Question 1
Figure 1 shows the triangle $ABC$
$\angle ABC = 90^\circ$
$AB = (2 + 4\sqrt{5}) \mathrm{cm}$
$BC = (a + b\sqrt{5}) \mathrm{cm}$ where $a$ and $b$ are integers.
The area of triangle $ABC = (34 + 11\sqrt{5}) \mathrm{cm}^2$
Without using a calculator, find the value of $a$ and the value of $b$ (4)
Solution
We are given a triangle $\triangle ABC$ with the following properties:
- $\angle ABC = 90^\circ$ (right triangle).
- $AB = (2 + 4\sqrt{5}) \ \mathrm{cm}$.
- $BC = (a + b\sqrt{5}) \ \mathrm{cm}$, where $a$ and $b$ are integers.
- The area of the triangle $\triangle ABC = (34 + 11\sqrt{5}) \ \mathrm{cm}^2$.
We aim to find the values of $a$ and $b$. Using the formula for the area of a right triangle:
$$ \text{Area} = \frac{1}{2} \times AB \times BC, $$we substitute the known values:
$$ (34 + 11\sqrt{5}) = \frac{1}{2} \times (2 + 4\sqrt{5}) \times (a + b\sqrt{5}). $$Step 1: Simplify the Equation
Multiply through by $2$ to eliminate the fraction:
$$ 68 + 22\sqrt{5} = (2 + 4\sqrt{5})(a + b\sqrt{5}). $$Expand the right-hand side:
$$ (2 + 4\sqrt{5})(a + b\sqrt{5}) = 2a + 2b\sqrt{5} + 4a\sqrt{5} + 20b. $$Combine like terms:
$$ = (2a + 20b) + (2b + 4a)\sqrt{5}. $$Equating the coefficients of $1$ and $\sqrt{5}$ on both sides:
$$ 2a + 20b = 68, $$ $$ 2b + 4a = 22. $$Step 2: Solve the System of Equations
From the first equation:
$$ 2a + 20b = 68 \implies a + 10b = 34. \tag{1} $$From the second equation:
$$ 2b + 4a = 22 \implies b + 2a = 11. \tag{2} $$Substitute $a$ from equation (1) into equation (2):
$$ a = 34 - 10b. \tag{3} $$Substitute equation (3) into equation (2):
$$ b + 2(34 - 10b) = 11. $$Expand and simplify:
$$ b + 68 - 20b = 11, $$ $$ -19b + 68 = 11 \implies -19b = -57 \implies b = 3. $$Substitute $b = 3$ into equation (3):
$$ a = 34 - 10(3) = 34 - 30 = 4. $$Step 3: Final Answer
The values of $a$ and $b$ are:
$$ a = 4, \quad b = 3. $$Question 2
$$ g(x)=2x^2+\frac{1}{2}x-3 $$(a) Express $g(x)$ in the form $p(x+q)^2+r$ where $p$, $q$ and $r$ are rational numbers to be found. (3)
(b) Find
(i) the minimum value of $g(x)$
(ii) the value of $x$ at which this minimum occurs.
(2)
(c) Hence, or otherwise, write down
(i) the minimum value of $h(x)$
(ii) the value of $x$ at which this minimum occurs.
(2)
Solution:
(a) Express $g(x)$ in the form $p(x+q)^2+r$
Given:
$$ g(x)=2x^2+\frac{1}{2}x-3 $$We complete the square:
$$ \begin{aligned} g(x) &= 2\left(x^2+\frac{1}{4}x\right)-3 \\ &= 2\left( x^2+\frac{1}{4}x+\frac{1}{64}-\frac{1}{64} \right)-3 \\ &= 2\left( \left(x+\frac{1}{8}\right)^2-\frac{1}{64} \right)-3 \\ &= 2\left(x+\frac{1}{8}\right)^2-\frac{2}{64}-3 \\ &= 2\left(x+\frac{1}{8}\right)^2-\frac{1}{32}-3 \\ &= 2\left(x+\frac{1}{8}\right)^2-\frac{97}{32}. \end{aligned} $$Thus, $g(x)$ in the form $p(x+q)^2+r$ is:
$$ g(x)=2\left(x+\frac{1}{8}\right)^2-\frac{97}{32}, $$where $p=2$, $q=\frac{1}{8}$, and $r=-\frac{97}{32}$.
(b) Find the minimum value of $g(x)$ and the value of $x$ at which it occurs
(i) The minimum value of $g(x)$ occurs when the square term is zero:
$$ \left(x+\frac{1}{8}\right)^2=0 \implies x=-\frac{1}{8}. $$Substituting $x=-\frac{1}{8}$ into $g(x)$:
$$ g\left(-\frac{1}{8}\right) = 2(0)^2-\frac{97}{32} = -\frac{97}{32}. $$Thus, the minimum value of $g(x)$ is $-\frac{97}{32}$, and it occurs at $x=-\frac{1}{8}$.
(c) Determine the minimum value of $h(x)$ and the value of $x$ at which it occurs
Given:
$$ h(x)=2x^6+\frac{1}{2}x^3-3. $$Notice that $h(x)$ is related to $g(x)$:
$$ h(x)=g(x^3), $$where
$$ g(x)=2x^2+\frac{1}{2}x-3. $$From part (b), the minimum value of $g(x)$ is $-\frac{97}{32}$, and it occurs at $x=-\frac{1}{8}$. Replacing $x$ in $g(x)$ with $x^3$ in $h(x)$, the minimum value of $h(x)$ occurs when
$$ x^3=-\frac{1}{8}. $$Solving for $x$:
$$ x=\sqrt[3]{-\frac{1}{8}}=-\frac{1}{2}. $$(i) The minimum value of $h(x)$ is:
$$ h(x)=-\frac{97}{32}. $$(ii) The value of $x$ at which this minimum occurs is:
$$ x=-\frac{1}{2}. $$
Post a Comment