The function with rule `f(x)` has derivative `f′(x) = cos\ 3x`.
If `f(pi/6) = 1,` find `f(x).` (3 marks)
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The function with rule `f(x)` has derivative `f′(x) = cos\ 3x`.
If `f(pi/6) = 1,` find `f(x).` (3 marks)
`f(x)= 1/3 sin\ 3x + 2/3`
`int f(x)\ dx` | `=int cos\ 3x\ dx` |
`= 1/3 sin\ 3x + c` | |
`f(pi/6)` | `= 1/3\ [sin\ (3 xx pi/6) + c` |
`1` | `= 1/3\ sin\ pi/2+c` |
`c` | `= 2/3` |
`:.f(x)= 1/3 sin\ 3x + 2/3`
The average value of the function `y = tan (2x)` over the interval `[0, pi/8]` is
`A`
`y_text(average)` | `= 1/(pi/8 – 0) int_0^(pi/8) (tan 2x)\ dx` |
`= 8/pi int_0^(pi/8) (tan 2x)\ dx` | |
`= (2 log_e (2))/pi` |
`=> A`
The heights of the children in a queue for an amusement park ride are normally distributed with mean 130 cm and standard deviation 2.7 cm. 35% of the children are not allowed to go on the ride because they are too short.
The minimum acceptable height correct to the nearest centimetre is
`D`
If a random variable `X` has probability density function
`f(x) = {(x/2, if x in[0,2]), (0,\ \ \ text(otherwise)):}`
then `text(E) (X)` is equal to
`C`
`text(E) (X)` | `= int_0^2 x (x/2)\ dx` |
`=[x^3/6]_0^2` | |
`= 4/3` |
`=> C`
For the graph of `y = 4x^3 + 27x^2 - 30x + 10` the subset of `R` for which the gradient is negative is given by the interval
`D`
Let `f: R -> R` be a differentiable function such that
When `x = 3`, the graph of `f` has a
`C`
Solve the equation `e^(4x) - 5e^(2x) + 4 = 0` for `x`
(A) `x= 1, 4`
(B) `x= – 4, – 1`
(C) `x= 0, log_e 2`
(D) `x= – log_e 2, 0, log_e 2 `
`C`
`e^(4x) – 5e^(2x) + 4 = 0`
`text(Let)\ \ X=e^(2x)`
`X^2-5X+4` | `=0` |
`(X-4)(X-1)` | `=0` |
`X` | `=4 or 1` |
`:.e^(2x)` | `=4` | `e^(2x)` | `=1` |
`2x` | `=log_e 4` | `x` | `=0` |
`x` | `=(2log_e 2)/2` | ||
`=log_e 2` |
`=> C`
The graph of `y = kx - 3` intersects the graph of `y = x^2 + 8x` at two distinct points for
A. `k = 11`
B. `k > 8 + 2 sqrt 3 or k < 8 - 2 sqrt 3`
C. `5 <= k <= 6`
D. `8 - 2 sqrt 3 <= k <= 8 + 2 sqrt 3`
E. `k = 5`
`B`
`text(Intersection occurs when:)`
`kx – 3` | `= x^2 + 8x` |
`x^2 + (8 – k)x + 3` | `= 0` |
`text(For 2 points of intersection:)`
`Delta` | `> 0` |
`(8 – k)^2 – 4 (3)` | `> 0` |
`:. k < 8 – 2 sqrt 3\ uu\ k > 8 + 2 sqrt 3`
`=> B`
Let `k = int_-2^-1 1/x\ dx`, then `e^k` is equal to
`E`
`text(The integral can be seen as the negative)`
`text(equivalent of the area under)\ \ y=1/x`
`text(between)\ \ x=1 and x=2.`
`k` | `= – [log_e x]_1^2` |
`=-(log_e(2)-log_e(1))` | |
`= – log_e (2)` | |
`= log_e (1/2)` | |
`:. e^k` | `= e^(log_e(1/2))` |
`= 1/2` |
`=> E`
The average rate of change of the function with rule `f(x) = x^3 - sqrt (x + 1)` between `x = 0` and `x = 3` is
A. `0`
B. `12`
C. `26/3`
D. `25/3`
E. `8`
`C`
`text(Define)\ \ f(x) = x^3 – sqrt (x + 1)\ \ text(on CAS)`
`text(Average ROC)` | `= (f(3) – f(0))/(3 – 0)` |
`= 26/3` |
`=> C`
The linear function `f: D -> R,\ f(x) = 6 - 2x` has range `text([−4, 12]).`
The domain `D` is
`A`
`12` | `= 6 – 2x` |
`x` | `= – 3` |
`- 4` | `= 6 – 2x` |
`x` | `= 5` |
`:. x in [– 3, 5]`
`=> A`
The polynomial `p(x) = x^3- ax + b` has a remainder of 2 when divided by `(x- 1)` and a remainder of 5 when divided by `(x + 2)`.
Find the values of `a` and `b`. (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
`a` | `= 4` |
`b` | `= 5` |
`p(x)` | `= x^3 – ax + b` |
`P(1)` | `= 2` |
`1\ – a + b` | `= 2` |
`b` | `= a+1\ \ \ …\ text{(1)}` |
`P (–2)` | `= 5` |
`-8 + 2a + b` | `= 5` |
`2a + b` | `= 13\ \ \ …\ text{(2)}` |
`text(Substitute)\ \ b = a+1\ \ text(into)\ \ text{(2)}`
`2a + a+1` | `= 13` |
`3a` | `= 12` |
`:. a` | `= 4` |
`:. b` | `= 5` |
The graph of `P(x) = x^2 + ax + b` cuts the `x`-axis when `x=2.` When `P(x)` is divided by `x + 1`, the remainder is 18.
