SPM 2021 Add Maths Paper 2, Question 8Differentiation
Solved on video in English and Bahasa Melayu · Form 5, Differentiation · 7:11 video
The question
Diagram 4 shows a trapezium PQRT (with point S on the bottom side TR). QS is parallel to PT, and there is a right angle at T. It is given that TR = 26 cm, QS = x cm and SR = 2x cm. (a)(i) State the area, in cm², of PQST in terms of x. (a)(ii) Hence, find the maximum area, in cm², of PQST.
For the trapezium in Diagram 4, the area of PQST is A = 26x − 2x² cm². (b) Given that the rate of change of the area of PQST is 36 cm² s⁻¹ when x = √5 cm, calculate the length, in cm, of QS after 3 seconds.
7:11Part (a)
Step 1
PQST is a rectangle: its height is QS = PT = x, and its width is TS = TR − SR = 26 − 2x.
Why this step
TR is the whole base, but the rectangle only uses TS. SR = 2x belongs to triangle QSR, so we subtract it.
4 more steps, each with the reason for it, in the full solution
Final answer
(i) A = 26x − 2x² cm²; (ii) maximum area = 84.5 cm²
Where students lose marks
Watch out: differentiate and set dA/dx = 0. Setting A = 0 only gives x = 0 or 13, not the maximum area.
Part (b)
4 more steps, each with the reason for it, in the full solution
Final answer
QS ≈ 8.57 cm
Where students lose marks
Watch out: after 3 seconds, QS = starting x + (dx/dt)(3). Don't forget the starting x = √5 or the ×3.
Part (c)
4 more steps, each with the reason for it, in the full solution
Final answer
δA ≈ −0.36 cm² (area decreases)
Where students lose marks
Watch out: 2x² differentiates to 4x, not 2x. Using 26 − 2x gives −0.44 instead of −0.36.
Watch every step of this question solved on video (7:11), with the reason behind each step and a box that checks your own answer.
No card. Every past-year question on video, in English and BM.
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