SPM 2019 Add Maths Paper 2, Question 2Indices, Surds & Logarithms
Solved on video in English and Bahasa Melayu · 6 marks · Form 4, Indices, Surds & Logarithms · 4:44 video
The question
Express 2^(n+2) - 2^(n+1) + 2^(n-1) in the form p(2^(n-1)), where p is a constant. Hence, solve the equation 8(2^(n+2) - 2^(n+1) + 2^(n-1)) = 5(2^(n²).
[6 marks]
4:44Step 1
Write every term with base 2^(n-1): 2^(n+2) = 2^(n-1)*2³, 2^(n+1) = 2^(n-1)*2², 2^(n-1) = 2^(n-1)*2⁰.
Why this step
We want a common factor. Writing 2^(n+2) as 2^(n-1) × 2³ (a product, not 2^n + 2² lets us pull 2^(n-1) out of every term.
7 more steps, each with the reason for it, in the full solution
Final answer
p = 5; n = 2 or n = -1
Where students lose marks
Watch out: 2^(n+2) is 2^n × 2², not 2^n + 2². Also keep both roots of the quadratic in n.
Watch every step of this question solved on video (4:44), 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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