Find the values of `a` and `b`. (3 marks)
`a = -7\ \ text(and)\ \ b = 10`
`P(x) = x^2 + ax + b`
`text(S)text(ince the graph cuts the)\ xtext(-axis at)\ \ x = 2,`
`P(2)` | `=0` | |
`2^2 + 2a + b` | `= 0` | |
`2a + b` | `= −4` | `…\ (1)` |
`P(−1) = 18,`
`(−1)^2 − a + b` | `= 18` | |
`−a + b` | `= 17` | `…\ (2)` |
`text(Subtract)\ \ (1) − (2),`
`3a` | `= −21` |
`a` | `= −7` |
`text(Substitute)\ \ a = −7\ \ text{into (1),}`
`2(−7) + b` | `= −4` |
`b` | `= 10` |
`:.a = −7\ \ text(and)\ \ b = 10`
The graph of the function `f` with domain `[0, 6]` is shown below.
Which one of the following is not true?
`C`
`f(x) > 0\ \ text(for)\ \ x = 2`
`:.\ text(Option)\ \ C\ \ text(is not true)`
`=> C`
The minimum number of times that a fair coin can be tossed so that the probability of obtaining a head on each trial is less than 0.0005 is
A. `8`
B. `9`
C. `10`
D. `11`
E. `12`
`D`
`text(Let)\ \ X = text(Number of heads,)`
`X ∼ text(Bi) (n, 1/2)`
`text(Pr) (X = n)` | `< 0.0005` |
`((n), (n)) (1/2)^n (1/2)^0` | `< 0.0005` |
`n` | `> 10.97` |
`:. n_min = 11`
`=> D`
According to a survey, 30% of employed women have never been married.
If 10 employed women are selected at random, the probability (correct to four decimal places) that at least 7 have never been married is
A. `0.0016`
B. `0.0090`
C. `0.0106`
D. `0.9894`
E. `0.9984`
`C`
`text(Let)\ \ X =\ text(Number who have not been married)`
`X∼\ text(Bi) (10, 0.3)`
`text(Pr) (X >= 7)\ \ \ [text(CAS: binomCdf) (10, 0.3, 7, 10)]`
`= 0.0106`
`=> C`
Let `X` be a discrete random variable with a binomial distribution. The mean of `X` is 1.2 and the variance of `X` is 0.72
The values of `n` (the number of independent trials) and `p` (the probability of success in each trial) are
A. `n = 4,\ \ \ \ \ p = 0.3`
B. `n = 3,\ \ \ \ \ p = 0.6`
C. `n = 2,\ \ \ \ \ p = 0.6`
D. `n = 2,\ \ \ \ \ p = 0.4`
E. `n = 3,\ \ \ \ \ p = 0.4`
`E`
`X∼\ text(Bi) (n, p)`
`text(E) (X)` | `= 1.2` |
`np` | `= 1.2\ …\ (1)` |
`text(Var) (X)` | `= 0.72` |
`np (1 – p)` | `= 0.72\ …\ (2)` |
`text(Solve simultaneous equations:)`
`:. n = 3,\ \ p = 0.4`
`=> E`
The average value of the function with rule `f(x) = log_e (3x + 1)` over the interval `[0, 2]` is
`D`
`y_text(avg)` | `= 1/(2 – 0) int_0^2 log_e (3x + 1)\ dx` |
`= (7 log_e(7) – 6)/6` |
`=> D`
The area under the curve `y = sin (x)` between `x = 0` and `x = pi/2` is approximated by two rectangles as shown.
This approximation to the area is
A. `1`
B. `pi/2`
C. `((sqrt 3 + 1) pi)/12`
D. `0.5`
E. `((sqrt 3 + 1) pi)/6`
`C`
`text(Rectangle width) = pi/6`
`text(Area)` | `~~ pi/6 [sin (pi/6) + sin (pi/3)]` |
`~~pi/6(sqrt3/2 + 1/2)` | |
`~~ (pi (sqrt 3 + 1))/12` |
`=> C`
Solve these simultaneous equations to find the values of `x` and `y`.
