SPM 2024 Add Maths Paper 1, Question 9Differentiation

Solved on video in English and Bahasa Melayu · Form 5, Differentiation · 5:55 video

The question

(a)

Diagram 6 shows the rectangle ABCD where AB = (5x − 1) cm and BC = 2x cm. At an instant, the value of x changes from 3 cm to (3 + p) cm, such that p is a small value. Using differentiation, express in terms of p, the small change in the area of ABCD, in cm², at that instant.

(b)(i)

Diagram 7 shows two lots of land, P and Q. Both lots will be planted completely with the same type of grass. Lot P is a rectangle with dimensions x m by (x − 1) m, and the perimeter of lot P is 54 m. Lot Q is a trapezium with parallel sides 3y m (top) and 5y m (bottom), and perpendicular height (38 − 8y) m; the area of lot Q is A m². (i) Using differentiation, find the value of y such that A is a maximum.

(b)(ii)

Lot Q is a trapezium with parallel sides 3y m (top) and 5y m (bottom), and perpendicular height (38 − 8y) m. (ii) Hence, determine the lot that will be planted with the most grass. Justify your answer.

Part (a)

Step 1

Express the area as a function of x: A = (5x − 1)(2x) = 10x² − 2x.

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

Final answer

δA ≈ 58p cm²

Where students lose marks

Watch out: substitute x = 3 into dA/dx before multiplying by δx. Leaving (20x − 2)p has no numerical value and loses marks.

Part (b)(i)

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

Final answer

y = 2.375

Where students lose marks

Watch out: for a maximum, set dA/dy = 0. Solving dA/dy < 0 only gives y < 2.375, not the value of y.

Part (b)(ii)

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

Final answer

Lot P, since 182 m² > 180.5 m²

Where students lose marks

Watch out: don't just say Lot P. Justify by comparing the two areas: 182 m² > 180.5 m².

Watch every step of this question solved on video (5:55), with the reason behind each step and a box that checks your own answer.

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