A curve, C, has equation
$$y = \frac{e^{3x-5}}{x^2}$$
Show that C has exactly one stationary point - AQA - A-Level Maths Pure - Question 13 - 2019 - Paper 1
Question 13
A curve, C, has equation
$$y = \frac{e^{3x-5}}{x^2}$$
Show that C has exactly one stationary point.
Fully justify your answer.
Worked Solution & Example Answer:A curve, C, has equation
$$y = \frac{e^{3x-5}}{x^2}$$
Show that C has exactly one stationary point - AQA - A-Level Maths Pure - Question 13 - 2019 - Paper 1
Step 1
Differentiate the curve
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Answer
To find the stationary points, we first need to differentiate the function using the quotient rule. The quotient rule states that if you have a function y=vu, then the derivative dxdy is given by:
dxdy=v2vdxdu−udxdv
For our function, let:
u=e3x−5
v=x2
We calculate:
dxdu=3e3x−5
dxdv=2x
Now applying the quotient rule:
dxdy=x4x2(3e3x−5)−e3x−5(2x)
This simplifies to:
dxdy=x3e3x−5(3x−2)
Step 2
Set the derivative to zero
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Answer
Next, to find the stationary points, we set the derivative equal to zero:
dxdy=0
Thus:
e3x−5(3x−2)=0
Since e3x−5=0 for any real x, we have:
3x−2=0
Solving for x gives:
x=32
Step 3
Determine the number of stationary points
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Answer
To confirm this is the only stationary point, we need to check the factor 3x−2 from the derived equation. This leads us to:
Since there are no other factors in the equation that could yield a solution, x=32 is indeed the only stationary point.
We also note that the expression e3x−5 is positive for all real x, hence there is no possibility of additional stationary points coming from that term.