`4x - 2y = 18`
`3x + 2y = 10` (3 marks)
`x = 4,\ \ y = -1`
`text(Solution 1 – Eliminations)`
`4x – 2y = 18\ …\ text{(i)}`
`3x +2y = 10\ …\ text{(ii)}`
`text{(i)} + text{(ii)}`
`7x` | `= 28` |
`x` | `= 4` |
`text(Substitute)\ \ x = 4\ \ text(into)\ text{(i)}`
`4 xx 4 – 2y` | `= 18` |
`- 2y` | `= 2` |
`y` | `= -1` |
`:.\ text(Solution is)\ \ x = 4\ \ text(and)\ y = -1`
`text(Alternative Solution – Substitution)`
`4x – 2y = 18\ …\ text{(i)}`
`3x + 2y = 10\ …\ text{(ii)}`
`text{Divide (i) by 2}`
`2x – y` | `= 9` |
`y` | `= 2x – 9` |
`text(Substitute)\ \ y = 2x – 9\ \ text(into)\ text{(ii)}`
`3x + 2(2x – 9)` | `= 10` |
`3x + 4x – 18` | `= 10` |
`7x` | `= 28` |
`x` | `= 4` |
`text(Substitute)\ \ x= 4\ \ text{into (i)}`
`4 xx 4 – 2y` | `= 18` |
`- 2y` | `= 2` |
`y` | `= -1` |
Solve these simultaneous equations to find the values of `x` and `y`. (3 marks)
`y = 2x + 1`
`x − 2y − 4 = 0`
`x = -2,\ y = -3`
`text(Solution 1 – Substitution)`
`y = 2x + 1\ \ \ \ \ …\ text{(i)}`
`x\ – 2y\ – 4 = 0\ \ \ \ \ …\ text{(ii)}`
`text(Substitute)\ \ y = 2x + 1\ \ text(into)\ text{(ii)}`
`x\ – 2(2x + 1)\ – 4` | `= 0` |
`x\ – 4x\ – 2\ – 4` | `= 0` |
`-3x\ – 6` | `= 0` |
`3x` | `= -6` |
`x` | `= -2` |
`text(Substitute)\ \ x = –2\ \ text(into)\ text{(i)}`
`y = 2(–2) + 1 = -3`
`:.\ text(Solution is)\ x = -2,\ y = -3`
`text(Alternative Solution – Elimination)`
`y = 2x + 1\ \ \ \ \ …\ text{(i)}`
`x\ – 2y\ – 4 = 0\ \ \ \ \ …\ text{(ii)}`
`text(Multiply)\ text{(i)} xx 2`
`2y` | `= 4x + 2` |
`-4x + 2y\ – 2` | `= 0\ \ \ \ \ …\ text{(iii)}` |
`text{Add (ii) + (iii)}`
`-3x\ – 6` | `= 0` |
`x` | `= -2` |
`y` | `= -3\ \ text{(see Solution 1)}` |
The function `f` has the rule `f(x) = 1 + 2 cos x`.
a. `f(x) = 1 + 2 cos x`
`f(x)\ text(cuts the)\ x text(-axis when)\ f(x) = 0`
`1 + 2 cos x` | `= 0` |
`2 cos x` | `=-1` |
`cos x` | `= -1/2` |
`:. x = (2 pi)/3\ …\ text(as required)`
b. |
c. `text(Area)` | `= int_(-pi/2)^((2 pi)/3) 1 + 2 cos x\ \ dx` |
`= [x + 2 sin x]_(-pi/2)^((2 pi)/3)` | |
`= [((2 pi)/3 + 2 sin (2 pi)/3) – ((-pi)/2 + 2 sin (-pi)/2)]` | |
`= ((2 pi)/3 + 2 xx sqrt 3/2) – ((-pi)/2 +2(- 1))` | |
`= (2 pi)/3 + sqrt(3) + pi/2 + 2` | |
`= ((7 pi)/6 + sqrt(3) + 2)\ text(u²)` |
The graph shown is `y = A sin bx`.
On the same set of axes, draw the graph `y = 3 sin x + 1` for `0 <= x <= pi`. (2 marks)
a. `A = 4`
b. `text(S)text(ince the graph passes through)\ \ (pi/4, 4)`
`text(Substituting into)\ \ y = 4 sin bx`
`4 sin (b xx pi/4)` | `=4` |
`sin (b xx pi/4)` | `= 1` |
`b xx pi/4` | `= pi/2` |
`:. b` | `= 2` |
c. |
Evaluate `int_0^(pi/4) cos 2x\ dx`. (2 marks)
`1/2`
`int_0^(pi/4) cos 2x`
`= [1/2 sin\ 2x]_0^(pi/4)`
`= [1/2 sin\ pi/2 – 1/2 sin\ 0]`
`= 1/2 – 0`
`= 1/2`
Find the equation of the tangent to the curve `y = cos 2x` at the point whose `x`-coordinate is `pi/6`. (3 marks)
`y = – sqrt 3 x + ((sqrt 3 pi)/6 + 1/2)`
`y = cos 2x`
`dy/dx = -2 sin 2x`
`text(When)\ \ x = pi/6,`
`y` | `= cos (2 xx pi/6)` |
`= cos (pi/3)` | |
`= 1/2` |
`dy/dx` | `= -2 sin (pi/3)` |
`= -2 xx sqrt 3 / 2` | |
`= – sqrt 3` |
`:. text(Equation of tangent,)\ \ m = – sqrt 3, text(through)\ \ (pi/6, 1/2) :`
`y – y_1` | `= m(x – x_1)` |
`y – 1/2` | `= – sqrt 3 ( x – pi/6)` |
`y – 1/2` | `= – sqrt 3 x + (sqrt 3 pi)/6` |
`y` | `= – sqrt 3 x + ((sqrt 3 pi)/6 + 1/2)` |
If `f(x)= 2 sin 3x - 3 tan x`, find `f′(0)`. (2 marks)
`3`
`y` | `= 2 sin 3x – 3 tan x` |
`(dy)/(dx)` | `= 6 cos 3x – 3 sec^2 x` |
`text(When)\ \ x = 0,`
`(dy)/(dx)` | `= 6 cos 0 – 3 sec^2 0` |
`= 6 (1) – 3/(cos^2 0)` | |
`= 6 – 3` | |
`= 3` |
The rule for `f` is `f(x) = e^x − e^(-x)`.
