9702 m17 ms_all

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Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/12
Paper 1 Multiple Choice March 2017
MARK SCHEME
Maximum Mark: 40
Published
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination.
Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.
Cambridge will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the March 2017 series for most Cambridge IGCSE®
,
Cambridge International A and AS Level components and some Cambridge O Level components.

9702/12 Cambridge International AS/A Level – Mark Scheme
PUBLISHED
March 2017
© UCLES 2017 Page 2 of 3
Question Answer Marks
1 C 1
2 B 1
3 D 1
4 C 1
5 D 1
6 A 1
7 B 1
8 A 1
9 D 1
10 B 1
11 B 1
12 B 1
13 C 1
14 C 1
15 B 1
16 A 1
17 A 1
18 C 1
19 C 1
20 B 1
21 A 1
22 C 1
23 A 1
24 C 1
25 D 1
26 A 1
27 C 1
28 C 1

PUBLISHED
March 2017
29 D 1
30 A 1
31 D 1
32 B 1
33 C 1
34 A 1
35 C 1
36 D 1
37 B 1
38 C 1
39 B 1
40 C 1

PHYSICS 9702/22
Paper 2 AS Level Structured Questions March 2017
MARK SCHEME
Maximum Mark: 60
Published
examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the
details of the discussions that took place at an Examiners’ meeting before marking began, which would have
considered the acceptability of alternative answers.
Teachers.
,

PUBLISHED
March 2017
1(a) scalars: kinetic energy, power, work A1
vectors: acceleration, force, momentum A1
1(b)(i) mass = volume × density or m = V × ρ
= 4/3 π (23 × 10–2
)3
× 82
C1
weight = 4/3 π (23 × 10–2
)3
× 82 × 9.8 = 41N A1
1(b)(ii) vertical component of tension = 290 sin75° or 290 cos15° (= 280) C1
upthrust = 290 sin75° + 41
= 320 (321)N
A1
1(b)(iii) the water pressure is greater than the air pressure
or
the pressure on lower surface (of sphere) is greater than the pressure on upper surface (of sphere)
B1

PUBLISHED
March 2017
2(a) sum/total momentum of bodies is constant
or
sum/total momentum of bodies before = sum/total momentum of bodies after
M1
for an isolated/closed system/no (resultant) external force A1
2(b)(i) EPE = area under graph or ½Fx or ½kx2
and F = kx C1
energy = ½ × 12.0 × 8.0 × 10–2
= 0.48J
or
energy = ½ × 150 × (8.0 × 10–2
)2
= 0.48J
A1
2(b)(ii)1 4.0 vA = 6.0 vB C1
EK = ½mv2
C1
×  
=  ×  
2
0.50 4.0 6.0
ratio
0.50 6.0 4.0
= 1.5 or ( )= ×
21
ratio 1.5
1.5
= 1.5
A1
2(b)(ii)2 0.48 = EK of A + EK of B
= EK of A + (EK of A / 1.5) = 5/3 × EK of A
C1
EK of A = 0.29 (0.288)J A1
2(b)(iii) curve starts from origin and has decreasing gradient M1
final gradient of graph line is zero A1

PUBLISHED
March 2017
3(a) change of displacement/time (taken) B1
3(b)(i) constant velocity, so resultant force is zero M1
(so car is) in (dynamic) equilibrium A1
3(b)(ii) FD = 0.40(kN) or 0.40 × 103
(N) C1
component of weight = 2.0 × 103
– 0.40 × 103
= 1.6 × 103
N
A1
3(b)(iii) P = Fv C1
= 2.0 ×103
× 9.0 = 1.8 × 104
W A1
3(b)(iv) (driving) force = 1.8 × 104
/15 (= 1.2 × 103
) C1
FD = 0.66(kN) or 0.66 × 103
(N) C1
acceleration = (1.2 × 103
– 0.66 × 103
)/850
= 0.64 (0.635)ms–2
A1

PUBLISHED
March 2017
4(a) change in frequency when source moves relative to observer M1
refers to ‘change in observed/apparent frequency’ A1
4(b)(i) f = (950 × 330)/(330 – 7.5) C1
= 970 (972)Hz A1
4(b)(ii) frequency decreases M1
from greater than 950Hz/from 970 (972)Hz/to less than 950Hz/to 930 (929)Hz/by 40 (43)Hz A1

