CAIE P1 2019 June — Question 10 13 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2019
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeEquation of tangent line
DifficultyStandard +0.3 This is a straightforward multi-part chain rule question requiring: (i) differentiation and tangent equation (standard), (ii) integration with substitution (routine), and (iii) related rates using dy/dx = 1/2 (direct application). All parts follow standard procedures with no novel insight required, making it slightly easier than average.
Spec1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08e Area between curve and x-axis: using definite integrals
\includegraphics{figure_10} The diagram shows part of the curve with equation \(y = (3x + 4)^{\frac{1}{3}}\) and the tangent to the curve at the point \(A\). The \(x\)-coordinate of \(A\) is 4.
  1. Find the equation of the tangent to the curve at \(A\). [5]
  2. Find, showing all necessary working, the area of the shaded region. [5]
  3. A point is moving along the curve. At the point \(P\) the \(y\)-coordinate is increasing at half the rate at which the \(x\)-coordinate is increasing. Find the \(x\)-coordinate of \(P\). [3]
Question 10:
AnswerMarks
10(i) 1( 3x+4 )− 1 
 2 
AnswerMarks Guidance
2 B1 oe
dy =   1( 3x+4 )− 1 2   × 3
AnswerMarks Guidance
dx 2 B1 ×3’
Must have ‘
dy 3
=
At x = 4, soi
AnswerMarks
dx 8B1
3
Line through (4, their4) with gradient their
AnswerMarks Guidance
8M1 If y ≠ 4 is used then clear evidence of substitution of x = 4 is needed
Equation of tangent is y−4= 3( x−4 ) or y= 3 x+ 5
AnswerMarks Guidance
8 8 2A1 oe
5
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
10(ii)1 5
Area under line =  4+  ×4=13
AnswerMarks Guidance
2 2B1 4
OR ∫ 3 x+ 5 =  3 x2 + 5 x  =[ 3+10 ]=13
 
8 2 16 2 
0
Area under curve: ∫ ( 3x+4 )1 2 = ( 3x+4 )3/2  [ ÷3 ]
 3/2 
AnswerMarks Guidance
 B1B1 Allow if seen as part of the difference of 2 integrals
[ ]
First B1 for integral without ÷3
[ ]
Second B1 must have ÷3
128 16 112 4
− = =12
AnswerMarks Guidance
9 9 9 9M1 Apply limits 0 → 4 to an integrated expression
4 5
Area = 13 ‒ 12 = (or 0.556)
AnswerMarks
9 9A1
Alternative method for question 10(ii)
AnswerMarks Guidance
Area for line = 1/2 × 4 × 3/2 = 3B1 4
OR ∫ 1( 8y−20 )= 1 4y2 −20= 1[ −16+25 ]=3
 
3 3 3
5/2
y3 4y
Area for curve = ∫⅓(y2 −4) =  −  
AnswerMarks
 9   3 B1B1
64 16 8 8 32
− − − =
   
AnswerMarks Guidance
 9 3  9 3 9M1 Apply limits 2 → 4 to an integrated expression for curve
32 5
Area = −3= (or 0.556)
AnswerMarks
9 9A1
5
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
10(iii)dy 1
=
AnswerMarks
dx 2B1
3( 3x+4 )− 1 2 = 1
AnswerMarks Guidance
2 2M1 Allow M1 for 3( 3x+4 )− 1 2 = 2.
2
( 3x+4 )1 2 =3 → 3x+4=9 → x= 5 oe
AnswerMarks
3A1
3
Question 10:
--- 10(i) ---
10(i) |  1( 3x+4 )− 1 
 2 
2  | B1 | oe
dy =   1( 3x+4 )− 1 2   × 3
dx 2  | B1 | ×3’
Must have ‘
dy 3
=
At x = 4, soi
dx 8 | B1
3
Line through (4, their4) with gradient their
8 | M1 | If y ≠ 4 is used then clear evidence of substitution of x = 4 is needed
Equation of tangent is y−4= 3( x−4 ) or y= 3 x+ 5
8 8 2 | A1 | oe
5
Question | Answer | Marks | Guidance
--- 10(ii) ---
10(ii) | 1 5
Area under line =  4+  ×4=13
2 2 | B1 | 4
OR ∫ 3 x+ 5 =  3 x2 + 5 x  =[ 3+10 ]=13
 
8 2 16 2 
0
Area under curve: ∫ ( 3x+4 )1 2 = ( 3x+4 )3/2  [ ÷3 ]
 3/2 
  | B1B1 | Allow if seen as part of the difference of 2 integrals
[ ]
First B1 for integral without ÷3
[ ]
Second B1 must have ÷3
128 16 112 4
− = =12
9 9 9 9 | M1 | Apply limits 0 → 4 to an integrated expression
4 5
Area = 13 ‒ 12 = (or 0.556)
9 9 | A1
Alternative method for question 10(ii)
Area for line = 1/2 × 4 × 3/2 = 3 | B1 | 4
OR ∫ 1( 8y−20 )= 1 4y2 −20= 1[ −16+25 ]=3
 
3 3 3
5/2
y3 4y
Area for curve = ∫⅓(y2 −4) =  −  
 9   3  | B1B1
64 16 8 8 32
− − − =
   
 9 3  9 3 9 | M1 | Apply limits 2 → 4 to an integrated expression for curve
32 5
Area = −3= (or 0.556)
9 9 | A1
5
Question | Answer | Marks | Guidance
--- 10(iii) ---
10(iii) | dy 1
=
dx 2 | B1
3( 3x+4 )− 1 2 = 1
2 2 | M1 | Allow M1 for 3( 3x+4 )− 1 2 = 2.
2
( 3x+4 )1 2 =3 → 3x+4=9 → x= 5 oe
3 | A1
3
\includegraphics{figure_10}

The diagram shows part of the curve with equation $y = (3x + 4)^{\frac{1}{3}}$ and the tangent to the curve at the point $A$. The $x$-coordinate of $A$ is 4.

\begin{enumerate}[label=(\roman*)]
\item Find the equation of the tangent to the curve at $A$. [5]

\item Find, showing all necessary working, the area of the shaded region. [5]

\item A point is moving along the curve. At the point $P$ the $y$-coordinate is increasing at half the rate at which the $x$-coordinate is increasing. Find the $x$-coordinate of $P$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2019 Q10 [13]}}
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Q1 5 Q2 5 Q3 6 Q4 5 Q5 7 Q6 7 Q7 9 Q8 8 Q9 10 Q10 13