Show that the inverse function is given by
`f^(-1)(x) = log_e((x + sqrt(x^2 + 4))/2)` (3 marks)
`text{Proof (See Worked Solutions.)}`
`y = e^x − e^(-x)`
`text(Inverse: swap)\ x harr y`
`x` | `= e^y − 1/(e^y)` |
`xe^y` | `= e^(2y) − 1` |
`e^(2y) − xe^y − 1` | `= 0` |
`text(Let)\ \ A = e^y`
`:.A^2 − xA − 1 = 0`
`text(Using the quadratic formula)`
`A` | `=(x ± sqrt((-x)^2 − 4 · 1 · (-1)))/(2 · 1)` |
`=(x ± sqrt(x^2 + 4))/2` |
`text(S)text(ince)\ \ (x – sqrt(x^2 + 4))/2<0\ \ text(and)\ \ e^y>0,`
`:.e^y` | `= (x + sqrt(x^2 + 4))/2` |
`log_e e^y` | ` = log_e((x + sqrt(x^2 + 4))/2)` |
`y` | `= log_e((x + sqrt(x^2 + 4))/2)` |
`:.f^(-1)(x)` | `= log_e((x + sqrt(x^2 + 4))/2)\ \ …\ text(as required)` |
The function `f: [0,oo) → R` with rule `f(x) = 1/(1 + x^2)` is drawn below.
a. |
b. `text(Range of)\ \ f(x):\ (0,1]`
`:.\ text(Domain of)\ \ f^(−1)(x): (0,1]`
c. `f(x) = 1/(1 + x^2)`
`text(Inverse: swap)\ x harr y`
`x` | `= 1/(1 + y^2)` |
`x(1 + y^2)` | `= 1` |
`1 + y^2` | `= 1/x` |
`y^2` | `= 1/x − 1` |
`= (1 − x)/x` | |
`y` | `= ± sqrt((1 − x)/x)` |
`:.y = sqrt((1 − x)/x), \ \ y >= 0`
d. `P\ \ text(occurs when)\ \ f(x)\ \ text(cuts)\ \ y = x`
`text(i.e. where)`
`1/(1 + x^2)` | `= x` |
`1` | `= x(1 + x^2)` |
`1` | `= x + x^3` |
`x^3 + x − 1` | `= 0` |
`=>\ text(S)text(ince)\ α\ text(is the)\ x\ text(-coordinate of)\ P,`
`text(it is a root of)\ \ \ x^3 + x − 1 = 0`
The rule for function `f` is `f(x) = e^(-x^2)`. The diagram shows the graph `y = f(x)`.
The graph has two points of inflection.
a. | `y` | `= e^(-x^2)` |
`dy/dx` | `= -2x * e^(-x^2)` | |
`(d^2y)/(dx^2)` | `= -2x (-2x * e^(-x^2)) + e ^(-x^2) (-2)` | |
`= 4x^2 e^(-x^2)\ – 2e^(-x^2)` | ||
`= 2e^(-x^2) (2x^2\ – 1)` |
`text(P.I. when)\ \ (d^2y)/(dx^2) = 0`
`2e^(-x^2) (2x^2\ – 1)` | `= 0` |
`2x^2\ – 1` | `= 0` |
`x^2` | `= 1/2` |
`x` | `= +- 1/sqrt2` |
`text(When)\ \ ` | `x < 1/sqrt2,` | `\ (d^2y)/(dx^2) < 0` |
`x > 1/sqrt2,` | `\ (d^2y)/(dx^2) > 0` |
`=>\ text(Change of concavity)`
`:.\ text(P.I. at)\ \ x = 1/sqrt2`
`text(When)\ \ ` | `x < – 1/sqrt2,` | `\ (d^2y)/(dx^2) > 0` |
`x > – 1/sqrt2,` | `\ (d^2y)/(dx^2) < 0` |
`=>\ text(Change of concavity)`
`:.\ text(P.I. at)\ \ x = – 1/sqrt2`
b. | `text(In)\ f(x), text(there are 2 values of)\ y\ text(for)` |
`text(each value of)\ x.` | |
`:.\ text(The domain of)\ f(x)\ text(must be restricted)` | |
`text(for)\ \ f^(-1) (x)\ text(to exist).` |
c. `y = e^(-x^2)`
`text(Inverse: swap)\ x harr y`
`x` | `= e^(-y^2),\ \ \ x >= 0` |
`lnx` | `= ln e^(-y^2)` |
`-y^2` | `= lnx` |
`y^2` | `= -lnx` |
`=ln(1/x)` | |
`y` | `= +- sqrt(ln (1/x))` |
`text(Restricting)\ \ x>=0,\ \ =>y>=0`
`:. f^(-1) (x)=sqrt(ln (1/x))`
d. | `f(0) = e^0 = 1` |
`:.\ text(Range of)\ \ f(x)\ \ text(is)\ \ 0 < y <= 1`
`:.\ text(Domain of)\ \ f^(-1) (x)\ \ text(is)\ \ 0 < x <= 1`
e. |
The rule for `f` is `f(x) = x - 1/2 x^2` for `x <= 1`. This function has an inverse, `f^(-1) (x)`.
a. |
b. | `y = x – 1/2 x^2,\ \ \ x <= 1` |
`text(For the inverse function, swap)\ \ x↔y,`
`x` | `= y – 1/2 y^2,\ \ \ y <= 1` |
`2x` | `= 2y – y^2` |
`y^2 – 2y + 2x` | `= 0` |
`text(Using quadratic formula,)`
`y` | `= (2 +- sqrt( (-2)^2 – 4 * 1 * 2x) )/2` |
`= (2 +- sqrt(4 – 8x))/2` | |
`= (2 +- 2 sqrt(1 – 2x))/2` | |
`= 1 +- sqrt (1 – 2x)` |
`:. y = 1 – sqrt(1 – 2x), \ \ (y <= 1)`
c. | `f^(-1) (3/8)` | `= 1 – sqrt(1 -2(3/8))` |
`= 1 – sqrt(1 – 6/8)` | ||
`= 1 – sqrt(1/4)` | ||
`= 1 – 1/2` | ||
`= 1/2` |
Let `f: R→R` where `f(x)= x^3 - 2`.
