Let $f(x) = ext{ln}(2x - 5) + 2x^2 - 30$, \ x > 2.5 - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 1
Question 9
Let $f(x) = ext{ln}(2x - 5) + 2x^2 - 30$, \ x > 2.5.
(a) Show that $f(x) = 0$ has a root $\alpha$ in the interval [3.5, 4].
A student takes 4 as the first approxi... show full transcript
Worked Solution & Example Answer:Let $f(x) = ext{ln}(2x - 5) + 2x^2 - 30$, \ x > 2.5 - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 1
Step 1
Show that $f(x) = 0$ has a root $\alpha$ in the interval [3.5, 4]
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Answer
First, evaluate the function at the endpoints of the interval:
Calculate f(3.5):
f(3.5)=ln(2×3.5−5)+2(3.5)2−30=ln(2)+24.5−30≈−4.8.
Calculate f(4):
f(4)=ln(2×4−5)+2(4)2−30=ln(3)+32−30≈3.1.
Since f(3.5)<0 and f(4)>0, there is a change of sign in the interval [3.5, 4], which implies by the Intermediate Value Theorem that there is at least one root α in the interval.
Step 2
apply the Newton-Raphson procedure once to obtain a second approximation for $\alpha$
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Rounding to 3 significant figures, the second approximation is:
x1≈3.81.
Step 3
Show that $\alpha$ is the only root of $f(x) = 0$
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Answer
To show that α is the only root, we must examine the behavior of f(x):
The function is defined for x>2.5.
Calculate the derivative:
f′(x)=2x−52+4x.
Identify critical points by setting f′(x)=0:
The only critical point occurs when f′(x) does not change sign. This can be shown graphically or by analyzing the sign of f′(x):
Since both terms are positive for x>2.5, f′(x)>0 for all x in (2.5,∞), indicating that the function is monotonically increasing in this interval.
With f(x) being continuous and strictly increasing, it can have at most one root. Since we found one root in [3.5, 4], we conclude that α is the only root of f(x)=0.
