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In the complex plane, the circle with equation `|\ z - (2 + 3i)\ | = 1` is intersected exactly twice by the curve with equation
A. `|\ z - 3i\ | = 1`
B. `|\ z + 3\ | = |\ z - 3i\ |`
C. `|\ z - 3\ | = |\ z - 3i\ |`
D. `text(Im)(z) = 4`
E. `text(Re)(z) = 3`
`text(Consider option C:)`
`|z – 3| = |z – 3i|`
`=> y = x\ text{(perpendicular bisector}`
`text{between (3, 0) and (0, 3)}`
`=> C`
The polynomial `P(z)` has real coefficients. Four of the roots of the equation `P(z) = 0` are `z = 0`, `z = 1 - 2i`, `z = 1 + 2i` and `z = 3i`.
The minimum number of roots that the equation `P(z) = 0` could have is
A. 4
B. 5
C. 6
D. 7
E. 8
A certain complex number `z`, where `|\ z\ | > 1`, is represented by the point on the following argand diagram.
All axes below have the same scale as those in the diagram above.
The complex number `1/barz` is best represented by
The region in the first quadrant enclosed by the curve `y = sin(x)`, the line `y = 0` and the line `x = pi/6` is rotated about the `x`-axis.
Find the volume of the resulting solid of revolution. (3 marks)
| `V` | `= pi int_(0)^(pi/6) y^2\ dx` |
| `= pi int_0^(pi/6) sin^2(x)\ dx` | |
| `= pi/2 int_0^(pi/6) (1 – cos 2x)\ dx` | |
| `= pi/2 [x – 1/2 sin (2x)]_0^(pi/6)` | |
| `=pi/2 [(pi/6 – 1/2 sin (pi/3)) – 0]` | |
| `=pi/2 (pi/6 – (sqrt 3)/4)` | |
| `=pi^2/12 – (sqrt3 pi)/8\ \ text(u³)` |
The velocity–time graph for the first 2 seconds of the motion of a particle that is moving in a straight line with respect to a fixed point is shown below.
The particle’s velocity `v` is measured in cm/s. Initially the particle is `x_0` cm from the fixed point.
The distance travelled by the particle in the first 2 seconds of its motion is given by
A. `int_0^2 v\ dt`
B. `int_0^2 v\ dt + x_0`
C. `int_1^2 v\ dt - int_0^1 v\ dt`
D. `|\ int_0^2 v\ dt\ |`
E. `int_1^2 v\ dt - int_0^1 v\ dt + x_0`
Consider the three vectors
`underset ~a = underset ~i - underset ~j + 2underset ~k,\ underset ~b = underset ~i + 2 underset ~j + m underset ~k` and `underset ~c = underset ~i + underset ~j - underset ~k`, where `m in R.`
A flowerpot of mass `m` kg is held in equilibrium by two light ropes, both of which are connected to a ceiling. The first rope makes an angle of 30° to the vertical and has tension `T_1` newtons. The second makes an angle of 60° to the vertical and has tension `T_2` newtons.
Find the maximum value of `m` for which the flowerpot will remain in equilibrium. (3 marks)
a. `text(Forces in equilibrium:)`
| `tan 30^@` | `=T_2/T_1` |
| `T_2` | `= T_1 xx tan30^@` |
| `T_2` | `= T_1/sqrt3\ \ text(.. as required)` |
| b. | `sin30^@` | `= (T_2)/(mg)` |
| `T_2` | `= (mg)/2` | |
| `98` | `=(m_text(max) xx 9.8)/2` | |
| `m_text(max)` | `=(2 xx 98)/9.8` | |
| `=20\ text(kg)` |
Evaluate `int_0^1 e^x cos (e^x)\ dx.` (2 marks)
For the curve with parametric equations
`x = 4 sin (t) - 1`
`y = 2 cos (t) + 3`
Find `(dy)/(dx)` at the point `(1, sqrt 3 + 3).` (3 marks)
If `z = a + bi`, where both `a` and `b` are non-zero real numbers and `z in C`, which of the following does not represent a real number?
