Band 3 – Page 49

Calculus, EXT2 C1 2009 HSC 1a

Find `int (ln x)/x\ dx.` (2 marks)

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`((ln x)^2)/2 + c`

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`text(Let)\ \ u=lnx,\ \ \ du=1/x\ dx`

`int (ln x)/x \ dx`	`=int u\ du`
	`=1/2 u^2 +c`
	`=1/2 (ln x)^2 +c`

Polynomials, EXT2 2010 HSC 6c

Expand `(cos theta + i sin theta)^5` using the binomial theorem. (1 mark)
Expand `(cos theta + i sin theta)^5` using de Moivre’s theorem, and hence show that
1. `sin 5theta = 16 sin^5 theta − 20sin^3 theta + 5 sin theta`. (3 marks)
Deduce that
`x = sin (pi/10)` is one of the solutions to
1. `16x^5 − 20x^3 + 5x − 1 = 0`. (1 mark)
Find the polynomial `p(x)` such that `(x − 1) p(x) = 16x^5 − 20x^3 + 5x − 1`. (1 mark)
Find the value of `a` such that `p(x) = (4x^2 + ax − 1)^2`. (1 mark)
Hence find an exact value for
1. `sin (pi/10)`. (1 mark)

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`cos^5 theta + 5i cos^4 theta sin theta − 10 cos^3 theta sin^2 theta − 10i cos^2 theta sin^3 theta`
`+ 5 cos theta sin^4 theta + i sin^5 theta`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`16x^4 + 16x^3 − 4x^2 − 4x + 1`
`a = 2`
`(-1 + sqrt5)/4`

Show Worked Solution

(i) `(cos theta + i sin theta)^5`

`=cos^5 theta + 5cos^4 theta (i sin theta) + 10 cos^3 theta (i sin theta)^2 + `

`10 cos^2 theta (isin theta)^3 + 5 cos theta (i sin theta)^4 + (i sin theta)^5`

`= cos^5 theta + 5i cos^4 theta sin theta − 10 cos^3 theta sin^2 theta − 10i cos^2 theta sin^3 theta +`

`5 cos theta sin^4 theta + i sin^5 theta`

(ii) `text(Using De Moivre)`

`(cos theta + i sin theta)^5 = cos 5theta + i sin 5theta`

`text(Equating imaginary parts)`

`sin 5theta`	`= 5 cos^4theta sin theta − 10cos^2 theta sin^3 theta + sin^5 theta`
	`= 5 sin theta (1 − sin^2 theta)^2 − 10 sin^3 theta (1 − sin^2 theta) + sin^5 theta`
	`= 5 sin theta (1 − 2 sin^2 theta + sin^4 theta) − 10 sin^3 theta + 10 sin^5 theta + sin^5 theta`
	`= 5 sin theta − 10 sin^3 theta + 5 sin^5 theta − 10 sin^3 theta + 11 sin^5 theta`
	`= 16 sin^5 theta − 20 sin^3 theta + 5 sin theta`

(iii) `text(If)\ x = sin (pi/10)`

`16 sin^5 (pi/10) − 20 sin^3 (pi/10) + 5 sin (pi/10)`	`=sin (5 xx pi/10)`
`16x^5 − 20x^3 + 5x`	`= sin\ pi/2`
`16x^5 − 20x^3 + 5x`	`=1`

`:. sin\ pi/10\ \ text(is one solution to)\ \ 16x^5 − 20x^3 + 5x − 1=0`

(iv) `16x^5 − 20x^3 + 5x − 1`

`=(x-1)(16x^4 + 16x^3 − 4x^2 − 4x + 1)`

`:.p(x) = 16x^4 + 16x^3 − 4x^2 − 4x + 1`

(v) `(4x^2 + ax − 1)^2`

`= 16x^4 + 4ax^3 − 4x^2 + 4ax^3 + a^2x^2 − ax − 4x^2 − ax + 1`

`= 16x^4 + 8ax^3 − 8x^2 + a^2x^2 − 2ax +1`

`text(By equating coefficients of)\ \ x^3`

`:.a = 2`

(vi) `4 x^2 + 2 x − 1 = 0`

♦♦ Mean mark part (vi) 27%.

`x`	`=(-2 ± sqrt(4 + 16))/8`
	`=(-1 ± sqrt5)/4`

`:. sin\ pi/10 = (-1 + sqrt5)/4\ \ \ \ (sin\ pi/10\ > 0)`

Conics, EXT2 2010 HSC 5a

The diagram shows two circles, `C_1` and `C_2`, centred at the origin with radii `a` and `b`, where `a > b`.

The point `A` lies on `C_1` and has coordinates `(a cos theta, a sin theta)`.

The point `B` is the intersection of `OA` and `C_2`.

The point `P` is the intersection of the horizontal line through `B` and the vertical line through `A`.

Write down the coordinates of `B`. (1 mark)
Show that `P` lies on the ellipse
`(x^2)/(a^2) + (y^2)/(b^2) = 1`. (1 mark)
Find the equation of the tangent to the ellipse
`(x^2)/(a^2) + (y^2)/(b^2) = 1` at `P`. (2 marks)
Assume that `A` is not on the `y`-axis.
Show that the tangent to the circle `C_1` at `A`, and the tangent to the ellipse
`(x^2)/(a^2) + (y^2)/(b^2) = 1` at `P`, intersect at a point on the `x`-axis. (2 marks)

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`B(b cos theta, b sin theta)`
`text{Proof (See Worked Solutions)}`
`b cos theta x + a sin theta y = ab, or`
`(x cos theta)/a + (y sin theta)/b = 1`
`text{Proof (See Worked Solutions)}`

Show Worked Solution

(i) `B(b cos theta, b sin theta)`

(ii) `text(Substitue)\ \ P(a cos theta, b sin theta)\ \ text(into)\ \ (x^2)/(a^2) + (y^2)/(b^2) = 1`

`text(LHS)`	`= (a^2 cos^2 theta)/(a^2) + (b^2 sin^2 theta)/(b^2)`
	`= cos^2 theta + sin^2 theta`
	`= 1`
	`=\ text(RHS)`

`:.P\ text(lies on the ellipse.)`

(iii) `text(Solution 1)`

`(x^2)/(a^2) + (y^2)/(b^2)`	`= 1`
`(2x)/(a^2) + (2y)/(b^2) xx (dy)/(dx)`	`= 0`
`(dy)/(dx)`	`= -(b^2x)/(a^2y)`

`text(At)\ \ P(a cos theta, b sin theta)`

`(dy)/(dx)`	`= (b^2a cos theta)/(a^2b sin theta)`
	`= -(b cos theta)/(a sin theta)`

`:.\ text(Equation of tangent at)\ P`

`y − b sin theta=`	` -(b cos theta)/(a sin theta)(x − a cos theta)`
`a sin theta y − ab sin^2 theta=`	` −b cos theta x + ab cos^2 theta`
`b cos theta x + a sin theta y=`	` ab(sin^2 theta + cos^2 theta)`
`:.b cos theta x + a sin theta y=`	`ab,\ \ \ text(or)`
`(x cos theta)/a + (y sin theta)/b=`	`1`

`text(Alternative Solution)`

`dy/dx`	`=(dy)/(d theta) xx (d theta)/(dx)`
	`=b cos theta xx 1/(-a sin theta)`
	`= -(bcos theta)/(a sin theta)`

`text{(then use the point-gradient formula as shown above)}`

(iv) `text(Finding the tangent to)\ x^2 + y^2 = 1\ text(at)\ \ A(a cos theta, a sin theta)`

`2x + 2y* (dy)/(dx)`	`=0`
`(dy)/(dx)`	`= (-x)/y`

`:.\ text(Equation of tangent)`

`y – a sin theta`	`=-(a cos theta)/(a sin theta)(x − a cos theta)`
`y sin theta – a sin^2 theta`	`=-x cos theta + a cos^2 theta`
`x cos theta + y sin theta`	`=a(sin^2 theta + cos^2 theta)`
`x cos theta + y sin theta`	`=a`
`:.y sin theta`	`=a- x cos theta`

`text(Substitute into the equation of the tangent at)\ \ P`

`b cos theta x + a sin theta y`	`=ab`
`b cos theta x+a(a- x cos theta)`	`=ab`
`bx cos theta + a^2 – ax cos theta`	`=ab`
`(b – a)x cos theta`	`=a(b – a)`
`x cos theta`	`= a`

`text(When)\ x cos theta = a,\ \ y sin theta = a − a = 0\ \ =>y=0,\ \ theta≠0`

`:.\ text(Intersection occurs on the)\ x text(-axis.)`

Mechanics, EXT2 2010 HSC 4b

A bend in a highway is part of a circle of radius `r`, centre `O`. Around the bend the highway is banked at an angle `α` to the horizontal.

A car is travelling around the bend at a constant speed `v`. Assume that the car is represented by a point `P` of mass `m`. The forces acting on the car are a lateral force `F`, the gravitational force `mg` and a normal reaction `N` to the road, as shown in the diagram.

By resolving forces, show that
`F = mg sin α − (mv^2)/r cos α`. (3 marks)
Find an expression for `v` such that the lateral force `F` is zero. (1 mark)

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`text{Proof (See Worked Solutions)}`
`sqrt(rg tan α)`

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(i)

`text(Resolving forces vertically)`
`N cos α + F sin α=`	`mg\ \ \ \ …\ (1)`
`text(Resolving forces horizontally)`
`N sin α − F cos α=`	`(mv^2)/r\ \ \ \ …\ (2)`
`text{Multiply (1) by sin α and (2) by cos α}`
`N sin α\ cos α + F sin^2 α=`	`mg sin α\ \ \ \ …\ (3)`
`N sin α\ cos α − F cos^2 α=`	`(mv^2)/r cos α\ \ \ \ …\ (4)`
`text{Subtract (3) – (4)}`
`F sin^2 α + F cos^2 α=`	`mg sin α − (mv^2)/r cos α`
`:.F=`	`mg sin α − (mv^2)/r cos α`

(ii) `F = mg sin α − (mv^2)/r cos α`

`mg sin α − (mv^2)/r cos α`	`=0`
`(v^2)/r cos α`	`=g sin α`
`v^2`	`=rg tan α`
`v`	`=sqrt(rg tan α)`

Graphs, EXT2 2010 HSC 4a

A curve is defined implicitly by `sqrtx + sqrty = 1`.
Use implicit differentiation to find `(dy)/(dx)`. (2 marks)
Sketch the curve `sqrtx + sqrty = 1`. (2 marks)
Sketch the curve `sqrt(|\ x\ |) + sqrt(|\ y\ |) = 1` (1 mark)

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`- sqrty/sqrtx`
`text(See Worked Solutions.)`
`text(See Worked Solutions.)`

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(i)	`sqrtx + sqrt y`	`= 1`
	`1/(2sqrtx) + 1/(2sqrty)*(dy)/(dx)`	`= 0`
	`:.(dy)/(dx)`	`= – sqrty/sqrtx`

(ii)

♦♦ Mean mark part (iii) 46%.

(iii)

Conics, EXT2 2010 HSC 3d

The diagram shows the rectangular hyperbola `xy = c^2`, with `c > 0`.

The points `A(c, c)`, `R(ct, c/t)` and `Q(-ct, -c/t)` are points on the hyperbola,
with `t ≠ ±1`.

