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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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