The set of points in the complex plane defined by `|\ z + 2i\ | = |\ z\ |` corresponds to
A. the point given by `z = −i`
B. the line `text(Im)(z) = −1`
C. the line `text(Im)(z) = −i`
D. the line `text(Re)(z) = −1`
E. the circle with centre `−2i` and radius `1`
If `z = sqrt 2 text(cis)(-(4 pi)/5)` and `w = z^9`, then
A. `w = 16 sqrt 2 text(cis)((36 pi)/5)`
B. `w = 16 sqrt 2 text(cis)(−pi/5)`
C. `w = 16 sqrt 2 text(cis)((4pi)/5)`
D. `w = 9 sqrt 2 text(cis)(-pi/5)`
E. `w = 9 sqrt 2 text(cis)((4pi)/5)`
Label any asymptotes with their equations and label any intercepts with the axes, writing them as coordinates. (3 marks)
| a. | `f(x)` | `= (2x^2 + 3)/(x^2 + 1)` |
| `= (2(x^2 + 1) + 1)/(x^2 + 1)` | ||
| `= 2 + 1/(x^2 + 1)` |
b. `underset (x→oo) (lim y) = 2`
`text(S)text(ince)\ \ x^2>0\ \ text(for all)\ x,`
`=> f(x)_text(max)\ \ text(occurs when)\ \ x=0\ \ text(at)\ \ (0,3)`
c. `f(x)\ \ text(is an even function.)`
| `:.\ text(Area)` | `= 2 int_0^1 2 + 1/(x^2 + 1)\ dx` | |
| `= 2[2x + tan^(−1)(x)]_0^1` | ||
| `=2[(2+pi/4) – 0]` | ||
| `= 4 + pi/2` |
Find the value of the real constant `k` given that `kxe^(2x)` is a solution of the differential equation
`(d^2y)/(dx^2)-2 (dy)/(dx) + 5y = e^(2x) (15x + 6).` (3 marks)
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Find an antiderivative of `(1 + x)/(9-x^2),\ \ x in R \ text(\{– 3, 3}).` (3 marks)
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The graph of `y = f(x)` is shown below.
All the axes below have the same scale as the axes in the diagram above.
The graph of `y = 1/(f(x))` is best represented by
| A. |  |
B. |  |
| C. |  |
D. |  |
| E. |  |
Consider the functions with rules `f(x) = arcsin (x/2) + 3/sqrt (25 x^2 - 1)` and `g(x) = arcsin (3x) - 3/sqrt (25x^2 - 1).`
Give your answer in the form `(a sqrt b)/c, \ a, b, c in Z.` (3 marks)
a.i. `text(Maximal Domain occurs when:)`
| `-1 <= x/2 <= 1` | |
| `{x: -2 <= x <= 2}` |
a.ii. `text(Maximal Domain occurs when:)`♦♦ Mean mark part (a)(ii) 33%.
| `25x^2 – 1 > 0` | |
| `x^2 > 1/25` | |
| `{x: x < -1/5 uu x > 1/5}` |
a.iii. `text(Max domain for which)\ \ f(x)\ \ text(is defined:)`♦♦ Mean mark part (a)(iii) 26%.
| `(-2 <= x <= 2) nn (x < -1/5 uu x > 1/5)` | |
| `{x: -2 <= x < -1/5 \ uu \ 1/5 < x <= 2}` |
| b. | `h(x)` | `= sin^(-1) (x/2) + 3/sqrt(25x^2 – 1) + sin^(-1)(3x) – 3/sqrt(25x^2 – 1)` |
| `= sin^(-1)(x/2) + sin^(-1)(3x)` |
`text(When)\ \ x=1/4,\ \ h(x)=theta`♦♦♦ Mean mark part (b) 20%.