Evaluate `f^(-1)(25),` where `f^(-1)` is the inverse function of `f`. (2 marks)
`3`
`text(Let)\ \ y = x^3 – 2`
`text(For inverse),\ \ x harr y`
`x` | `= y^3 – 2` |
`y^3` | `= x + 2` |
`y` | `= (x + 2)^(1/3)` |
`:. f^(-1)(25)` | `=(25+2)^(1/3)` |
`=3` |
If `f′(x)= (x^2)/(x^3 + 1)` and `f(1)= log_e 2,` find `f(x)`. (3 marks)
`f(x) = 1/3 log_e(x^3 + 1) + 2/3 log_e 2`
`f(x)` | `=int f′(x)\ dx` |
`=int (x^2)/(x^3 + 1)\ dx` | |
`= 1/3 int (3x^2)/(x^3 + 1)\ dx` | |
`=1/3 log_e |(x^3 + 1)|+c` |
`text(S)text(ince)\ \ f(1)= log_e 2,`
`1/3 log_e 2+c` | `= log_e 2` |
`c` | `=2/3 log_e 2` |
`:.\ f(x) = 1/3 log_e(x^3 + 1) + 2/3 log_e 2`
Find the value of `int_e^(e^3) 5/x` with respect to `x` (2 marks)
`10`
`int_e^(e^3) 5/x\ dx`
`= 5 int_e^(e^3) 1/x\ dx`
`= 5[lnx]_e^(e^3)`
`= 5(ln e^3 – ln e)`
`= 5(3 – 1)`
`= 10`
a. | `y` | `= log_e(cos x)` |
`(dy)/(dx)` | `= (−sin x)/(cos x)` | |
`= −tan x` |
b. `int_0^(pi/4) tan x\ dx`
`= −[log_e(cos x)]_0^(pi/4)`
`= −[log_e(cos(pi/4)) – log_e(cos 0)]`
`= −[log_e(1/sqrt2) – log_e 1]`
`= −[log_e(1/sqrt2) – 0]`
`= −log_e(1/sqrt2)`
`= 0.346…`
`= 0.35\ \ (text(2 d.p.))`
a. `y = xe^(3x)`
`text(Using product rule,)`
`(dy)/(dx)` | `= x · 3e^(3x) + 1 · e^(3x)` |
`= e^(3x) (1 + 3x)` |
b. `int_0^2 e^(3x) (3 + 9x)\ dx`
`= 3 int_0^2 e^(3x) (1 + 3x)\ dx`
`= 3 [x e^(3x)]_0^2`
`= 3 (2e^6 – 0)`
`= 6e^6`
Differentiate `(e^x + x)^5`. (2 marks)
`5(e^x + 1)(e^x + x)^4`
`y` | `= (e^5 + x)^5` |
`(dy)/(dx)` | `= 5(e^x + x)^4 xx d/(dx)(e^x + x)` |
`= 5(e^x + x)^4 xx (e^x + 1)` | |
`= 5(e^x + 1)(e^x + x)^4` |
A bag contains three white balls and seven yellow balls. Three balls are drawn, one at a time, from the bag, without replacement.
The probability that they are all yellow is
`D`
`text{Pr(YYY)}` | `= 7/10 xx 6/9 xx 5/8` |
`= 7/24` |
`=> D`
The diagram shows the graph of the function `f: (0,oo) →R,` where `f(x) = 1/x`.
The area under `f(x)` between `x = a` and `x = 1` is `A_1`. The area under the curve between `x = 1` and `x = b` is `A_2`.
The areas `A_1` and `A_2` are each equal to 1 square unit.
Find the values of `a` and `b`. (3 marks)
`a = 1/e`
`b = e`
`int_a^1 1/x dx` | `= 1` |
`[ln x]_a^1` | `= 1` |
`ln 1 – ln a` | `= 1` |
`ln a` | `= −1` |
`:. a` | `= e^(−1) = 1/e` |
`int_1^b 1/x dx` | `= 1` |
`[ln x]_1^b` | `= 1` |
`ln b – ln 1` | `= 1` |
`ln b` | `= 1` |
`:. b` | `= e` |
The curves `y = e^(2x)` and `y = e^(−x)` intersect at the point `(0,1)` as shown in the diagram.