The line `l_1` is the line through `R` perpendicular to `QA`.
Show that the equation of `l_1` is
1. `y = -tx + c(t^2 + 1/t)`. (2 marks)
The line `l_2` is the line through `Q` perpendicular to `RA`.
Write down the equation of `l_2`. (1 mark)
Let `P` be the point of intersection of the lines `l_1` and `l_2`.
Show that `P` is the point `(c/(t^2), ct^2)`. (2 marks)
Give a geometric description of the locus of `P`. (1 mark)

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`text{Proof (Show Worked Solutions)}`
`y = tx + c(t^2 − 1/t)`
`text{Proof (Show Worked Solutions)}`
`text{Proof (Show Worked Solutions)}`

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(i)

`m_(QA)`	`= (c + c/t)/(c + ct)`
	`= (1 + 1/t)/(1 + t)`
	`= (t + 1)/(t(1 + t))`
	`= 1/t`

`:.\ text(Gradient of perpendicular to)\ QA\ \ (l_1) = -t`

`:.text(Equation of)\ \ l_1`

`y − c/t`	`= -t(x − ct)`
`y`	`= -tx + ct^2 + c/t`
	`= -tx + c(t^2 + 1/t)\ \ \ …\ (1)`

(ii)	`m_(RA)`	`= (c − c/t)/(c − ct)`
		`= (1 − 1/t)/(1 − t)`
		`= (t − 1)/(t(1 − t))`
		`= -1/t`

`:.m\ text(of)\ l_2 = t`

`:.\ text(Equation of)\ l_2`

`y + c/t`	`= t(x + ct)`
`y`	`= tx + ct^2 – c/t`
	`= tx + c(t^2 – 1/t)\ \ \ …\ (2)`

(iii) `text{Solving (1) and (2) simultaneously}`

`-tx + c(t^2 + 1/t)`	`= tx + c(t^2 − 1/t)`
`ct^2+c/t-ct^2+c/t`	`=2tx`
`(2c)/t`	`= 2tx`
`x`	`= c/(t^2)`

`text{Substitute into (2)}`

`y`	`= (ct)/(t^2) + ct^2 − c/t`
	` = ct^2`

`:.P\ text(has coordinates)\ (c/(t^2), ct^2)`

(iv) `text(Using the coordinates of)\ P`

♦♦ Mean mark part (iv) 1%.
MARKER’S COMMENT: Most students found the correct locus, although very few indicated that it was only the branch in the 1st quadrant.

`=>t^2 = c/x,\ \ t^2 = y/c`

`y/c`	`= c/x`
`:.xy`	`=c^2`

`text(S)text(ince the parameter of)\ P\ text(is)\ t^2`

`=>\ text(the locus is in the first quadrant.)`

`:.\ text(The locus of)\ P\ text(is the branch of the rectangular hyperbola)`

`xy = c^2\ text{in the first quadrant (excluding}\ A, t≠±1 text{)}`

Complex Numbers, EXT2 N2 2010 HSC 2c

Sketch the region in the complex plane where the inequalities `1 ≤ |\ z\ | ≤ 2` and `0 ≤ z + bar z ≤ 3` hold simultaneously. (2 marks)

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`text(See Worked Solutions.)`

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`text(Consider)\ \ \ \ `	`1≤\|\ z\ \|≤2`
	`1≤x^2+y^2≤4`

`z + bar z = (x+iy)+(x-iy)=2x`

`text(Consider)\ \ \ \ `	`0≤z + bar z≤3`
	`0≤2x≤3`
	`0≤x≤3/2`

Complex Numbers, EXT2 N1 2010 HSC 2b

Express `-sqrt3 − i` in modulus–argument form. (2 marks)

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Show that `(-sqrt3 − i)^6` is a real number. (2 marks)

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`2text(cis)(-(5pi)/6)`
`-64`

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i.	`\|-sqrt3 − i\ \|`	`=sqrt((-sqrt3)^2+sqrt((-1)^2))`
		`=2`

`text(From the graph)`

`text{arg}(-sqrt3-i)=- (5pi)/6\ \ \ \ text{(for}\ –pi<theta<pi text{)}`

`:.-sqrt3 − i= 2text(cis)(-(5pi)/6)`

`text{Alternative Solution (to find the argument)}`

`-sqrt3-i`	`= 2(- sqrt3/2 − 1/2 i)`
	`=2text(cis)(-(5pi)/6)`

ii.	`(-sqrt3 − i)^6`	`= [2text(cis)(-(5pi)/6)]^6`
		`=2^6[cos((-5pi)/6 xx6) +i sin((-5 pi)/6 xx6)]\ \ \ \ text{(De Moivre)}`
		`= 2^6[cos(-5pi) + i sin(-5pi)]`
		`= 64(-1 + 0i)`
		`= -64`

Calculus, EXT2 C1 2010 HSC 1d

Using the substitution `t = tan\ x/2`, or otherwise, evaluate `int_0^(pi/2) (dx)/(1 + sin\ x)`. (4 marks)

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`1`

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`t = tan\ x/2, \ dx = (2\ dt)/(1 + t^2), \ sin\ x = (2t)/(1 + t^2)`

`text(When)\ \ x=pi/2,\ \ t=tan\ pi/4=1`

`text(When)\ \ x=0,\ \ t=tan0=0`

`:.int_0^(pi/2) (dx)/(1 + sin\ x)`	`=int_0^1 ((2\ dt)/(1 + t^2))/(1 + (2t)/(1 + t^2))`
	`=int_0^1 2/(1 + t^2 + 2t)\ dt`
	`=int_0^1 (2)/((1 + t)^2)\ dt`
	`=[(-2)/(1 + t)]_0^1`
	`=(-2)/2 − (-2)/1`
	`=1`

Calculus, EXT2 C1 2010 HSC 1b

Evaluate `int_0^(pi/4) tan\ x\ dx`. (3 marks)

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`1/2 ln 2 \ \ text(or)\ \ ln\ sqrt2`

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`int_0^(pi/4) tan\ x\ dx`	`=int_0^(pi/4) (sin\ x)/(cos\ x)\ dx`
	`=[-ln\ cos\ x]_0^(pi/4)`
	`=[-ln\ cos\ pi/4 – (-ln cos 0)]`
	`=-ln\ 1/sqrt2 + ln\ 1`
	`=ln sqrt2`
	`=1/2 ln 2`

Calculus, EXT2 C1 2010 HSC 1a

Find `int x/(sqrt(1 + 3x^2))\ dx`. (2 marks)

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`1/3 sqrt(1 + 3x^2) + c`

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`text(Let)\ u = 1 + 3x^2, \ du = 6x\ dx`

`:.int x/(sqrt(1 + 3x^2))\ dx`	`=1/6 int u^(-1/2)\ du`
	`=1/6 xx 2 xx sqrtu + c`
	`=1/3 sqrt(1 + 3x^2) + c`

Calculus, EXT2 C1 2011 HSC 7b

Let `I = int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx.`

Use the substitution `u = 4-x` to show that

`I = int_1^3 (sin^2(pi/8 u))/(u(4-u))\ du.` (2 marks)

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Hence, find the value of `I`. (3 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`1/4 log_e 3`

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i. `text(Let)\ \ u = 4-x,\ \ du = -dx,\ \ x = 4-u`

`text(When)\ \ x = 1,\ \ u = 3`

`text(When)\ \ x = 3,\ \ u = 1`

`I`	`=int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx`
	`=int_3^1 (cos^2 (pi/8(4-u)))/((4-u)u) (-du)`
	`=-int_3^1 (cos^2(pi/2-pi/8 u))/((4-u)u)\ du`
	`=int_1^3 (sin^2(pi/8 u))/(u (4-u))\ du`

ii. `text{S}text{ince}\ \ int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx = int_1^3 (sin^2(pi/8 x))/(x(4-x))\ dx`

`text(We can add the integrals such that)`

`2I`	`=int_1^3 (cos^2(pi/8 x))/(x(4-x)) dx + int_1^3 (sin^2(pi/8 x))/(x(4-x)) dx`
	`=int_1^3 1/(x(4-x)) dx`

`text(Using partial fractions:)`

♦ Mean mark 36%.

`1/(x(4-x))`	`=A/x+B/(4-x)`
`1`	`=A(4-x)+Bx`

`text(When)\ \ x=0,\ \ A=1/4`

`text(When)\ \ x=0,\ \ B=1/4`

`2I`	`=1/4 int_1^3 (1/x + 1/(4-x))\ dx`
	`=1/4 [log_e x-log_e (4-x)]_1^3`
	`=1/4 [log_e 3-log_e 1-(log_e 1-log_e 3)]`
	`=1/2 log_e 3`
`:.I`	`=1/4 log_e 3`

Mechanics, EXT2 M1 2011 HSC 6a

Jac jumps out of an aeroplane and falls vertically. His velocity at time `t` after his parachute is opened is given by `v(t)`, where `v(0) = v_0` and `v(t)` is positive in the downwards direction. The magnitude of the resistive force provided by the parachute is `kv^2`, where `k` is a positive constant. Let `m` be Jac’s mass and `g` the acceleration due to gravity. Jac’s terminal velocity with the parachute open is `v_T.`

Jac’s equation of motion with the parachute open is

`m (dv)/(dt) = mg - kv^2.` (Do NOT prove this.)

Explain why Jac’s terminal velocity `v_T` is given by

`sqrt ((mg)/k).` (1 mark)

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By integrating the equation of motion, show that `t` and `v` are related by the equation

`t = (v_T)/(2g) ln[((v_T + v)(v_T - v_0))/((v_T - v)(v_T + v_0))].` (3 marks)

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Jac’s friend Gil also jumps out of the aeroplane and falls vertically. Jac and Gil have the same mass and identical parachutes.

Jac opens his parachute when his speed is `1/3 v_T.` Gil opens her parachute when her speed is `3v_T.` Jac’s speed increases and Gil’s speed decreases, both towards `v_T.`

Show that in the time taken for Jac's speed to double, Gil's speed has halved. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

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`text(See Worked Solutions)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

i. `m (dv)/(dt) = mg – kv^2`

`text(As)\ \ v ->text(terminal velocity,)\ \ (dv)/(dt) -> 0`

`mg – kv_T^2`	`= 0`
`v_T^2`	`= (mg)/k`
`v_T`	`= sqrt ((mg)/k)`

ii. `m (dv)/(dt) = mg – kv^2`

`int_0^t dt`	`=int_(v_0)^v m/(mg – kv^2)\ dv`
`t`	`= m/k int_(v_0)^v (dv)/((mg)/k – v^2)\ \ \ \ text{(using}\ \ v_T^2= (mg)/k text{)}`
	`= (v_T^2)/g int_(v_0)^v (dv)/(v_T^2 – v^2)`

♦ Mean mark part (ii) 39%.

