CAIE P1 2016 November — Question 7 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2016
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeArea of triangle or parallelogram using vectors
DifficultyModerate -0.3 This is a straightforward vector question requiring standard techniques: dot products to verify perpendicularity (routine calculation), then area via cross product magnitude and volume formula application. While it has multiple parts and requires careful arithmetic, it involves only direct application of well-practiced methods with no conceptual challenges or novel insights needed.
Spec1.10c Magnitude and direction: of vectors1.10g Problem solving with vectors: in geometry
\includegraphics{figure_2} The diagram shows a triangular pyramid \(ABCD\). It is given that $$\overrightarrow{AB} = 3\mathbf{i} + \mathbf{j} + \mathbf{k}, \quad \overrightarrow{AC} = \mathbf{i} - 2\mathbf{j} - \mathbf{k} \quad \text{and} \quad \overrightarrow{AD} = \mathbf{i} + 4\mathbf{j} - 7\mathbf{k}.$$
  1. Verify, showing all necessary working, that each of the angles \(DAB\), \(DAC\) and \(CAB\) is \(90°\). [3]
  2. Find the exact value of the area of the triangle \(ABC\), and hence find the exact value of the volume of the pyramid. [4]
[The volume \(V\) of a pyramid of base area \(A\) and vertical height \(h\) is given by \(V = \frac{1}{3}Ah\).]
Question 7:
(ii) ---
7 (i)
AnswerMarks
(ii)AB.AC =3−2−1=0 hence perpendicular or 90˚
AB.AD =3+4−7=0 hence perpendicular or 90˚
AC.AD =1−8+7=0 hence perpendicular or 90˚ AG
Area ABC =(½) 32 +12 +12 × 12 +(−2)2 +(−1)2
=½ 11× 6
Vol. =⅓×their ∆ABC× 12 +42 +(−7)2
1
= 66× 66 = 1 1
AnswerMarks
6B1
B1
B1
M1
A1
M1
AnswerMarks
A1[3]
[4]3 ‒ 2 ‒ 1 or sum of prods etc
must be seen
Or single statement: mutually
perpendicular or 90˚ seen at least
once .
Expect ½ 66
Not 11.0
8 (i)
(ii)
AnswerMarks
(iii)(2x+3)2
+1 Cannot score retrospectively in (iii)
g(x)=2x+3 cao
y=(2x+3)2 +1⇒ 2x+3=(±) y−1 or ft from (i)
1 3
x=(±) y−1− or ft from (i)
2 2
1 3
(fg)−1(x)= x−1− cao Note alt. method g−1f−1
2 2
Domain is (x)>10
ALT. method for first 3 marks:
Trying to obtain g−1f−1(x)
 
g−1 =½(x−3), f−1 = x−1
1 3
A1 for x−1−
AnswerMarks
2 2B1B1B1
B1
M1
M1
A1
B1
*M1
DM1
AnswerMarks
A1[3]
[1]
AnswerMarks
[4]For a=2, b=3, c=1
In (ii),(iii) Allow if from
2
 3
4 x+ +1
 
 2
Or with x/y transposed.
Or with x/y transposed Allow
sign errors.
Must be a function of x. Allow y
= ....
Allow (10, ∞), 10 < x < ∞ etc.
but not with y or f or g involved.
Not ⩾10
Both required
AnswerMarks Guidance
Page 6Mark Scheme Syllabus
Cambridge International AS/A Level – October/November 20169709 13
9 (a)
AnswerMarks
(b)6 12
=
1−r 1+r
1
r =
3
S =9
13
2cosθ+12sin2θ=52
 
