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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 function
Answer
To find the stationary points, we first differentiate the given function:\n\nLet ( y = \frac{e^{3x-5}}{x^2} ). We use the quotient rule for differentiation, which states that if ( y = \frac{u}{v} ), then ( \frac{dy}{dx} = \frac{u'v - uv'}{v^2} ).\n\nHere, ( u = e^{3x-5} ) and ( v = x^2 ).\n\nCalculating the derivatives, we have:\n( u' = 3e^{3x-5} ) (using the chain rule) and ( v' = 2x ).\n\nApplying the quotient rule: [ \frac{dy}{dx} = \frac{(3e^{3x-5})(x^2) - (e^{3x-5})(2x)}{(x^2)^2} = \frac{e^{3x-5}(3x^2 - 2x)}{x^4} ]
Step 2
Set the derivative to zero
Answer
Next, we set the derivative equal to zero to find the stationary points:\n[ \frac{dy}{dx} = 0 \Rightarrow e^{3x-5}(3x^2 - 2x) = 0 ]\nSince ( e^{3x-5} ) is never zero, we focus on the factor ( 3x^2 - 2x = 0 ). This can be factored:\n[ x(3x - 2) = 0 ] \nThus, the stationary points occur at ( x = 0 ) and ( x = \frac{2}{3} ).\nHowever, ( x = 0 ) is not valid since it would make the original function undefined.
Step 3
Analyze the stationary points
Answer
The only valid stationary point is at ( x = \frac{2}{3} ). To confirm that there is exactly one stationary point, we analyze the second derivative or the behavior of the first derivative near the stationary point.
Since we found only one valid stationary point with the first derivative of the function considered over the valid domain, we can conclude that ( C ) has exactly one stationary point.