HKDSE M2 Question Review (20250615)

Question 17: Explain why the leading coefficient of a polynomial cannot be zero.

Solution:

The leading coefficient of a polynomial f(x)can be rewritten as the amplitude of a polynomial, i.e.:

f(x) = a (f(x)/a) = a g(x)

Put a = 0, then g(x) will be undefined.

Even without factor out the amplitude a, just recall a fact that for any number n:

0 • n = 0

The value of n cannot be zero.

Now put n=f(x), then f(x) cannot be zero and thus have no root such that it can no longer be a polynomial by the fundamental theorem of algebra, i.e. no degree at all.

Even if you argue that f(x) of even order can have complex roots, you will face the inequalities f(x) > 0 or f(x) < 0. So if you atempt to extract the complex roots, you will be comparing complex numbers which is nonsense.

Even if you argue that f(x) of even degrees can have complex roots, you will face the inequalities f(x) > 0 or f(x) < 0. So if you attempt to extract the complex roots, you will be comparing complex numbers which is nonsense.

Even without factoring out the amplitude a, just recall a fact that for any number n:

0 • n = 0

Try writing f(x) = x^2 + 1 <0,

if you attempt to extract the comple x roots:

-i < x < i

You will be comparing non-real values dude. Remember inequalities can be used to compare real numbers only.

By the same token, if y=mx + c

y is a polynomial of degree 1 if and only if the leading coefficient m is non-zero, otherwise we have 2 cases that defy the fundamental theorem of algebra that a polynomial of n degrees have exactly n roots:

Case 1: m = 0

Then y has no roots (y=c and c is non-zero real number) or infinite roots (y=0)

Case 2: m=infinity

Consider the inverse of y=c, i.e. x=c, the reasoning is similar to case 1.

Question 18:

Given -2 ≤ x ≤ 1, the following function has the maximum value 4:

y = - ( x - m)^2 + m^2 + 1

Find the value(s) of m.

Solution:

The vertex is (m, m^2 + 1).

Since the leading coefficient is negative, the vertex represents global maximum.

The axis of symmetry is x=m.

Here are 3 cases:

Case 1: m < -2

Then y is maximum when x = -2

m = -7/4 > -2

This case is rejected.

Case 2: -2≤m≤1

Then y is maximun when x = m

i.e. m^2 + 1 = 4

m = √3 or -√3

We reject √3 as √3 > 1.

Case 3: m > 1

Then y is maximum when x = 1

m = 2

So the answer is 2 or -√3.

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Question 19:

Solve 1/x > -2

Some will say:

-1 > -2x

1 < 2x

x >1/2

Is it valid?

Solution:

We don't know the sign of x,

to keep both side having the signs intact,

we do the following trick:

1/x • x^2 > -2 • x^2

x > -2 x^2

2x^2 > -x

2x^2 + x > 0

x(2x+1) >0

x< -1/2 or x >0

Alternatively, take cases.

Case 1: x=0

Go fuck yourself.

The expression is undefined then.

Case 1: x > 0

1 > -2x

2x > -1

x > -1/2

Notice that x > 0 > -1/2

Combining 2 inequality above using AND,

we have x > 0

Case 2: x< 0

1 < -2x

x < -1/2

Notice that x < -1/2 < 0

Combining 2 inequality above using AND,

we have x < -1/2