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Published by artsculturebiro, 2022-08-07 20:35:36

15a Differentiation and integration

15a Differentiation and integration

AlvinAADccaiadfdefAememlryvyeinAnAltAIlvivicGainantdCiAAeomcSncaaydEdaeAemnmlPdyvyiIanAnlsAtvUeictnagndYrietamet1yia5orna

0606 Additional Mathematics Unit 15a Differentiation and integration

2
Mathematical Formulae

cademyQuadratic Equation 1.  ALGEBRA
in A AlvinFor the equation ax2 + bx + c = 0, x = −b b2 − 4ac

2a
emy( ) Alv( ) cadem( )y cademyBinomial Theorem
cad ( ) in A in Awhere n is a positive integer and(a + b)n = an +nan–1b+nan–2b2+…+nan–rbr + … + bn,
1 2 r

n n!
r – r)!r!
AlvinAAcadeAmlyvinAlAvcademy AlvIdentities = (n

2.  TRIGONOMETRY

sin2 A + cos2 A = 1
sec2 A = 1 + tan2 A
cosec2 A = 1 + cot2 A

Formulae for ∆ABC

a = b = c
sin A sin B sin C

a2 = b2 + c2 – 2bc cos A

∆ = 1 bc sin A
2

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0606 Additional Mathematics Unit 15a Differentiation and integration

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5  (i)  Given that y = e x2 , find dy . [2]
dx [2]
[2]
  ((iAiiii))   lHUvesiencnyeoeuAvrAaalncuscawateaedryd0t2oexAepmeaxmr2ldt x(yvi.y)itonAfAinldAlvyvicxeinaxn2ddxA.AemccaayddeAemmlyvyinAlAvicnademy


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0606 Additional Mathematics Unit 15a Differentiation and integration

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7  A curve is such that dy = 4x + ^x 1 for x 2 0. The curve passes through the point KJ1 , 5NO.
dx + 1h2 L2 6P

  (i)  Find the equation of the curve.AlvinAAccaaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy  (ii)  Find the equation of the normal to the curve at the point where x = 1.[4]
[4]

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0606 Additional Mathematics Unit 15a Differentiation and integration

0606/21/M/J/14 8

7  Given that a curve has equation y = 1 + 2 x , where x 2 0, find
x
dy
  (i)  dx , [2]
in AcademyAlvin  [2]
cademy Alvin Academyin Academy  Hence, or otherwise, find(ii) d2y. [4]
AlvinAAcadeAmlyvinAlAvcademy Alv  (iii)  the coordinates and nature of the stationary point of the curve.dx2

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0606 Additional Mathematics Unit 15a Differentiation and integration

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10  Find dy when
dx

  ((Aiii))   lvyy ==inclot+asAnl2nAxxxcs.icnaLKJa3xdNPOd,eAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy [4]
  [4]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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4  The region enclosed by the curve y = 2 sin 3x, the x-axis and the line x = a , where

  0 1 a 1 1 radian, lies entirely above the x-axis. Given that the area of this region is 1 square unit,
3
AlvinAAccaaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademyfind the value of a.
[6]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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9  A solid circular cylinder has a base radius of r cm and a volume of 4000 cm3.

  (i)  Show that the total surface area, A cm2, of the cylinder is given by A = 8000 + 2rr2. [3]
r
emy Alvin AccaaddeemmyyAlvicnademy  (ii)  Given that r can vary, find the minimum total surface area of the cylinder, justifying that this area
AlvinAAccaaddeAmlyvinAlAvicnadAemy Alvin Ais a minimum. [6]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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7  Differentiate with respect to x

  ((i(Aiiiii)))    llx1vn4s+^eii2n3nxx+,xcA.oAsxchc,aaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy [2]
  [2]

  [3]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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9  A curve is such that dy = ^2x + 1 The curve passes through the point (4, 10).
dx
1h2 .

