Let this be a reminder to you to show your workings properly during the exam. Calculators should only be used to confirm your answer.

BTW, if you didn't press wrongly, time to fix that calculator

Instead of using f(x) = 0, since this question specifically says f(x) = x, try thinking along the line of:

f(x) - x = 0 when x = 1, x = 2

Is it clearer now?

ps:sry for spammin juz tat everytime i come bak i see my comments gone thus asummin it was a glich or some kind

Read part 1 again. It says: The function f(x) = x³ + ax + b is such that the equation f(x) = x has solutions x = 1 and x = 2.

Don't always assume f(x) = 0!

P.S. Miss Loi have deleted your other two comments, since they are similar to your latest one.

Find the values of a and b

err i was under de impression that "has solutions x=1 and x =2 " means that f(x) = 0 when x is 1 or 2 thus by equating them i got 7..

anyway de last part i did de samemethod but i couldnt solve due to my ans in de first part being incorrect..

As mentioned, once you know what to do it's pretty simple.

The range of f^-1 = domain of f, and vice versa.

So y = f(x) maps to x = f^-1(y)

From this you can simply let

f^-1(15) = x

and you'll get:

15 = f(x) = x³ - 6x + 6
⇒ x³ - 6x -9 = 0

Classic textbook stuff, but unfortunately there will be stubborn students who are so intent on finding the full expression for f^-1 first before doing anything else.

The tedious part, as Kiroii and 123 pointed out, is to solve this cubic equation where you'll need to do a bit of trial and error for your root (see your Factor Theorem chapter).

When you look at the constant of the cubic equation which is 9, the possible values of the root can only be ±1, ±3 and ± 9.

So let's see ... f(1) ≠ 0 ... f(-1) ≠ 0 ... f(3) = 0 (eureka!)

Factorizing the cubic function with (x -3) with your super-duper long division skills:

And you'll get:

(x - 3)(x² + 3x + 3) = 0

To find the other roots from the quadratic equation, and using your

You'll discover that,

b² - 4ac < 0

So there is only one solution for f^-1(15) = 3

Yeah it's a little tedious but you do need to know all the possible roots of the cubic function in order to satisfy your final answer.

You just have to show that you understand the following concept:

A one-to-one function cannot have any turning point. (see diagram above if you are not sure)

So since the question talks about x, first you'll have to find the value of x at the turning point(s), which kiroii and 123 rightly showed us the way:

df(x)/dx = 0
3x² - 6 = 0

Solving the above,
x = ±√2

This means there are two turning points, one at x = √2 and another at x = -√2, which implies that f(x) is NOT one-to-one within this range.

But the question asks for x ≥ 2. Sadly this is where the insecure, under-confident, panic-stricken Singaporean student takes centrestage. (S)he will say "%#$*! The question says x = 2, how come I keep getting √2???!!", before proceeding redo and redo the question again.

Taking a page off that imaginary textbook called common sense, 2 is GREATER than √2 for goodness sake! Since the 2 turning points are contained within -√2 ≤ x ≤ √2, there will be no more turning points for x > √2 which also imply that there is no turning point for x ≥ 2!

So to answer this question in full, simply repeat the following. Everybody now!

The turning points of f(x) are found at x = ±√2. As f(x) has no turning points at x ≥ 2, it is a one-to-one function at x ≥ 2 and therefore f^-1 exists.

Look ma, no diagrams required!

Young and handsome 123 has summarized the steps correctly. Miss Loi is stating the obvious here but sometimes students don't really catch the keywords has solutions, which means the x=1 or 2 to satisfy f(x) = x. Duh.

So simply sub in these values of x:

(1)³ + a(1) + b = 1

⇒ a + b = 0 ----- (1)

(2)³ + a(2) + b = 2

⇒ 2a + b = -6 ----- (2)

Sub a = -b from (1) into (2) to solve the simultaneous equations,

You'll get a = -6 and b = 6

So simple right? How did young and handsome kiroii get a = 7????