A-Level Edexcel Maths Pure Trigonometric Equations: Figure 2 shows part of the graph

Figure 2 shows part of the graph with equation $y = f(x)$, where $f(x) = 2|5 - x| + 3, \, x > 0$ Given that the equation $f(x) = k$, where $k$ is a constant, has exactly one root, (a) state the set of possible values of $k$; (b) Solve the equation $f(x) = \frac{1}{2} + 10$ The graph with equation $y = f(x)$ is transformed onto the graph with equation $y = 4f(x - 1)$ - Edexcel - A-Level Maths Pure - Question 6 - 2018 - Paper 5

Question 6

$Figure-2-shows-part-of-the-graph-with-equation-$y-=-f(x)$,-where---$f(x)-=-2|5---x|-+-3,-\,-x->-0$---Given-that-the-equation-$f(x)-=-k$,-where-$k$-is-a-constant,-has-exactly-one-root,--(a)-state-the-set-of-possible-values-of-$k$;----(b)-Solve-the-equation-$f(x)-=-\frac{1}{2}-+-10$----The-graph-with-equation-$y-=-f(x)$-is-transformed-onto-the-graph-with-equation-$y-=-4f(x---1)$-Edexcel-A-Level Maths Pure-Question 6-2018-Paper 5.png$

Figure 2 shows part of the graph with equation $y = f(x)$, where $f(x) = 2|5 - x| + 3, \, x > 0$ Given that the equation $f(x) = k$, where $k$ is a constant, has... show full transcript

Worked Solution & Example Answer:Figure 2 shows part of the graph with equation $y = f(x)$, where $f(x) = 2|5 - x| + 3, \, x > 0$ Given that the equation $f(x) = k$, where $k$ is a constant, has exactly one root, (a) state the set of possible values of $k$; (b) Solve the equation $f(x) = \frac{1}{2} + 10$ The graph with equation $y = f(x)$ is transformed onto the graph with equation $y = 4f(x - 1)$ - Edexcel - A-Level Maths Pure - Question 6 - 2018 - Paper 5

Step 1

state the set of possible values of k

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Answer

For the function to have exactly one root, the value of $k$ must be equal to the minimum value of $f(x)$ .

Calculating the minimum: Since the vertex of the function occurs at $x = 5$ , we find $f(5) = 2|5-5| + 3 = 3$ . Thus, the set of possible values for $k$ is $k = 3.$
In addition, $k$ must be greater than or equal to this minimum, so we have:
$k \geq 3 \text{ and } k \neq 3 \implies k > 3.$

Step 2

Solve the equation f(x) = 1/2 + 10

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Answer

We need to solve for $x$ in the equation:
$f(x) = \frac{1}{2} + 10 = \frac{21}{2}.$
Setting up the equation:
$2|5 - x| + 3 = \frac{21}{2}$
Subtracting $3$ from both sides:
$2|5 - x| = \frac{21}{2} - 3 = \frac{15}{2}.$
Dividing by $2$ :
$|5 - x| = \frac{15}{4}.$
This gives us two scenarios:

$5 - x = \frac{15}{4}$
\Rightarrow \quad x = 5 - \frac{15}{4} = \frac{20 - 15}{4} = \frac{5}{4}$
$5 - x = -\frac{15}{4}$
\Rightarrow \quad x = 5 + \frac{15}{4} = \frac{20 + 15}{4} = \frac{35}{4}$
Thus, the solutions are:
$x = \frac{5}{4} \text{ and } x = \frac{35}{4}.$

Step 3

State the value of p and the value of q

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Answer

To find the new vertex after the transformation $y = 4f(x - 1)$ , we need to consider that the transformation shifts the graph to the right by 1 unit and scales the function vertically by a factor of 4.

The original vertex of $f(x)$ is at $(5, 3)$ . After the horizontal shift, the new vertex is at $(5 + 1, 3) = (6, 3)$ . Now applying the vertical scaling, $q = 4 \cdot 3 = 12.$
Thus, the coordinates of the vertex $(p, q)$ are:
$(6, 12)$ So, $p = 6$ and $q = 12.$

state the set of possible values of k

Solve the equation f(x) = 1/2 + 10

State the value of p and the value of q

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