Find the exact area enclosed by the curves and the line `x = 2`. (3 marks)
`1/2 e^4 + e^(−2) – 3/2\ \ \ text(u²)`
Note: there’s marker’s comment in this part
`text(Area)` | `= int_0^2 e^(2x)\ \ dx – int_0^2 e^(−x)\ \ dx` |
`= int_0^2 (e^(2x) – e^(−x))dx` | |
`= [1/2 e^(2x) + e^(−x)]_0^2` | |
`= [(1/2 e^4 – e^(−2)) – (1/2 e^0 + e^0)]` | |
`= 1/2 e^4 + e^(−2) – 3/2\ \ \ text(u²)` |
For the function `f:R→R,\ \ f(x)= 2e^x + 3x`, determine the coordinates of the point `P` at which the tangent to `f(x)` is parallel to the line `y = 5x - 3`. (3 marks)
`(0, 2)`
`y = 5x − 3\ \ =>\ m=5`
`y` | `= 2e^x + 3x` |
`(dy)/(dx)` | `= 2e^x + 3` |
`text(Find)\ \ x\ \ text(when)\ \ (dy)/(dx) = 5,`
`5` | `= 2e^x + 3` |
`2e^x` | `= 2` |
`e^x` | `= 1` |
`x` | `= 0` |
`text(When)\ \ x = 0,`
`y` | `= 2e^0 + (3 xx 0)=2` |
`:.P\ \ text{has coordinates (0, 2)}`
Solve the following equation for `x`:
`2e^(2x) - e^x = 0`. (2 marks)
`x = ln\ 1/2`
`text(Solution 1)`
`2e^(2x) – e^x = 0`
`text(Let)\ \ X = e^x`
`2X^2 – X` | `= 0` |
`X (2X – 1)` | `= 0` |
`X = 0 or 1/2`
`text(When)\ \ e^x = 0\ =>\ text(no solution)`
`text(When)\ \ e^x = 1/2`
`ln e^x` | `= ln\ 1/2` |
`:. x` | `= ln\ 1/2` |
`text(Solution 2)`
`2e^(2x)-e^x` | `=0` |
`2e^(2x)` | `=e^x` |
`ln 2e^(2x)` | `=ln e^x` |
`ln 2 +ln e^(2x)` | `=x` |
`ln 2 + 2x` | `=x` |
`x` | `=-ln2` |
`=ln\ 1/2` |
Write `log2 + log4 + log8 + log16 + … + log128` in the form `a logb` where `a` and `b` are integers greater than 1. (2 marks)
`28log2`
`log2 + log4 + log8 + … + log 128`
`= log2^1 + log2^2 + log2^3 + … + log2^7`
`= log2 + 2log2 + 3log2 + … + 7log2`
`= 28log2`
Solve `5^x = 4` for `x`. (2 marks)
`(log_e 4)/(log_e 5)`
`text(Given)\ \ 5^x` | `= 4` | |
`log_e 5^x` | `= log_e 4` | |
`x xx log_e 5` | `= log_e 4` | |
`:. x` | `= (log_e 4)/(log_e 5)` |
Let `a = e^x`
Which expression is equal to `log_e(a^2)`?
`C`
`log_e(a^2)` | `= log_e(e^x)^2` |
`= log_e(e^(2x))` | |
`= 2xlog_e e` | |
`= 2x` |
`=> C`
A company produces motors for refrigerators. There are two assembly lines, Line A and Line B. 5% of the motors assembled on Line A are faulty and 8% of the motors assembled on Line B are faulty. In one hour, 40 motors are produced from Line A and 50 motors are produced from Line B. At the end of an hour, one motor is selected at random from all the motors that have been produced during that hour.
What is the probability that it was assembled on Line A? Express your answer in the form `1/c`, where `c` is a positive integer. (1 mark)
Let `f : (0, ∞) → R`, where `f(x) = log_e(x)` and `g: R → R`, where `g (x) = x^2 + 1`.
a.i. | `h(x)` | `= f(x^2 + 1)` |
`= log_e(x^2 + 1)` |
a.ii. `text(Domain)\ (h) =\ text(Domain)\ (g) = R`
`text(For)\ x ∈ R` | `-> x^2 + 1 >= 1` |
`-> log_e(x^2 + 1) >= 0` |
`:.\ text(Range)\ (h) = [0,∞)`
a.iii. | `text(LHS)` | `= h(x) + h(−x)` |
`= log_e(x^2 _ 1) + log_e((−x)^2 + 1)` | ||
`= log_e(x^2 + 1) + log_e(x^2 + 1)` | ||
`= 2log_e(x^2 + 1)` |
`text(RHS)` | `= f((x^2 + 1)^2)` |
`= 2log_e(x^2 + 1)` |
`:. h(x) + h(−x) = f((g(x))^2)\ \ text(… as required)`
a.iv. `text(Stationary points when)\ \ h′(x) = 0`
`text(Using Chain Rule:)`
`h′(x)` | `= (2x)/(x^2 + 1)` |
`:.\ text(S.P. when)\ \ x=0`
`text(Find nature using 1st derivative test:)`
`:.\ text{Minimum stationary point at (0, 0)}.`
b.i. `text(Let)\ \ y = k(x)`
`text(Inverse: swap)\ x ↔ y`
`x` | `= log_e(y^2 + 1)` |
`e^x` | `= y^2 + 1` |
`y^2` | `= e^x – 1` |
`y` | `= ±sqrt(e^x – 1)` |
`text(But range)\ \ (k^(−1)) =\ text(domain)\ (k)`
`:.k^(−1)(x) = −sqrt(e^x – 1)`
b.ii. `text(Range)\ (k^(−1)) =\ text(Domain)\ (k) = (−∞,0]`
`text(Domain)\ (k^(−1)) =\ text(Range)\ (k) = [0,∞)`
A paddock contains 10 tagged sheep and 20 untagged sheep. Four times each day, one sheep is selected at random from the paddock, placed in an observation area and studied, and then returned to the paddock.
