HKDSE Mathematics Question Review (2024-11-17)

Question 1: Try evaluate this determinant

Hints (out of syllabus):

Question 2: Try solve this system of linear equations

Question 3: Given 2 random events A and B, which of the following statements is correct?

(a) P(AB) ≤ P(A) P(B)

(b) P(AB) ≥ P(A) P(B)

(c) 2 P(AB) ≤ P(A) + P(B)

(d) 2 P(AB) ≥ P(A) + P(B)

Question 4:

(a) Prove that ln(x+1)/x = 1 when x -> 0+

(b) Hence, show that when x -> 0+:

ln ((1+x)/(1-√x)) ~ √x

Hint: Use small value aproximation.

答案唔提供

鑑於香港蠢人甚多，本人明天會給出題解

Solution:

(a) Intuitively, sketch y=x and y=ln(x+1).

Then denote f(x) = ln(x-1) - x.

Clearly f"(x) < 0,

and the global extremum is at (0,0)

i.e. ln(x-1) = x if and only if x=0

Formally, using the power series:

ln(x+1) = x - x^2/2 + x^3/3 ...

Divide both sides by x, and put x = 0,

then ln(x+1)/x = 1

QED

(b) Note that:

(1+x)/(1-√x) = (1-√x+-√x+x)/(1-√x)

= 1 + (√x+x)/(1-√x)

For x -> O+, 1-√x ~ 1

Then (√x+x)/(1-√x) ~ 0

Also, ln(1+x) ~ 1

Now ln(1 + (√x+x)/(1-√x)) ~ (√x+x)/(1-√x)

~ √x+x = √x (1+√x) ~ √x

For the last step, consider 1 ±√x.

(1+x)/(1-√x) = (1-√x+√x+x)/(1-√x)

Answer: C

Using the method of exclusion:

When AB is an empty set,

P(AB) = 0,

but it is not sufficient to say:

P(A) = 0 or P(B) = 0

So options B and D are wrong.

When A is a subset of B,

P(AB) = P(A) ≥ P(A)P(B)

So option A is also wrong.

For option C,

note that AB is a subset of A∪B

So P(AB) ≤ P(A∪B)

= P(A) + P(B) - P(AB)

Rearranging, we obtain the desired result.

Solution:

Denote the given augmented matrix as A.

Note that r(A) = 3

Now, using elementary row operations:

Now r(B) = 3, i.e. the system is solvable:

x_1 = 3 - x_3

x_2 = -8 + 2 x_3

x_4 = 6

Denote x_3 = k as an arbitrary real constant,

then the solution set will be:

{x_1, x_2, x_3, x_4} = {(3, -8, 0, 6) + k(-1, 2, 1, 0)}

Solution: Straightforward.

We simply split the given determinant into 2,

and sum them up by separating the common factors,

each of them a and b respectively.

We will then obtain 2 matrics X and Y of (n-1) x (n-1) dimensions:

aX + [(-1)^(n+1) b]Y

Note that:

| a 0|

| 0 a|

= a^2 E

where E =

| 1 0|

| 0 1|

i.e.

det(aX) = a^2 det(X)

The desired result will be:

a^n + [(-1)^(n+1) b^n]

det(aX) = a^(n-1) det(X)

This is X:

This is Y:

Intuitively, sketch y=x and y=ln(x+1).

Then denote f(x) = ln(x+1) - x.

Clearly f"(x) < 0,

and the global extremum is at (0,0)

i.e. ln(x+1) = x if and only if x=0

The gap is the smallest when x=0.