`theta=sin^(-1) (1/8) + sin^(-1) (3/4)`
`text(Let)\ \ theta_1 = sin^(-1) (1/8),\ \ theta_2 = sin^(-1) (3/4)`
`text(Using)\ \ sin(theta_1 + theta_2)=sin theta_1 cos theta_2 + cos theta_1 sin theta_2:`
| `sin(theta)` | `= 1/8 * sqrt7/4 + sqrt63/8 * 3/4` |
| `=sqrt7/32 + (9sqrt7)/32` | |
| `=(5 sqrt7)/16` |
The position of a particle at time `t` is given by
`underset ~r (t) = (2 sqrt (t^2 + 2) - t^2) underset ~i + (2 sqrt (t^2 + 2) + 2t) underset ~j,\ \ t >= 0.`
a. `underset ~r (t) = (2 sqrt (t^2 + 2) – t^2) underset ~i + (2 sqrt (t^2 + 2) +2t) underset ~j`
| `underset ~v(t)` | `=dot underset ~r(t)` | |
| `= (2(2t)(1/2)(t^2 + 2)^(-1/2) – 2t) underset ~i + (2(2t)(1/2)(t^2 + 2)^(-1/2) + 2)underset ~j` | ||
| `= ((2t)/sqrt(t^2 + 2) – 2t) underset ~i + ((2t)/sqrt(t^2 + 2) + 2) underset ~j` |
| b. | `underset ~v(1)` | `= (2/sqrt(1 + 2) – 2)underset ~i + (2/sqrt(1 + 2) + 2) underset ~j` |
| `= (2/sqrt 3 – 2) underset ~i + (2/sqrt 3 + 2)underset ~j` | ||
| `|\ underset ~v(1)\ |` | `= sqrt((2/sqrt 3 – 2)^2 + (2/sqrt 3 + 2)^2)` | |
| `= sqrt(4/3 – 8/sqrt 3 + 4 + 4/3 + 8/sqrt 3 + 4)` | ||
| `= sqrt(8/3 + 8)` | ||
| `= sqrt(32/3)` | ||
| `= (4sqrt2)/sqrt3` | ||
| `= (4 sqrt 6)/3` |
♦ Mean mark part (c) 41%.
| c. | `(dy)/(dt)` | `= (dy)/(dx) xx (dx)/(dt)` |
| `(dy)/(dx)` | `=((dy)/(dt))/((dx)/(dt))` | |
| `:. (dy)/(dx)|_(t=1)` | `= (2/sqrt 3 + 2)/(2/sqrt 3 – 2)` | |
| `= ((2 + 2 sqrt 3)/sqrt 3) xx (sqrt 3/(2 – 2 sqrt 3))` | ||
| `= (2 + 2 sqrt 3)/(2 – 2 sqrt 3)` | ||
| `= (1 + sqrt 3)/(1 – sqrt 3)` |
d. `text(At)\ \ t=0:` ♦♦ Mean mark part (d) 32%.
`underset ~r(0) = 2 sqrt 2 underset ~i + 2 sqrt 2 underset ~j`
| `theta_1` | `= tan^(-1)((2 sqrt 2)/(2 sqrt 2)) = pi/4` |
| `theta_2` | `= tan^(-1)(1/sqrt 3)=pi/6` |
`text(Let)\ \ theta=\ text(angle between the vectors:)`
| `theta` | `= pi – theta_1 – theta_2` |
| `= pi – pi/4 – pi/6` | |
| `= (12 pi – 3 pi – 2 pi)/12` | |
| `= (7 pi)/12` |
The velocity, `v` m/s, of a body when it is `x` metres from a fixed point `O` is given by
`v = (2x)/sqrt(1 + x^2).`
Find an expression for the acceleration of the body in terms of `x` in simplest form. (3 marks)
Consider the curve with equation `y = (x - 1) sqrt (2 - x),\ \ 1 <= x <= 2.`
Calculate the area of the region enclosed by the curve and the `x`-axis. (3 marks)
Find the gradient of the tangent to the curve `xy^2 + y + (log_e (x - 2))^2 = 14` at the point `(3, 2).` (3 marks)
Let `y = arctan (2x).`
Find the value of `a` given that `(d^2y)/(dx^2) = ax ((dy)/(dx))^2`, where `a` is a real constant. (3 marks)
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Find `int(3x^2 + 8)/(x(x^2 +4))\ dx`. (3 marks)
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Consider the equation `z^3 - z^2 - 2z - 12 = 0, \ z in C.`
a. `text(Given)\ \ z = 2 text(cis)((2 pi)/3)\ \ text(is a root, and)`
`z^3 – z^2 – 2z – 12 = 0\ \ text(has real coefficients,)`
`=> z= 2 text(cis)(-(2 pi)/3)\ \ text{is a root (conjugate).}`
`=> text(Roots:)\ \ -1+sqrt3i,\ \ -1-sqrt3i`
`(z – alpha) (z – beta) (z – gamma) = z^3 – z^2 – 2z – 12`
`text(Using)\ \ alpha beta gamma = 12:`
| `(-1 + i sqrt 3) (-1 – i sqrt 3) gamma` | `=12` | |
| `((-1)^2-(sqrt3i)^2)gamma` | `=12` | |
| `4gamma` | `=12` | |
| `gamma` | `=3` |
`:.\ text(Other two roots:)`♦♦ Mean mark part (b) 33%.
| `z_2` | `= -1 – i sqrt 3` |
| `z_3` | `= 3 + 0i` |
| b. |  |
Find an antiderivative of `(6 + x)/(x^2 + 4).` (2 marks)
A 5 kg parcel is on the floor of a lift that is accelerating downwards at 3 m/s².