  ((Aiii))   lFFviinniddntyheAydAexqcuacanatdiaodhndeoenfcAemethmeelvcyvaulyurivanetAe.Ayl0A1lv.5vyicdinxan. dAAemccaayddeAemmlyvyinAlAvicnademy [4]
  [5]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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12  A curve has equation y = x3 - 9x2 + 24x.

  (i)  Find the set of values of x for which dy H 0. [4]
lvin AcademyyAAllvviinnAAccaayddeAemmlyvyinAlAvicnademy  The normal to the curve at the point on the curve where x = 3 cuts the y-axis at the point P.dx[5]
A AcadeAmlvin Academ  (ii)  Find the equation of the normal and the coordinates of P.

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0606 Additional Mathematics Unit 15a Differentiation and integration

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4 The diagram shows a thin square sheet of metal measuring 24 cm by 24 cm. A square of side x cm is

cut off from each corner. The remainder is then folded to form an open box, x cm deep, whose square

ybase is shown shaded in the diagram.
Academ lvin24cm24 cm
Alvin emy A emyxcm
emy cad cadxcm [2]
AlvinAAccaaddeAmlyvinAlAvicnadAemy Alvin A(i) Show that the volume, Vcm3, of the box is given by V = 4x3 - 96x2 + 576x.
(ii) Given that x can vary, find the maximum volume of the box. [4]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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6 Find the equation of the normal to the curve y = x (x2 - 12) 1 at the point on the curve
3

where x = 2. [6]

AlvinAAccaaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy

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0606 Additional Mathematics Unit 15a Differentiation and integration

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11 The diagram shows the graph of y = cos 3x + 3 sin 3x, which crosses the x-axis at A and has a
maximum point at B.

y y = cos 3x + 3 sin 3x

emyB A
lvin Acad y Alvin yO x
demy A Academ Academ(i) Find the x-coordinate of A.
AlvinAAccaadeAmlyvinAlAvicnademy Alvin(ii) [3]
Find dy and hence find the x-coordinate of B. [4]
dx

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0606 Additional Mathematics Unit 15a Differentiation and integration

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11 (iii) Showing all your working, find the area of the shaded region bounded by the curve, the x-axis

and the line through B parallel to the y-axis. [5]

AlvinAAccaaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy

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0606 Additional Mathematics Unit 15a Differentiation and integration

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8  A curve is such that dy = 6x2 - 8x + 3.
dx

  (i)  Show that the curve has no stationary points. [2]
[2]
emy Alvin AccaaddeemmyyAlvicnademy  Given that the curve passes through the point P(2,10), [4]
d A A  (ii)  find the equation of the tangent to the curve at the point P,
AlvinAAccaadeAmlyvinAlAvicnademy Alvin  (iii)  find the equation of the curve.

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0606 Additional Mathematics Unit 15a Differentiation and integration

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10  (i)  Given that y = 2x , show that dy = k , where k is a constant to be found. [5]
x2 + 21 dx ^x2 + 21h3

AlvinAAccaaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy

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0606 Additional Mathematics Unit 15a Differentiation and integration

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1 0 (Aii)  lHveincne fAinAdceddccaa^dxd2e6+Aem2m1lh3yvdyxinaAnAdleAlvvavilcuinaantedddecA21Ae0 mc^caxa2yd6+d2e1Aemh3mdlxyv.yinAlAvicnademy [3]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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1  Find the coordinates of the stationary point on the curve y = x2 + 16 .  [4]
x [3]

2  (Aa) lOvnitnheOaAy642xAesccbaealodwde, Asemkmetlc4yvh5y°tihenAcAulrAvlvevicinayn9=0d°A3Aecmcosca2axyd-de1Aem1f3mo5lr°yv0y°iGnAx GlAv181i0c8°n0a.°dexmy

–2

–4

–6

  (b)  (i)  State the amplitude of 1 - 4 sin 2x. [1]

    (ii)  State the period of 5 tan 3x + 1. [1]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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3  A curve is such that dy = 2 for x 2-3. The curve passes through the point (6, 10).
dx x+3

  (i)  Find the equation of the curve. [4]
[1]
AlvinAAccaaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy  (ii)  Find the x-coordinate of the point on the curve where y = 6.