`text(If)\ \ v_0 < v_T,\ \ text(then)\ \ v < v_T\ \ text(at all times.)`

`:. t`	`=(v_T^2)/g int_(v_0)^v (dv)/(v_T^2 – v^2)`
	`=(v_T^2)/g int_(v_0)^v (dv)/((v_T – v)(v_T + v))`
	`=(v_T^2)/(2 g v_T) int_(v_0)^v (1/((v_T – v)) + 1/((v_T + v))) dv`
	`=v_T/(2g)[-ln (v_T – v) + ln (v_T + v)]_(v_0)^v`
	`=v_T/(2g)[ln(v_T + v)-ln(v_T – v)-ln(v_T + v_0)+ln(v_T – v_0)]`
	`=v_T/(2g) ln[((v_T + v)(v_T – v_0))/((v_T – v)(v_T + v_0))]`

`text(If)\ \ v_0 > v_T,\ \ text(then)\ \ v > v_T\ \ text(at all times.)`

`text(Replace)\ \ v_T – v\ \ text(by) – (v – v_T)\ \ text(in the preceeding)`

`text(calculation, leading to the same result.)`

iii. `text{From (ii) we have}:\ \ t = v_T/(2g) ln [((v_T + v)(v_T – v_0))/((v_T – v)(v_T + v_0))]`

`text(For Jac,)\ \ v_0 = V_T/3\ \ text(and we have to find the time)`

`text(taken for his speed to be)\ \ v = (2v_T)/3.`

`t`	`=v_T/(2g) ln[((v_T + (2v_T)/3)(v_T – v_T/3))/((v_T – (2v_T)/3)(v_T + v_T/3))]`
	`=v_T/(2g) ln [(((5v_T)/3)((2v_T)/3))/((v_T/3)((4v_T)/3))]`
	`=v_T/(2g) ln[(10/9)/(4/9)]`
	`=v_T/(2g) ln (5/2)`

`text(For Gil),\ \ v_0 = 3v_T\ \ text(and we have to find)`

♦♦♦ Mean mark part (iii) 16%.

`text(the time taken for her speed to be)\ \ v = (3v_T)/2.`

`t`	`=v_T/(2g) ln [((v_T + (3v_T)/2)(v_T – 3v_T))/((v_T – (3v_T)/2)(v_T + 3v_T))]`
	`=v_T/(2g) ln [(((5v_T)/2)(-2v_T))/((-v_T/2)(4v_T))]`
	`=v_T/(2g) ln [(-5)/-2]`
	`=v_T/(2g) ln (5/2)`

`:.\ text(The time taken for Jac’s speed to double is)`

`text(the same as it takes for Gil’s speed to halve.)`

Mechanics, EXT2 2011 HSC 5a

A small bead of mass `m` is attached to one end of a light string of length `R`. The other end of the string is fixed at height `2h` above the centre of a sphere of radius `R`, as shown in the diagram. The bead moves in a circle of radius `r` on the surface of the sphere and has constant angular velocity `omega > 0`. The string makes an angle of `theta` with the vertical.

Three forces act on the bead: the tension force `F` of the string, the normal reaction force `N` to the surface of the sphere, and the gravitational force `mg`.

By resolving the forces horizontally and vertically on a diagram, show that
1. `F sin theta - N sin theta = m omega^2 r`
and
1. `F cos theta + N cos theta = mg.` (2 marks)
Show that
1. `N = 1/2 mg sec theta - 1/2 m omega^2 r\ text(cosec)\ theta.` (2 marks)
Show that the bead remains in contact with the sphere if
1. `omega <= sqrt (g/h).` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i)

`text(Resolving forces horizontally)`

`text(Net force)`	`= m r omega^2\ \ \ text{(towards the centre of the circle)}`
`F sin theta – N sin theta`	`= mr omega^2\ \ \ \ text{… (1)}`

♦ Mean mark part (i) 50%.

`text(Resolving forces vertically)`

`text(Net force)`	`= mg\ \ \ \ text{(gravitational force)}`
`F cos theta + N cos theta`	`= mg\ \ \ \ text{… (2)}`

(ii) `text{Divide (1) by}\ \ sin theta`

`F – N = mr omega^2\ text(cosec)\ theta\ \ \ \ text{… (3)}`

`text{Divide (2) by}\ \ cos theta`

`F+N = mg sec theta\ \ \ \ text{… (4)}`

`text{Subtract (4) – (3)}`

`2N`	`= mg sec theta – mr omega^2\ text(cosec)\ theta`
`:.N`	`= 1/2 mg sec theta – 1/2 mr omega^2\ text(cosec)\ theta`

(iii) `text(When in contact with the sphere,)\ \ N >= 0.`

`1/2 mg sec theta – 1/2 mr omega^2`	`>= 0`
`1/2 mg sec theta`	`>= 1/2 mr omega^2\ text(cosec)\ theta`
`g sec theta`	`>= r omega^2\ text(cosec)\ theta`
`omega^2`	`<= (g sec theta)/(r\ text(cosec)\ theta)`
	`<= g/r tan theta,\ \ \ \ (text{since}\ \ tan theta = r/h)`
	`<= g/r xx r/h`
	`<=g/h`
`:. omega`	`<= sqrt (g/h)`

Harder Ext1 Topics, EXT2 2011 HSC 4b

In the diagram, `ABCD` is a cyclic quadrilateral. The point `E` lies on the circle through the points `A, B, C` and `D` such that `AE\ text(||)\ BC`. The line `ED` meets the line `BA` at the point `F`. The point `G` lies on the line `CD` such that `FG\ text(||)\ BC.`

Copy or trace the diagram into your writing booklet.

Prove that `FADG` is a cyclic quadrilateral. (2 marks)
Explain why `/_ GFD =/_ AED.` (1 mark)
Prove that `GA` is a tangent to the circle through the points `A, B, C` and `D.` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`/_ GFD = /_ AED\ \ text{(alternate angles,}\ \ FG\ text(||)\ AE text{)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i)

`text(Let)\ /_ BCD`	`= alpha`
`/_ FAD`	`= alpha\ \ text{(exterior angle of a cyclic quadrilateral}\ ABCD text{)}`
`/_ FGC`	`= pi – alpha\ \ text{(cointerior angles,}\ \ FG\ text(\|\|)\ BC text{)}`

`:.\ \ /_ FAD + /_ FGD = pi`

`:.\ FADG\ \ text{is a cyclic quadrilateral (opposite angles are supplementary)}`

(ii) `/_ GFD = /_ AED\ \ text{(alternate angles,}\ \ FG\ text(||)\ AE text{)}`

(iii) `text(Join)\ GA`

`/_ GAD = /_ GFD\ \ text{(angles in the same segment on arc}\ GDtext{)}`

`text(S)text(ince)\ /_ GFD`	`= /_ AED\ \ \ text{(part (ii))}`
`/_ GAD`	`= /_ AED`

`:.GA\ \ text(is a tangent to the circle through)\ \ A, B, C and D.`

`text{(angle in the alternate segment equals the angle}`

`text(between)\ GA\ text(and chord)\ AD text{).}`

Conics, EXT2 2011 HSC 3d

The equation `x^2/16 - y^2/9 = 1` represents a hyperbola.

Find the eccentricity `e.` (1 mark)
Find the coordinates of the foci. (1 mark)
State the equations of the asymptotes. (1 mark)
Sketch the hyperbola. (1 mark)
For the general hyperbola
`x^2/a^2 - y^2/b^2 = 1`,
describe the effect on the hyperbola as `e -> oo.` (1 mark)

Show Answers Only

`e = 5/4`
`text(Foci) (+-5, 0)`
`y = +- 3/4 x`
`text(The hyperbola approaches the)\ \ y text(-axis from both sides.)`

Show Worked Solution

(i) `x^2/16 – y^2/9 = 1,\ \ \ =>a^2 = 16,\ \ \ b^2 = 9`

`b^2`	`= a^2 (e^2 – 1)`
`9`	`= 16 (e^2 – 1)`
`e^2`	`= 1 + 9/16 `
	`= 25/16`
`:. e`	`= 5/4`

(ii) `S (ae, 0),\ \ S prime (-ae, 0)`

`S(5, 0),\ \ S prime (–5, 0)`

(iii) `y = +- b/a x`

`y = +- 3/4 x`

(iv)

♦♦♦ Mean mark part (v) 24%.

(v) `text(As)\ \ e -> oo,\ \ b -> oo\ \ text(and the asymptotes approach the)`

`y text{-axis from both sides (i.e. the gradients of the asymptotes →∞)}`

` text(Thus the hyperbola approaches the)\ \ y text(-axis from both sides.)`

Functions, EXT1′ F1 2011 HSC 3a

Draw a sketch of the graph

`quad y = sin\ pi/2 x` for `0 < x < 4.` (1 mark)

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Find `lim_(x -> 0) x/(sin\ pi/2 x).` (1 mark)

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Draw a sketch of the graph

`quad y = x/(sin\ pi/2 x)` for `0 < x < 4.` (2 marks)
(Do NOT calculate the coordinates of any turning points.)

--- 10 WORK AREA LINES (style=lined) ---

Show Answers Only

ii. `2/pi`

iii.

Show Worked Solution

ii. `lim_(x->0) x/(sin\ pi/2 x)`	`= 2/pi lim_(x->0) (pi/2 x)/(sin\ pi/2 x)`
	`= 2/pi xx 1`
	`= 2/pi`

iii.

Complex Numbers, EXT2 N2 2011 HSC 2d

Use the binomial theorem to expand `(cos theta + i sin theta)^3.` (1 mark)

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Use de Moivre’s theorem and your result from part (i) to prove that

`cos^3 theta = 1/4 cos 3 theta + 3/4 cos theta.` (3 marks)

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Hence, or otherwise, find the smallest positive solution of

`4 cos^3 theta - 3 cos theta = 1.` (2 marks)

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`cos^3 theta + 3 i cos^2 theta sin theta – 3 cos theta sin^2 theta – i sin^3 theta`
`text(Proof)\ \ text{(See Worked Solutions)}`
`theta = (2 pi)/3`

Show Worked Solution

i. `(cos theta + i sin theta)^3`

`=sum_(k=0)^3 \ ^3C_k (cos theta)^(3-k) (i sin theta)^k`

`= cos^3 theta + 3 cos^2 theta (i sin theta)+ 3 cos theta (i sin theta)^2 + (i sin theta)^3`

`= cos^3 theta + 3 i cos^2 theta sin theta- 3 cos theta sin^2 theta – i sin^3 theta`

ii. `text(Using De Moivre’s Theorem)`

`(cos theta + i sin theta)^3 = cos 3 theta + i sin 3 theta`

`text(Equate real parts)`

`cos 3 theta`	`= cos^3 theta – 3 cos theta sin^2 theta`
`cos 3 theta`	`= cos^3 theta – 3 cos theta (1 – cos^2 theta)`
`cos 3 theta`	`= 4 cos^3 theta – 3 cos theta`
`4 cos^3 theta`	`=cos 3 theta+3cos theta`
`:.cos^3 theta`	`= 1/4 cos 3 theta + 3/4 cos theta\ \ \ text(… as required)`

iii. `text(If)\ \ \ 4 cos^3 theta – 3 cos theta = 1`

`=>cos 3 theta = 1\ \ \ \ text{(from part (ii))}`

`3 theta`	`= 2 k pi`
`:. theta`	`= (2 k pi)/3`

`:.\ text(Smallest positive solution occurs when)`

`theta = (2 pi)/3\ \ \ \ text{(i.e. when}\ k = 1 text{)}`

Complex Numbers, EXT2 N2 2011 HSC 2c

Find, in modulus-argument form, all solutions of `z^3 = 8.` (2 marks)

--- 7 WORK AREA LINES (style=lined) ---

Show Answers Only

`2text(cis)\ 0,\ 2text(cis)\ (2pi)/3,\ 2text(cis)((-2pi)/3)`

Show Worked Solution

`z^3 = 8`

`z^3 = 8 [cos (2 k pi) + i sin (2 k pi)],\ \ k=0, +-1, +-2`

`z = 2[cos ((2 k pi)/3) + i sin ((2 k pi)/3)]\ \ \ text{(De Moivre)}`

`text(When)\ \ k = 0,`

`z`	`= 2 (cos 0 + i sin 0)`
	`= 2text(cis)\ 0`

`text(When)\ \ k = 1,`

`z`	`= 2 [cos ((2 pi)/3) + i sin((2 pi)/3)]`
	`=2text(cis)\ (2pi)/3`

`text(When)\ \ k = -1,`

`z`	`= 2 [cos ((-2 pi)/3) + i sin ((-2 pi)/3)]`
	`=2 text(cis)((-2 pi)/3)`

Complex Numbers, EXT2 N2 2011 HSC 2b

On the Argand diagram, the complex numbers `0, 1 + i sqrt 3 , sqrt 3 + i` and `z` form a rhombus.

