CAIE P1 2024 November — Question 11 12 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind second derivative
DifficultyStandard +0.3 This is a standard multi-part differentiation question requiring routine techniques: basic differentiation of power functions, finding stationary points, and using tangent equations. Part (c) involves more algebraic manipulation but follows a predictable method of equating two tangent line equations. All techniques are standard P1 content with no novel insight required, making it slightly easier than average.
Spec1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives
The equation of a curve is \(y = kx^{\frac{1}{2}} - 4x^2 + 2\), where \(k\) is a constant.
  1. Find \(\frac{\text{d}y}{\text{d}x}\) and \(\frac{\text{d}^2y}{\text{d}x^2}\) in terms of \(k\). [2]
  2. It is given that \(k = 2\). Find the coordinates of the stationary point and determine its nature. [4]
  3. Points \(A\) and \(B\) on the curve have \(x\)-coordinates 0.25 and 1 respectively. For a different value of \(k\), the tangents to the curve at the points \(A\) and \(B\) meet at a point with \(x\)-coordinate 0.6. Find this value of \(k\). [6]
Question 11:
AnswerMarks
11(a)dy 1 − 1
= kx 2 −8x
AnswerMarks
dx 2B1
d2y 1 − 3
=− kx 2 −8
AnswerMarks
dx2 4B1
2
AnswerMarks
11(b)1 3  3 
− 1
x 2 −8x=0 ⇒ 1−8x2 =0 or x−1 =64x2 x3 = or 8x2 =1 
 64 
dy
Setting their to zero and solving, providing their only error(s) are
dx
AnswerMarks Guidance
incorrect coefficientsM1 OE
1
Award if working leads to x= WWW.
4
1
−
Squaring x 2 −8x2 = 0 to x−1−64x2 =0 gets M0.
1
x= only
AnswerMarks Guidance
4A1 If x=0 included, A0 and max of 3/4.
1
1 −
SC B1 only for x= only from squaring x 2 −8x2 = 0
4
directly to x−1−64x2 =0 (SC B1 replacing the M1A1).
11
y=
AnswerMarks Guidance
4A1 1
11 −
SC B1 for y= from squaring x 2 −8x2 = 0 to
4
x−1−64x2 =0.
d2y 1 − 3
=− x 2 −8 which is negative, so maximum
AnswerMarks Guidance
dx2 2B1 FT WWW
d2y
FT their x-value and their .
dx2
No FT if x=0 is the only solution.
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
11(c)1
When x=1,attempting to find y=k−2and gradient = k−8
AnswerMarks Guidance
2M1* OE
SC B1 if both correct gradients only, or both correct
y-coordinates only.
1 
Equation of tangent is y−k+2= k−8  (x−1)
AnswerMarks Guidance
2 A1 k  k k k
OE, e.g. y= −8x+ +6 or y= x−8x+ +6.
2  2 2 2
1 1
When x= ,attempting to find y= k+1.75 and gradient = k−2
AnswerMarks Guidance
4 2M1* OE
1
Equation of tangent is y− k−1.75 =(k−2 )(x−0.25)
AnswerMarks Guidance
2A1 k 9 k 9
OE, e.g. y=(k−2)x+ + or y=kx−2x+ + .
4 4 4 4
1  1
Meet at  k−8  (0.6−1)+k−2=(k−2 )(0.6−0.25)+ k+1.75
2  2
AnswerMarks Guidance
Equate two tangent equations and substitute x=0.6DM1 k  k k 9
OE, e.g.  −80.6+ +6 =(k−2)0.6+ + .
2  2 4 4
M0 if constants in both equations are the same.
⇒ −0.2k+k+3.2−2=0.35k−0.7+0.5k+1.75 
⇒ 0.05k= 0.15
AnswerMarks
k=3A1
6
Question 11:
--- 11(a) ---
11(a) | dy 1 − 1
= kx 2 −8x
dx 2 | B1
d2y 1 − 3
=− kx 2 −8
dx2 4 | B1
2
--- 11(b) ---
11(b) | 1 3  3 
− 1
x 2 −8x=0 ⇒ 1−8x2 =0 or x−1 =64x2 x3 = or 8x2 =1 
 64 
dy
Setting their to zero and solving, providing their only error(s) are
dx
incorrect coefficients | M1 | OE
1
Award if working leads to x= WWW.
4
1
−
Squaring x 2 −8x2 = 0 to x−1−64x2 =0 gets M0.
1
x= only
4 | A1 | If x=0 included, A0 and max of 3/4.
1
1 −
SC B1 only for x= only from squaring x 2 −8x2 = 0
4
directly to x−1−64x2 =0 (SC B1 replacing the M1A1).
11
y=
4 | A1 | 1
11 −
SC B1 for y= from squaring x 2 −8x2 = 0 to
4
x−1−64x2 =0.
d2y 1 − 3
=− x 2 −8 which is negative, so maximum
dx2 2 | B1 FT | WWW
d2y
FT their x-value and their .
dx2
No FT if x=0 is the only solution.
4
Question | Answer | Marks | Guidance
--- 11(c) ---
11(c) | 1
When x=1,attempting to find y=k−2and gradient = k−8
2 | M1* | OE
SC B1 if both correct gradients only, or both correct
y-coordinates only.
1 
Equation of tangent is y−k+2= k−8  (x−1)
2  | A1 | k  k k k
OE, e.g. y= −8x+ +6 or y= x−8x+ +6.
2  2 2 2
1 1
When x= ,attempting to find y= k+1.75 and gradient = k−2
4 2 | M1* | OE
1
Equation of tangent is y− k−1.75 =(k−2 )(x−0.25)
2 | A1 | k 9 k 9
OE, e.g. y=(k−2)x+ + or y=kx−2x+ + .
4 4 4 4
1  1
Meet at  k−8  (0.6−1)+k−2=(k−2 )(0.6−0.25)+ k+1.75
2  2
Equate two tangent equations and substitute x=0.6 | DM1 | k  k k 9
OE, e.g.  −80.6+ +6 =(k−2)0.6+ + .
2  2 4 4
M0 if constants in both equations are the same.
⇒ −0.2k+k+3.2−2=0.35k−0.7+0.5k+1.75 
⇒ 0.05k= 0.15
k=3 | A1
6
The equation of a curve is $y = kx^{\frac{1}{2}} - 4x^2 + 2$, where $k$ is a constant.

\begin{enumerate}[label=(\alph*)]
\item Find $\frac{\text{d}y}{\text{d}x}$ and $\frac{\text{d}^2y}{\text{d}x^2}$ in terms of $k$. [2]

\item It is given that $k = 2$.

Find the coordinates of the stationary point and determine its nature. [4]

\item Points $A$ and $B$ on the curve have $x$-coordinates 0.25 and 1 respectively. For a different value of $k$, the tangents to the curve at the points $A$ and $B$ meet at a point with $x$-coordinate 0.6.

Find this value of $k$. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2024 Q11 [12]}}
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