2009 Soln Promo Binomial MI Diffn Integn Maclaurin SOLN

2009 Promotion Examination H2 MathematicsCompilation by Topics (Binomial Series 2009)

1. DHS/2009 Promo/Q2

2. NYJC/2009 Promo/Q3

=

Combining

-

= +…

=

Given that the second term is ,

=

a = 8

Range of x for which this series expansion is valid:

3. SRJC/2009 Promo/Q6

=

=

Condition for validity:

Set of values of x :

When ,

4. TPJC/2009 Promo/Q11ii

(ii)

5. AJC/2009 Promo/Q2

i)

= = …. = 2 34 2 11 8...

3 9 108 81x x x

ii)

Expansion is valid for and

iii) = =

6. CJC/2009 Promo/Q1

(i)

(ii) The expansion is valid for ,

i.e., (OR OR any equivalent forms involving x or |x|)

(iii) Putting in the expansion, we have

7. HCI/2009 Promo/Q3

8. IJC/2009 Promo/Q1

9. JJC/2009 Promo/Q1

The series is valid for

Coefficient of

10. MI/2009 Promo/Q9

11. SAJC/2009 Promo/Q2

= =

= =

Expansion is valid for < 1

Range of values of x for which expansion is valid is for − < x < , x .

Put x = 0.4

= 1.8945833

37.89167 (5 decimal places)

12. TJC/2009 Promo/Q1

= (1 – x) [1 + ]

=

=

Expansion is valid for or

13. VJC/2009 Promo/Q9

2009 Promotion Examination H2 MathematicsCompilation by Topics (Mathematical Induction 2009)

1. DHS/2009 Promo/Q8Let P(n) be the proposition:

for

When n = 1, LHS of

RHS of =

P(1) is true.

Assume P(k) is true for some ,

i.e.

To show P(k+1) is also true, i.e.

LHS =

=

=

=

=

= = RHS

P(k) is true P(k+1) is true

Since P(1) is true and P(k) is true P(k+1) is true, hence by mathematical induction, P(n) is true for n +.

2. NYJC/2009 Promo/Q2(i) Using the recurrence relation, we find .

= 6 – 1, 26 – 1, 126 – 1, 626 – 1 = 5, 52, 53, 54 for n = 1, 2, 3, 4 respectively.

Thus we conjecture that .

(ii) Let Pn denotes for n = 1, 2, 3, ….

For n = 1, LHS = RHS = 5 + 1 = 6 = LHS. So P1 is true. Assume Pk is true for some k = 1, 2, 3, ….

That is, ------ (IH)

For n = k + 1, LHS = by the recurrence relation

= by (IH)

= = RHS.

Hence Pk+1 is true.

Hence by MI, Pn is true for n = 1, 2, 3, …. 3. SRJC/2009 Promo/Q1

(i) Let Pn be the statement .

n = 1, LHS

RHS . Therefore, P1 is true.

Assume Pk is true for some i.e.

To show Pk+1 is true i.e.

Therefore Pk+1 is true if Pk is true.Since P1 is true and Pk+1 is true if Pk is true, Pn is true if Pk is true .

4. TPJC/2009 Promo/Q8(i) As

we have,

(rejected since ) or 2

(ii) Let be the statement “un = ” where n is a positive integer

When n = 1, ,

LHS = RHS is true

Assume that Pk is true for some positive integer k, ie. uk = for some

positive integer k.

To prove that Pk+1 is true, i.e. to prove that uk+1 =

is true is true.

Since is true and is true is true, by mathematical induction,

is true for all positive integers n. i.e. un = for all positive integers n.


To prove for n Z .

When n = 1,

LHS =

RHS = 1Since LHS = RHS Statement is true for n = 1.

Assume the statement is true for some n = k, where k Z ,

i.e.

Require to prove that the statement is true for n = k+1,

i.e. to prove

If the statement is true for n k , it is also true for 1n k .Since the statement is true for n =1, By mathematical induction, the statement is true for n Z .


Let Pn be the statement for .

When n = 0, LHS = a

RHS = = LHS

P0 is true and this forms the basis of induction.

Assume Pk is true for some , i.e

Required to prove Pk+1 is true, i.e

LHS =

Since P0 is true and Pk true Pk+1 is also true, hence by Mathematical Induction Pn is true for .


,

8. IJC/2009 Promo/Q11

Let denote the statement,

Consider : LHS =

RHS =

Thus, is true

Assume that is true for some : i.e.,

Want to show that is true, : i.e.,

= RHS

Thus, true true.