The reaction, in newtons, of the floor of the lift on the parcel is
A. `−15 + 5g`
B. `15 + 5g`
C. `−15 + 3g`
D. `−15 − 5g`
E. `15 + 3g`
A tourist in a hot air balloon, which is rising at 2 m/s, accidentally drops a camera over the side and it falls 100 m to the ground.
Neglecting the effect of air resistance on the camera, the time taken for the camera to hit the ground, correct to the nearest tenth of a second, is
A. 4.3 s
B. 4.5 s
C. 4.7 s
D. 4.9 s
E. 5.0 s
A particle moves in a straight line such that its acceleration is given by `a = sqrt(v^2 - 1)` , where `v` is its velocity and `x` is its displacement from a fixed point.
Given that `v = sqrt2` when `x = 0`, the velocity `v` in terms of `x` is
A. `v = sqrt(2 + x)`
B. `v = 1 + |\ x + 1\ |`
C. `v = sqrt(2 + x^2)`
D. `v = sqrt(1 + (1 + x)^2)`
E. `v = sqrt(1 + (x - 1)^2)`
Consider the four vectors `underset~a = underset~j + 3underset~k`, `underset~b = underset~i - 4underset~k`, `underset~c = 3underset~j - k` and `underset~d = 2underset~j + underset~k`.
Which one of the following is a linearly dependent set of vectors?
Forces of magnitude 5 N, 7 N and `Q` N act on a particle that is in equilibrium, as shown in the diagram below.
The magnitude of `Q`, in newtons, can be found by evaluating
A. `sqrt(5^2 + 7^2 - 2 xx 5 xx 7 cos(70^@))`
B. `5^2 + 7^2 - 2 xx 5 xx 7 cos(110^@)`
C. `sqrt(5^2 + 7^2 - 2 xx 5 xx 7 cos(110^@))`
D. `5^2 + 7^2 - 2 xx 5 xx 7 cos(70^@)`
E. `sqrt(5^2 + 7^2 - 2 xx 5 xx 7 cos(20^@))`
`text(Using cosine rule:)`
| `Q^2` | `= 5^2 + 7^2 – 2(5)(7)cos(70^@)` |
| `Q` | `= sqrt(5^2 + 7^2 – (5)(7)cos(70^@))` |
`=> A`
Let `underset~u = 4underset~i - underset~j + underset~k`, `underset~v = 3underset~j + 3underset~k` and `underset~w = −4underset~i + underset~j + underset~k`.
Which one of the following statements is not true?
Water containing 2 grams of salt per litre flows at the rate of 10 litres per minute into a tank that initially contained 50 litres of pure water. The concentration of salt in the tank is kept uniform by stirring and the mixture flows out of the tank at the rate of 6 litres per minute.
If `Q` grams is the amount of salt in the tank `t` minutes after the water begins to flow, the differential equation relating `Q` to `t` is
A. `(dQ)/(dt) = 20 - (3Q)/(25 + 2t)`
B. `(dQ)/(dt) = 10 - (3Q)/(25 + 2t)`
C. `(dQ)/(dt) = 20 - (3Q)/(25 - 2t)`
D. `(dQ)/(dt) = 10 - (3Q)/(25 - 2t)`
E. `(dQ)/(dt) = 20 - (3Q)/25`
The differential equation that best represents the above direction field is
A. `(dy)/(dx) = x^2 - y^2`
B. `(dy)/(dx) = y^2 - x^2`
C. `(dy)/(dx) = y/x`
D. `(dy)/(dx) = −x/y`
E. `(dy)/(dx) = x/y`
Consider the differential equation `(dy)/(dx) = 1/(3 + 3x + x^2)`, with `y_0 = 1` when `x_0 = 0`.
