Question 7
(2+5+2=9)Two numbers $x$ and $y$ are such that $3x - y = 4$, \[S = 5x^3 + y^2.\]
(a) Show that $S = 5x^3 + 9x^2 - 24x + 16$. (2)
Given that $x$ can vary,
(b) use calculus to find the value of $x$ for which $S$ is a minimum, justifying that this value of $x$ gives a minimum value of $S$. (5)
(c) Find the minimum value of $S$. (2)
Solution
Step 1: Express $y$ in terms of $x$
We are given the equation:
$$ 3x - y = 4. $$Rearranging for $y$, we have:
$$ y = 3x - 4. $$Step 2: Substitute $y$ into $S$
The expression for $S$ is given as:
$$ S = 5x^3 + y^2. $$Substituting $y = 3x - 4$ into $y^2$, we get:
$$ y^2 = (3x - 4)^2 = 9x^2 - 24x + 16. $$Thus, the expression for $S$ becomes:
$$ S = 5x^3 + 9x^2 - 24x + 16. $$This verifies that:
$$ S = 5x^3 + 9x^2 - 24x + 16. $$Step 3: Differentiate $S$ with respect to $x$
To find the critical points, we differentiate $S$ with respect to $x$:
$$ \frac{dS}{dx} = 15x^2 + 18x - 24. $$Step 4: Solve $\frac{dS}{dx} = 0$
Setting $\frac{dS}{dx} = 0$:
$$ 15x^2 + 18x - 24 = 0. $$Divide through by 3:
$$ 5x^2 + 6x - 8 = 0. $$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 5$, $b = 6$, and $c = -8$:
$$ x = \frac{-6 \pm \sqrt{6^2 - 4(5)(-8)}}{2(5)} $$ $$ x = \frac{-6 \pm \sqrt{36 + 160}}{10} $$ $$ x = \frac{-6 \pm \sqrt{196}}{10} $$ $$ x = \frac{-6 \pm 14}{10}. $$This gives two solutions:
$$ x = \frac{-6 + 14}{10} = 0.8, \quad x = \frac{-6 - 14}{10} = -2. $$Step 5: Determine the nature of the critical points
The second derivative of $S$ is:
$$ \frac{d^2S}{dx^2} = 30x + 18. $$Substitute $x = 0.8$:
$$ \frac{d^2S}{dx^2} = 30(0.8) + 18 = 24 + 18 = 42 > 0 $$Since $\frac{d^2S}{dx^2} > 0$, $S$ has a local minimum at $x = 0.8$.
Substitute $x = -2$:
$$ \frac{d^2S}{dx^2} = 30(-2) + 18 = -60 + 18 = -42 < 0 $$Since $\frac{d^2S}{dx^2} < 0$, $S$ has a local maximum at $x = -2$.
Step 6: Find the minimum value of $S$
Substitute $x = 0.8$ into the expression for $S$:
$$ S = 5x^3 + 9x^2 - 24x + 16 $$ $$ S = 5(0.8)^3 + 9(0.8)^2 - 24(0.8) + 16 $$ $$ S = 5(0.512) + 9(0.64) - 19.2 + 16 $$ $$ S = 2.56 + 5.76 - 19.2 + 16 $$ $$ S = 5.12 $$Note: The final line in the working is inconsistent with the stated answer.
Thus, the minimum value of $S$ (from correct substitution) is:
$$ \boxed{5.12} $$Question 8
(6+4=10)The sum to $n$ terms of an arithmetic series $A$ is $S_n$.
The sum of the first four terms of $A$ is 42 and the fifth term of $A$ is 23.
(a) Show that $S_n = \sum_{r=1}^n (P r - Q)$ where $P$ and $Q$ are prime numbers. (6)
$S_{2n} - 3 U_n = 1062$ where $U_n$ is the $n$th term of $A$.
(b) Find the value of $n$. (4)
Solution
Step 1: Expressing the general terms of the arithmetic series
The sum to $n$ terms of an arithmetic series $A$ is given by
$$ S_n = \frac{n}{2} \left( 2a + (n-1)d \right), $$where $a$ is the first term and $d$ is the common difference.
The $n$th term of the arithmetic series is given by
$$ U_n = a + (n-1)d. $$Step 2: Using the given information
The sum of the first four terms is given as $S_4 = 42$. Substituting $n = 4$ into the formula for $S_n$:
$$ S_4 = \frac{4}{2} \left( 2a + 3d \right) = 42. $$Simplify:
$$ 2a + 3d = 21. \tag{1} $$The fifth term is given as $U_5 = 23$. Substituting $n = 5$ into the formula for $U_n$:
$$ U_5 = a + 4d = 23. $$Simplify:
$$ a + 4d = 23. \tag{2} $$Step 3: Solving the system of equations
From equations (1) and (2), we solve for $a$ and $d$. Subtract equation (1) from twice equation (2):
$$ 2(a + 4d) - (2a + 3d) = 2 \times 23 - 21 $$ $$ 8d - 3d = 25 $$ $$ 5d = 25 $$ $$ d = 5. \tag{3} $$Substitute $d = 5$ into equation (2):
$$ a + 4 \times 5 = 23 $$ $$ a + 20 = 23 $$ $$ a = 3. \tag{4} $$Thus, $a = 3$ and $d = 5$.
Step 4: Verifying $S_n = \sum_{r=1}^n \left(Pr - Q\right)$
The $r$th term of the series is $U_r = a + (r-1)d$. Substituting $a = 3$ and $d = 5$:
$$ U_r = 3 + 5(r-1) = 5r - 2. $$The sum of the first $n$ terms is:
$$ S_n = \sum_{r=1}^n U_r = \sum_{r=1}^n \left( 5r - 2 \right) $$ $$ = \sum_{r=1}^n 5r - \sum_{r=1}^n 2 $$ $$ = 5 \sum_{r=1}^n r - 2n $$ $$ = 5 \cdot \frac{n(n+1)}{2} - 2n $$ $$ = \frac{5n(n+1)}{2} - 2n. \tag{5} $$This verifies the form $S_n = \sum_{r=1}^n \left( Pr - Q \right)$, where $P = 5$ and $Q = 2$.
Step 5: Solving for $n$ given $S_{2n} - 3U_n = 1062$
The sum of $2n$ terms is:
$$ S_{2n} = \frac{5(2n)((2n)+1)}{2} - 2(2n) $$ $$ = \frac{5 \cdot 2n \cdot (2n + 1)}{2} - 4n $$ $$ = 5n(2n+1) - 4n. \tag{6} $$The $n$th term is:
$$ U_n = a + (n-1)d $$ $$ = 3 + (n-1)5 $$ $$ = 5n - 2. \tag{7} $$Substitute into the given equation:
$$ S_{2n} - 3U_n = 1062 $$ $$ 5n(2n+1) - 4n - 3(5n - 2) = 1062 $$ $$ 10n^2 + 5n - 4n - 15n + 6 = 1062 $$ $$ 10n^2 - 14n + 6 = 1062 $$ $$ 10n^2 - 14n - 1056 = 0 $$ $$ 5n^2 - 7n - 528 = 0. \tag{8} $$Solve this quadratic equation using the quadratic formula:
$$ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ $$ n = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(5)(-528)}}{2(5)} $$ $$ n = \frac{7 \pm \sqrt{49 + 10560}}{10} $$ $$ n = \frac{7 \pm \sqrt{10609}}{10} $$ $$ n = \frac{7 \pm 103}{10} $$ $$ n = 11 \text{ (positive solution)}. \tag{9} $$Final Answer
The value of $n$ is
$$ \boxed{11}. $$
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