PUBLISHED
March 2017
5(a) to the right/from the left/from A to B/in the same direction as electron velocity B1
5(b) v 2
= u 2
+ 2as
a = (1.5 × 107
)2
/(2 × 2.0 × 10–2
)
Other alternative calculations for the C1 mark:
e.g. a = 1.5×107
/2.67×10–9
e.g. a = [(1.5×107
× 2.67×10–9
) – 2.0×10–2
] × [2/(2.67×10–9
)2
]
e.g. a = (2.0×10–2
× 2)/(2.67×10–9
)2
C1
= 5.6 × 1015
m s–2
A1
5(c) E = F/Q C1
= (9.1 × 10–31
× 5.6 × 1015
)/1.6 × 10–19
C1
= 3.2 × 104
V m–1
A1
5(d) straight line with negative gradient starting at an intercept on the v-axis and ending at an intercept on the t-axis. B1

PUBLISHED
March 2017
6(a) I = I1 + I2 + I3 B1
(V / R) = (V / R1) + (V / R2) + (V / R3) or (I / V) = (I1 / V) + (I2 / V) + (I3 / V)
and (so) 1 / R = 1 / R1 + 1 / R2 + 1 / R3
A1
6(b)(i) e.m.f. is total energy available per unit charge B1
energy is dissipated in the internal resistance/resistor/r B1
6(b)(ii)1 Energy = EQ C1
= 6.0 × 2.5 × 103
= 1.5 × 104
J
A1
6(b)(ii)2 number = 2.5 × 103
/1.6 × 10–19
= 1.6 × 1022
(1.56 × 1022
)
A1
6(b)(iii) 1 / 4.8 = 1 / 12 + 1 / RX
RX = 8.0 Ω
A1
6(b)(iv) P = V2
/R
or
P = VI and V = IR
C1
ratio = (V 2
/ 8)/(V 2
/ 12) = 12/8
= 1.5
A1
6(b)(v) (total) current, or I, increases and P = EI or P = 6I or P ∝ I
or
total (circuit) resistance decreases and P = E 2
/R or P = 36/R or P ∝ 1/R
B1

PUBLISHED
March 2017
7(a) number of protons = 83 and number of neutrons = 129 A1
7(b) λ = 3.8 × 10–12
C1
f = 3.0 × 108
/ 3.8 × 10–12
C1
f = 7.9 × 1019
(7.89 × 1019
)Hz A1
7(c) use an electric field (at an angle to the beam) M1
α is deflected and γ is undeflected A1
7(d) either
energy = 9.3 × 10–13
/1.8 × 105
(= 5.17 × 10–18
J) C1
= 5.17 × 10–18
/1.6 × 10–19
= 32 (32.3)eV
A1
or
energy = 9.3 × 10–13
/1.6 × 10–19
(= 5.81 × 106
eV) (C1)
= 5.81 × 106
/1.8 × 105
= 32 (32.3)eV
(A1)

PHYSICS 9702/33
Paper 3 Advanced Practical Skills 1 March 2017
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
,

PUBLISHED
March 2017
1(a)(iii) Values of l 1 and l 2, both with units, and l 2 > l 1 1
1(b)(iii) Value of m in range 30 to 70g, with unit. 1
1(c) Six sets of readings of l 1, l 2 and m with correct trend and without help scores 5 marks, five sets scores 4 marks etc. 5
Range:
mmax ⩾ 80.0g and mmin ⩽ 30.0g.
1
Column headings:
Each column heading must contain a quantity and a unit where appropriate.
The presentation of quantity and unit must conform to accepted scientific convention e.g. (l2–l1)/mm.
1
Consistency:
All values of l1 and l2 must be given to the nearest mm.
1
1(d)(i) Axes:
Sensible scales must be used, no awkward scales (e.g. 3:10).
Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions.
Scales must be labelled with the quantity which is being plotted.
Scale markings should be no more than 3 large squares apart.
1
Plotting of points:
All observations must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no blobs).
Plots must be accurate to within half a small square in both x and y directions.
1
Quality:
All points in the table must be plotted (at least 5) for this mark to be awarded. Scatter of points must be no more than
±10.0g in the m direction from a straight line.
1