Since is true, and true true, by mathematical induction, is true for

all (Shown)


Let be the statement for all .

LHS =

RHS =

Hence is true.

Assume is true for some , i.e.

To prove is true , i.e.

LHS =

= RHS

is true is true.

Since P1 is true and true true, by Mathematical Induction, Pn is true for

.

(i)

(ii)

10. MI/2009 Promo/Q4

Let for all positive integers .

For

LHS:

RHS: Since LHS=RHS, therefore is true.Now, assume that is true for some positive integer .Then,

Therefore, is true if is true.And so by Mathematical Induction , is true for all positive integers

11. MJC/2009 Promo/Q5(i)

(ii)

12. NJC/2009 Promo/Q4i)

Let Pn be the statement that for all ( ).

For ,

;

Therefore, is true.

Assume that Pk is true for some ( ). ie.

To prove that is true, ie.

Thus, is true is also true.

Since is true, and that is true is also true, by Mathematical Induction,

is true for all ( ).

ii)

iii)

13. PJC/2009 Promo/Q7(i) Observe the pattern of the

sequence: 2,2,4,4,6,6,….

(ii) Let be the statement ,

When n = 1,LHS = 2

RHS =

Thus, is true.

Assume that is true for some k, i.e.

To prove that is also true, i.e.

LHS =2

=

=

= RHS

Thus, is also true.

Since is true and is true is also

true, then by induction, is true for

.

Learn how to manipulate the numbers to arrive at the required statement. Don’t try to “cheat”.

14. RVHS/2009 Promo/Q11

(a)Let be the statement for .

When ,

LHS

RHS

Since LHS = RHS, therefore is true.

Assume that is true for some

i.e.

Need to prove that is true

i.e.

Consider ,

LHS

= RHS

Thus is true implies is true.

Since ( is true) and ( is true implies is true), by mathematical induction, is true for .

(b)

(ii)Since series is convergent.

15.Q Let P(n) be the statement for

When n = 1, LHS = = ; RHS =

So P(1) is true.

Assume P(k) is true for some k ≥ 1, i.e.

To show P(k + 1) is true i.e.

LHS = = + = =

= = =

= = RHS

So P ( k+1) is true.Since P(1) is true, and P(k) is true P(k + 1) is true. By mathematical

induction, P(n) is true for all n Ζ+, n 1 ie. for

Since

Therefore (deduced)


Let Pn be the statement for all positive integers n.

When n = 1, and

P1 is true.

Assume Pk is true for some positive integer k. i.e. .

To prove that Pk+1 is true, i.e. .

Hence, Pk is true Pk+1 is true.

Since P1 is true and Pk is true Pk+1 is true, by Mathematical Induction,Pn is true for all positive integers n.

ie. for all positive integers n.

17.VJC/2009 Promo/Q2

2009 Promotion Examination H2 Mathematics

Compilation by Topics (Differentiation & Applications 2009)

1. DHS/2009 Promo/Q4i)

ii)

2.

SRJC/2009 Promo/Q2


Differentiating w.r.t. x,

At (2,3),

Thus equation of tangent at (2,3),


a)

b)


a. The vertical height of cone, (Shown)

y

x0a

b

b. Let h = vertical height of cone.

then R = and l =

Thus area of metal sheet, A =

(Shown)

iii) Let

Since

then

For stationary point,

or (Rejected)

Using sign test:s 0.4 0.5

-ve 0 +ve

Therefore, A is minimum when (Shown)

6. TPJC/2009 Promo/Q2


(i)

(ii) Since f is strictly decreasing,

since

Method 1:

Method 2: Discriminant =


−−−−−− (1)

−−−−−− (2)

Substitute (1) into (2):

Differentiating A w.r.t x

When , meters

Since , A will be maximum when meters

When meters,

Hence, when A is maximum, the cross−sectional shape of the drain will be a

semicircle with radius meters.


,

When x = 11,

Equation of tangent:

so therefore

Assuming tangent cuts the curve again:

Substituting and into equation of tangent,

Discriminant (Alternatively, show that )

Since there is only 1 point of intersection between the curve and the tangent at P, the tangent at P does not cut the curve again.

