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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 y with respect to x
Answer
To find the stationary points, we first need to differentiate the function. We can use the quotient rule, which states that if we have a function ( y = \frac{u}{v} ), then:
[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} ]
Here, let:
- ( u = e^{3x-5} )
- ( v = x^2 )
Calculating derivatives:
- ( \frac{du}{dx} = 3e^{3x-5} )
- ( \frac{dv}{dx} = 2x )
Applying the quotient rule: [ \frac{dy}{dx} = \frac{x^2 (3e^{3x-5}) - e^{3x-5} (2x)}{(x^2)^2} = \frac{e^{3x-5} (3x^2 - 2x)}{x^4} ]
Step 2
Find stationary points
Answer
Stationary points occur where the derivative is equal to zero: [ \frac{dy}{dx} = 0 \implies e^{3x-5}(3x^2 - 2x) = 0 ] Since ( e^{3x-5} ) is never zero, we have: [ 3x^2 - 2x = 0 ] Factoring gives: [ x(3x - 2) = 0 ] Thus, ( x = 0 ) or ( x = \frac{2}{3} ).
However, ( x = 0 ) is not valid in the context of our original equation since it would lead to division by zero.
Step 3
Justify the number of stationary points
Answer
The only valid stationary point is therefore at ( x = \frac{2}{3} ). To confirm that this is indeed the only stationary point, we analyze the function further:
- The factor ( e^{3x-5} ) ensures the function is always positive for real values of x.
- The quadratic ( 3x^2 - 2x ) indicates that the function can only cross the x-axis once, confirming one stationary point.
Hence, we conclude that the curve C has exactly one stationary point at ( x = \frac{2}{3} ), and it doesn't allow for any other stationary points.