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0606 Additional Mathematics Unit 15a Differentiation and integration

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5  (i)  Find the equation of the tangent to the curve y = x3 - ln x at the point on the curve [4]
where x = 1. [2]

  (Aii)  lSvhoiwnthAaAt tchciasatdandgeeAemntmbliyvseycitsnAtAhelAllvinviecinjaonindiAnAeg mtchceaapydodinetAsem (m-l2yv,y1i6)nA anldAv(1i2c,n2a).demy

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0606 Additional Mathematics Unit 15a Differentiation and integration

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8  (i)  Given that y = 2 x2 , show that dy = (2 kx , where k is a constant to be found. [3]
+ x2 dx + x2) 2 [2]

  (Aii)  lHveincne fAinAdcedcdca(2ad+dxxe2Aem)2mdlxyv.yinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy

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0606 Additional Mathematics Unit 15a Differentiation and integration

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12  (i)  Show that x - 2 is a factor of 3x3 - 14x2 + 32. [1]
[4]
  (Aii)  lHveincne fAaActocrciaseadde3Aexm3m-lyv1y4ix2nA+A3lA2lvviccinaonmdpAlAeetmeclcya.ayddeAemmlyvyinAlAvicnademy

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0606 Additional Mathematics Unit 15a Differentiation and integration

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1 2 The diagram below shows part of the curve y = 3x - 14 + 32 cutting the x-axis at the points P and Q.
x2

y y = 3x – 14 + 32
x2
Academy lvinO P
emy Alvin cademy A cademy  (iii)  State the x-coordinates of P and Q.
Qx

AlvinAAccaaddeAmlyvinAlAvicnadAemy Alvin Ay  (iv)  Find [1]

(3x - 14 + 32 ) dx and hence determine the area of the shaded region. [4]
x2

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0606 Additional Mathematics Unit 15a Differentiation and integration

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8  (i) Given that f ^xh = x ln x3, show that f l^xh = 3^1 + ln xh. [3]
[2]
  (Aii)  lHveincne fAinAdcyca^1ad+delnAemxhmdlxyv.yinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy

y  (iii)  Hence find 2 lnx dx in the form p + ln q, where p and q are integers. [3]
1

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0606 Additional Mathematics Unit 15a Differentiation and integration

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11  The diagram shows part of the curve y = ^x + 5h^x - 1h2.

  (Ai)  lFvinidntheAAx-ccocaoraddidneatAemesmolfyvtyhienAstAatliAolvnvaicryinanpodOiAnAetsmcyocfatahydedceuAermvmel. yvyinAlAvyic=na(xd+ e5)m(xx –y1)2 [5]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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y1 1 (ii)  Find ^x + 5h^x - 1h2 dx. [3]
[2]
  (iAii)  lHveincne fAinAdcthcaeaadrdeeaAeemnmclloyvsyeidnAbAy ltAhlvevcicuinravnedaAnAedmcthcaeaxyd-daxeiAesm.mlyvyinAlAvicnademy

  (iv)  Find the set of positive values of k for which the equation ^x + 5h^x - 1h2 = k has only one real
solution. [2]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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9  (i)  Determine the coordinates and nature of each of the two turning points on the

  A  lcvurivne AAy =cca4axd+dexAem-1m2l.yvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy [6]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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9  (ii)  Find the equation of the normal to the curve at the point (3, 13) and find the x-coordinate of the

point where this normal cuts the curve again. [6]

AlvinAAccaaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy

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0606 Additional Mathematics Unit 15a Differentiation and integration

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6  (i)  Given that y = tan 2x , find dy . [3]
x dx