Find `z` in the form `a + ib`, where `a` and `b` are real numbers. (1 mark)

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An interior angle, `theta`, of the rhombus is marked on the diagram.

Find the value of `theta.` (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

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`(1 + sqrt 3) + i(1 + sqrt 3)`
`(5 pi)/6`

Show Worked Solution

i. `z`	`= 1 + i sqrt 3 + sqrt 3 + i`
	`= (1 + sqrt 3) + i (1 + sqrt 3)`

ii. `text(arg)\ z = tan^-1 ((1 + sqrt 3)/(1 + sqrt 3)) = pi/4`

`text(arg)\ (sqrt 3 + i) = tan^-1 (1/sqrt 3) = pi/6`

`text(Difference) = pi/4 – pi/6 = pi/12`

`=>\ text(Opposite angles of a rhombus are equal)`

`=>\ text(The diagonals of a rhombus bisect the angles)`

`:.theta`	`= pi – 2 xx pi/12`
	`= (5 pi)/6\ \ text{(angle sum of triangle)`

Calculus, EXT2 C1 2011 HSC 1e

Evaluate `int_-1^1 1/(5 - 2t + t^2) \ dt.` (3 marks)

Show Answers Only

`pi/8`

Show Worked Solution

`int_-1^1 1/{(5 – 2t + t^2)}dt`	`= int_-1^1 1/{(4 + 1 – 2t + t^2)dt}`
	`= int_-1^1 1/(4 + (t – 1)^2)dt`
	`= 1/2[tan^-1 ((t – 1)/2)]_-1^1`
	`= 1/2 [tan^-1 0 – tan^(-1)(-1)]`
	`= 1/2 [0 – (-pi/4)]`
	`= pi/8`

Calculus, EXT2 C1 2011 HSC 1d

Find `int cos^3 theta\ d theta` (3 marks)

Show Answers Only

`sin theta-(sin^3 theta)/3 + c`

Show Worked Solution

`int cos^3 theta\ d theta`	`= int cos^2 theta cos theta\ d theta`
	`= int (1-sin^2 theta) cos theta\ d theta`
	`= int (cos theta-sin^2 theta cos theta) d theta`
	`= sin theta-(sin^3 theta)/3 + c`

Calculus, EXT2 C1 2011 HSC 1c

Find real numbers `a, b` and `c` such that

`1/(x^2 (x - 1)) = a/x + b/x^2 + c/(x - 1).` (2 marks)

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Hence, find `int 1/(x^2 (x - 1))\ dx` (2 marks)

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

`a = −1,\ \ \ b = −1,\ \ \ c = 1`
`log_e\ ((x – 1)/x) + 1/x + c`

Show Worked Solution

i. `1/(x^2 (x – 1)) = a/x + b/x^2 + c/(x – 1)`

`1 = ax (x – 1) + b (x – 1) + cx^2`

`1 = ax^2 + cx^2 – ax + bx – b`

`1=(a+c)x^2+(b-a)x-b`

`text(Equating coefficients:)`

`=>0 = a + c,\ \ \ b – a = 0,\ \ \ -b = 1`

`:.a = -1,\ \ \ b = -1,\ \ \ c = 1`

ii. `int 1/(x^2 (x – 1)) \ dx`	`= int(-1/x – 1/x^2 + 1/(x – 1))\ dx`
	`= -log_e x + 1/x + log_e (x – 1) + c`
	`= log_e\ ((x – 1)/x) + 1/x + c`

Polynomials, EXT2 2012 HSC 15b

Let `P(z) = z^4 − 2kz^3 + 2k^2z^2 − 2kz + 1`, where `k` is real.

Let `α = x + iy`, where `x` and `y` are real.

Suppose that `α` and `iα` are zeros of `P(z)`, where `bar α ≠ iα`.

Explain why `bar α` and `-i bar α` are zeros of `P(z)`. (1 mark)
Show that `P(z) = z^2(z − k)^2 + (kz − 1)^2`. (1 mark)
Hence show that if `P(z)` has a real zero then
1. `P(z) = (z^2 + 1)(z+ 1)^2` or `P(z) = (z^2 + 1)(z − 1)^2.` (2 marks)
Show that all zeros of `P(z)` have modulus `1`. (2 marks)
Show that `k = x − y`. (1 mark)
Hence show that `-sqrt2 ≤ k ≤ sqrt2`. (2 marks)

Show Answers Only

`text{Proof (See Worked Solutions.)}`
`text{Proof (See Worked Solutions.)}`
`text{Proof (See Worked Solutions.)}`
`text{Proof (See Worked Solutions.)}`
`text{Proof (See Worked Solutions.)}`
`text{Proof (See Worked Solutions.)}`

Show Worked Solution

(i) `P(z) = z^4 − 2kz^3 + 2k^2z^2 − 2kz +1`

`P(α) = P(iα) = 0, \ \ bar α ≠ iα.`

`text(S)text(ince)\ \ P(z)\ text(has real coefficients,)`

`=>\ text(Its zeros occur in conjugate pairs.)`

`text(i.e.)\ \ P(α) = 0\ text(and)\ P(bar α) = 0`

`bar α`	`= x − iy`
`bar(i α)`	`=bar (i(x+iy))`
	`=bar (ix-y)`
	`=-y-ix`
	`=-i(x-iy)`
	`=-i barα`

`:. -i bar α\ text(is a zero of)\ P(z)\ text(as it is the conjugate of)\ iα.`

(ii)	`H(z)`	`= z^2(z − k)^2 + (kz − 1)^2`
		`= z^2 (z^2 − 2kz + k^2) + (k^2z^2 − 2kz + 1)`
		`= z^4 − 2kz^3 + k^2z^2 + k^2z^2 − 2kz +1`
		`= z^4 − 2kz^3 + 2k^2z^2 − 2kz + 1`
		`= P(z)`

(iii) `text(If)\ z\ text(is real then)\ z^2(z − k)^2\ text(and)\ (kz −1)^2\ text(are real and)`

`text(either positive or zero.)`

♦♦♦ Mean mark part (iii) 11%.

`:.text(If)\ \ P(z) = z^2(z − k)^2 + (kz − 1)^2 = 0`

`=>(z − k)^2 = 0\ text(and)\ (kz − 1)^2 = 0`

`text(When)\ \ z = k,\ \ \ k^2 − 1 = 0\ \ text(or)\ k = ±1.`

`text(If)\ k = 1`

`P(z)`	`= z^2(z − 1)^2 + (z − 1)^2`
	`= (z^2 + 1)(z − 1)^2.`

`text(If)\ k = -1`

`P(z)`	`= z^2(z + 1)^2 + (-z − 1)^2`
	`= (z^2 + 1)(z + 1)^2`

(iv) `text(Product of the roots) = e/a=1`

♦♦♦ Mean mark part (iv) 20%.

`:. α bar α iα *(-i barα)`	`=1`
`(α bar α)^2`	`=1`
`(\|α\|)^4`	`=1`
`\|α\|`	`=1`

`:.\ text(All zeros have modulus 1.)`

(v) `text(Sum of zeroes)\ = -b/a=-(-2k)/1 = 2k`

♦♦♦ Mean mark part (v) 20%.

`:. 2k`	`=α + bar α + iα + (-i bar α )`
	`=x + iy + x − iy + (-y + ix) − i(x − iy)`
	`=2x − y + ix − ix − y`
	`=2x − 2y`
`:. k`	`=x-y`

(vi) `text(S)text(ince)\ \ |α| = 1\ \ \ text{(part (iv))}`

♦♦♦ Mean mark part (vi) 2%.

`=>x^2 + y^2`	`=1`
`text(Substitute)\ \ y=x-k\ \ \ text{(part (v))}`
`x^2 + (x − k)^2`	`=1`
`2x^2 − 2kx + k^2 − 1`	`=0`

`text(For a real solution to exist), Δ ≥ 0`

`4k^2 − 8(k^2 − 1) `	`≥ 0`
`-4k^2 + 8`	`≥ 0`
`k^2`	`≥ 2`

`:. -sqrt2 ≤ k ≤ sqrt2`

Integration, EXT2 2012 HSC 14a

Find `int(3x^2 + 8)/(x(x^2 +4))\ dx`. (3 marks)

Show Answers Only

` 2log_e x + 1/2log_e(x^2 + 4) + c`

Show Worked Solution

`(3x^2 + 8)/(x(x^2 +4))`	`= a/x + (bx + c)/(x^2 + 4)`
`3x^2 +8`	`=ax^2 + 4a+ bx^2 + cx`
	`=(a+b)x^2+cx+4a`

`a + b = 3, \ c = 0, \ 4a = 8`

`:.a = 2, b = 1, c = 0`

`int(3x^2 + 8)/(x(x^2 +4))\ dx`	`= int2/x\ dx + int x/(x^2 + 4)\ dx`
	`= 2log_e x + 1/2log_e(x^2 + 4) + c`

Harder Ext1 Topics, EXT2 2012 HSC 13b

The diagram shows `ΔS′SP`. The point `Q` is on `S′S` so that `PQ` bisects `∠S′PS`. The point `R` is on `S′P` produced so that `PQ\ text(||)\ RS`.

Show that `PS = PR`. (1 mark)
Show that `(PS)/(QS) = (PS′)/(QS′)`. (2 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`

Show Worked Solution

(i)	`∠S′PQ`	`= ∠PRS = α`	`text{(corresponding angles,}\ QP\ text(\|\|)\ SR)`
	`∠QPS`	`= ∠PSR = α`	`text{(alternate angles,}\ QP\ text(\|\|)\ SR)`
	`:. ∠PSR`	`= ∠PRS = α`

`:. ΔPSR\ text{is isosceles (two angle equal)}`

`:. PS = PR\ \ \ text{(opposite equal angles in}\ Delta PSR text{)}`

(ii) `text(Draw a line through)\ S′\ text(parallel to)\ QP`

`:.(PS′)/(PR)`	`= (QS′)/(QS)\ \ \ text{(parallel lines preserve ratios)}`
`(PS′)/(QS′)`	`= (PR)/(QS)`
`text(S)text(ince)\ \ PS = PR\ \ \ text{(from part (i))}`
`(PS′)/(QS′)`	`= (PS)/(QS)\ \ \ text(… as required)`

Mechanics, EXT2 M1 2012 HSC 13a

An object on the surface of a liquid is released at time `t = 0` and immediately sinks. Let `x` be its displacement in metres in a downward direction from the surface at time `t` seconds.

