CAIE P1 2018 November — Question 11 12 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2018
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferentiating Transcendental Functions
TypeFind stationary points - trigonometric functions
DifficultyModerate -0.3 This is a straightforward multi-part calculus question requiring standard techniques: differentiating a sum involving a square root (chain rule), integrating using reverse chain rule, finding a stationary point by setting dy/dx = 0, and computing a definite integral for area. All steps are routine A-level procedures with no novel problem-solving required, making it slightly easier than average.
Spec1.07i Differentiate x^n: for rational n and sums1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08e Area between curve and x-axis: using definite integrals
\includegraphics{figure_11} The diagram shows part of the curve \(y = 3\sqrt{(4x + 1)} - 2x\). The curve crosses the \(y\)-axis at \(A\) and the stationary point on the curve is \(M\).
  1. Obtain expressions for \(\frac{\text{d}y}{\text{d}x}\) and \(\int y \text{d}x\). [5]
  2. Find the coordinates of \(M\). [3]
  3. Find, showing all necessary working, the area of the shaded region. [4]
Question 11:
AnswerMarks Guidance
11(i)dy =   3 ×( 4x+1 )− 1 2   [×4] [− 2]   6 −2  
dx 2   4x+1 B2,1,0 Looking for 3 components
∫ydx =   3 ( 4x+1 )3 2 ÷ 3  [ ÷ 4 ] [ − 2x² ] (+ C)
 2 2
 ( 4x+1 )3 
= 2 
−x2
 2 
 
AnswerMarks Guidance
 B1 B1 B1 B1 for 3 ( 4x+1 )3 2 ÷ 3 B1 for ‘÷4’. B1 for ‘− 2x2 ’.
2 2
Ignore omission of + C. If included isw any attempt at
evaluating.
5
AnswerMarks
11(ii)dy 6
At M, = 0 → = 2
AnswerMarks Guidance
dx 4x+1M1 dy
Sets their 2 term to 0 and attempts to solve
dx
(as far as x = k)
AnswerMarks
x = 2, y = 5A1 A1
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
11(iii)2
 1( )3 
Area under the curve =  4x+1 2 − x² 
2 
AnswerMarks Guidance
0M1 Uses their integral and their ‘2’ and 0 correctly
(13.5 – 4) – 0.5 or 9.5 – 0.5 = 9A1 No working implies use of integration function on calculator
M0A0.
Area under the chord = trapezium = ½ × 2 × (3 + 5) = 8
2
x2 
Or  +3x =8
2
 
AnswerMarks Guidance
0M1 Either using the area of a trapezium with their 2, 3 and 5 or
∫ ( theirx+3 ) dx using their ‘2’ and 0 correctly.
AnswerMarks Guidance
(Shaded area = 9 – 8) = 1A1 Dependent on both method marks,
OR Area between the chord and the curve is:
2
∫3 4x+1−2x− ( x+3 ) dx
0
2
=∫3 4x+1−3x−3dx
AnswerMarks Guidance
0M1 Subtracts their line from given curve and uses their ‘2’ and 0
correctly.
2
 1( )3 x2 
=3  4x+1 2 − −x 
6 2
 
AnswerMarks Guidance
0A1 All integration correct and limits 2 and 0.
27  1
=3 −2−2  −  
AnswerMarks Guidance
 6  6M1 Evidence of substituting their ‘2’ and 0 into their integral.
1 1 1
=3 −  =3  =1
AnswerMarks Guidance
2 6 3A1 No working implies use of a calculator M0A0.
[4]
Question 11:
--- 11(i) ---
11(i) | dy =   3 ×( 4x+1 )− 1 2   [×4] [− 2]   6 −2  
dx 2   4x+1  | B2,1,0 | Looking for 3 components
∫ydx =   3 ( 4x+1 )3 2 ÷ 3  [ ÷ 4 ] [ − 2x² ] (+ C)
 2 2
 ( 4x+1 )3 
= 2 
−x2
 2 
 
  | B1 B1 B1 | B1 for 3 ( 4x+1 )3 2 ÷ 3 B1 for ‘÷4’. B1 for ‘− 2x2 ’.
2 2
Ignore omission of + C. If included isw any attempt at
evaluating.
5
--- 11(ii) ---
11(ii) | dy 6
At M, = 0 → = 2
dx 4x+1 | M1 | dy
Sets their 2 term to 0 and attempts to solve
dx
(as far as x = k)
x = 2, y = 5 | A1 A1
3
Question | Answer | Marks | Guidance
--- 11(iii) ---
11(iii) | 2
 1( )3 
Area under the curve =  4x+1 2 − x² 
2 
0 | M1 | Uses their integral and their ‘2’ and 0 correctly
(13.5 – 4) – 0.5 or 9.5 – 0.5 = 9 | A1 | No working implies use of integration function on calculator
M0A0.
Area under the chord = trapezium = ½ × 2 × (3 + 5) = 8
2
x2 
Or  +3x =8
2
 
0 | M1 | Either using the area of a trapezium with their 2, 3 and 5 or
∫ ( theirx+3 ) dx using their ‘2’ and 0 correctly.
(Shaded area = 9 – 8) = 1 | A1 | Dependent on both method marks,
OR Area between the chord and the curve is:
2
∫3 4x+1−2x− ( x+3 ) dx
0
2
=∫3 4x+1−3x−3dx
0 | M1 | Subtracts their line from given curve and uses their ‘2’ and 0
correctly.
2
 1( )3 x2 
=3  4x+1 2 − −x 
6 2
 
0 | A1 | All integration correct and limits 2 and 0.
27  1
=3 −2−2  −  
 6  6 | M1 | Evidence of substituting their ‘2’ and 0 into their integral.
1 1 1
=3 −  =3  =1
2 6 3 | A1 | No working implies use of a calculator M0A0.
[4]
\includegraphics{figure_11}

The diagram shows part of the curve $y = 3\sqrt{(4x + 1)} - 2x$. The curve crosses the $y$-axis at $A$ and the stationary point on the curve is $M$.

\begin{enumerate}[label=(\roman*)]
\item Obtain expressions for $\frac{\text{d}y}{\text{d}x}$ and $\int y \text{d}x$. [5]
\item Find the coordinates of $M$. [3]
\item Find, showing all necessary working, the area of the shaded region. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2018 Q11 [12]}}
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Q1 4 Q2 4 Q3 5 Q4 6 Q5 7 Q6 7 Q7 7 Q8 7 Q9 7 Q10 9 Q11 12