Revision Test
This Revision Test covers the material contained in Chapters 1 to 4.The marks for each question are shown in
brackets at the end of each question.
- Factorisex^3 + 4 x^2 +x−6 using the factor theo-
rem. Hence solve the equation
x^3 + 4 x^2 +x− 6 =0(6) - Use the remainder theorem to find the remainder
when 2x^3 +x^2 − 7 x−6isdividedby
(a) (x− 2 ) (b) (x+ 1 )
Hence factorise the cubic expression (7)
- Simplify
6 x^2 + 7 x− 5
2 x− 1
by dividing out (4)
- Resolve the following into partial fractions
(a)
x− 11
x^2 −x− 2
(b)
3 −x
(x^2 + 3 )(x+ 3 )
(c)
x^3 − 6 x+ 9
x^2 +x− 2
(24)
- Evaluate, correct to 3 decimal places,
5e−^0.^982
3ln0. 0173
(2)
- Solve the following equations, each correct to 4
significant figures:
(a) lnx= 2 .40 (b) 3x−^1 = 5 x−^2
(c) 5= 8 ( 1 −e−
x
(^2) ) (10)
- (a) The pressurepat heighthabove groundlevel is
given by:p=p 0 e−khwherep 0 is the pressure
at ground level andkis a constant. Whenp 0
is 101 kilopascals and the pressure at a height
of 1500m is 100 kilopascals, determine the
value ofk.
(b) Sketch a graph ofpagainsth(pthe vertical
axis andhthe horizontal axis) for values of
height from zero to 12000m whenp 0 is 101
kilopascals.
(c) If pressurep=95kPa, ground level pressure
p 0 =101kPa, constant k= 5 × 10 −^6 , deter-
mine the height above ground level,h,in
kilometres correct to 2 decimal places. (13) - Solve the following equations:
(a) log
(
x^2 + 8
)
−log( 2 x)=log3
(b) lnx+ln(x–3)=ln6x–ln(x–2) (13)
- If θf−θi=
R
J
ln
(
U 2
U 1
)
find the value of U 2
given thatθf= 3. 5 ,θi= 2. 5 ,R= 0. 315 ,J= 0. 4 ,
U 1 = 50 (6)
- Solve, correct to 4 significant figures:
(a) 13e^2 x−^1 =7ex
(b) ln(x+ 1 )^2 =ln(x+ 1 )–ln(x+ 2 )+2 (15)