AQA Further Paper 2 2023 June — Question 14 10 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2023
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeImproper integral to infinity with inverse trig
DifficultyChallenging +1.3 This is a Further Maths question requiring completion of the square, recognition of arctan derivative, and improper integral evaluation. Part (a) is straightforward algebra, part (b) is a standard arctan integration with careful arithmetic, and part (c) uses symmetry. While multi-step, these are well-practiced techniques for Further Maths students with no novel insight required, placing it moderately above average difficulty.
Spec4.08c Improper integrals: infinite limits or discontinuous integrands4.08h Integration: inverse trig/hyperbolic substitutions
The function \(f\) is defined by $$f(x) = \frac{1}{4x^2 + 16x + 19} \quad (x \in \mathbb{R})$$
  1. Show, without using calculus, that the graph of \(y = f(x)\) has a stationary point at \(\left(-2, \frac{1}{3}\right)\) [3 marks]
  2. Show that \(\int_{-2}^{-\frac{1}{2}} f(x) \, dx = \frac{\pi\sqrt{3}}{18}\) [5 marks]
  3. Find the value of \(\int_{-2}^{\infty} f(x) \, dx\) Fully justify your answer. [2 marks]
Question 14:
AnswerMarks
14(a)Completes the square for
denominator.
Or
( )
Sets f x =k and forms a
AnswerMarks Guidance
quadratic equation in x3.1a M1
f(x)=
( )
4 x2 +4x+ 19
4
1
=
( )
( )2 3
4 x+2 +
4
f is maximum when the denominator
is minimum, that is when x = −2
1 1
and y= =
( )
4 3 3
4
So the graph of y = f(x) has a
( )
stationary point at −2, 1
3
Explains that f has a stationary
point when x=−2
Or
Equates the discriminant of the
quadratic equation to 0 and
AnswerMarks Guidance
solves for k2.4 E1
Completes a reasoned
argument, without using
calculus, to show that the
 1
stationary point is at  −2,  to
 3
AnswerMarks Guidance
obtain the required result.2.1 R1
Subtotal3
QMarking Instructions AO
AnswerMarks
14(b)Expresses the denominator of
the integrand in completed
AnswerMarks Guidance
square form.3.1a M1
2 2
1 1
∫ f(x)dx= ∫ dx
4 ( x+2 )2 + 3
−2 −2
4
1
−
1 2  2 ( x+2 ) 2
−1
= × tan  
4 3   3 
−2
1 ( )
= tan −1 3−tan −1 0
2 3
1 π  π π 3
=  −0= =
2 3 3  6 3 18
Uses inverse tan to integrate
their integrand of the form
1
( x+k )2 +a2
Or
AnswerMarks Guidance
Makes a correct substitution3.1a M1
Integrates to obtain
2 ( x+2 )
Atan−1
AnswerMarks Guidance
31.1b A1
Substitutes the upper limit
correctly into their integrated
tan−1
AnswerMarks Guidance
expression which includes1.1a M1
Completes a reasoned
argument to obtain the required
result.
Must see substitution of -2 in the
AnswerMarks Guidance
integrated expression.2.1 R1
Subtotal5
QMarking Instructions AO
AnswerMarks
14(c)Replaces ∞by a letter (N)
and considers lim in the
N→∞
integral or integrated
AnswerMarks Guidance
expression.3.1a E1
−2 2 3 N→∞  3  −2
1 π 
=  −0
2 32 
π 3
=
12
Obtains the correct exact
value.
AnswerMarks Guidance
ACF2.2a B1
Subtotal2
Question total10
QMarking Instructions AO 
Question 14:
--- 14(a) ---
14(a) | Completes the square for
denominator.
Or
( )
Sets f x =k and forms a
quadratic equation in x | 3.1a | M1 | 1
f(x)=
( )
4 x2 +4x+ 19
4
1
=
( )
( )2 3
4 x+2 +
4
f is maximum when the denominator
is minimum, that is when x = −2
1 1
and y= =
( )
4 3 3
4
So the graph of y = f(x) has a
( )
stationary point at −2, 1
3
Explains that f has a stationary
point when x=−2
Or
Equates the discriminant of the
quadratic equation to 0 and
solves for k | 2.4 | E1
Completes a reasoned
argument, without using
calculus, to show that the
 1
stationary point is at  −2,  to
 3
obtain the required result. | 2.1 | R1
Subtotal | 3
Q | Marking Instructions | AO | Marks | Typical solution
--- 14(b) ---
14(b) | Expresses the denominator of
the integrand in completed
square form. | 3.1a | M1 | −1 −1
2 2
1 1
∫ f(x)dx= ∫ dx
4 ( x+2 )2 + 3
−2 −2
4
1
−
1 2  2 ( x+2 ) 2
−1
= × tan  
4 3   3 
−2
1 ( )
= tan −1 3−tan −1 0
2 3
1 π  π π 3
=  −0= =
2 3 3  6 3 18
Uses inverse tan to integrate
their integrand of the form
1
( x+k )2 +a2
Or
Makes a correct substitution | 3.1a | M1
Integrates to obtain
2 ( x+2 )
Atan−1
3 | 1.1b | A1
Substitutes the upper limit
correctly into their integrated
tan−1
expression which includes | 1.1a | M1
Completes a reasoned
argument to obtain the required
result.
Must see substitution of -2 in the
integrated expression. | 2.1 | R1
Subtotal | 5
Q | Marking Instructions | AO | Marks | Typical solution
--- 14(c) ---
14(c) | Replaces ∞by a letter (N)
and considers lim in the
N→∞
integral or integrated
expression. | 3.1a | E1 | ∞ ∫ f(x)dx= 1 lim  tan −1   2 (x+2 )    N
−2 2 3 N→∞  3  −2
1 π 
=  −0
2 32 
π 3
=
12
Obtains the correct exact
value.
ACF | 2.2a | B1
Subtotal | 2
Question total | 10
Q | Marking Instructions | AO | Marks | Typical solution
The function $f$ is defined by
$$f(x) = \frac{1}{4x^2 + 16x + 19} \quad (x \in \mathbb{R})$$

\begin{enumerate}[label=(\alph*)]
\item Show, without using calculus, that the graph of $y = f(x)$ has a stationary point at $\left(-2, \frac{1}{3}\right)$
[3 marks]

\item Show that $\int_{-2}^{-\frac{1}{2}} f(x) \, dx = \frac{\pi\sqrt{3}}{18}$
[5 marks]

\item Find the value of $\int_{-2}^{\infty} f(x) \, dx$

Fully justify your answer.
[2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2 2023 Q14 [10]}}
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