The equation of motion is given by

`(dv)/(dt) = 10 − (v^2)/40`,

where `v` is the velocity of the object.

Show that `v = (20(e^t − 1))/(e^t + 1)`. (4 marks)

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Use `(dv)/(dt) = v (dv)/(dx)` to show that

`x = 20\ log_e(400/(400 − v^2))` (2 marks)

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How far does the object sink in the first 4 seconds? (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(See Worked Solutions.)`
`text(See Worked Solutions.)`
`40log_e((e^4 + 1)/(2e^2))\ text(m)`

Show Worked Solution

i.	`(dv)/(dt)`	`= 10 − (v^2)/40`
	`(dv)/(dt)`	`= (400 − v^2)/40`
	`dt`	`= 40/(400 −v^2)\ dv`
	`int dt`	`=int 40/(400 −v^2)\ dv`
	`t`	`= int (1/(20 + v) + 1/(20 − v))\ dv`
		`= log_e(20 + v) − log_e(20 − v) + c`

`text(When)\ \ t = 0, v = 0\ \ => \ c = 0`

`t=`	` log_e((20 + v)/(20 − v))`
`e^t=`	` (20 + v)/(20 − v)`
`20e^t-ve^t=`	` 20 + v`
`v+ ve^t=`	` 20e^t − 20`
`v(1+e^t)=`	`20(e^t − 1)`
`v=`	` (20(e^t − 1))/(e^t + 1)\ \ \ \ text(… as required)`

ii.	`v (dv)/(dx)`	`= 10 − (v^2)/40`
	`(40v\ dv)/(400 − v^2)`	`= dx`

`int dx`	`= int (40v)/(400 − v^2)\ dv`
`x`	`= -20log_e(400 − v^2) + c`

`text(When)\ \ x = 0, v = 0\ \ \ => c = 20log_e 400`

`:.x`	`= 20log_e400 − 20log_e(400 − v^2)`
	`= 20log_e((400)/(400 − v^2))\ \ \ \ text(… as required)`

iii. `text(When)\ \ t = 4,\ v = (20(e^4 − 1))/(e^4 + 1)`

`x`	`= 20log_e[400/(400 −((20(e^4 − 1))/(e^4 + 1))^2)]`
	`= 20log_e[((e^4 + 1)^2)/((e^4 + 1)^2 − (e^4 − 1)^2)]`
	`= 20log_e(((e^4 + 1)^2)/((e^4+1 + e^4 − 1)(e^4 + 1 − e^4 + 1)))`
	`= 20log_e(((e^4 + 1)^2)/(4e^4))`
	`= 40log_e\ (e^4 + 1)/(2e^2)\ \ text(metres)`

Calculus, EXT2 C1 2012 HSC 12c

For every integer `n ≥ 0` let `I_n = int_1^(e^2)(log_e x)^n\ dx`.

Show that for `n ≥ 1,`

`I_n = e^2 2^n − nI_(n − 1)`. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`I_n = int_1^(e^2)(log_e x)^n\ dx`

`u`	`= (log_e x)^n,`	`v′`	`= 1`
`u′`	`= n (log_e x)^(n − 1) xx 1/x`	`v`	`= x`

`I_n`

`=int_1^(e^2) 1*(ln x)^n\ dx`

`= [x (log_e x)^n]_1^(e^2) − int_1^(e^2)n/x(log_e x)^(n − 1) xx x\ dx`

`= [e^2(log_e e^2)^n − 1*ln\ 1] − n int_1^(e^2)(log_e x)^(n − 1)\ dx`

`= [e^2(2 log_ee)^n-0 ]− nI_(n − 1)`

`= e^2 2^n − nI_(n − 1)`

Conics, EXT2 2012 HSC 12b

The diagram shows the ellipse `(x^2)/(a^2) + (y^2)/(b^2) = 1` with `a > b`. The ellipse has focus `S` and eccentricity `e`. The tangent to the ellipse at `P(x_0, y_0)` meets the `x`-axis at `T`. The normal at `P` meets the `x`-axis at `N`.

Show that the tangent to the ellipse at `P` is given by the equation
1. `y − y_0 = -(b^2x_0)/(a^2y_0)(x − x_0)`. (2 marks)
Show that the `x`-coordinate of `N` is `x_0e^2`. (2 marks)
Show that `ON xx OT = OS^2` (2 marks)

Show Answers Only

(i) `text(See Worked Solutions.)`

(ii) `text(See Worked Solutions.)`

(iii) `text(See Worked Solutions.)`

Show Worked Solution

(i) `(x^2)/(a^2) + (y^2)/(b^2) = 1`

`(2x)/(a^2) + (2y)/(b^2) *(dy)/(dx)`	`=0`
`(dy)/(dx)`	`=-(2x)/(a^2) xx (b^2)/(2y)`
	`= -(b^2x)/(a^2y)`

`text(At)\ P(x_0, y_0),\ \ m_(tan)= -(b^2x_0)/(a^2y_0)`

`:.text(Equation of tangent at)\ P\ text(is)`

`y − y_0 = -(b^2x_0)/(a^2y_0)(x − x_0)`

(ii) `text(At)\ P(x_0, y_0),\ \ m_(norm) = (a^2y_0)/(b^2x_0)`

`:.text(Equation of normal at)\ P\ text(is)`

`y − y_0 = (a^2y_0)/(b^2x_0)(x − x_0)`

`N\ \ text(occurs when)\ \ y=0`

`0-y_0=`	` (a^2y_0)/(b^2x_0)(x − x_0)`
`-b^2x_0y_0=`	` a^2y_0x − a^2x_0y_0`
`a^2y_0x=`	` a^2x_0y_0 − b^2x_0y_0`
`a^2x=`	` x_0(a^2 − b^2)`
`x=`	` (x_0(a^2 − b^2))/(a^2)\ \ \ \ \ text{(using}\ \ a^2e^2=a^2-b^2 text{)}`
`=`	`x_0e^2`

(iii) `ON = x_0e^2`

`OS=ae\ \ \ \ text{(given}\ \ S(ae,0) text{)}`

`T\ text(is the)\ x text(-axis intercept of the tangent)\ PT`

`0-y_0`	`= -(b^2x_0)/(a^2y_0)(x − x_0)`
`a^2y_0^2`	`=b^2x_0x-b^2x_0^2`
`b^2x_0x`	`=b^2x_0^2+a^2y_0^2`
`x`	`=(b^2x_0^2+a^2y_0^2)/(b^2x_0)`

`text(S)text(ince)\ P(x_0,y_0)\ text(lies on the ellipse,)`

`=> (x_0^2)/(a^2) + (y_0^2)/(b^2)`	` = 1`
`b^2 x_0^2+a^2y_0^2`	`=a^2b^2`

`:.x=OT`

`=(a^2b^2)/(b^2x_0)=a^2/x_0`

`:.ON xx OT`	`= x_0e^2 xx a^2/x_0`
	`=a^2e^2`
	`=OS^2`

Calculus, EXT2 C1 2012 HSC 12a

Using the substitution `t = tan\ theta/2`, or otherwise, find `int(d theta)/(1 − cos\ theta)`. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`-cot\ theta/2 + c`

Show Worked Solution

`t = tan\ theta/2`

`dt=1/2 sec^2 (theta/2)\ d theta,\ \ \ d theta=(2\ dt)/sec^2 (theta/2)=2/(1+t^2)\ dt`

`cos\ theta = (1 − t^2)/(1 + t^2)`

`int(d theta)/(1 − cos\ theta)`	`= int1/(1 − ((1 − t^2)/(1 + t^2))) xx 2/(1 + t^2)\ dt`
	`= int 2/(1 + t^2 − (1 − t^2))`
	`= int(dt)/(t^2)`
	`= -1/t + c`
	`= -1/(tan\ theta/2) + c`
	`= -cot\ theta/2 + c`

Calculus, EXT2 C1 2012 HSC 11e

Evaluate `int_0^1 (e^(2x))/(e^(2x) + 1)\ dx`. (3 marks)

Show Answers Only

`1/2log_e((e^2 + 1)/2)`

Show Worked Solution

`int_0^1 (e^(2x))/(e^(2x) + 1)\ dx`	`= 1/2[log_e(e^(2x) + 1)]_0^1`
	`= 1/2[log_e(e^2 + 1) − log_e2]`
	`= 1/2log_e((e^2 + 1)/2)`

Complex Numbers, EXT2 N2 2012 HSC 11b

Shade the region on the Argand diagram where the two inequalities

`|\ z + 2\ | ≥ 2` and `|\ z − i\ | ≤ 1`

both hold. (2 marks)

Show Answers Only

Show Worked Solution

`|\ z + 2\ | ≥ 2 and |\ z − i\ | ≤ 1`

`(x + 2)^2 + y^2`	`≥ 4`
`x^2 + (y − 1)^2`	`≤ 1`

Mechanics, EXT2 M1 2013 HSC 15d

A ball of mass `m` is projected vertically into the air from the ground with initial velocity `u`. After reaching the maximum height `H` it falls back to the ground. While in the air, the ball experiences a resistive force `kv^2`, where `v` is the velocity of the ball and `k` is a constant.

The equation of motion when the ball falls can be written as

`m dot v = mg-kv^2.` (Do NOT prove this.)

Show that the terminal velocity `v_T` of the ball when it falls is
`sqrt ((mg)/k).` (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Show that when the ball goes up, the maximum height `H` is
`H = (v_T^2)/(2g) ln (1 + u^2/(v_T^2)).` (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
When the ball falls from height `H` it hits the ground with velocity `w`.
Show that `1/w^2 = 1/u^2 + 1/(v_T^2).` (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

i. `m dot v = mg-kv^2`

`t = 0,\ \ \ v = 0,\ \ \ x = 0\ \ \ text(Ball falling)`

`text{For terminal velocity}\ \(v_T),\ \ \ dot v = 0`

`v_T^2`	`= (mg)/k`
`:.v_T`	`= sqrt ((mg)/k)`

ii. `text(When the ball rises),\ \ m dot v = -mg-kv^2`

♦ Mean mark 43%.
 MARKER’S COMMENT: More than half of students incorrectly wrote the equation to solve as `m dot v=mg-kv^2!`

`text(Using)\ \ dot v`	`= v (dv)/(dx)`
`mv (dv)/(dx)`	`= -mg-kv^2`
`dx`	`=(-mv)/(mg + kv^2) dv`

`int_0^H dx`	`= -int_u^0 (mv)/(mg + kv^2) dv`
`[x]_0^H`	`= -m/(2k) [log_e (mg + kv^2)]_u^0`
`H`	`= -m/(2k) (log_e (mg)-log_e (mg + ku^2))`
	`= m/(2k) log_e ((mg + ku^2)/(mg))`
	`= m/(2k) log_e (1 + (ku^2)/(mg))\ \ \ \ text{(using}\ \ k = (mg)/v_T^2 text{)}`
`:.H`	`=(v_T^2)/(2g) log_e (1 + u^2/(v_T^2))`

iii. `text(When the ball falls),\ \ m dot v = mg-kv^2`

♦♦ Mean mark 14%.