Express your answer in the form `(a/b)^c`, where `a`, `b` and `c` are positive integers. (1 mark)
a. `text(Let)\ \ X =\ text(Number of tagged sheep,)`
`X ~\ text(Bi)(4,1/3)`
`text(Pr)(X = 0)` | `= ((4),(0)) xx (1/3)^0 xx (2/3)^4` |
`= 16/81` |
b. | `text(Pr)(X >= 1)` | `= 1 – text(Pr)(X = 0)` |
`= 1 – 16/81` | ||
`= 65/81` |
c. `text(Let)\ \ Y =\ text(Days that no tagged sheep selected,)`
`Y ~\ text(Bi)(6,16/81)`
`text(Pr)(Y = 6)` | `= ((6),(6)) xx (16/81)^6 xx (65/81)^0` |
`= (16/81)^6 = (2/3)^24` |
Let `f: R text{\}{1} -> R` where `f(x) = 2 + 3/(x - 1)`.
Let `f: (−∞,1/2] -> R`, where `f(x) = sqrt(1 - 2x)`.
a. | `f(x)` | `= (1 – 2x)^(1/2)` |
`f′(x)` | `= 1/2(1 – 2x)^(−1/2) (−2)qquadtext([Using Chain Rule])` | |
`= (−1)/(sqrt(1 – 2x))` |
b. | `tan theta` | `= f′(−1)` |
`= (−1)/(sqrt(1 – 2(−1)))` | ||
`= (−1)/(sqrt3)` |
`text(S)text(ince)\ theta ∈ [0,pi],`
`=> theta = (5pi)/6`
A train is travelling at a constant speed of `w` km/h along a straight level track from `M` towards `Q.`
The train will travel along a section of track `MNPQ.`
Section `MN` passes along a bridge over a valley.
Section `NP` passes through a tunnel in a mountain.
Section `PQ` is 6.2 km long.
From `M` to `P`, the curve of the valley and the mountain, directly below and above the train track, is modelled by the graph of
`y = 1/200 (ax^3 + bx^2 + c)` where `a, b` and `c` are real numbers.
All measurements are in kilometres.
The driver sees a large rock on the track at a point `Q`, 6.2 km from `P`. The driver puts on the brakes at the instant that the front of the train comes out of the tunnel at `P`.
From its initial speed of `w` km/h, the train slows down from point `P` so that its speed `v` km/h is given by
`v = k log_e ({(d + 1)}/7)`,
where `d` km is the distance of the front of the train from `P` and `k` is a real constant.
a.i. `N (2, 0),`
`1/200 (8a + 4b + c) = 0\ \ text{… (1)}`
`{:frac (dy) (dx)|:}_(x = 2) = (– 3)/50,`
`1/200 (12a + 4b) = (– 3)/50\ \ text{… (2)}`
`{:frac (dy) (dx)|:}_(x = 4) = 0,`
`1/200 (48a + 8b) = 0\ \ text{… (3)}`
ii. `text(Using the above equations,)`
`12a + 4b` | `=-12` | `\ \ \ …\ (2′)` |
`24 a + 4b` | `=0` | `\ \ \ …\ (3′)` |
`text(Solve simultaneous equations:)`
`(3′) – (2′):`
`12a` | `= 12` |
`:. a` | `= 1` |
`text(Substitute)\ \ a = 1\ \ text(into)\ \ (3′):`
`24(1) + 4b` | `= 0` |
`:. b` | `= – 6` |
`text(Substitute)\ \ a = 1,\ \ b = – 6\ \ text(into)\ \ (1),`
`8(1) + 4 (– 6) + c` | `= 0` |
`:. c` | `= 16` |
b.i. `text(For)\ \ x text(-intercepts),`
`text(Solve:)\ \ x^3 + -6x^2 + 16` | `= 0\ \ text(for)\ x,` |
`x= 2 – 2 sqrt 3, 2, 2 + 2 sqrt 3` |
`:. M (2 + 2 sqrt 3, 0),\ \ P (2 – 2 sqrt 3, 0)`
ii. `N (2, 0)`
`bar (NP)` | `=\ text(Tunnel length)` |
`= 2 – (2 – 2 sqrt 3)` | |
`= 2 sqrt 3\ text(km)` |
iii. `text(Solve)\ \ frac (dy) (dx) = 0\ \ text(for)\ \ x in (2, 2 + 2 sqrt 3)`
`x = 4`
`text(When)\ \ x=4,\ \ y = – 2/25\ text(km) = 80\ text{m (below track)}`
`:.\ text(Max depth below is)\ \ 80\ text(m.)`
c. `text(Solution 1)`
`text(Let)\ \ v(d) = k log_e ({(d + 1)}/7)`
`text(S)text(ince)\ \ v=w\ \ text(when)\ \ d=0\ \ text{(given),}`
`k log_e ({(0 + 1)}/7)` | `=w` |
`k` | `=w/log_e (1/7)` |
`=(-w)/log_e7` |
`text(Solution 2)`
`text(Solve:)\ \ v (0)` | `= w\ \ text(for)\ \ k` |
`:. k` | `= (– w)/(log_e (7))` |
d. `v (2.5) = k log_e(1/2)`
`text(Solve)\ \ v(2.5)` | `= (120 log_e (2))/(log_e (7))\ \ text(for)\ \ k,` |
`:. k` | `= (– 120)/(log_e (7))` |
`(– w)/(log_e (7))` | `= (– 120)/(log_e (7))` |
`:. w` | `= 120` |
e. `text(Define)\ \ v (d) = (– 120)/(log_e (7)) log_e ((d + 1)/7)`
`text(Solve:)\ \ v(d)` | `= 0\ \ text(for)\ d,` |
`:. d` | `= 6\ text(km from)\ \ P` |
`:.\ text(Distance between train and)\ \ Q`
`= 6.2 – 6`
`= 0.2\ text(km)`
Let `f: R^+ uu {0} -> R,\ f(x) = 6 sqrt x - x - 5.`
The graph of `y = f (x)` is shown below.
