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 $y = 4f(x - 1)$. The vertex on the graph with equation $y = 4(f(x - 1))$ has coordinates $(p, q)$. (c) State the value of $p$ and the value of $q$.

A-Level Edexcel Maths Pure Compound & Double Angle Formulae: 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 $f(x) = \frac{1}{2} + 10$ - 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-$f(x)-=-\frac{1}{2}-+-10$-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 $f(x) = \frac{1}{2} + 10$ - Edexcel - A-Level Maths Pure - Question 6 - 2018 - Paper 5

Step 1

state the set of possible values of k

96%

114 rated

Answer

To determine the set of possible values of $k$ for which the equation $f(x) = k$ has exactly one root, we must consider the properties of the function $f(x)$ . The function is a V-shaped graph due to the absolute value, opening upwards and having a vertex at $(5, 3)$ . Therefore, the minimum value of $f(x)$ is 3 when $x = 5$ . For $f(x) = k$ to have exactly one solution, $k$ must equal this minimum value, leading us to:

$k = 3$

Thus, the correct answer is:

$k = 3 \text{ or } k > 13$ .

Step 2

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

99%

104 rated

Answer

We start by rewriting the equation as:

$f(x) = 10.5$

Substituting the expression of $f(x)$ into the equation, we have:

$2|5 - x| + 3 = 10.5$

Simplifying this gives:

$2|5 - x| = 7.5$

which simplifies to:

$|5 - x| = 3.75$

This leads to two cases:

$5 - x = 3.75 \implies x = 1.25$
$5 - x = -3.75 \implies x = 8.75$

Thus, the solutions to the equation are $x = 1.25$ and $x = 8.75$ .

Step 3

State the value of p and the value of q

96%

101 rated

Answer

The function is transformed to $y = 4f(x - 1)$ . The vertex of the original function is at $(5, 3)$ . By applying the transformation, we translate the vertex 1 unit to the right:

$x = 5 + 1 = 6$

The y-coordinate now scales by a factor of 4:

$y = 4 * 3 = 12$

Hence, the coordinates of the vertex on the transformed graph are $(p, q) = (6, 12)$ . Thus,

$p = 6 \, \text{ and } \, q = 12$ .

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

Join the A-Level students using SimpleStudy...