Using Euler's method with a step size of 0.1, the value of `y_2` correct to three decimal places, is
A. 1.033
B. 1.063
C. 1.064
D. 1.065
E. 1.066
The region bounded by the lines `x = 0`, `y = 3` and the graph of `y = x^(4/3)` where `x ≥ 0` is rotated about the `y`-axis to form a solid of revolution.
The volume of this solid is
A. `(81pi3^(2/3))/11`
B. `(12pi3^(3/4))/7`
C. `(27pi3^(1/3))/7`
D. `(18pi3^(1/2))/5`
E. `(6pi3^(1/2))/5`
The principal arguments of the solutions to the equation `z^2 = 1 + i` are
The region in the complex plane that is outside the circle of radius `b` centred at the origin is given by the set of points `z`, where `z ∈ C`, such that
A. `|\ z\ | < b`
B. `|\ z\ | > b`
C. `|\ z\ | > b^2`
D. `|\ z\ | = b`
E. `|\ z\ | < b^2`
The shaded region below is enclosed by the graph of `y = sin(x)` and the lines `y = 3x` and `x = pi/3.`
This region is rotated about the `x`-axis.
Find the volume of the resulting solid of revolution. (4 marks)
A container of water is heated to boiling point (100°C) and then placed in a room that has a constant temperature of 20°C. After five minutes the temperature of the water is 80°C.
a. `text(Maximal domain:)`
| `1 – 2x` | `∈ [−1, 1]` |
| `-2x` | `∈ [−2, 0]` |
| `x` | `∈ [0, 1]` |
`text(When)\ \ x = 0,\ \ y = cos^-1 1= 0;`
`text(When)\ \ x = 1,\ \ y = cos^-1 (-1)= pi`
`:.\ text(Range is)\ \ [0, pi]`
| b. | |
c. `y = cos^-1 (1 – 2x),`
| `(dy)/(dx)` | `= (-1(-2))/sqrt(1 – (1 – 2x)^2)` |
| `= 2/sqrt(1 – (1 – 2x)^2)` |
`text(At)\ \ x = 1/4,`
| `m_T` | `= 2/sqrt(1 – (1 – 1/2)^2)` |
| `=2/sqrt(3/4)` | |
| `= 4/sqrt 3` | |
| `= (4 sqrt 3)/3` |
The coordinates of three points are `A (– 1, 2, 4), \ B(1, 0, 5) and C(3, 5, 2).`
Prove that the triangle has a right angle at `A.` (2 marks)
Evaluate `int_0^1 (x-5)/(x^2-5x + 6)\ dx.` (4 marks)
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Find the value of `c`, where `c in R`, such that the curve defined by
`y^2 + (3e^{(x - 1)})/(x - 2) = c`
has a gradient of 2 where `x = 1.` (4 marks)
The acceleration, in `text(ms)^(-2)`, of a particle moving in a straight line is given by `–4x`, where `x` metres is its displacement from a fixed origin `O`.
If the particle is at rest where `x = 5`, the speed of the particle, in `text(ms)^(−1)`, where `x = 3` is
A. `8`
B. `8 sqrt 2`
C. `12`
D. `4 sqrt 2`
E. `2 sqrt 34`
Particles of mass 3 kg and 5 kg are attached to the ends of a light inextensible string that passes over a fixed smooth pulley, as shown above. The system is released from rest.
Assuming the system remains connected, the speed of the 5 kg mass after two seconds is
A. 4.0 m/s
B. 4.9 m/s
C. 9.8 m/s
D. 10.0 m/s
E. 19.6 m/s
The acceleration vector of a particle that starts from rest is given by
`underset ~a(t) = −4 sin(2t) underset ~i + 20 cos (2t) underset ~j - 20 e^(−2t) underset ~k`, where `t >= 0`.
The velocity vector of the particle, `underset ~v(t)`, is given by
If `theta` is the angle between `underset ~a = sqrt 3 underset ~i + 4 underset ~j - underset ~k` and `underset ~b = underset ~i - 4 underset ~j + sqrt 3 underset ~k`, then `cos(2 theta)` is
The differential equation that is best represented by the above direction field is
A. `(dy)/(dx) = 1/(x - y)`
B. `(dy)/(dx) = y - x`
C. `(dy)/(dx) = 1/(y - x)`
D. `(dy)/(dx) = x - y`
E. `(dy)/(dx) = 1/(y + x)`
An object is moving in a straight line, initially at `5\ text(ms)^(–1)`. Sixteen seconds later, it is moving at `11\ text(ms)^(–1)` in the opposite direction to its initial velocity.
