Summation of geometric sequence, logarithm, quadratic inequality

Question Sample Titled 'Summation of geometric sequence, logarithm, quadratic inequality'

Briana is studying about the number of mobile phones in a city. It is given that the total number of mobile phones at the end of the first year is ${90000}$ , and in subsequent years, the total number of mobile phones imported each year is ${r}\%$ of the total number of mobile phones at the end of the previous year, where ${r}$ is a constant, and the total number mobile phones abandoned each year is ${3000}$ . It is found that the total number of mobile phones at the end of the third year is ${1.026}\times{10}^{{5}}$.

 (a) (i) Express, in terms of ${r}$, the total number of mobile phones at the end of the second year. (ii) Find ${r}$ .  (4 marks) (b) (i) Express, in terms of ${n}$, the total number of mobile phones at the end of the ${n}^{{{t}{h}}}$ year. (ii) At the end of which year will the total number of mobile phones first exceed ${4.3}\times{10}^{{5}}$ ?  (5 marks) (c) It is assumed that the total number of mobile phones needed at the end of the ${n}^{{{t}{h}}}$ year is ${\left({a}{\left({1.21}\right)}^{{n}}+{b}\right)}$ , where ${a}$ and ${b}$ are constants. Some research results reveal that the total number of mobile phones needed at the end of the first year and the end of the second year are ${1.384}\times{10}^{{5}}$ and ${1.48564}\times{10}^{{5}}$ respectively. Briana claims that based on the assumption, the total number of mobile phones will be greater than the total number of mobile phones needed at the end of a certain year. Is the claim correct? Explain your answer. (4 marks)

 (ai) Required number of mobile phones $={90000}{\left({1}+{r}\%\right)}-{3000}$ 1A $={900}{r}+{87000}$  (aii) ${\left({90000}{\left({1}+{r}\%\right)}-{3000}\right)}{\left({1}+{r}\%\right)}-{3000}$ $={1.026}\times{10}^{{5}}$ 1M ${150}{\left({1}+{r}\%\right)}^{{2}}-{5}{\left({1}+{r}\%\right)}-{176}$ $={0}$ 1M ${\left({1}+{r}\%\right)}={1.1}$ or ${\left({1}+{r}\%\right)}=-\dfrac{{16}}{{15}}$ (rejected) Thus, we have ${r}={10}$ . 1A  (bi) Required number of mobile phones $={90000}{\left({1.1}\right)}^{{{n}-{1}}}-{3000}{\left({1.1}\right)}^{{{n}-{2}}}-{3000}{\left({1.1}\right)}^{{{n}-{3}}}-\ldots-{3000}$ 1M $={90000}{\left({1.1}\right)}^{{{n}-{1}}}-{3000}{\left(\dfrac{{{1.1}^{{{n}-{1}}}-{1}}}{{{1.1}-{1}}}\right)}$ 1M for sum of G.S. $={90000}{\left({1.1}\right)}^{{{n}-{1}}}-{30000}{\left({1.1}^{{{n}-{1}}}-{1}\right)}$ $={\left({60000}{\left({1.1}\right)}^{{{n}-{1}}}+{30000}\right)}$ 1A  (bii) Put ${60000}{\left({1.1}\right)}^{{{n}-{1}}}+{30000}>{4.3}\times{10}^{{5}}$ ${1.1}^{{{n}-{1}}}>\dfrac{{20}}{{3}}$ ${{\log{{\left({1.1}\right)}}}^{{{n}-{1}}}>}{\log{{\left(\dfrac{{20}}{{3}}\right)}}}$ 1M ${\left({n}-{1}\right)}{\log{{\left({1.1}\right)}}}>{\log{{\left(\dfrac{{20}}{{3}}\right)}}}$ ${n}-{1}>\dfrac{{{\log{{\left(\dfrac{{20}}{{3}}\right)}}}}}{{{\log{{\left({1.1}\right)}}}}}$ ${n}>{20.904694218190894}$ Thus, the total number of mobile phones will first exceed ${4.3}\times{10}^{{5}}$ at the end of the ${21}^{{{s}{t}}}$ year. 1A  (c) Note that ${a}{\left({1.21}\right)}^{{1}}+{b}={1.384}\times{10}^{{5}}$ and ${a}{\left({1.21}\right)}^{{2}}+{b}={1.48564}\times{10}^{{5}}$ . Solving, we have ${a}={40000}$ and ${b}={90000}$ . 1M Consider ${60000}{\left({1.1}\right)}^{{{n}-{1}}}+{30000}>{\left({40000}{\left({1.21}\right)}^{{n}}+{90000}\right)}$ $\ldots{\left(\ast\right)}$ ${22}{\left({\left({1.1}\right)}^{{{n}-{1}}}\right)}^{{2}}-{30}{\left({1.1}\right)}^{{{n}-{1}}}+{33}<{0}$ 1M $\Delta={\left(-{30}\right)}^{{2}}-{4}{\left({22}\right)}{\left({33}\right)}$ $=-{2004}$ $<{0}$ Since ${22}>{0}$ , for all ${n}$ , we have ${22}{\left({\left({1.1}\right)}^{{{n}-{1}}}\right)}^{{2}}-{30}{\left({1.1}\right)}^{{{n}-{1}}}+{33}>{0}$ . So, ${\left(\ast\right)}$ does not have real solution. 1M Thus, the claim is incorrect. 1A

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