  (Aii)  lHveincne fAinAdcthcaeaedqdueaAetmiomnlyovfyithnAeAnolArlvmvaiclintaontdhAeAecmucrcvaeayddeyAem=mtlanyvxy2ixnAaltAtvheicpnaoindt wehmere yx = r . [3]
8

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0606 Additional Mathematics Unit 15a Differentiation and integration

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8    (iii)  Solve gl(x) = hl(x).
[3]

AlvinAAccaaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy

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0606 Additional Mathematics Unit 15a Differentiation and integration

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9
x cm

yx cm
cadem inx cm
A lvy cm
lvin y A y  The diagram shows an empty container in the form of an open triangular prism. The triangular faces
A em emare equilateral with a side of x cm and the length of each rectangular face is y cm. The container is
y d dmade from thin sheet metal. When full, the container holds 200 3 cm3.
AlvinAAccaaddeAemmlyvinAlAvicnadAemcay Alvin Aca  3x2 1600
(i)  Show that A cm2 , the total area of the thin sheet metal used, is given by A = 2 + x . [5]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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9  (ii)  Given that x and y can vary, find the stationary value of A and determine its nature. [6]

AlvinAAccaaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy

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0606 Additional Mathematics Unit 15a Differentiation and integration

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4   (i)  Differentiate sin x cos x with respect to x, giving your answer in terms of sin x. [3]
[3]
  (Aii)  lHveincne fAinAdcycasaind2dxedAemx.mlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy

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0606 Additional Mathematics Unit 15a Differentiation and integration

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9 
y

y = 4x
in AcademyAlvinA
y Alv demy demyO y= 4 + 2x
(2x + 1)2

Academ lvin Aca lvin Aca  x

The diagram shows part of the curve y = 4 + 2x and the line y = 4x.
(2x + 1) 2
AlvinAcadeAmlyvinA Academy A  (i)  Find the coordinates of A, the stationary point of the curve.
[5]

  (ii)  Verify that A is also the point of intersection of the curve y = (2x 4 1) 2 + 2x and the line y = 4x.
+ [1]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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9  (iii)  Without using a calculator, find the area of the shaded region enclosed by the line y = 4x, the

curve and the y-axis. [6]

AlvinAAccaaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy

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0606 Additional Mathematics Unit 15a Differentiation and integration

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y8  (i)  Find ^10e2x + e-2xhdx. [2]
[2]
lvin AcademyyAlvin yy  (ii)  Hence find k (10e2x + e-2x)dx in terms of the constant k. [2]
-k

Alvin AcadeemmyyAAlvin AemcaydeAmlvin Academy  (iii)  Given that k (10e2x + e-2x)dx =-60, show that 11e2k - 11e-2k + 120 = 0.
Acad Alvin Acad-k

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0606 Additional Mathematics Unit 15a Differentiation and integration

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8  (iv)  Using a substitution of y = e2k or otherwise, find the value of k in the form a ln b, where [3]
a and b are constants.

AlvinAAccaaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy

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0606 Additional Mathematics Unit 15a Differentiation and integration

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9  A curve has equation y = 4x + 3 cos 2x. The normal to the curve at the point where x = r meets
4
the x- and y-axes at the points A and B respectively. Find the exact area of the triangle AOB,

wAhelrevOiins thAeAocricgaiand.deAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy [8]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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8 y y = x + 10

B
y = x2– 6x + 10

demyA
y Alvin Academy AlvindemyO C x
em ca ca  The graph of y = x2 - 6x + 10 cuts the y-axis at A. The graphs of y = x2 - 6x + 10 and y = x + 10
d A Acut one another at A and B. The line BC is perpendicular to the x-axis. Calculate the area of the shaded
AlvinAAccaadeAmlyvinAlAvicnademy Alvinregion enclosed by the curve and the line AB, showing all your working. [8]

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0606 Additional Mathematics Unit 15a Differentiation and integration

