2010 set b paper 3 rvhs promo (solutions)
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2010 RVHS Year 5 (JC1) H2 Maths Promo Exam Solutions
1Given , for
Let P(n) be the statement that , for
For P(1),
L.H.S. = (given)
R.H.S. =
Assume that P(k) is true for some ,
i.e.
To show P(k + 1) is true,
i.e.
.
2
Since the above equation has real solutions
Discriminant ( 0
Using GC or linear line method:
(+) (() (+) (()
0
Therefore the solution set is
3(i)Let , and be the number of batches of candies X, Y and Z produced respectively.
Using GC to solve the augmented matrix,
, ,
(ii)
4(i)
(ii)Let f(r) =,
then f(r+1) = .
Alternatively,
= (3.5.7 2.3.5)
+(4.7.9 3.5.7)
+(5.9.11 4.7.9)
+
+
(proved)
4(iii)
= 1357477276
5
Differentiating (1) wrt x again:
Then when x = 0,
Therefore,
Next,
6(i)
The unit vector is .
6(ii)
6(iii)
Since OB is parallel to OP, and O is a common point,
O, B and P are collinear.
6(iv)
7Let the radius and height of the cylinder be r and h respectively.r = , h =
Total surface area S =
=
Therefore,
(shown)Then
For stationary value, let = 0
Next,
When, ,
So, [In fact, .]
8(i)
8(ii)Any horizontal line y = k, where , cuts the graph of g at most once.
g is oneone. Hence g 1 exists.
g-1 :
8(iii)For x < 0, hg(x) =
8(iv)Rhg =
9(a)Sketch of curve defined parametrically by and by GC:
Converting parametric equation to Cartesian equation,
We have
Solving and ,
We have the intersection points (1, 0) and (4, (3)
So exact area of R =
=
= units2 Alternatively:
=
+ = units2
9(b)Required Vol.
=
Letting &
Then &
=
=
=
=
= or (units3)
Alternatively
Required Vol.
=
=
=
=
=
= or (units3)
10(i)
10(ii)
is a G..P. , with first term, and common ratio, .
Given
Min.
Alternative MethodGiven
Min.
11a.. Then:
11b.
12(a)The graph is
shifted 1 unit to the left (translated -1 unit along the x-axis)
then reflected in the y-axis.
OR
reflected in the y-axis
shifted 1 unit to the right (translated 1 unit along the x-axis)
12(b)(i)
12(b)(ii)
12(b)(iii)
13(a)(i)
Then, by substitution,
We have:
So, (general soln.)
(ii) Given that when x = 1, y = (1
(1 = 1ln1+c
c = ( 1
Thus, the particular solution is
(iii)
The stationary points lie on the line .
13(b)(i) The differential equation is
(ii) Solving the differential equation:
Given that when t = 0 h, x = 100 grams
We have C = ( 0.01
Also, when when t = 1 h, x = 25 grams
We have , so k = (0.03
Thus,
So, after 3 hrs, x = = 10 grams
(iii) As x decreases, the absolute value of also decreases. Thus, there is slowing down of rate of conversion of substance A to substance B in the chemical reaction.
EMBED Equation.DSMT4 (c = 1)
EMBED Equation.DSMT4 (c = 0)
EMBED Equation.DSMT4 (c = 1)
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