2011 A-level H2 Mathematics (9740) Paper 1 Question 9 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Depth drilled on the 10th day = 256 +9(-7) = 193

T_n \textless 10

256 + (n-1)(-7) \textless 10

n > 36.1

Total depth = S_{37} = \frac{37}{2}[2(256) + 36(-7))] = 4810

(ii)
S_n > 0.99 s_{\infty}

\frac{256(1-(\frac{8}{9})^n)}{1-\frac{8}{9}} > 0.99 (\frac{256}{1-\frac{8}{9}})

(\frac{8}{9})^n \textless 0.01

n \mathrm{ln}(\frac{8}{9}) \textless \mathrm{ln} 0.01

n > \frac{\mathrm{ln} 0.01}{\mathrm{ln}(\frac{8}{9})} = 39.1

Thus, it takes 40 days

KS Comments:

I feel that (i) is fine, student just needs to read really carefully and beware if they are using the term or sum formulas.
For (ii), it is worth noting that some students still have no clue how to deal with inequalities when there is \mathrm{ln} involved.

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