10. TPJC/2009 Promo/Q11(i)

Differentiate wrt x

At ,

Maclaurin’s series for

(ii)

(iii)

Equation of tangent to at x = 0 is


(a)

(b)

(c)Shape + Minimum Point x-intercept at -2: Asymptotes: x = 2, y = 0


(i)

–2

x = 2y

x 0

y = f ’ (x)

(ii)

Check when , hence maximum volume

Alternative first derivative test accepted

(iii)

m3


=

=

=


Area of base (hexagon) =

Volume of prism, V =

(since )

V =

V =

Hence, maximum volume occurs when .

15. HCI/2009 Promo/Q9i)

3

x

hx

d

ii)a.

At t = 1,

ii)b.When t = 1 and k = 2,

Equation of normal:

ii)c.

Required angle =

OR

Required angle =

16. MJC/2009 Promo/Q9(a)(i)

(ii)

(b) Consider rectangle ABCD where A is vertex on origin, B on positive x-axis and D on positive y-axis. Then vertex C has to be the point (x,y).


Total surface area of hemisphere and circular base =

Total surface area of cylinder =

ii)

(Since r > 0)

OR

Hence T is a minimum.

(nearest $)

18. NJC/2009 Promo/Q10a

i)

ii) From part (a)(i),

19. NJC/2009 Promo/Q12(i),(ii),(iii)i)

For stationary points, .

Since a > 0, thus, there are no stationary points.

ii)

For graph to concave downwards, .

Since a > 0, we need

Therefore, when , the curve is concave downwards.

iiia)

Vertical asymptote :

Horizontal Asymptote:

x-intercept:

y-intercept:

3

2

2y

13

2 2

a

23

a

y

x

iiib)

iiic)

20. PJC/2009 Promo/Q1

3

2

1

2y

13

2 2

a

1

23

a

y

x

3

2

y

x

2

9

a

21. PJC/2009 Promo/Q3Solution Remarks

Given ,

Given gradient at P is 5,

Coordinates of P is .

Equation of normal to the curve at P:

Need to equate to 5 to obtain the x-

coordinate of P.Need to substitute x-coordinate into the equation of tangent (not the equation of the curve) to get the y-coordinate of P.

22. PJC/2009 Promo/Q14(i) By Pythagoras’ Theorem,

(ii) By Pythagoras’ Theorem,

Curved surface area, A

From (i)

(iii)

Solving ,

Using 1st derivative test,

h

+ 0 –

The curved surface area A is a maximum when .

Maximum

It is easier to do 1st derivative test here as it is already stated in the question that the value of A is a maximum. Also, it is tedious to obtain the 2nd derivative.



i. Let denote the height of the cylinder in .

by similar triangle or

ii.

Let then

(Reject) or

When ,

occurs when .


(a)

(b)

26. SAJC/2009 Promo/Q4Let the depth of the water be h and the radius of the water level at that point be R.

at h = 12 cm

27. SAJC/2009 Promo/Q6(i) Base circumference of cone = rθ

To find base radius R,

(shown)

Height of cone

EITHER

Volume of cone =

Treat r as a number (say r = 1) and sketch the graph of

Using GC to calculate the maximum point (students need to resize window first e.g. using ZoomFit) givesx = 5.130, y = 0.403

[OR

Volume of cone =

Treat r as a number (say r = 1) and sketch the graph of

Then use to find θ.]

Angle of sector he should cut off= 2 – 5.130= 1.15 radians (3 s.f.)


(i)

(ii)

(iii)

At , t = 2

Gradient of curve at =

Equation of normal at t = 2 is

-- (1)

Equation of normal at t = -2 is

-- (2)

Solving,

The line is y = x.

29. TJC/2009 Promo/Q12By Pythagoras’ Theorem,

Let T be the time taken.

= 0 x = 55.279 55.3

Using G.C.,

xve 0 ve

By 1st Derivative Test or any other valid testT is minimum when (3 s.f.). Least T is 86.6 s



55.279

2009 Promotion Examination H2 MathematicsCompilation by Topics (Integration & Applications 2009)


i) using GC

ii) Using 5 s.f. answer:Volume of solid formed

units3 (to 3 s.f) (Using GC)

Using 3 s.f. answer:Volume of solid formed

units3 (to 3 s.f) (Using GC)


i) = 2cos 2x

ii)

iii)


a)

b)

c)


(i)

To find x-intercepts:

When .

To find y-intercepts:

When

(ii)

y

x O 1

π23e

(iii) Gradient of normal at P(1, 0) is by (ii).