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11

O
cademy inB
lvin A y Alv yA C
A em emQ D
8
Rh
in AcademyAlvin Acad Alvin AcadP 4 S
Alv emy emy  The diagram shows a cuboid of height h units inside a right pyramid OPQRS of height 8 units and with
d dsquare base of side 4 units. The base of the cuboid sits on the square base PQRS of the pyramid. The
ca capoints A, B, C and D are corners of the cuboid and lie on the edges OP, OQ, OR and OS, respectively,

of the pyramid OPQRS. The pyramids OPQRS and OABCD are similar.
A in A  (i)  Find an expression for AD in terms of h and hence show that the volume V of the cuboid is given
Alv   byV = h3 - 4h2 + 16h units3. [4]
4

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0606 Additional Mathematics Unit 15a Differentiation and integration

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  (ii)  Given that h can vary, find the value of h for which V is a maximum. [4]

AlvinAAccaaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy

Question 12 is printed on the next page.

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0606 Additional Mathematics Unit 15a Differentiation and integration

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7  The point A, where x = 0, lies on the curve y= ln ^4x 2 + 3h . The normal to the curve at A meets the
x-axis at the point B. x- 1

  (i)  Find the equation of this normal. [7]
[2]
AlvinAAccaaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy  (ii)  Find the area of the triangle AOB, where O is the origin.

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0606 Additional Mathematics Unit 15a Differentiation and integration

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9

y D y = 3x + 10
cademyA
inB
in A Alv y = x3 – 5x2 + 3x + 10
lv y yC
A em emO x
demy Acad Acad  The diagram shows parts of the line y = 3x + 10 and the curve y = x3 - 5x2 + 3x + 10.
The line and the curve both pass through the point A on the y-axis. The curve has a maximum at the
ca in inpoint B and a minimum at the point C. The line through C, parallel to the y-axis, intersects the line
A lv lvy = 3x + 10 at the point D.
AlvinAcadeAmlyvinA Academy A  (i)  Show that the line AD is a tangent to the curve at A. [2]

  (ii)  Find the x-coordinate of B and of C. [3]

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0606 Additional Mathematics Unit 15a Differentiation and integration

0606/13/M/J/15 13

  (iii)  Find the area of the shaded region ABCD, showing all your working. [5]

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0606 Additional Mathematics Unit 15a Differentiation and integration

0606/23/M/J/15 10

y8  (a)  (i)  Find e4x + 3dx. [2]
[2]
lvin AcademyyAlvin yy    (ii)  Hence evaluate 3 e4x+3dx. [2]
2.5 [2]
lvin AcademyyAAlvin AcaydeAmlvin Academy  (b)  (i)  Find cosLKJ3xPONdx.
A AcadeAmlvin Academy 
  (ii)  Hence evaluate r cos KJL3xPNOdx .
6

0

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0606 Additional Mathematics Unit 15a Differentiation and integration

0606/23/M/J/15 11

y8  ^x-1 2 dx [4]
(c)  Find + .
xh

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0606 Additional Mathematics Unit 15a Differentiation and integration

0606/12/O/N/15 4

2  A curve, showing the relationship between two variables x and y, passes through the point P^-1, 3h.

  TAhelcvurvienhaAsAacgcraaaddiednetAemofm2lyvatyiPn.AGAivlAelvnvitchinaant dAddAex2mcy2c=aa-yd5d,eAemmfilnydvytihenAeqluAavtioicnnaofdthee cmurvey. [4]

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0606 Additional Mathematics Unit 15a Differentiation and integration

0606/12/O/N/15 7

5  Variables x and y are such that y = ^x - 3hln^2x2 + 1h.

    (i)  Find the value of dy when x = 2. [4]
AlvinAAccaaddeAemmlyvyinAAlAlvvicinandAAemccaayddeAemmlyvyinAlAvicnademy    (ii)   Hence find the approximate change in y when x changes from 2 to 2.03.dx[2]

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0606 Additional Mathematics Unit 15a Differentiation and integration

0606/12/O/N/15 10

8  Find the equation of the tangent to the curve y = 2x - 1 at the point where x = 2. [7]
x2 + 5

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