PUBLISHED
March 2017
1(d)(ii) Line of best fit:
Judged by balance of all points on the grid (at least 5) about the candidate’s line. There must be an even distribution of
points either side of the line along the full length.
One anomalous point is allowed only if clearly indicated (i.e. circled or labelled) by the candidate.
Lines must not be kinked or thicker than half a small square.
1
1(d)(iii) Gradient:
The hypotenuse of the triangle used must be greater than half the length of the drawn line.
Method of calculation must be correct.
Both read-offs must be accurate to half a small square in both the x and y directions.
1
y-intercept:
Either
Correct read-off from a point on the line substituted into y = mx + c or an equivalent expression, with read-off accurate
to half a small square in both x and y directions.
Or
Intercept read directly from the graph, with read-off at x = zero accurate to half a small square in y direction.
1
1(e) Value of a equal to candidate’s gradient. Value of b equal to candidate’s intercept.
The values must not be fractions and must be to at least 2 sig. fig.
1
Unit for a is dimensionally correct.
Unit for b is dimensionally correct.
1
1(f) k calculated correctly. 1
k given to 2 or 3 s.f. 1
Total: 20

PUBLISHED
March 2017
2(a)(ii) All raw values of D to nearest 0.1cm. 1
Evidence of repeat readings. 1
2(b)(ii) Correct calculation of W. 1
2(b)(iii) Justification for s.f. in W linked to s.f. in m and g. 1
2(c)(iii) Raw values of F to nearest 0.1N. 1
Evidence of repeat readings of F. 1
2(d) Absolute uncertainty in F of 0.2 to 0.5N and correct method of calculation to obtain percentage uncertainty.
If repeated readings have been taken, then the absolute uncertainty can be half the range (but not zero if values are
equal).
1
2(e) D for second pipe. 1
F for second pipe. 1
Quality:
Both F > 3N and < 10N.
1
2(f)(i) Two values of k calculated correctly. 1
2(f)(ii) Sensible comment relating to the calculated values of k, testing against a criterion specified by the candidate. 1

PUBLISHED
March 2017
2(g)(i) • Two readings are not enough to draw a valid conclusion.
• Difficult to measure diameter with reason, e.g. diameter varies /e.g. ends not square.
• Large uncertainty in D.
• Difficult to pull hook down steadily.
• Difficult to estimate average force during movement/F changes during movement.
• Newton-meter collides with mass/observation time too short.
• String slides off pipe/slides down pipe.
1 mark for each point up to a maximum of 4.
4
2(g)(ii) • Take more readings and plot a graph/calculate more k values and compare.
• Use (vernier) calliper/measure between two blocks/use micrometer.
• Improved pulling method to give constant speed, e.g. use a motor/e.g. use a known weight.
• Record images (or video) of the reading (or measurement) and play back.
• Use longer string with taller stand.
• Use two stands/support pipe on rod/use spirit level to ensure pipe is horizontal/use groove around pipe (to guide
string).
1 mark for each point up to a maximum of 4.
4
Total: 20

PHYSICS 9702/42
Paper 4 A Level Structured Questions March 2017
MARK SCHEME
Maximum Mark: 100
Published
Teachers.
,

PUBLISHED
March 2017
1(a) work done per unit mass M1
bringing (small test) mass from infinity (to the point) A1
1(b)(i) ∆φ = (GM / 2R) – (GM / 5R) = 3GM /10R A1
1(b)(ii) change in GPE = (3 × 4.0 × 1014
/ 10 R) × 4.7 × 104
C1
(3 × 4.0 × 1014
/ 10 R) × 4.7 × 104
= (1.70 – 0.88) × 1012
R = 6.88 ×106
C1
distance = 3 × 6.88 ×106
= 2.1 × 107
m
A1
2(a) +∆U increase in internal energy
+q heat (energy) transferred to the system/heating of system
+w work done on system
B2
2(b)(i) W = p∆V
= 5.2 × 105
× (5.0 – 1.6) × 10–4
(=177J)
B1
∆U = q + w
= 442 – 177 = 265J
A1
2(b)(ii) no (molecular) potential energy B1
internal energy decreases so (total molecular) kinetic energy decreases B1
(mean molecular) kinetic energy decreases so temperature decreases B1