2
2cosθ+12(1−cos²θ)=8→6cos2θ−cosθ−2(=0)
cosθ=2/3 or −1/2 soi
AnswerMarks
θ=0.841 , 2.09 Dep on previous A1M1
A1
A1
M1*
DM1
A1
AnswerMarks
A1A1[3]
[5]Use of correct formula for sum of
AP
Use s2 =1−c2 & simplify to 3-
term quad
Accept 0.268π, 2π/3. SRA1 for
48.2˚, 120˚ Extra solutions in
range –1
10 (i)
(ii)
AnswerMarks
(iii)dy 2 1  3 
at x=a2, = + or 2a−2 +a−2 = or 3a−2 
dx a2 a2  a2 
y−3= 3 ( x−a2) or y= 3 x+c →3= 3 a2 +c
a2 a2 a2
3
y= x or 3a−2x cao
a2
2 x½ ax−½
(y)= + (+ c)
a ½ −½
sub x=a2, y=3 into ∫dy/dx
4x½
c = 1 (y= −2ax−½ +1)
a
4 1
sub x=16, y=8 → 8= ×4−2a× +1
a 4
a2 +14a−32 (=0)
a=2
AnswerMarks
A = (4, 3), B = (16, 8) AB2 =122 +52 → AB=13B1
M1
A1
B1B1
M1
A1
*M1
A1
A1
AnswerMarks
DM1A1[3]
[4]
AnswerMarks
[5]2 1
+ or 2a−2 +a−2 seen
a2 a2
anywhere in (i)
Through (a2,3) & with their
grad as f(a)
c must be present. Expect
3 = 4 ‒ 2 + c
Sub into their y
Allow ‒16 in addition
AnswerMarks Guidance
Page 7Mark Scheme Syllabus
Cambridge International AS/A Level – October/November 20169709 13 
Question 7:
--- 7 (i)
(ii) ---
7 (i)
(ii) | AB.AC =3−2−1=0 hence perpendicular or 90˚
AB.AD =3+4−7=0 hence perpendicular or 90˚
AC.AD =1−8+7=0 hence perpendicular or 90˚ AG
Area ABC =(½) 32 +12 +12 × 12 +(−2)2 +(−1)2
=½ 11× 6
Vol. =⅓×their ∆ABC× 12 +42 +(−7)2
1
= 66× 66 = 1 1
6 | B1
B1
B1
M1
A1
M1
A1 | [3]
[4] | 3 ‒ 2 ‒ 1 or sum of prods etc
must be seen
Or single statement: mutually
perpendicular or 90˚ seen at least
once .
Expect ½ 66
Not 11.0
8 (i)
(ii)
(iii) | (2x+3)2
+1 Cannot score retrospectively in (iii)
g(x)=2x+3 cao
y=(2x+3)2 +1⇒ 2x+3=(±) y−1 or ft from (i)
1 3
x=(±) y−1− or ft from (i)
2 2
1 3
(fg)−1(x)= x−1− cao Note alt. method g−1f−1
2 2
Domain is (x)>10
ALT. method for first 3 marks:
Trying to obtain g−1f−1(x)
 
g−1 =½(x−3), f−1 = x−1
1 3
A1 for x−1−
2 2 | B1B1B1
B1
M1
M1
A1
B1
*M1
DM1
A1 | [3]
[1]
[4] | For a=2, b=3, c=1
In (ii),(iii) Allow if from
2
 3
4 x+ +1
 
 2
Or with x/y transposed.
Or with x/y transposed Allow
sign errors.
Must be a function of x. Allow y
= ....
Allow (10, ∞), 10 < x < ∞ etc.
but not with y or f or g involved.
Not ⩾10
Both required
Page 6 | Mark Scheme | Syllabus | Paper
Cambridge International AS/A Level – October/November 2016 | 9709 | 13
9 (a)
(b) | 6 12
=
1−r 1+r
1
r =
3
S =9
13
2cosθ+12sin2θ=52
 
2
2cosθ+12(1−cos²θ)=8→6cos2θ−cosθ−2(=0)
cosθ=2/3 or −1/2 soi
θ=0.841 , 2.09 Dep on previous A1 | M1
A1
A1
M1*
DM1
A1
A1A1 | [3]
[5] | Use of correct formula for sum of
AP
Use s2 =1−c2 & simplify to 3-
term quad
Accept 0.268π, 2π/3. SRA1 for
48.2˚, 120˚ Extra solutions in
range –1
10 (i)
(ii)
(iii) | dy 2 1  3 
at x=a2, = + or 2a−2 +a−2 = or 3a−2 
dx a2 a2  a2 
y−3= 3 ( x−a2) or y= 3 x+c →3= 3 a2 +c
a2 a2 a2
3
y= x or 3a−2x cao
a2
2 x½ ax−½
(y)= + (+ c)
a ½ −½
sub x=a2, y=3 into ∫dy/dx
4x½
c = 1 (y= −2ax−½ +1)
a
4 1
sub x=16, y=8 → 8= ×4−2a× +1
a 4
a2 +14a−32 (=0)
a=2
A = (4, 3), B = (16, 8) AB2 =122 +52 → AB=13 | B1
M1
A1
B1B1
M1
A1
*M1
A1
A1
DM1A1 | [3]
[4]
[5] | 2 1
+ or 2a−2 +a−2 seen
a2 a2
anywhere in (i)
Through (a2,3) & with their
grad as f(a)
c must be present. Expect
3 = 4 ‒ 2 + c
Sub into their y
Allow ‒16 in addition
Page 7 | Mark Scheme | Syllabus | Paper
Cambridge International AS/A Level – October/November 2016 | 9709 | 13
\includegraphics{figure_2}

The diagram shows a triangular pyramid $ABCD$. It is given that
$$\overrightarrow{AB} = 3\mathbf{i} + \mathbf{j} + \mathbf{k}, \quad \overrightarrow{AC} = \mathbf{i} - 2\mathbf{j} - \mathbf{k} \quad \text{and} \quad \overrightarrow{AD} = \mathbf{i} + 4\mathbf{j} - 7\mathbf{k}.$$

\begin{enumerate}[label=(\roman*)]
\item Verify, showing all necessary working, that each of the angles $DAB$, $DAC$ and $CAB$ is $90°$. [3]
\item Find the exact value of the area of the triangle $ABC$, and hence find the exact value of the volume of the pyramid. [4]
\end{enumerate}

[The volume $V$ of a pyramid of base area $A$ and vertical height $h$ is given by $V = \frac{1}{3}Ah$.]

\hfill \mbox{\textit{CAIE P1 2016 Q7 [7]}}
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