Assuming that the acceleration of the object is constant, after 16 seconds the distance, in metres, of the object from its starting point is
A. 24
B. 48
C. 73
D. 96
E. 128
A light inextensible string passes over a smooth pulley, as shown below, with particles of mass 1 kg and `m` kg attached to the ends of the string.
If the acceleration of the 1 kg particle is 4.9 `text(ms)^(-2)` upwards, then `m` is equal to
A. 1
B. 2
C. 3
D. 4
E. 5
`T – (9.8 xx 1) = 4.9 xx 1`
`T=14.7`
| `m xx 9.8 – T` | `=m xx 4.9` |
| `4.9m` | `= 14.7“ |
| `:. m` | `= 14.7/4.9` |
| `= 3` |
`=> C`
The position vectors of two moving particles are given by `underset~r_1(t) = (2 + 4t^2)underset~i + (3t + 2)underset~j` and `underset~r(t) = (6t)underset~i + (4 + t)underset~j`, where `t >=0`.
The particles will collide at
The velocity–time graph for a body moving along a straight line is shown below.
The body first returns to its initial position within the time interval
A. (0, 0.5)
B. (0.5, 1.5)
C. (1.5, 2.5)
D. (2.5, 3.5)
E. (3.5, 5)
The lifetime of a certain brand of batteries is normally distributed with a mean lifetime of 20 hours and a standard deviation of two hours. A random sample of 25 batteries is selected.
The probability that the mean lifetime of this sample of 25 batteries exceeds 19.3 hours is
A. 0.0401
B. 0.1368
C. 0.6103
D. 0.8632
E. 0.9599
Oranges grown on a citrus farm have a mean mass of 204 grams with a standard deviation of 9 grams
Lemons grown on the same farm have a mean mass of 76 grams with a standard deviation of 3 grams.
The masses of the lemons are independent of the masses of the oranges.
The mean mass and standard deviation, in grams, respectively of a set of three of these oranges and two of these lemons are
A body of mass 3 kg is moving to the left in a straight line at 2 `text(ms)^(-1)`. It experiences a force for a period of time, after which it is then moving to the right at 2 `text(ms)^(-1)`.
The change in momentum of the particle, in kg `text(ms)^(-1)`, in the direction of the final motion is
A. − 6
B. 0
C. 4
D. 6
E. 12
A variable force of `F` newtons acts on a 3 kg mass so that it moves in a straight line. At time `t` seconds, `t >= 0`, its velocity `v` metres per second and position `x` metres from the origin are given by `v = 3 - x^2`.
It follows that
A. `F = -2x`
B. `F = -6x`
C. `F = 2x^3 - 6x`
D. `F = 6x^3 - 18x`
E. `F = 9x - 3x^3`
A particle of mass 5 kg is subject to forces `12 underset ~i` newtons and `9 underset ~j` newtons.
If no other forces act on the particle, the magnitude of the particle’s acceleration, in ms¯², is
If `underset ~a = -2 underset ~i - underset ~j + 3 underset ~k` and `underset ~b = -m underset ~i + underset ~j + 2 underset ~k`, where `m` is a real constant, the vector `underset ~a - underset ~b` will be perpendicular to vector `underset ~b` where `m` equals
A. 0 only
B. 2 only
C. 0 or 2
D. 4.5
E. 0 or −2
Let `underset ~a = 3 underset ~i + 2 underset ~j + alpha underset ~k` and `underset ~b = 4 underset ~i - underset ~j + alpha^2 underset ~k`, where `alpha` is a real constant.
If the scalar resolute of `underset ~a` in the direction of `underset ~b` is `74/sqrt 273`, then `alpha` equals
A. 1
B. 2
C. 3
D. 4
E. 5
The direction field for the differential equation `(dy)/(dx) + x + y = 0` is shown above.
A solution to this differential equation that includes `(0, -1)` could also include
A. `(3, –1)`
B. `(3.5, –2.5)`
C. `(–1.5, –2)`
D. `(2.5, –1)`
E. `(2.5, 1)`
`=> B`