PUBLISHED
March 2017
2(b)(iii) ∆U + 265 – 313 = 0
∆U = 48J
A1
2(b)(iv) pV = NkT or pV = nRT and N = nNA C1
5.2×105
× 1.6×10–4
= N × 1.38×10–23
× (273 + 227)
or
5.2×105
× 1.6×10–4
= n × 8.31 × (273 + 227) and n = N / 6.02×1023
N = 1.2 × 1022
A1
3(a) m is constant or k / m is constant and so acceleration/a proportional to displacement/x B1
negative sign shows that acceleration/a is in opposite direction to displacement/x
or
negative sign shows acceleration/a is towards fixed point
B1
3(b) evidence of comparison to expression to a = – ω2
x B1
ω2
= k/m or ω2
= 4.0/m hence ω = 2.0/√m A1
3(c) EK = ½ m ω2
x0
2
or EK = ½mv 2
and v = ωx0 C1
= ½m (4.0/m) (3.0 × 10–2
)2
C1
= 1.8 × 10–3
J A1

PUBLISHED
March 2017
3(d) new x0 = –3
[( ) (1.8 10 / 2 2 / ( / 4.0))]m m×××
or
(EK ∝ x0
2
so) new x0 = –2 2
[½ 3.0 10( ) ]× ×
C1
= 2.12 × 10–2
m A1
3(e) flux linked to block changes/flux is cut by block which induces an e.m.f. in block B1
(eddy) currents induced in block cause heating B1
thermal/heat energy comes from (kinetic/potential) energy of oscillations/block B1
4 piezo-electric/quartz crystal/transducer B1
alternating p.d. applied across crystal/transducer B1
causes crystal to vibrate/resonate B1
crystal resonates at ultrasound frequencies/ crystal’s natural frequency is in the ultrasound range/alternating p.d. is in
ultrasound frequency range
B1

PUBLISHED
March 2017
5(a) any three from:
• greater bandwidth
• does not suffer from (e.m.) interference/can be used in (e.m.) ‘noisy’ environments
• no/less power/energy radiated/better security/less cross-talk
• less attenuation/fewer repeaters/amplifiers needed
• less weight/easier to handle/cheaper/occupy less space
B3
5(b)(i) attenuation/gain = 10 log P1 / P2 C1
0.50 × 57 = 10 log (15 × 10–3
/P) so P = 2.1 × 10–5
W
or
– (0.50 × 57) = 10 log (P/15 × 10–3
) so P = 2.1 × 10–5
W
A1
5(b)(ii) either
(calculation of S/N ratio at receiver)
S/N ratio = 10 log(2.1 × 10–5
/ 9.0 × 10–7
) or S/N ratio = 14
M1
14 < 24 or S/N ratio < minimum S/N ratio A1
so not able to distinguish signal from noise A1
or
(calculation of minimum acceptable power at receiver)
24 = 10 log (P / 9.0 × 10–7
) or P = 2.3 × 10–4
(M1)
2.1 × 10–5
< 2.3 × 10–4
or power < minimum power (A1)
so not able to distinguish signal from noise (A1)

PUBLISHED
March 2017
6(a) similarity: lines are radial/greater separation of lines with increased distance from the sphere B1
difference: gravitational lines directed towards sphere and electric lines directed away from sphere B1
6(b)(i) E = Q / 4πε0r 2
or E = kQ / r 2
with k defined/substituted in C1
4.1 × 10–5
= [Q / (4π × 8.85 ×10–12
× 0.0252
)] – [Q / (4π × 8.85 × 10–12
× 0.0752
)] C1
Q = 3.2 × 10–18
C A1
6(b)(ii) smooth curve with gradient decreasing starting at (0, 4.1 × 10–5
) to d-axis at (2.5, 0) B1
smooth curve with gradient increasing from (2.5, 0) ending at (5, – 4.1 × 10–5
) B1
6(b)(iii) acceleration decreases (to zero at mid-point) B1
then acceleration increases in the opposite direction/increasing negative acceleration B1
7(a) correct grid shape (of wire) B1
fine wire/foil strip B1
plastic/insulating envelope containing the wire B1
7(b)(i) 2.00 / 6.00 = 153.0 / (R + 153.0)
or
4.00 / 6.00 = R / (R + 153.0) (so R = 306.0)
C1
∆R = 306.0 – 300.0 = 6.0 (Ω) C1
so ∆L = 8(.0) × 10–5
m A1

