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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.
5. AJC/2009 Promo/Q5
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 .
6. CJC/2009 Promo/Q6
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 .
7. HCI/2009 Promo/Q6
,
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)
9. JJC/2009 Promo/Q2
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)
16. TJC/2009 Promo/Q4
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
3. SRJC/2009 Promo/Q3
Differentiating w.r.t. x,
At (2,3),
Thus equation of tangent at (2,3),
4. SRJC/2009 Promo/Q4
a)
b)
5. SRJC/2009 Promo/Q9
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
7. TPJC/2009 Promo/Q3
(i)
(ii) Since f is strictly decreasing,
since
Method 1:
Method 2: Discriminant =
8. TPJC/2009 Promo/Q7
−−−−−− (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.
9. TPJC/2009 Promo/Q10
,
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
11. CJC/2009 Promo/Q10
(a)
(b)
(c)Shape + Minimum Point x-intercept at -2: Asymptotes: x = 2, y = 0
12. CJC/2009 Promo/Q12
(i)
–2
x = 2y
x 0
y = f ’ (x)
(ii)
Check when , hence maximum volume
Alternative first derivative test accepted
(iii)
m3
13. HCI/2009 Promo/Q1
=
=
=
=
=
14. HCI/2009 Promo/Q4
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).
17. NJC/2009 Promo/Q3i)
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.
23. RVHS/2009 Promo/Q1
24. RVHS/2009 Promo/Q8
i. Let denote the height of the cylinder in .
by similar triangle or
ii.
Let then
(Reject) or
When ,
occurs when .
25. SAJC/2009 Promo/Q1
(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.)
28. SAJC/2009 Promo/Q8
(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
30. VJC/2009 Promo/Q4
31. VJC/2009 Promo/Q11
55.279
2009 Promotion Examination H2 MathematicsCompilation by Topics (Integration & Applications 2009)
1. DHS/2009 Promo/Q5
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)
2. DHS/2009 Promo/Q9
i) = 2cos 2x
ii)
iii)
3. DHS/2009 Promo/Q11
a)
b)
c)
4. DHS/2009 Promo/Q13
(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)
5. TPJC/2009 Promo/Q9
(a)
(b) Let
6. AJC/2009 Promo/Q11
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
10. JJC/2009 Promo/Q10
(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
12. JJC/2009 Promo/Q12
(i) Let
(ii) Area
(shown)
(iii) Using (ii) and GC, we have
Using (i),
Hence,
13. MJC/2009 Promo/Q6(i)
+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
15. NJC/2009 Promo/Q6i)
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
17. NJC/2009 Promo/Q11i)
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
20. RVHS/2009 Promo/Q10
(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)
22. TJC/2009 Promo/Q2
(a)
(b)
23. TJC/2009 Promo/Q6
Hence,
24.T
JC
TJC/2009 Promo/Q13ii
25. TJC/2009 Promo/Q15
(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
3. CJC/2009 Promo/Q11
when x = 0,
by Maclaurin’s Theorem,
When
4. MJC/2009 Promo/Q5(i)
(ii)
5. NJC/2009 Promo/Q7
(ans in MF15)
(shown)
Differentiating implicitly wrt x,
When x = 0, , , ,
Maclaurin’s series for y is
;
6. PJC/2009 Promo/Q4(i)
(ii)
7. PJC/2009 Promo/Q11(i)
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.
8. RVHS/2009 Promo/Q2
or (Reject, since x is numerically small.)9. RVHS/2009 Promo/Q13
(a)
.
Using ,
10. SAJC/2009 Promo/Q5
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 +
12. VJC/2009 Promo/Q7