----~~
u =~ "=-- - cAV)- =3 ~-+S~l::"l-)'2- x- ( \ +~as2:x..:}_____ --
---
---
-- ~~
'\\} _ :x_ :::
-~
-
_C()l:_~ -=---~
-------------~0~
~---
-
6.
(i) Without using a calculator, find the exact value of
(sin 22.5° + cos 22.5°) 2
You must show each stage of your working.
(5)
(ii) (a) Show that cos 28 + sin 8 = 1 may be written in the form
k sin2 8- sin 8 = 0, stating the value of k.
(2)
(b) Hence solve, for 0 :::;; 8 < 360°, the equation
cos 28 + sin 8 = 1
(4)
__·,) - s ' n::_u._.s _±___2___S_, o.ll___:S_Cos 2.2~ s
=
Stn'2.1.2·S +(~22·S) + Sn'V~~s
:::
\ +
+ ( o5 '2. 2 '2."---"'·.S..,____
\ri_
Collft:-T_S,cij_?_\_ _ _ _ _ _ _ __
\\)
~)
/_
~StY\&.;? c..) 2s,"?fr- s,~a =O_
I
- - - -- - - -- -- - - - - - - - - - - - - - - - - -
X- 15,¥"1
- ::-:- t(:: ~
_6_~$ td:l -l~G- _- _,_0"'------- - -- - ____,S,"'G = o =) 9; 0 1 \KQ_ _ _ _ __
---"''-"
# --
- . _ ~,oe~
i
~flo;
y),,so
h(x)=-2- + _
4__
18
x+2 x 2 +5 (x 2 +5)(x+2)'
7.
x~O
(a) Show that h(x) =--f.£x +5
(4)
(b) Hence, or otherwise, find h'(x) in its simplest form.
(3)
y
y = h(x)
0
X
Figure 2
Figure 2 shows a graph of the curve with equation y = h(x).
(c) Calculate the range of h(x) .
(5)
-------
=-)
_b)
lA..~'2..-x...__ y ~ ~-z-t:-_S_
\A I :; k
\{ f
::.
£..X...
_b~(~ ~ '21...'2+\0-~x.,"2.
(:L'2.+-J )..,_
-b-~ ~ -L-y:X:.
'---'2-.+r-\\--1Q
H-- - - -
------------'-.
cxz-+-s'r
________________
3f ~~ ~~l~--~o
-----
j:. ~JS
-
=5-t5
- s
8.
The value of Bob's car can be calculated from the formula
V = 17000e-0.251 + 2000e-O.st + 500
where Vis the value of the car in pounds ( £) and t is the age in years.
(a) Find the value of the car when t = 0
(1)
(b) Calculate the exact value oft when V
= 9500
(4)
(c) Find the rate at which the value of the car is decreasing at the instant when t = 8.
Give your answer in pounds per year to the nearest pound.
·
(4)
l=t<X)O<i* 0 ·'2Sf""±2£X:De: 0 ·St ±SOD ::-9SC5D
b)
~ '20Db@:-o·Br)z +\~o&Sf") -L)O{)Q :;Q
:. 2tx:nx?- + \-=tOCDL -C) crt) =0 x:e-o·2Sl=
~
27-'2. +\3-'X..-9 :: 0