`mv (dv)/(dx)`	`= mg-kv^2`
`dx`	`=(mv)/(mg-kv^2) dv`

`int_0^H dx`	`= int_0^w (mv)/(mg-kv^2)dv`
`[x]_0^H`	` =-m/(2k)[log_e (mg-kv^2)]_0^w`
`H`	`= -m/(2k) (log_e (mg-kw^2)-log_e (mg))`
	`= -m/(2k) log_e ((mg-kw^2)/(mg))`
	`= -m/(2k) log_e (1-(kw^2)/(mg))\ \ \ \ text{(using}\ \ k = (mg)/v_T^2 text{)}`
`H`	`=-(v_T^2)/(2g) log_e (1-w^2/(v_T^2))`
	`=-(v_T^2)/(2g) log_e ((v_T^2-w^2)/(v_T^2))`
	`=(v_T^2)/(2g) log_e ((v_T^2)/(v_T^2-w^2))`

`text{Using part (ii):}`

`(v_T^2)/(2g) log_e ((v_T^2)/(v_T^2-w^2))`	`=(v_T^2)/(2g) log_e (1 + u^2/(v_T^2))`
`(v_T^2)/(v_T^2-w^2)`	`=(v_T^2 + u^2)/(v_T^2)`
`(v_T^2 + u^2)(v_T^2-w^2)`	`=v_T^4`
`v_T^4-v_T^2 w^2+v_T^2 u^2-w^2 u^2`	`=v_T^4`
`v_T^2 w^2+w^2 u^2`	`=v_T^2 u^2`
`(v_T^2 w^2)/(v_T^2w^2u^2)+(w^2 u^2)/(v_T^2w^2u^2)`	`=(v_T^2 u^2)/(v_T^2w^2u^2)`
`:.1/u^2 + 1/(v_T^2)`	`=1/w^2`

Harder Ext1 Topics, EXT2 2013 HSC 15c

Eight cars participate in a competition that lasts for four days. The probability that a car completes a day is `0.7`. Cars that do not complete a day are eliminated.

Find the probability that a car completes all four days of the competition. (1 mark)
Find an expression for the probability that at least three cars complete all four days of the competition. (2 marks)

Show Answers Only

`0.2401`
`1-(0.76^8 + \ ^8C_1 * 0.76^7 * 0.24 + \ ^8C_2 * 0.76^6 * 0.24^2)`

Show Worked Solution

(i) `P text{(Completes 1 day)} = 0.7`

`:.P text{(Completes 4 days)}`	`= 0.7^4`
	`~~ 0.2401`

(ii) `P text{(Does not complete all 4 days)`	`= 1 – 0.7^4`
	`= 0.7599`
	`~~ 0.76`

♦♦ Mean mark 26%. 
STRATEGY: The use of complementary events reduced the required calculations and subsequent errors in this part.

`text(Let)\ \ C =\ text(number of cars that complete 4 days)`

`:. P text{(At least 3 cars complete all 4 days)}`

`= 1 – [P (C=0) + P (C = 1) + P (C = 2)]`

`= 1 – (\ ^8C_0 *0.76^8 + \ ^8C_1 * 0.76^7 * 0.24 + \ ^8C_2 * 0.76^6 * 0.24^2)`

`=1-(0.76^8 + \ ^8C_1 * 0.76^7 * 0.24 + \ ^8C_2 * 0.76^6 * 0.24^2)`

Functions, EXT1′ F2 2013 HSC 15b

The polynomial `P(x) = ax^4 + bx^3 + cx^2 + e` has remainder `-3` when divided by `x - 1`. The polynomial has a double root at `x = -1.`

Show that `4a + 2c = -9/2.` (2 marks)

--- 7 WORK AREA LINES (style=lined) ---
Hence, or otherwise, find the slope of the tangent to the graph `y = P(x)` when `x = 1.` (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`-9`

Show Worked Solution

i.	`P(x)`	`= ax^4 + bx^3 + cx^2 + e`
	`P prime (x)`	`=4ax^3 + 3bx^2 + 2cx`

`P(1)=-3`
`a+b+c+e`	`=-3\ \ \ \ …\ (1)`
`P(-1)=0`
`a-b+c+e`	`=0\ \ \ \ …\ (2)`
`P′(-1)=0`
`-4a+3b-2c`	`=0\ \ \ \ …\ (3)`

`(1)-(2)`

`2b`	`=-3`
`b`	`=-3/2`

`text(Substitute into)\ \ (3)`

`:.4a+2c`	`=3b`
	`=-9/2\ \ \ \ text(… as required)`

ii. `P prime (1)`	`= 4a + 3b + 2c`
	`= -9/2 – 9/2`
	`=-9`

Functions, EXT1′ F1 2013 HSC 13b

The diagram shows the graph of a function `f(x).`

Sketch the following curves on separate half-page diagrams.

`y^2 = f(x).` (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
`y = 1/(1 - f(x)).` (3 marks)

--- 12 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(See Worked Solutions)`
`text(See Worked Solutions)`

Show Worked Solution

i. `y^2 = f(x)`

`text(Only exist for)\ \ f(x) >= 0`

`y^2 = 1,\ \ \ y = +- 1`

ii. `y = 1/(1 – f(x))`

MARKER’S COMMENT: Correct working sketches such as `y=-f(x)` and `y=1-f(x)` meant that students could obtain some marks, even if their final sketch was wrong. Note this important advice.

`f(x) = 1,\ \ \ y\ text(undefined.)`

`f(x) > 1,\ \ \ y < 0`

`f(x) <= 0, \ \ \ y <= 1`

Conics, EXT2 2013 HSC 12d

The points `P (cp, c/p)` and `Q (cq, c/q)`, where `|\ p\ | ≠ |\ q\ |`, lie on the rectangular hyperbola with equation `xy = c^2.`

The tangent to the hyperbola at `P` intersects the `x`-axis at `A` and the `y`-axis at `B`. Similarly, the tangent to the hyperbola at `Q` intersects the `x`-axis at `C` and the `y`- axis at `D`.

Show that the equation of the tangent at `P` is `x + p^2 y = 2cp.` (2 marks)
Show that `A, B and O` are on a circle with centre `P.` (2 marks)
Prove that `BC` is parallel to `PQ.` (1 mark)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `P (cp, c/p),\ \ Q (cq, c/q),\ \ xy = c^2,\ \ |\ p\ | ≠ |\ q\ |`

`xy = c^2`

`y + x*(dy)/(dx) = 0,\ \ (dy)/(dx) = -y/x`

`text(At)\ \ (cp, c/p)`

`(dy)/(dx) = (-c/p)/(cp) = -1/p^2`

`text(Equation of tangent at)\ \ P`

`y – c/p`	`= -1/p^2 (x – cp)`
`p^2 y – cp`	`= -x + cp`
`:. x + p^2 y`	`= 2cp\ \ \ text(… as required)`

(ii) `text(Solution 1)`

`text(At)\ \ A, y = 0,\ \ \ x = 2cp`

`text(At)\ \ B, x = 0,\ \ \ y = (2c)/p`

`OP`	`= sqrt (c^2 p^2 + c^2/p^2) = c/p sqrt(p^4 + 1)`
`PA`	`= sqrt {(2cp – cp)^2 + c^2/p^2} = c/p sqrt (p^4 + 1)`
`PB`	`= sqrt{c^2 p^2 + ((2c)/p – c/p)^2} = c/p sqrt (p^4 + 1)`

`OP = PA = PB`

`:. A, B and O\ \ text(are on the circle centre)\ \ P.`

`text(Alternate Solution)`

`text(Mid-point of)\ \ AB`

`=((2cp+0)/2,\ (0+(2c)/p)/2)`

`=(cp,\ c/p)`

`=>P\ \ text(is the midpoint of)\ \ AB`

`text(S)text(ince)\ \ ∠AOB=90^@`

`:. A, B and O\ \ text(are on the circle centre)\ \ P.`

(iii) `text(Equation of tangent at)\ \ Q\ \ \ text{(using part (i))}`

`x + q^2y = 2cq`

`=> C\ \ text(is)\ \ (2cq, 0)`

`m_(BC)`	`= {(2c)/p – 0}/(0 – 2cq)`
	`= – 1/(pq)`
`m_(PQ)`	`= (c/p – c/q)/(cp – cq)`
	`=((cq-cp)/(pq))/(c(p-q))`
	`= {-c(p – q)}/{cpq (p – q)}`
	`= -1/(pq)`

`m_(BC) = m_(PQ)`

`:. BC\ text(||)\ PQ`

Graphs, EXT2 2013 HSC 12b

The equation `log_e y - log_e (1000 - y) = x/50 - log_e 3` implicitly defines `y` as a function of `x`.

Show that `y` satisfies the differential equation

`(dy)/(dx) = y/50 (1 - y/1000).` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`log_e y – log_e (1000 – y) = x/50 – log_e 3`

`1/y* (dy)/(dx) – (-1)/((1000 – y))*(dy)/(dx)`	`= 1/50 – 0`
`(dy)/(dx)(1/y + 1/(1000 – y))`	`= 1/50`
`(dy)/(dx) ((1000 – y + y)/(y (1000 – y)))`	`= 1/50`
`(dy)/(dx) (1000/(y (1000 – y)))`	`= 1/50`
`(dy)/(dx)`	`= (y (1000 – y))/(50 xx 1000)`
`(dy)/(dx)`	`= y/50 (1 – y/1000)`

Calculus, EXT2 C1 2013 HSC 12a

Using the substitution ` t = tan\ x/2`, or otherwise, evaluate

`int_0^(pi/2) 1/(4 + 5 cos x)\ dx.` (4 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`1/3 ln 2`

Show Worked Solution

`t = tan\ x/2`

`=>cos x = (1 – t^2)/(1 + t^2)\ ,\ \ \ dx = (2\ dt)/(1 + t^2)`

`text(When)\ \ x = 0\ ,\ t = 0\ ;\ \ x = pi/2\ ,\ t = 1`

`int_0^(pi/2) (dx)/(4 + 5 cos x)`	`= int_0^1 1/{4 + (5(1 – t^2))/(1 + t^2)} xx (2\ dt)/(1 + t^2)`
	`= int_0^1 (2\ dt)/(4 + 4t^2 + 5 – 5t^2)`
	`= int_0^1 (2\ dt)/(9 – t^2)`
	`=2 int_0^1 1/((3-t)(3+t))`
	`= 1/3 int_0^1 (1/(3 – t) + 1/(3 + t))\ dt`
	`= 1/3 [-ln (3 – t) + ln (3 + t)]_0^1`
	`= 1/3 [(-ln 2 + ln 4) – (-ln 3 + ln 3)]`
	`= 1/3 ln2`

Complex Numbers, EXT2 N1 2013 HSC 11c

Factorise `z^2 + 4iz + 5.` (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`(z – i) (z + 5i)`

Show Worked Solution

`text(Solution 1)`

`alpha + beta = 4i,\ \ \ alpha beta = 5 = -5i^2`

`:. z^2 + 4iz + 5 = (z – i) (z + 5i)`

`(alpha + beta = 4i,\ \ \ alpha beta = 5 = -5i^2)`

MARKER’S COMMENT: Best practice includes stating the general quadratic formula before substituting in values.