Equation of normal at P(1,0) :

Therefore coordinates of

Thus area of triangle POQ =

(iv)


(a)

(b) Let


a)

Area = area of Triangle

= = … =

b) Volume = =… =

7. CJC/2009 Promo/Q13(a) =

= ln + C = ln + C

OR = = + By cover-up rule, A = =

B = = =

= [ ln |3 + 2x| ln |3 2x| ] + C = ln + C(b) Let = cos

Differentiate with respect to := sin dx = 2 sin d = 2 cos sin d dx = ( 2 cos sin ) d

= [2 cos sin ] d = 2 cos2 d

= cos 2 + 1 d = sin 2 + C= sin cos + C= cos + C= cos1 + C

= cos1 + C(c) Let u = (ln x)2 and dv = x

du = 2(ln x)( ) and v = x2

= [ x2 (ln x)2 ]

Let u = ln x and dv = xdu = and v = x2

= e2 { [x2 ln x ] x dx }= e2 e2 + [ x2 ]

= (e2 1)8. HCI/2009 Promo/Q10

i)

ii)

=

= …… (A)

or ……(B)

A B

=

=

= =

=

=

= =

iii)

=

= = =

or

9. JJC/2009 Promo/Q9(i) Substitute into :

(ii) Differentiate throughout w.r.t x.

At

Equation of tangent at P:

Area of OQR


(a)

(b) Let

When , .

When , .

(c)

11. JJC/2009 Promo/Q11(a)

Volume

(to 2 dec pl)

(b)

When ,

, or ,

When ,

, or ,

Since

y

x

R1 R2

0

Area


(i) Let

(ii) Area

(shown)

(iii) Using (ii) and GC, we have

Using (i),

Hence,


+C

(ii)

14. MJC/2009 Promo/Q10

(i)Area of R = dx

dx

R0 1

y

x

xy

22 xy

21

units2

(ii) Required Volume

= dy dy

=

units3


When , .

When ,

Equation of tangent l is

ii)

When k = 1, ; .

Using GC, the other point of intersection is (1, 0).

16. NJC/2009 Promo/Q10b

1 1

2 2y x

2

11

1

xy

x


ii)

Area of S =

=

= (shown)

Given

Area of S =

=

=

=

=

ln6x

O

y

(ln3, 0)

=18. PJC/2009 Promo/Q5

LetNote that it is easier and neater if we square both sides and make x the subject before differentiating.

Note that the substitutions are done all at one go.

19. PJC/2009 Promo/Q8(i)

(ii)

Let


(a)

(b)

Let

Comparing coefficient of x,

21. RVHS/2009 Promo/Q12(i)

Since ,

x-coordinate of P is

(ii) Area of A

(iii)

(iv)


(a)

(b)


Hence,

24.T

JC

TJC/2009 Promo/Q13ii


(i)

Let

Since x is positive, we get and hence .

Therefore, coordinates of P is . (Shown)

(ii) Area of shaded region A =

x

y

a3a 2a

x=a/2

=

=

=

(iii) Volume generated = .

= 0.813 cubic units (correct to 3 s.f.)26.V

JC

VJC/2009 Promo/Q10

2009 Promotion Examination H2 MathematicsCompilation by Topics (Maclaurin’s Series 2009)

1. DHS/2009 Promo/Q6Method 1

Method 2

Equation of tangent is y = x.2. TPJC/2009 Promo/Q11i

(i)

Differentiate wrt x

At ,

Maclaurin’s series for


when x = 0,

by Maclaurin’s Theorem,

When


(ii)

5. NJC/2009 Promo/Q7

(ans in MF15)

(shown)

Differentiating implicitly wrt x,

When x = 0, , , ,

Maclaurin’s series for y is

;


(ii)


Differentiate implicitly,

It is easier to do implicit differentiation here.

(i) OR

Differentiate wrt x,

(ii) Differentiate implicitly,

When , , , .

Thus, the Maclaurin’s series for y is

.

(iii)The Maclaurin’s series for is

.

Note that the Maclaurin’s

series of is

obtained through differentiating the Maclaurin’s series of y.


or (Reject, since x is numerically small.)9. RVHS/2009 Promo/Q13

(a)

.

Using ,


Differentiating (1) further wrt x,

Differentiating (2) further wrt x,

When x = 0, y = 2,

(1):

(2):

(3):

Maclaurin’s series for y is

11. TJC/2009 Promo/Q14 y2 = 3 +

Differentiating wrt x :

Differentiating again :

When x = 0, y = 2 , , 2 =

By Maclaurin’s Theorem : y = 2 +

= 2 +


2009 Soln Promo Binomial MI Diffn Integn Maclaurin SOLN

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