Question 7
A geometric series $G$ with common ratio $r$, has first term $16$ and third term $\frac{2704}{625}$
(a) Find the two possible values of $r$ (2)
Given that $r \gt 0$
(b) find the sum to infinity of $G$ (2)
The sum to $n$ terms of $G$ is greater than $33$
(c) Find, using logarithms, the least possible value of $n$
Show your working clearly.
(5)
Step-by-step Solution
A geometric series $G$ with common ratio $r$, has first term 16 and third term $\frac{2704}{625}$.
Given:
- First term $a = 16$
- Third term $ar^2 = \frac{2704}{625}$
(a) Find the two possible values of $r$
From the formula for the $n$-th term of a geometric series:
$$ T_n = ar^{n-1} $$For the third term, $T_3$:
$$ ar^2 = \frac{2704}{625} $$ $$ 16r^2 = \frac{2704}{625} $$ $$ r^2 = \frac{2704}{625 \times 16} $$ $$ r^2 = \frac{169}{625} $$ $$ r = \pm \frac{13}{25} $$Thus, the two possible values of $r$ are:
$$ r = \frac{13}{25} \quad \text{or} \quad r = -\frac{13}{25} $$(b) Find the sum to infinity of $G$ (Given that $r > 0$)
The sum to infinity of a geometric series is given by:
$$ S_\infty = \frac{a}{1-r}, \quad \text{for } |r| \lt 1 $$Substituting $a = 16$ and $r = \frac{13}{25}$:
$$ S_\infty = \frac{16}{1 - \frac{13}{25}} $$ $$ = \frac{16}{\frac{25 - 13}{25}} $$ $$ = \frac{16 \times 25}{12} $$ $$ = \frac{400}{12} $$ $$ = \frac{100}{3} $$Thus, the sum to infinity is:
$$ S_\infty = \frac{100}{3} $$(c) Find the least possible value of $n$ (using logarithms)
The sum to $n$ terms of a geometric series is given by:
$$ S_n = \frac{a(1 - r^n)}{1 - r}, \quad \text{for } |r| \lt 1 $$We are given that $S_n > 33$, and $r = \frac{13}{25}$. Substituting into the formula:
$$ \frac{16(1 - r^n)}{1 - \frac{13}{25}} > 33 $$ $$ \frac{16(1 - r^n)}{\frac{12}{25}} > 33 $$ $$ 16(1 - r^n) > 33 \times \frac{12}{25} $$ $$ 16(1 - r^n) > \frac{396}{25} $$ $$ 1 - r^n > \frac{396}{25 \times 16} $$ $$ 1 - r^n > \frac{99}{100} $$ $$ r^n < \frac{1}{100} $$Taking logarithms (base 10) on both sides:
$$ \log_{10}(r^n) < \log_{10}\left(\frac{1}{100}\right) $$ $$ n \log_{10}\left(\frac{13}{25}\right) < -2 $$ $$ n > \frac{-2}{\log_{10}\left(\frac{13}{25}\right)} $$Using $\log_{10}\left(\frac{13}{25}\right) = \log_{10}(13) - \log_{10}(25)$:
$$ \log_{10}(13) \approx 1.11394, \quad \log_{10}(25) \approx 1.39794 $$ $$ \log_{10}\left(\frac{13}{25}\right) \approx 1.11394 - 1.39794 = -0.284 $$ $$ n > \frac{-2}{-0.284} $$ $$ n > 7.04 $$Thus, the least possible value of $n$ is:
$$ n = 8 $$Question 8
$$ y=\frac{2e^{3x+1}}{5x^2} $$(a) Find $\frac{dy}{dx}$
Give your answer in the form $\frac{Ae^{3x+1}(Bx-A)}{Cx^3}$ where $A$, $B$ and $C$ are prime numbers to be found. (5)
The value of $x$ increases by $2\%$
(b) Use your answer to part (a) to find an estimate, in terms of $x$, for the percentage change in $y$
Give your answer in the form $(Px-Q)$ where $P$ and $Q$ are integers. (3)
Step-by-Step Solution
We are given:
$$ y = \frac{2 e^{3x+1}}{5x^2} $$(a) Find \( \frac{\mathrm{d}y}{\mathrm{d}x} \)
The function \( y \) is a product of \( 2e^{3x+1} \) and \( \frac{1}{5x^2} \). To differentiate, we use the product rule:
$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}\left(2e^{3x+1}\right) \cdot \frac{1}{5x^2} + 2e^{3x+1} \cdot \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{5x^2}\right). $$Step 1: Differentiate \( 2e^{3x+1} \):
$$ \frac{\mathrm{d}}{\mathrm{d}x}\left(2e^{3x+1}\right) = 2 \cdot e^{3x+1} \cdot \frac{\mathrm{d}}{\mathrm{d}x}(3x+1) = 6e^{3x+1}. $$Step 2: Differentiate \( \frac{1}{5x^2} \):
$$ \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{5x^2}\right) = \frac{1}{5} \cdot \frac{\mathrm{d}}{\mathrm{d}x}(x^{-2}) = \frac{1}{5} \cdot (-2x^{-3}) = -\frac{2}{5x^3}. $$Step 3: Substitute into the product rule:
$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \left(6e^{3x+1}\right) \cdot \frac{1}{5x^2} + \left(2e^{3x+1}\right) \cdot \frac{-2}{5x^3}. $$Simplify each term:
$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6e^{3x+1}}{5x^2} - \frac{4e^{3x+1}}{5x^3}. $$Combine terms over a common denominator:
$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6e^{3x+1}x - 4e^{3x+1}}{5x^3}. $$Factorize the numerator:
$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2e^{3x+1}(3x-2)}{5x^3}. $$Thus, the answer is:
$$ \boxed{\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2e^{3x+1}(3x-2)}{5x^3}}. $$Here, \( A = 2 \), \( B = 3 \), and \( C = 5 \) are the prime numbers.
(b) Estimate the percentage change in \( y \)
We know that the percentage change in \( y \) can be approximated by:
$$ \text{Percentage change in } y \approx \frac{\mathrm{d}y}{\mathrm{d}x} \cdot \frac{\Delta x}{y} \times 100\%, $$where \( \Delta x \) is the percentage change in \( x \) (in decimal form). For a \( 2\% \) increase, \( \Delta x = 0.02x \).
Substitute \( \frac{\mathrm{d}y}{\mathrm{d}x} \):
$$ \text{Percentage change in } y \approx \frac{2e^{3x+1}(3x-2)}{5x^3} \cdot \frac{0.02x}{\frac{2e^{3x+1}}{5x^2}} \times 100\%. $$Simplify the expression:
$$ \text{Percentage change in } y \approx \frac{2e^{3x+1}(3x-2)}{5x^3} \cdot \frac{0.02x \cdot 5x^2}{2e^{3x+1}} \times 100\%. $$ $$ \text{Percentage change in } y \approx \frac{(3x-2) \cdot 0.02 \cdot 5x^3}{5x^3} \times 100\%. $$ $$ \text{Percentage change in } y \approx (3x-2) \cdot 0.02 \cdot \frac{x^3}{x^3} \times 100\%. $$ $$ \text{Percentage change in } y \approx (3x-2) \cdot 0.02 \times 100\%. $$ $$ \text{Percentage change in } y \approx 2(3x-2)\%. $$Expand and simplify:
$$ \text{Percentage change in } y \approx 6x - 4\%. $$Thus, the answer is:
$$ \boxed{6x - 4\%} $$where \( P = 6 \) and \( Q = 4 \).
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