Figure 2 shows a sketch of part of the graph $y = f(x)$, where
$f(x) = 2/3 - |x| + 5, \, x \geq 0$
(a) State the range of $f$ - Edexcel - A-Level Maths Pure - Question 13 - 2017 - Paper 2
Question 13
Figure 2 shows a sketch of part of the graph $y = f(x)$, where
$f(x) = 2/3 - |x| + 5, \, x \geq 0$
(a) State the range of $f$.
(b) Solve the equation
$f(x) = \fra... show full transcript
Worked Solution & Example Answer:Figure 2 shows a sketch of part of the graph $y = f(x)$, where
$f(x) = 2/3 - |x| + 5, \, x \geq 0$
(a) State the range of $f$ - Edexcel - A-Level Maths Pure - Question 13 - 2017 - Paper 2
Step 1
State the range of f
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Answer
To determine the range of the function f(x)=2/3−∣x∣+5, we first note that since x is non-negative (x≥0), ∣x∣=x. Thus, the function simplifies to:
f(x)=2/3−x+5=17/3−x.
As x increases starting from 0, the maximum value occurs when x=0, giving us:
f(0)=17/3.
As x approaches infinity, f(x) approaches negative infinity. Therefore, the range of f is:
f(x)≥5⇒f(x)∈[5,∞).
Step 2
Solve the equation f(x) = 1/2 x + 30
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Answer
To solve the equation f(x)=21x+30, we set:
2/3−x+5=21x+30.
Rearranging gives:
−2/3+5−30=(1/2+1)x
Multiplying through by 6 to eliminate fractions:
−4+30=(3+6)x,26=9x.
Thus, we find:
x=926.
Step 3
Given that the equation f(x) = k has two distinct roots, state the set of possible values for k
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Answer
For the equation f(x)=k to have two distinct roots, the horizontal line y=k must intersect the graph of f(x) at two points. From the graph of f(x), it can be concluded that k must be such that 5<k<11. Hence, the set of possible values for k is:
{k:5<k<11}.
