PoLeungKukCelineHoYamTongCollege
Form Seven PureMathematics
(2009-2010)
HKALE
Paper One
Marking Scheme
Solution / Marks1. (a)
(b)
By comparing the coefficients of in both sides, we have
(c)
(by (a))
(by (b))
/ 1M
1
1M
1
1M for using (a)
1M for using (b)
1A
(7)
2. (a) Let
Putting ,,, we have ,,.
Thus, we have .
(b)
(by (a))
(c)
(by (b))
(which is also true for m = 1)
For
We have
Therefore, we have
So we have
As m is appositive integer, we have .
Thus, the required greatest positive integer m is 15. / 1A
1A for all correct
1M
1A or equivalent
1A
1A or equivalent
1A
(7)
3. (a) By (2), we have
By (1), we have .
So, we have .
Thus, we have .
(b) (i) Let for some polynomial .
So, we have .
By (1), we have .
Thus, with the help of (a), we have.
(ii) Let for some polynomial .
By (b)(i), we hve .
Note that.
By (1), we have.
So, with the help of (a), we have .
Therefore, we have .
Solving, we have and .
Thus, the required remainder is. / 1M for either one
1A
1M
1A
1M accept using long division
1A
4 (a)
(b) (i) Note that B = , where .
Thus, B represents the reflection in the straight line .
(ii)
Thus, the matrix AB represents a rotation. / 1A
1M can be absorbed
1A for reflection +1A for the details
1A
1M
1A f.t.
(7)
5. (a) Since , the statement is true for n = 1.
Assume that for some positive integer k.
By, mathematical induction, we have
(b) Note that for all .
Since , we have .
(c) Since for all n, the sequence is strictly increasing.
Note that
< 3
Therefore, the sequence is bounded above by3.
Thus, exists. / 1M for using induction assumption
1
1M accept
1A
1M
1M
1Af.t.
(7)
6. (a) (i) By A.M. G.M., we have
Since , we have
Thus, we have .
(ii) By A.M. G.M., we have
So, we have .
Since , we have .
Therefore, we have .
Thus, we have .
(b) Note that ( by (a)(i) ).
Also note that ( by (a)(ii) ).
Thus, we have . / 1M
1
1M
1
1M
1
(6)
7. (a) (i) (E) has a unique solution
and
, or
When (E) has a unique solution,
(ii) When , the augmented matrix of (E) becomes
~
(E) is consistent when .
Solving, the solution set is .
(b) Putting and in (a), we have, by(a)(i), the solution of the first
three linear equations of the system is , and .
Note that .
Thus, the system of linear equations is inconsistent.
(c) Putting and in (a), we have, by (a)(ii), the solution of the
system of linear equations is , and , where .
Putting, and in ,
We have .
Solving, we have or .
Thus, the required solutions are and . / 1M
1A
1A
1M for Cramer’s rule
1A+1A(1A for any one, 1A for all)
1A
1A
(8)
1A
1M
1A f.t.
(3)
1A
1M
1A
1A for all correct
(4)
8. (a) Since , we have and.
Note that and .
Thus, we have and .
(b)
is a 2X2 diagonal matrix if and only if .
As n is a positive integer, we have n = 3, 6 , 9,… .
When n = 3, 6 , 9,… , we have .
(c)
(d) Since , we have .
So, we have
Thus, is not a singular matrix.
(e) Note that .
Putting in (c), we have
.
Note that .
By (b), we have .
Therefore, we have
By (d), is a non-singular matrix.
Hence, the inverse of exists.
So, we have
Therefore, we have
Thus, we have . / 1M
1A
(2)
1M
1A
1A
(3)
1M
1
1M
1A f.t.
(2)
1M
1M
1M for using (b)
1A
1M
1A
(6)
9. (a)
If (*) can be written as (**), then , and .
So, we have , and .
Thus, we have .
Assume that .
By letting , and , (*) can be written as (**).
(b) By (a), we have .
So, we have .
Solving, we have .
or
, , or
(c) By (a), we have .
With the help of (b), satisfies the equation .
Simplifying, we have .
Therefore, we have .
Thus, we have , or . / 1M
1
1M
1
1
(5)
1M
1A
1M+1A
1M+1A
1A
1M
1M
1A
10. (a)
Since , we have .
Note that .
Thus, we have for all t > 0.
(b) Define for all x > 0.
Since , we have ( by(a) ).
Thus, we have for all x > 0.
(c) (i) By (b), is a strictly decreasing function.
Since , we have .
So, we have .
Note that is a strictly increasing function.
Thus, we have for all a > 0.
(ii) Note that for any 2 positive real numbers and .
Putting in (c)(i), we have .
Therefore, we have .
Thus, we have .
(iii) Since , the statement is true for n = 1.
Assume that the statement is true for n = m.
Then, for any (m+1) positive real numbers ,
( by induction assumption )
( by (c)(ii) )
By mathematical induction, we have for any
N positive real numbers .
(iv) Note that .
By (c)(iii), we have .
So, we have .
Thus, we have . / 1M
1
(2)
1M
1A
1M
1
(4)
1M
1
1M accept
1
1M
1M
1
1M
1
(9)
11. (a) (i) by solving the equation ,
the roots are , where .
(ii)
(b)
(i)
(ii)
(c) (i)
= 5
(ii) Putting in (a)(ii) and using (b), we have
.
Hence, we have .
So, we have for all .
By (c)(i), we have .
Therefore, we have .
So, we have .
Hence, we have .
Therefore, we have .
So, we have .
Note that for all k =1, 2, 3, 4.
Thus, we have . / 1A
1M
1M
1
1
(5)
1A
1A
1A
(3)
1M
1A
1
1M
1M
1M
1A
(7)
END OF MARKING SCHEME
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