`text(Solution 2)`

`z`	`=(-b+- sqrt(b^2 – 4ac))/(2a)`
	`=(-4i +- sqrt((4i)^2-4 xx 5))/2`
	`=(-4i+-sqrt(-16-20))/2`
	`=(-4i +- 6i)/2`
	`=i\ \ text(or)\ \ -5i`

`:.z^2 + 4iz + 5 = (z – i) (z + 5i)`

Harder Ext1 Topics, EXT2 2014 HSC 16a

The diagram shows two circles `C_1` and `C_2` . The point `P` is one of their points of intersection. The tangent to `C_2` at `P` meets `C_1` at `Q`, and the tangent to `C_1` at `P` meets `C_2` at `R`.

The points `A` and `D` are chosen on `C_1` so that `AD` is a diameter of `C_1` and parallel to `PQ`. Likewise, points `B` and `C` are chosen on `C_2` so that `BC` is a diameter of `C_2` and parallel to `PR`.

The points `X` and `Y` lie on the tangents `PR` and `PQ`, respectively, as shown in the diagram.

Copy or trace the diagram into your writing booklet.

Show that `∠APX = ∠DPQ`. (2 marks)
Show that `A`, `P` and `C` are collinear. (3 marks)
Show that `ABCD` is a cyclic quadrilateral. (1 mark)

Show Answers Only

`text(See Worked Solutions.)`
`text(See Worked Solutions.)`
`text(See Worked Solutions.)`

Show Worked Solution

(i)

`∠APX`	`= ∠ADP\ \ \ text{(angle in alternate segment)`βαγ
`∠ADP`	`= ∠DPQ\ \ \ text{(alternate angles,}\ AD\ text(\|\|)\ PQ)`
`:.∠APX`	`= ∠DPQ`

(ii) `text(Join)\ PC\ text(and)\ PB.`

♦♦ Mean mark 31%.
MARKER’S COMMENT: Be vigilant that your reasoning does not implicitly assume `APC` or `DPB` are straight lines.

`text{Let}\ \ ∠APX = ∠DPQ=α\ \ \ text{(from part (i))}`

`text{Similarly,}\ \ ∠YPB = ∠CPR=β`

`∠XPY = ∠RPQ=γ\ \ \ text{(vertically opposite angles.)}`

`∠APD`	`= 90^@\ \ \ text{(angle in a semicircle in}\ C_1)`
`∠CPB`	`= 90^@\ \ \ text{(angle in a semicircle in}\ C_2)`

`90^@ + 90^@ + 2(α+β+γ)`	`= 360^@ \ \ text{(sum of angles at a point}\ Ptext{)}`
`:. (α+β+γ)`	`= 90^@`

`∠APC`

`=90+(α+β+γ)=180^@`

`:.\ A, P\ text(and)\ C\ text(are collinear.)`

♦ Mean mark 44%.

(iii) `:.∠APX = ∠CPR\ text{(vertically opposite angles,}\ APC\ text{is a straight line)}`

`:.α=β\ \ =>∠BCA = ∠BDA`

`text(S)text(ince chord)\ AB\ text(subtends a pair of equal angles at)\ C and D`

`=>ABCD\ text(is a cyclic quadrilateral.)`

Functions, EXT1′ F2 2014 HSC 14a

Let `P(x) =x^5-10x^2 +15x-6`.

Show that `x = 1` is a root of `P(x)` of multiplicity three. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`text{See Worked Solutions}`

Show Worked Solution

`P(x) =x^5-10x^2 +15x-6`

`P(1) = 1-10 + 15-6 = 0`

`P^{′}(x)`	`= 5x^4-20x + 15`
`P^{′}(1)`	`= 5-20 + 15 = 0`

`P^{″}(x)`	`= 20x^3-20`
`P^{″}(1)`	`= 20-20 = 0`

`P^{‴}(x)`	`= 60x^2`
`P^{‴}(1)`	`= 60 ≠ 0`

`:.x = 1\ text(is a root of)\ P(x),text(of multiplicity 3.)`

Conics, EXT2 2014 HSC 13c

The point `S(ae, 0)` is the focus of the hyperbola `(x^2)/(a^2)-(y^2)/(b^2) = 1` on the positive `x`-axis.

The points `P(at, bt)` and `Q(a/t, −b/t)` lie on the asymptotes of the hyperbola, where `t > 0`.

The point `M((a(t^2 + 1))/(2t), (b(t^2 – 1))/(2t))` is the midpoint of `PQ`.

Show that `M` lies on the hyperbola. (1 mark)
Prove that the line through `P` and `Q` is a tangent to the hyperbola at `M`. (3 marks)
Show that `OP xx OQ = OS^2`. (2 marks)
If `P` and `S` have the same `x`-coordinate, show that `MS` is parallel to one of the asymptotes of the hyperbola. (2 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`

Show Worked Solution

(i) `text(Substitute)\ \ \ M((a(t^2 + 1))/(2t), (b(t^2 − 1))/(2t))\ \ text(into)`

`(x^2)/(a^2)-(y^2)/(b^2) = 1`

`text(LHS)`	`=1/(a^2) xx (a^2(t^2 + 1)^2)/(4t^2) − 1/(b^2) xx (b^2(t^2 − 1)^2)/(4t^2)`
	`=1/(4t^2)[t^4 + 2t^2 + 1 – (t^4 − 2t^2 + 1)]`
	`=1/(4t^2) xx 4t^2`
	`=1`
	`=\ text(RHS)`

`:.M\ text(lies on the hyperbola).`

(ii) `P(at, bt), \ \ Q(a/t, −b/t)`

`m_(PQ)`	`= (bt − (-b/t))/(at − a/t) xx t/t`
	`=(bt^2+b)/(at^2-a)`
	`= b/a ((t^2 + 1)/(t^2 − 1))`

`text(Differentiate the hyperbola)`

`(2x)/(a^2) − (2y*dy/dx)/(b^2)`	`= 0`
`(dy)/(dx)`	`= (b^2x)/(a^2y)`

`text(At)\ \ M((a(t^2 + 1))/(2t), (b(t^2 − 1))/(2t)),`

`(dy)/(dx)`	`= (b^2)/(a^2) xx (a(t^2 + 1))/(2t) xx (2t)/(b(t^2 – 1))`
	`= b/a ((t^2 + 1)/(t^2 -1))`

`:. m_(PQ)=m_(at\ M)`

`:.text(S)text(ince the tangent to the hyperbola has the same gradient as)\ \ PQ`

`text(and both pass through)\ M,text(they are the same line.)`

(iii)	`OP`	`= sqrt(a^2t^2 + b^2t^2) = t sqrt(a^2 + b^2)`
	`OQ`	`= sqrt((a^2)/(t^2) + (b^2)/(t^2)) = 1/tsqrt(a^2 + b^2)`

`OP xx OQ`	`= tsqrt(a^2 + b^2) xx 1/tsqrt(a^2 + b^2)`
	`= a^2 + b^2`
	`=a^2+a^2(e^2 − 1)`
	`= a^2e^2`
	`=OS^2\ \ \ …\ text(as required)`

(iv) `text(Given)\ \ P and S\ \ text(have the same)\ x text(-coordinate)`

`at = ae\ \ => t = e`

`S(ae,0),\ \ M((a(e^2 + 1))/(2e), (b(e^2 − 1))/(2e))`

`m_(MS)`	`= ((b(e^2 − 1))/(2e) − 0)/((a(e^2 + 1))/(2e) − ae)`
	`= (b(e^2 − 1))/(ae^2 + a − 2ae^2)`
	`= (b(e^2 − 1))/(a − ae^2)`
	`= (b(e^2 − 1))/(-a(e^2 − 1))`
	`= – b/a`

`:.text(S)text(ince the gradients of the hyperbola asymptotes are)\ ±b/a.`

`=>MS\ text(is parallel to one … as required)`

Graphs, EXT2 2014 HSC 12c

The point `P(x_0, y_0)` lies on the curves `x^2 − y^2 = 5` and `xy = 6`. Prove that the tangents to these curves at `P` are perpendicular to one another. (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Differentiating)\ \ \ x^2 − y^2 = 5`

`2x − 2y*(dy)/(dx)`	`= 0`
`(dy)/(dx)`	`=x/y`
`text(At)\ P(x_0, y_0),\ \ (dy)/(dx)`	`= x_0/y_0`

`text(Differentiating)\ \ \ xy = 6`

`y + x*(dy)/(dx)`	`= 0`
`(dy)/(dx)`	`= -y/x`
`text(At)\ P(x_0, y_0),\ \ (dy)/(dx)`	`= -y_0/x_0`

`text(S)text(ince)\ \ \ x_0/y_0 xx (-y_0/x_0) = -1`

`:.\ text(The tangents to the curves at)\ P(x_0, y_0)\ text(are)`

`text(perpendicular to one another.)`

Functions, EXT1′ F1 2014 HSC 12a

The diagram shows the graph of a function `f(x)`.

Draw a separate half-page graph for each of the following, showing all asymptotes and intercepts.

`y = f(|\ x\ |)` (2 marks)

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`y = 1/(f(x))` (2 marks)

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Show Answers Only

`text(See Worked Solutions.)`
`text(See Worked Solutions.)`

Show Worked Solution

ii.

Volumes, EXT2 2014 HSC 11e

The region enclosed by the curve `x = y(6 − y)` and the `y`-axis is rotated about the `x`-axis to form a solid.

Using the method of cylindrical shells, or otherwise, find the volume of the solid. (3 marks)

Show Answers Only

`216pi\ \ text(u³)`

Show Worked Solution

`x = y(6 − y)\ \ =>text(Rotate about)\ xtext(-axis.)`

MARKER’S COMMENT: The best responses included a sketch that clearly showed the axis of rotation and an understanding of the use of cylindrical shells, as per the worked solution.

`δV`	`=2 pi r h delta y`
	`= 2 pi y*y (6-y) delta y`
	`=2 pi(6y^2 – y^3) delta y`

`:.V`	`= 2pi int_0^6(6y^2 − y^3)\ dy`
	`= 2pi[2y^3 −(y^4)/4]_0^6`
	`= 2pi[(432 − 324) − 0]`
	`= 216pi\ \ text(u³)`

Calculus, EXT2 C1 2014 HSC 11b

Evaluate `int_0^(1/2)(3x-1)\ cos\ (pix)\ dx`. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`(pi-6)/(2pi²)`

Show Worked Solution

`text(Integrating by parts:)`

`u`	`=3x-1`	`u^{′}`	`=3`
`v^{′}`	`=cos(pi x)`	`v`	`=1/pi sin(pi x)`

`int_0^(1/2)(3x-1)\ cos\ (pix)\ dx`

`= [(3x-1)(sin\ (pix))/pi]_0^(1/2)-int_0^(1/2) 3 xx (sin\ (pix))/pi\ dx`

`= (1/(2pi)\ sin\ pi/2-0) − 3/pi[(-cos\ (pix))/pi]_0^(1/2)`

`= 1/(2pi) + 3/(pi^2)(cos\ pi/2-cos\ 0)`

`= 1/(2pi) + 3/(pi^2)(0-1)`

`= 1/(2pi)-3/(pi^2)`

Complex Numbers, EXT2 N1 2014 HSC 11a

Consider the complex numbers `z = -2- 2i` and `w = 3 + i`.