PUBLISHED
March 2017
7(b)(ii) R or ∆R increases B1
V +
< V –
or VA < 2.00 or V +
/ VA decreases M1
output is negative/–5V A1
diode X emits light/is ‘on’ A1
8(a) region (of space) where there is a force M1
produced by/on a magnet/magnetic pole/moving charge/current-carrying conductor A1
8(b)(i) out of (the plane of) the paper/page B1
8(b)(ii) the force on the particle is (always) perpendicular to the velocity/perpendicular to the direction of travel/towards the
centre of path
B1
no work is done by the force on the particle/there is no acceleration in the direction of the velocity/the acceleration is
(always) perpendicular to the velocity
B1
8(b)(iii) F = Bqv or F = mv2
/r C1
mv2
/ (d / 2) = Bqv so d = 2mv / Bq A1
8(b)(iv) time = distance / speed
T(F) = πd / 2v
C1
T(F) = (π / 2v) × (2mv / Bq)
T(F) = πm/ Bq and so T(F) independent of v
A1

PUBLISHED
March 2017
9(a)(i) increase flux linkage (with secondary coil)/to reduce flux loss B1
9(a)(ii) e.m.f. (induced only) when flux (in core/coil) is changing B1
constant/direct voltage gives constant flux/field B1
9(b)(i) NS / NP = VS / VP C1
NS = (52 / 150) × 1200
= 416 turns
A1
9(b)(ii) 0ms or 7.5ms or 15.0ms or 22.5ms A1
9(c)(i) either
mean power = V 2
/ 2R and V = 52 (V) C1
R = 522
/ (2 × 1.2)
= 1100 (1127)Ω
A1
or
mean power = V 2
/ R and V = 52 / 2 (= 36.8V) (C1)
R = 36.82
/ 1.2
= 1100Ω
(A1)
9(c)(ii) sinusoidal shape with troughs at zero power B1
only 3 ‘cycles’ B1
each ‘cycle’ is 2.4W high and zero power at correct times B1

PUBLISHED
March 2017
10(a) packet/quantum of energy M1
of electromagnetic radiation A1
10(b)(i) light is re-emitted in all directions/only part of the re-emitted light is in the direction of the beam B1
10(b)(ii) an arrow between –3.40eV and –1.51eV and an arrow between –3.40eV and –0.85eV B1
all arrows shown point ‘upwards’ B1
10(b)(iii) E = hc / λ or E = hf and c = fλ C1
2.60 × 1.60 × 10–19
= (6.63 × 10–34
× 3.00 × 108
) / λ C1
λ = 4.8 × 10–7
m A1

PUBLISHED
March 2017
11 any five from:
• electrons need energy to enter conduction band (from valence band)
• (positively-charged) holes are left in valence band
• moving charge carriers/holes/electrons are current
• (increase of temperature leads to) more (positive and negative) charge carriers/more holes/more electrons so
more current
• more charge carriers/holes/electrons gives rise to less resistance
• (increase of temperature causes) greater (amplitude of) vibrations of atoms/ions/lattice
• effect of more charge carriers/holes/electrons is greater than effect of greater vibrations (and so resistance
decreases)
B5

PUBLISHED
March 2017
12(a) either
(minimum) energy required/work done to separate the nucleons (in a nucleus) M1
to infinity A1
or
energy released when nucleons come together (to form a nucleus) (M1)
from infinity (A1)
12(b)(i) (total) binding energy of thorium and helium (nuclei) greater than binding energy of uranium (nucleus) B1
12(b)(ii)1 change in mass = 238.05076 – (234.04357 + 4.00260)
= 4.59 × 10–3
u
A1
12(b)(ii)2 either
E = mc 2
= 4.59 × 10–3
× 1.66 × 10–27
× (3.00 × 108
)2
C1
= 6.9 × 10–13
J A1
or
1u = 931MeV
E = 4.59 × 10–3
× 931 × 106
× 1.6 × 10–19
(C1)
= 6.8 × 10–13
J (A1)
12(b)(iii) Th nucleus/He nucleus/product nucleus has kinetic energy M1
energy of gamma photon must be less than energy released A1

