Kuala Lumpur Trial 2025 Add Maths Paper 2, Question 8Differentiation

Solved on video in English · Form 5, Differentiation · 8:30 video

The question

(a)(i)

It is given that the equation of a curve is y = 4x⁵ − kx + 2, where k is a constant, and the curve passes through the point (1, 2).

(a)(i) Find dy/dx.

(a)(ii)

(a)(ii) Hence, find the equation of the tangent at the point (1, 2).

(b)

It is given that the gradient of the normal to the curve y = ax² + b/x² at the point (1/2, 5/2) is 1/6. Find the value of a and of b.

Step 1

First find k using the point (1, 2): 2 = 4(1)⁵ − k(1) + 2 → 2 = 4 − k + 2 → 2 = 6 − k → k = 4.

Why this step

Any point on a curve fits its equation, so putting in (1, 2) pins down k. Students who skip this step lose marks.

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

Final answer

dy/dx = 20x⁴ − 4

Where students lose marks

Watch out: find k first using the point (1, 2). The constant +2 differentiates to 0, and 4x⁵ becomes 20x⁴.

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

Final answer

y = 16x − 14

Where students lose marks

Watch out: the gradient is dy/dx at x = 1, not the y-value. Also −16 + 2 = −14, not −18.

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

Final answer

a = 2, b = 1/2

Where students lose marks

Watch out: 1/6 is the normal's gradient. Convert it with m₁m₂ = −1 to get −6, and check 2b ÷ (1/8) = 16b.

Watch every step of this question solved on video (8:30), with the reason behind each step and a box that checks your own answer.

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