Question 5
DO NOT USE A CALCULATOR IN THIS QUESTION.
(a) Show that \(x-1\) is a factor of the expression \(x^3-2x^2-19x+20\). [1]
(b) Hence write \(x^3-2x^2-19x+20\) as a product of its linear factors. [3]
(c) Hence find the exact solutions of the equation \(\mathrm{e}^{3y}-2\mathrm{e}^{2y}-19\mathrm{e}^y+20=0\). [2]
Solution
Step (a): Show that \(x-1\) is a factor of the expression \(x^3-2x^2-19x+20\)
To show that \(x-1\) is a factor of the expression \(x^3 - 2x^2 - 19x + 20\), we can use the Factor Theorem. According to the Factor Theorem, if \(x-1\) is a factor of the polynomial, then substituting \(x=1\) into the polynomial should result in zero.
We evaluate the polynomial at \(x=1\):
\[ P(1) = (1)^3 - 2(1)^2 - 19(1) + 20 = 1 - 2 - 19 + 20 = 0. \]
Since \(P(1)=0\), we can conclude that \(x-1\) is indeed a factor of the polynomial \(x^3 - 2x^2 - 19x + 20\).
Step (b): Write \(x^3 - 2x^2 - 19x + 20\) as a product of its linear factors
Now that we know \(x-1\) is a factor, we can perform polynomial division to divide \(x^3 - 2x^2 - 19x + 20\) by \(x-1\).
We divide \(x^3 - 2x^2 - 19x + 20\) by \(x-1\) using synthetic division:
\[ \begin{array}{r|rrrr} 1 & 1 & -2 & -19 & 20 \\ & & 1 & -1 & -20 \\ \hline & 1 & -1 & -20 & 0 \\ \end{array} \]
The quotient is \(x^2 - x - 20\), and the remainder is \(0\).
Now, we factor \(x^2 - x - 20\). To factor this quadratic, we need two numbers that multiply to \(-20\) and add to \(-1\). These numbers are \(-5\) and \(4\), so:
\[ x^2 - x - 20 = (x - 5)(x + 4). \]
Thus, we can express the original cubic polynomial as:
\[ x^3 - 2x^2 - 19x + 20 = (x - 1)(x - 5)(x + 4). \]
Step (c): Find the exact solutions of the equation \(\mathrm{e}^{3y} - 2\mathrm{e}^{2y} - 19\mathrm{e}^y + 20 = 0\)
To solve the equation \(\mathrm{e}^{3y} - 2\mathrm{e}^{2y} - 19\mathrm{e}^y + 20 = 0\), we make the substitution:
\[ z = \mathrm{e}^y. \]
This transforms the equation into:
\[ z^3 - 2z^2 - 19z + 20 = 0. \]
From part (b), we already know that:
\[ z^3 - 2z^2 - 19z + 20 = (z - 1)(z - 5)(z + 4). \]
Thus:
\[ (z - 1)(z - 5)(z + 4) = 0. \]
This gives:
\[ z = 1, \quad z = 5, \quad z = -4. \]
Recall that \(z=\mathrm{e}^y\), so:
For \(z=1\):
\[ \mathrm{e}^y = 1 \]
\[ y = 0. \]
For \(z=5\):
\[ \mathrm{e}^y = 5 \]
\[ y = \ln 5. \]
For \(z=-4\):
\[ \mathrm{e}^y = -4. \]
Since \(\mathrm{e}^y\) is always positive, there is no real solution in this case.
Conclusion
\[ y = 0 \quad \text{or} \quad y = \ln 5. \]
Question 6
(a) A geometric progression has first term 64 and common ratio 0.5.
(i) Find the 10th term. [2]
(ii) Find the sum of the first 10 terms. [2]
(iii) Find the sum to infinity. [1]
(b) An arithmetic progression is such that $S_{20} - 400 = 2S_{10}$ and $u_1 : u_6$ is $1 : 5$. Find the sum of the first 3 terms of this progression. [6]
Solution
Step (a): Geometric Progression
A geometric progression has first term 64 and common ratio 0.5.
(i) Find the 10th term.
The general formula for the $n$-th term of a geometric progression is:
$u_n = a r^{n-1}$
Given $a = 64$, $r = 0.5$, and $n = 10$:
$u_{10} = 64 \times (0.5)^{10-1} = 64 \times (0.5)^9 = 64 \times \frac{1}{512} = \frac{64}{512} = \frac{1}{8}$
Thus, the 10th term is $u_{10} = \frac{1}{8}$.
(ii) Find the sum of the first 10 terms.
The sum of the first $n$ terms of a geometric progression is:
$S_n = \frac{a(1 - r^n)}{1 - r} \quad (r \lt 1)$
For $n=10$:
$S_{10} = \frac{64(1 - (0.5)^{10})}{1 - 0.5} = \frac{64(1 - \frac{1}{1024})}{0.5} = 128 \left(1 - \frac{1}{1024}\right)$
$S_{10} = 128 \times \frac{1023}{1024} = \frac{1023}{8} = 127.875$
Thus, $S_{10} = 127.875$.
(iii) Find the sum to infinity.
The sum to infinity of a geometric progression is:
$S_{\infty} = \frac{a}{1 - r} \quad (|r| \lt 1)$
$S_{\infty} = \frac{64}{1 - 0.5} = \frac{64}{0.5} = 128$
Thus, $S_{\infty} = 128$.
Step (b): Arithmetic Progression
An arithmetic progression is such that $S_{20} - 400 = 2S_{10}$ and $u_1 : u_6 = 1 : 5$.
Find the sum of the first 3 terms.
The sum of the first $n$ terms of an arithmetic progression is:
$S_n = \frac{n}{2}(2a + (n-1)d)$
From $u_1 : u_6 = 1 : 5$:
$\frac{a + 5d}{a} = 5 \Rightarrow a + 5d = 5a \Rightarrow 5d = 4a \Rightarrow d = \frac{4a}{5}$
Now:
$S_{20} = 10(2a + 19d), \quad S_{10} = 5(2a + 9d)$
Substitute $d = \frac{4a}{5}$:
$S_{20} = 10\left(2a + \frac{76a}{5}\right) = \frac{860a}{5}$
$S_{10} = 5\left(2a + \frac{36a}{5}\right) = 46a$
Given $S_{20} - 400 = 2S_{10}$:
$\frac{860a}{5} - 400 = 92a$
$\frac{860a - 460a}{5} = 400 \Rightarrow \frac{400a}{5} = 400$
$400a = 2000 \Rightarrow a = 5$
Then:
$d = \frac{4a}{5} = 4$
Now the first 3 terms sum:
$S_3 = \frac{3}{2}(2a + 2d) = \frac{3}{2}(10 + 8) = 27$
Thus, $S_3 = 27$.
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