Find the length of `AD` such that the area of rectangle `ABCD` is equal to the area of the shaded region. (2 marks)
a. `x in [9,oo)`
b. `text(Not in syllabus.)`
c. `8/3`
d.i. `m=-1/3`
d.ii. `a=81/4`
a. |
`text(Stationary point when)\ \ f prime (x)` | `= 0` |
`x` | `= 9` |
`:.\ text(Strictly decreasing for)\ \ x in [9, oo)`
b. `text(No longer in syllabus.)`
c. |
`AD` | `= y_text(average)` |
`= 1/(25 – 1) int_1^25\ f(x)\ dx` | |
`= 8/3\ \ text{[by CAS]}` |
d.i. | `m_(PB)` | `= (3 – 0)/(16 – 25)` |
`:. m_(PB)` | `= – 1/3` |
d.ii. | `text(Solve)\ \ f prime (a)` | `= -1/3\ \ text(for)\ \ a in [16, 25]` |
`:. a` | `= 81/4` |
The graph of a function `f`, with domain `R`, is as shown.
The graph which best represents `1 - f (2x)` is
`E`
`text(Determine transformations)`
`text(from)\ \ f -> – f (2x) + 1`
`text(- Dilate by factor)\ 1/2\ text(from)\ y text(-axis.)`
`text(- Reflect in)\ x text(-axis.)`
`text(- Shift up 1 unit.)`
`=> E`
The average value of the function `f: R\ text(\){text(−)1/2} -> R,\ f(x) = 1/(2x + 1)` over the interval `[0, k]` is `1/6 log_e (7).`
The value of `k` is
`B`
`text(Solve:)\ \ 1/(k-0) int_0^k 1/(2x + 1)\ dx` | `= 1/6 log_e (7)` |
`text(for)\ \ k` | `> 0` |
`:. k = 3`
`=> B`
The inverse of the function `f: R^+ -> R,\ f(x) = e^(2x + 3)` is
`C`
`text(Let)\ \ y = f(x)`
`text(Inverse: swap)\ \ x harr y`
`x` | `=e^(2y + 3)` |
`log_e x` | `=2y+3` |
`2y` | `=log_e x -3` |
`y` | `=1/2 log_e (x) – 3/2` |
`:. f^-1 (x) = log_e (sqrt x) – 3/2`
`=> C`
For `y = sqrt (1 - f(x)),\ \ (dy)/(dx)` is equal to
A. `(2 f prime (x))/(sqrt(1 - f(x))`
B. `(-1)/(2 sqrt (1 - f prime (x)))`
C. `1/2 sqrt (1 - f prime (x))`
D. `3/(2(1 - f prime(x)))`
E. `(-f prime (x))/(2 sqrt (1 - f (x)))`
`E`
`text(Using Chain Rule,)`
`dy/dx` | `= – f′(x) xx 1/2 xx (1-f(x))^(- 1/2)` |
`=(- f′(x))/(2 sqrt (1 – f (x)))` |
`=> E`
A fair coin is tossed twelve times.
The probability (correct to four decimal places) that at most 4 heads are obtained is
A. `0.0730`
B. `0.1209`
C. `0.1938`
D. `0.8062`
E. `0.9270`
`C`
`text(Let)\ \ X = text(Number of heads),`
`X∼\ text(Bi) (12, 1/2)`
`text(Pr) (X <= 4) ~~ 0.1938\ \ [text(CAS: binomCdf) (12, 1/2, 0, 4)]`
`=> C`
A transformation `T: R^2 -> R^2` that maps the curve with equation `y = sin (x)` onto the curve with equation `y = 1 - 3 sin(2x + pi)` is given by
`D`
`text(Let)\ \ f(x) = sin (x)`
`text(Let)\ \ g(x) = – 3 sin (2 (x + pi/2)) + 1`
`text(Find transformations taking)\ \ f -> g`
`3 f (2x) = 3 sin (2x) = h(x)`
`– h(x) = – 3 sin (2x) = k(x)`
`k (x + pi/2) + 1 = – 3 sin (2 (x + pi/2)) + 1 = g(x)`
`text(Dilate by factor 3 from)\ \ x text(-axis)`
`text(Dilate by factor)\ \ 1/2\ \ text(from)\ \ y text(-axis)`
`text(Reflect in)\ \ x text(-axis)`
`text(Translate left)\ \ pi/2\ \ text(up 1)`
`=> D`
The continuous random variable `X` has a probability density function given by
`f(x) = {(pi sin (2 pi x), text(if)\ \ 0 <= x <= 1/2), (0, text(elsewhere)):}`
The value of `a` such that `text(Pr) (X > a) = 0.2` is closest to
`D`
`text(Solve)\ \ int_a^(1/2) f (x)\ dx` | `= 0.2,\ \ a in [0, 1/2]` |
`:. a` | `~~ 0.35` |
`=> D`
The continuous random variable `X` has a normal distribution with mean 14 and standard deviation 2.
If the random variable `Z` has the standard normal distribution, then the probability that `X` is greater than 17 is equal to
`D`
`text(Pr) (X > 17)` | `= text(Pr) (Z > (17 – 14)/2)` |
`= text(Pr) (Z > 1.5)` | |
`= text(Pr) (Z < – 1.5)` |
`=> D`