Kuala Lumpur Trial 2025 Add Maths Paper 1, Question 15Functions

Solved on video in English · Form 4, Functions · 5:10 video

The question

(a)

Given fg(x) = 3/(2x − 3), x ≠ k, and f(x) = 3/(−2x + 9), x ≠ 9/2.

(a) Find the value of k.

(b)

(b) Find g(x).

(c)(i)

(c)(i) Hence, find the value of x if g(x) maps onto itself.

(c)(ii)

(c)(ii) Find g²n+6)(x).

Kuala Lumpur Trial 2025 Add Maths Paper 1, Question 15: the question as drawn in Eduly's video5:10

Part (a)

Step 1

fg(x) = 3/(2x − 3) is undefined when its denominator is zero.

Why this step

We can't divide by zero, so the excluded value comes from the denominator, not the numerator. Setting the numerator to 0 is a common slip.

2 more steps, each with the reason for it, in the full solution

Final answer

k = 3/2

Where students lose marks

Watch out: the excluded value comes from the denominator being zero, not the numerator. Then solve 2x − 3 = 0 carefully.

Part (b)

3 more steps, each with the reason for it, in the full solution

Final answer

g(x) = −x + 6

Where students lose marks

Watch out: fg(x) means f(g(x)), so substitute g(x) into f, and take care with the sign when dividing by −2.

Part (c)(i)

3 more steps, each with the reason for it, in the full solution

Final answer

x = 3

Where students lose marks

Watch out: 'maps onto itself' means g(x) = x, not g(x) = 0. Then solve −x + 6 = x carefully.

Part (c)(ii)

3 more steps, each with the reason for it, in the full solution

Final answer

g²n+6)(x) = x

Where students lose marks

Watch out: g applied twice gives x. The exponent 2n + 6 is even, so the answer is x, not −x + 6.

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