Question 3
Find the coefficient of $x^8$ in the expansion of $\left(1-x^2\right)\left(2x - \frac{1}{x}\right)^{10}$. [5]
Solution
Step 1: Expand the binomial $\left(2x - \frac{1}{x}\right)^{10}$
Using the binomial theorem, we have:
$$ \left(2x - \frac{1}{x}\right)^{10} = \sum_{k=0}^{10} \binom{10}{k} (2x)^k \left(-\frac{1}{x}\right)^{10-k}. $$Simplify each term in the summation:
$$ (2x)^k \left(-\frac{1}{x}\right)^{10-k} = 2^k x^k \cdot (-1)^{10-k} x^{-(10-k)} = 2^k (-1)^{10-k} x^{2k - 10}. $$Thus, the expansion becomes:
$$ \left(2x - \frac{1}{x}\right)^{10} = \sum_{k=0}^{10} \binom{10}{k} 2^k (-1)^{10-k} x^{2k - 10}. $$We need the coefficients of $x^6$ and $x^8$ in the expansion of $\left(2x - \frac{1}{x}\right)^{10}$.
For $x^6$:
$$ 2k - 10 = 6 \implies k = 8. $$Substitute $k=8$:
$$ \binom{10}{8} 2^8 (-1)^{10-8} = \binom{10}{8} 2^8 (-1)^2 = 45 \cdot 256 = 11520. $$For $x^8$:
$$ 2k - 10 = 8 \implies k = 9. $$Substitute $k=9$:
$$ \binom{10}{9} 2^9 (-1)^{10-9} = \binom{10}{9} 2^9 (-1)^1 = -10 \cdot 512 = -5120. $$Thus,
$$ \left(1 - x^2\right)\left(2x - \frac{1}{x}\right)^{10} = (1-x^2)(\cdots +11520 x^6 -5120 x^8 +\cdots) $$Step 3: Collect coefficients for $x^6$ and $x^8$
When multiplying by $1 - x^2$, the coefficient of $x^8$ comes from:
$$ (1-x^2) $$ $$ (-1)\cdot (\text{coefficient of } x^6) + (1)\cdot (\text{coefficient of } x^8). $$Substitute the values:
$$ (-1)(11520) + (1)(-5120) = -11520 - 5120 = -16640. $$Final Answer
The coefficient of $x^8$ is $\boxed{-16640}$.
Question 4
(a) Write $3 \lg x-2 \lg y^2-3$ as a single logarithm to base 10. [3]
(b) Solve the equation $\log _3 x+\log _x 3=\frac{5}{2}$. [5]
Solution
Step 1: Simplify $3 \lg x - 2 \lg y^2 - 3$ as a single logarithm to base 10
We start with the given expression:
$$ 3 \lg x - 2 \lg y^2 - 3. $$
Simplify each term:
1. Recall the logarithmic power rule:
$ a \lg b = \lg b^a $.
$ 3 \lg x = \lg x^3 $
$ 2 \lg y^2 = \lg (y^2)^2 = \lg y^4 $
2. Substitute the simplified terms:
$$ \lg x^3 - \lg y^4 - 3. $$
Combine using logarithmic properties:
1. Recall the logarithmic subtraction rule:
$ \lg a - \lg b = \lg \frac{a}{b} $.
$ \lg x^3 - \lg y^4 = \lg \frac{x^3}{y^4} $.
2. Introduce the constant $-3$:
$ \lg \frac{x^3}{y^4} - 3 = \lg \frac{x^3}{y^4} + \lg 10^{-3} $, because $-3 = \lg 10^{-3}$.
3. Combine using the addition rule:
$ \lg a + \lg b = \lg (ab) $:
$ \lg \frac{x^3}{y^4} + \lg 10^{-3} = \lg \left( \frac{x^3}{y^4} \cdot 10^{-3} \right) $.
Final expression:
$$ \lg \left( \frac{x^3}{10^3 y^4} \right). $$
Step 2: Solve the equation $\log_3 x + \log_x 3 = \frac{5}{2}$
Given the equation:
$$ \log_3 x + \log_x 3 = \frac{5}{2}. $$
Simplify $\log_x 3$:
Using the change of base formula, $ \log_x 3 = \frac{1}{\log_3 x} $. Substitute this into the equation:
$$ \log_3 x + \frac{1}{\log_3 x} = \frac{5}{2}. $$
Let $ y = \log_3 x $. Then the equation becomes:
$$ y + \frac{1}{y} = \frac{5}{2}. $$
Multiply through by $y$ to eliminate the fraction:
$$ y^2 + 1 = \frac{5}{2} y. $$
Rearrange into a standard quadratic form:
$$ 2y^2 - 5y + 2 = 0. $$
Solve the quadratic equation:
Use the quadratic formula $ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $:
$$ y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4}. $$
Simplify the solutions:
$$ y = \frac{5 + 3}{4} = 2 \quad \text{or} \quad y = \frac{5 - 3}{4} = \frac{1}{2}. $$
Interpret the solutions:
Recall $ y = \log_3 x $:
1. If $ y = 2 $, then $ \log_3 x = 2 $, so $ x = 3^2 = 9 $.
2. If $ y = \frac{1}{2} $, then $ \log_3 x = \frac{1}{2} $, so $ x = 3^{1/2} = \sqrt{3} $.
Final solutions:
$$ x = 9 \quad \text{or} \quad x = \sqrt{3}. $$
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