Challenging +1.2 This question requires finding dy/dx using the quotient rule with an exponential function, setting it to zero, and proving uniqueness of the solution. While it involves multiple techniques (quotient rule, exponential differentiation, algebraic manipulation) and requires a rigorous justification of uniqueness (likely showing the numerator equation has exactly one solution), it follows a standard calculus workflow without requiring deep insight. The 7 marks reflect the extended working needed, but the problem-solving approach is straightforward for a competent A-level student.
Question 13:
13 | Chooses an appropriate
technique to differentiate accept
any evidence of product rule or
quotient rule | 3.1a | M1 | x≠0 as y is undefined
dy 3e3x−5x2 −2xe3x−5
=
dx x4
dy
At a turning point =0
dx
3e3x−5x2 −2xe3x−5
⇒ =0
x4
⇒3e3x−5x2 −2xe3x−5 =0
⇒( 3x−2 ) xe3x−5 =0
2
⇒ x= , e3x−5 =0
3
e3x−5 ≠0
∴there is only one stationary
point.
e3x−5
Differentiates correctly | 1.1b | B1
dy
Obtains correct
dx
ACF | 1.1b | A1
Explains that stationary points
dy
occur when =0
dx | 2.4 | E1
dy
Equates their to zero and
dx
solves their equation with at least
one correct line of correct
rearrangement. Resulting in a
value for x. | 1.1a | M1
Deduces their factor e3x−5 ≠0 | 2.2a | B1F
Completes argument to show
exactly one
2
stationary point at x= .
3
Must include consideration of
x≠0somewhere. | 2.1 | R1
Total | 7
Q | Marking instructions | AO | Mark | Typical solution
A curve, C, has equation
$$y = \frac{e^{3x-5}}{x^2}$$
Show that C has exactly one stationary point.
Fully justify your answer.
[7 marks]
\hfill \mbox{\textit{AQA Paper 1 2019 Q13 [7]}}