HKDSE M2 Question Review (20250615)

Question 1:

Given a curve y = x ln(e + 1/(x-1) ).

(a) By Googling the power series for ln(x+1), show that there exists a small value approximation for x that:

ln (1/x + 1) ~ 1/x

(b) Hence ompute the oblique asymptote for y.

Hint:

Take the required equation as y=ax + b

(b) Hence compute the oblique asymptote for y.

Answer:

y= x + 1/e

題解自己搵

Reference:

This is to find the oblique asymptote for polynomials, for the general method:

Question 2:

烏大龜係點樣自學higher math? 🐢

Solution:

The Feynman Technique

Question 3:

Show that for 0 < x < 𝝿:

2x/𝝿 < sin(x)

Show that for 0 < x < 𝝿/2

Solution:

Gweilos from Quora attemps to use calculus to solve it,

i.e. differentiate sin(x) - 2x/𝝿 = 0 w.r.t. x.

This is stupid and time-consuming.

What method I use is simply intuition:

For the first quadrant, 0 < sin(x) < 1

As 1 = 2/𝝿 ‧ 𝝿/2, it is clear that 0 < 2x/𝝿 < 1

By observing the graph of sin(x) for the first quadrant,

clearly sin(x) is concave and sin'(𝝿/2) = 0

i.e. y=sin(x) pass through y=2x/𝝿 at the 2 ends of the domain.

This sufficiently shows that sin(x) > 2x/𝝿. Q.E.D.

Gweilos from Quora attempt to use calculus to solve it

A even more quicker method without calculus:

We know that from first quadrant, 𝝿/2 > x > sin(x) > 0

and

(𝝿/2) > (𝝿/2) / x > x /sin(x) > x > 1 > 0

Rearranging the terms, we have

sin (x) > x/(𝝿/2) = 2x/𝝿

Simple as fuck.

Remember when x -> 0, sin(x)/x --> 1

From the graph in the first quadrant, sin x < x < tan x

Remark: This method implies that within the given domain, sin(x) is bijective. A bijective function cut horizontal line and verticle line once only, so When sin'(x) = 0, the point must intersect with another given bijective function, i.e. a straight line that is neither horizontal nor vertical.

A bijective function cuts horizontal line and verticle line once only,

Quesion 4:

Rewrite the integration by substitution as a chain principle of differentiation.

Soluion:

f(x) = ∫ f'(g(x)) g'(x) dx

Solution

Question 5:

Explain graphically the direction of cross product such that i x j = k, j x k = i, k x i = j and j x i = -k, k × j = -i , i x k = -j

where i is on x-axis j is on y-axis and k is on z-axis in the 3D coordinate geometry.

Solution:

To make an 1D vector 2D, we rotate anticlockwisely such that the added y axis is linearly independent with x axis. Likewise, to make an 2D vector 3D, we rotate anticlockwisely su h that the added z axis is linearly independent with y axis.

Interestingly, by rotating z-axis antiockwisely we will obtain an x-axis again instead of forming a 4D axis, so overall the cross product implies anticlockwise rotation with z-axis (the direction of angular acceleration) pointing out of page. Look at the typhoons on the Northern hemisphere: they are all rotating anticlockwisely with moments of inertia pointing out of page.

The prototype of Question 1 with the suggested solution:

Enrichment: Questions that are not included in HKALE Pure Mathematics from 1994 onwards

Topic 1: Relation

Topic 2: Combinatorics - Counting Principles

Topic 3: Number Theory - Modulus

Question 6 with the suggested solution:

It is indefinite integration.

Question 7: Sequences

Question 8: Improper Integral

By checking the condition(s) of the final answer, you earn 1 more method mark,

Question 9:

Given f(x) = (x^2 + a) e^x. If f(x) has no extremum but one point of inflextion, what is the range of a?

Solution:

f'(x) = (x^2 + 2x + a) e^x

f"(x) = (x^2 + 4x + 2 + a) e^x

Since f(x) has no extremum, we have f'(x) ≥ 0,

because e^x > 0 and (x^2 + a) has to be equal or greater than zero.

(Otherwise f'(x) can have more than 1 root).

Now x^2 + 2x + a ≥ 0,

i.e. ∆ = 4 - 4a ≤ 0

a ≥ 1

Since f(x) has point of inflexion, f"(x) has 2 distinct roots.

For x^2 + 4x + 2 + a = 0,

∆ = 16 - 4(a+2) > 0

i.e. a < 2

So the range of a is [1, 2).

Question 10: Now we blend both M1 and M2

Given a random variable X observe Poisson distribution with the parameter = 1, compute the expected value E(|X-E(X)|).

observes*

Solution:

Since X ~ P(1), then E(X) = 1,

P{X = k} = (1/e)(1/k!). k = 0, 1, 2, ...

E(|X-1|) = 1 P{X = 0} + 0 P{X = 1} + 1P{X = 2} + ... + k P{X = k+1} + ...

= 1/e + 0 + ∑ (k/e)(1/((k+1)!))

= 1/e + 1/e [∑ (k+1-1)/((k+1)!)]

= 1/e + 1/e [∑ (1/(k!)) - ∑ (1/(k+1)!)]

= 1/e + 1/e [(1/1 + 1/2 + .... ) - (1/2 + 1/3 + ...)]

= 1/e + 1/e = 2/e.

= 1/e + 1/e [(1/1! + 1/2! + .... ) - (1/2! + 1/3! + ...)]

= 1/e + 1/e = 2/e.

Question 11:

Compute this power series:

Σ (x)^2n / (2n)!

Answers: straightforward

Recall e^x, e^r and put r= -x

Then the given expression is:

(e^x + e^r)/2

Question 12:

Given (1 + x)^n, using binomial expression,

find the greatest coefficient(s) of x^n.

This can be Googled directly.

Using Pascal triangle, the answer is obvious.

binomial expansion

Solution:

The greatest coefficient is in the middle.

For n being odd, take n=5,

then C2 and C3 are greatest.

For n being even, take n=8,

then C4 is the greatest.

Question 13:

Without loss of generality, outline how to compute the greatest term in binomial expansion of (ax + b)^n where a and b are any real numbers.

Solution:

Question 14:

Explain how to use inverse matrix, and otherwise, to solve a system of simultaneous kinear equations.

Solution:

In Singaporean GCE O Level additional mathematics, the square matrix is confined to 2x2 instead of 3x3.

Solution (Continued):

In HKDSE or Singaporean A Level, we can preview the method below apart from using conventional methods:

Solution (Continued):

We can use the method above to compute any inverse matrix we like:

linear* equations

Question 15: I am going to introduce mean value theorem (MVT). First, do the following reading comprehension for DSERs:

Consult Google for everything you get stuck with.

Now I show a typical question type for MVT:

Below is a harder example:

上面睇唔明？

嗱，一早話中大數學系係難過科大