PHYSICS 9702/52
Paper 5 Planning, Analysis and Evaluation March 2017
MARK SCHEME
Maximum Mark: 30
Published
Teachers.
,

PUBLISHED
March 2017
1 Defining the problem
M is the independent variable and v is the dependent variable, or vary M and measure v 1
keep x/compression of spring constant 1
Methods of data collection
labelled diagram including horizontal spring in line with vehicle attached to wall /retort stand 1
use a ruler/calliper to determine compression of spring 1
use of stopwatch/use of light gate connected to a timer/motion sensor correctly positioned 1
use of balance to measure mass of vehicle M 1
Method of Analysis
plots a graph of 1/v2
against M
[Do not allow lg-lg graphs]
1
relationship valid if a straight line produced 1
2
1
k
gradient x
=
×
or 2
b
k
y intercept x
=
− ×
1

PUBLISHED
March 2017
Additional detail including safety considerations Max 6
D1 use safety screen; use goggles to avoid ball/spring hitting eye
D2 add masses to the vehicle to change M
D3 repeat experiment for each M and average v
D4 use of ruler to measure an appropriate distance for the time taken in stopwatch/light gate methods
D5 method to determine speed of vehicle, e.g. time vehicle over a measured distance and use speed = distance/time
D6 method to release ball with guide or support for spring /ball
D7 release the ball close to the vehicle
D8 detail on determining x e.g. difference between compressed length and original length
D9 method to ensure constant speed along track, e.g. friction compensate track/use of air track
D10 (relationship valid if a straight line produced) with (y-)intercept = 2
b
kx

PUBLISHED
March 2017
2(a) gradient = Q/E
y-intercept = 1/E
1
2(b) 4.0 or 4.00 or 4.000 1.5 or 1.52
3.0 or 3.03 or 3.030 1.2 or 1.16
2.1 or 2.13 or 2.128 0.870 or 0.8696
1.8 or 1.79 or 1.786 0.769 or 07692
1.5 or 1.47 or 1.471 0.671 or 0.6711
1.2 or 1.19 or 1.190 0.610 or 0.6098
First mark for all first column correct either 2 and 3 significant figures or 3 and 4 significant figures.
Second mark for all second column correct.
2
absolute uncertainties from 0.4 to 0.1 1
2(c)(i) six points plotted correctly
must be within half a small square
1
error bars in 1/P plotted correctly
all error bars to be plotted
1
2(c)(ii) line of best fit drawn
If points are plotted correctly then lower end of line should pass between (1.50, 0.70) and (1.65, 0.70) and upper end of
line should pass between (3.60, 1.40) and (3.80, 1.40).
1
worst acceptable line drawn
steepest or shallowest possible line
mark scored only if all error bars are plotted
1

PUBLISHED
March 2017
2(c)(iii) gradient determined with a triangle that is at least half the length of the drawn line 1
uncertainty = gradient of line of best fit – gradient of worst acceptable line
or
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
1
2(c)(iv) y-intercept determined by substitution into y = mx + c 1
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
or
uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept).
1
2(d)(i) E determined with correct unit using y-intercept
1
E
y intercept
=
−
1
Q determined with correct unit using gradient and given to two or three significant figures
penalise power of ten errors
correct substitution of numbers must be seen
gradient
Q E gradient
y intercept
= × =
−
1
2(d)(ii) percentage uncertainty in Q
correct substitution of numbers must be seen
%uncertainty E + %uncertainty in gradient or %uncertainty in y-intercept + %uncertainty in gradient
Maximum/minimum methods
max
max max
min
gradient
MaxQ gradient E
y intercept
= × =
−
min
min min
max
gradient
MinQ gradient E
y intercept
= × =
−
1

9702 m17 ms_all

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9702 m17 ms_all