Express `z + w` in modulus–argument form. (2 marks)

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Express `z/w` in the form `x + iy`, where `x` and `y` are real numbers. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`sqrt2\ text(cis)(-pi/4)`
`−4/5 −2/5i`

Show Worked Solution

i.	`z + w`	`= −2 − 2i + 3 + i`
		`= 1 − i`
	`\|\ z+w\ \|`	`= sqrt2`
	`text(arg)\ (z+w)`	`=- pi/4`
	`:. z+w`	`= sqrt2\ text(cis)(-pi/4)`

ii.	`z/w`	`= (−2 − 2i)/(3 + i)`
		`= ((−2 − 2i)(3 − i))/((3 + i)(3 − i))`
		`= (−6 − 6i + 2i − 2)/(9 + 1)`
		`= −8/10 −4/10i`
		`= −4/5 − 2/5i`

Statistics, EXT1 S1 2007 HSC 4a

In a large city, 10% of the population has green eyes.

What is the probability that two randomly chosen people both have green eyes? (1 mark)

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What is the probability that exactly two of a group of 20 randomly chosen people have green eyes? Give your answer correct to three decimal places. (1 mark)

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What is the probability that more than two of a group of 20 randomly chosen people have green eyes? Give your answer correct to two decimal places. (2 marks)

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Show Answers Only

`0.01`
`0.285\ \ \ text{(to 3 d.p.)}`
`0.32\ \ \ text{(to 2 d.p.)}`

Show Worked Solution

i.	`P(text(G))`	`= 0.1`
	`P(text(GG))`	`= 0.1 xx 0.1`
		`= 0.01`

ii. `P(text(not G)) = 1 − 0.1 = 0.9`

`:. P(text(2 out of 20 have green eyes))`

`= \ ^(20)C_2 · (0.1)^2 · (0.9)^(18)`

`= 0.2851…`

`= 0.285\ \ \ text{(to 3 d.p.)}`

iii. `P(text(more than 2 have green eyes))`

`= 1 − [P(0) + P(1) + P(2)]`

`= 1 − [0.9^20 + \ ^20C_1(0.1)^1(0.9)^19 + \ ^20C_2(0.1)^2(0.9)^(18)]`

`= 1 − [0.1215… + 0.2701… + 0.2851…]`

`= 1 − 0.6769…`

`= 0.3230`

`= 0.32\ \ \ text{(to 2 d.p.)}`

Mechanics, EXT2* M1 2007 HSC 2d

A skydiver jumps from a hot air balloon which is 2000 metres above the ground. The velocity, `v` metres per second, at which she is falling `t` seconds after jumping is given by `v =50(1 - e^(-0.2t))`.

Find her acceleration ten seconds after she jumps. Give your answer correct to one decimal place. (2 marks)

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Find the distance that she has fallen in the first ten seconds. Give your answer correct to the nearest metre. (2 marks)

--- 7 WORK AREA LINES (style=lined) ---

Show Answers Only

`1.4\ text(ms)^(−2)\ \ text{(to 1 d.p.)}`
`284\ text{m (nearest m)}`

Show Worked Solution

i.	`v`	`= 50(1 − e^(-0.2t))`
		`=50-50e^(-0.2t)`
	`ddot x`	`= (dv)/(dt)`
		`= −0.2 xx 50 xx −e^(−0.2t)`
		`= 10 e^(−0.2t)`

`text(When)\ \ t = 10:`

`ddot x`	`= 10 e^(−0.2 xx 10)`
	`= 10 e ^(−2)`
	`= 1.353…`
	`= 1.4\ text(ms)^(−2)\ \ \ text{(to 1 d.p.)}`

ii. `text(Distance travelled)`

`= int_0^10 v\ dt`

`= 50 int_0^10 1 − e^(−0.2t) \ dt`

`= 50 [t + 1/0.2 · e^(−0.2t)]_0^10`

`= 50 [t + 5e^(−0.2t)]_0^10`

`= 50 [(10 + 5e^(−2)) − (0 + 5e^0)]`

`= 50 [10 + 5e^(−2) − 5]`

`= 50 [5 + 5e^(−2)]`

`= 283.833…`

`= 284\ text{m (nearest m)}`

Functions, EXT1 F2 2007 HSC 2c

The polynomial `P(x) = x^2 + ax + b` has a zero at `x = 2`. When `P(x)` is divided by `x + 1`, the remainder is `18`.

Find the values of `a` and `b`. (3 marks)

--- 7 WORK AREA LINES (style=lined) ---

Show Answers Only

`a = -7\ \ text(and)\ \ b = 10`

Show Worked Solution

`P(x) = x^2 + ax + b`

`text(S)text(ince there is a zero 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`

Trigonometry, EXT1 T1 2007 HSC 2b

Let `f(x) = 2 cos^(-1)x`.

Sketch the graph of `y = f(x)`, indicating clearly the coordinates of the endpoints of the graph. (2 marks)

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State the range of `f(x)`. (1 mark)

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Show Answers Only

`text(See Worked Solutions.)`
`0 ≤ y ≤ 2pi`

Show Worked Solution

i. `y= 2\ cos^(-1)x`

`text(Domain:)\ -1 ≤ x ≤ 1`

`text{Range:}\ \0 ≤`	`y/2 ≤ pi`
`0 ≤`	`y ≤ 2pi`

ii. `text(Range:)\ \ 0 ≤ y ≤ 2pi`

Linear Functions, EXT1 2007 HSC 1d

The graphs of the line `x - 2y + 3= 0` and the curve `y = x^3+ 1` intersect at `(1, 2)`. Find the exact value, in radians, of the acute angle between the line and the tangent to the curve at the point of intersection. (3 marks)

Show Answers Only

`pi/4\ text(radians)`

Show Worked Solution

`x − 2y + 3`	`= 0`
`2y`	`= x + 3`
`y`	`= 1/2x + 3/2`

`:. m_1 = 1/2`

`y`	`= x^3 + 1`
`(dy)/(dx)`	`= 3x^2`

`text(When)\ x = 1,`

`(dy)/(dx) = 3`

`:. m_2 = 3`

`tan\ theta`	`= \|\ (m_1 − m_2)/(1 + m_1m_2)\ \|`
	`= \|\ (1/2 − 3)/(1 + 1/2 xx 3)\ \|`
	`= \|\ (−2 1/2)/(2 1/2)\|`
	`= 1`
`:. theta`	`= pi/4\ text(radians)`

Calculus, EXT1 C2 2007 HSC 1c

Differentiate `tan^(–1)(x^4)` with respect to `x`. (2 marks)

Show Answers Only

`(4x^3)/(1 + x^8)`

Show Worked Solution

`y`	`= tan^(−1)(x^4)`
`(dy)/(dx)`	`= 1/(1 + (x^4)^2) xx d/(dx) (x^4)`
	`= (4x^3)/(1 + x^8)`

Linear Functions, EXT1 2007 HSC 1b

The interval `AB`, where `A` is `(4, 5)` and `B` is `(19, text(−5))`, is divided internally in the ratio `2\ :\ 3` by the point `P(x,y)`. Find the values of `x` and `y`. (2 marks)

Show Answers Only

`(10, 1)`

Show Worked Solution

`A(4, 5), \ B(19, text(−5))`

`text(Internal division in ratio 2 : 3 at)\ \ P(x, y)`

`P`	`= ((nx_1 + mx_2)/(m + n) , (ny_1 + my_2)/(m + n))`
	`= ((3 xx 4 + 2 xx 19)/(2 + 3) , (3 xx 5 + 2 xx (text(−5)))/(2+ 3))`
	`= ((12 + 38)/5 , (15 − 10)/5)`
	`= (10, 1)`

Mechanics, EXT2* M1 2004 HSC 6b

A fire hose is at ground level on a horizontal plane. Water is projected from the hose. The angle of projection, `theta`, is allowed to vary. The speed of the water as it leaves the hose, `v` metres per second, remains constant. You may assume that if the origin is taken to be the point of projection, the path of the water is given by the parametric equations

`x = vt\ cos\ theta`

`y = vt\ sin\ theta − 1/2 g t^2`

where `g\ text(ms)^(−2)` is the acceleration due to gravity. (Do NOT prove this.)

Show that the water returns to ground level at a distance`(v^2\ sin\ 2theta)/g` metres from the point of projection. (2 marks)

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This fire hose is now aimed at a 20 metre high thin wall from a point of projection at ground level 40 metres from the base of the wall. It is known that when the angle `theta` is 15°, the water just reaches the base of the wall.

Show that `v^2 = 80g`. (1 mark)

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Show that the cartesian equation of the path of the water is given by

`y = x\ tan\ theta − (x^2\ sec^2\ theta)/160`. (2 marks)

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Show that the water just clears the top of the wall if

`tan^2\ theta − 4\ tan\ theta + 3 = 0`. (2 marks)

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Find all values of `theta` for which the water hits the front of the wall. (2 marks)

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Show Answers Only

`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`
`15^@ ≤ theta ≤ 45^@ \ \ text(and)\ \ 71.6^@ ≤ theta ≤ 75^@`

Show Worked Solution

i.	`x`	`= vt\ cos\ theta`
	`y`	`= vt\ sin\ theta − 1/2 g t^2`

`text(Find)\ \ t\ \ text(when)\ \ y = 0`

`vt\ sin\ theta − 1/2 g t^2`	`= 0`
`t(v\ sin\ theta − 1/2 g t)`	`= 0`
`v\ sin\ theta − 1/2 g t`	`= 0, \ \ t ≠ 0`
`1/2 g t`	`= v\ sin\ theta`
`t`	`= (2v\ sin\ theta)/g`

`text(Find)\ \ x\ \ text(when)\ \ t = (2v\ sin\ theta)/g`

`x`	`= v · (2v\ sin\ theta)/g\ cos\ theta`
	`= (v^2 · \ 2\ sin\ theta\ cos\ theta)/g`
	`= (v^2\ sin\ 2theta)/g`

`:.\ text(When)\ \ x = (v^2\ sin\ 2theta)/g,\ \ \text(the water returns)`

`text(to ground level … as required.)`

ii. `text(Show)\ \ v^2 = 80\ text(g)`

`text(When)\ \ theta = 15^@, \ x = 40`

`text{From part (i)}`

`40`	`= (v^2\ sin\ 30^@)/g`
`v^2 xx 1/2`	`= 40g`
`v^2`	`= 80g\ \ \ …text(as required)`

iii. `text(Show)\ \ y = x\ tan\ theta − (x^2\ sec^2\ theta)/160`

`x`	`= vt\ cos\ theta`
`:.t`	`= x/(v\ cos\ theta)`

`text(Substitute into)`

`y`	`= vt\ sin\ theta − 1/2 g t^2`
	`= v · x/(v\ cos\ theta) · sin\ theta −1/2 g (x/(v\ cos\ theta))^2`
	`= x\ tan\ theta − 1/2 g (x^2/(v^2\ cos^2\ theta))`
	`= x\ tan\ theta − 1/2 g · x^2/(80g\ cos^2\ theta)`
	`= x\ tan\ theta − (x^2\ sec^2\ theta)/160\ \ \ …text(as required.)`

iv. `text(Water clears the top if)\ \ y = 20\ \ text(when)\ \ x = 40`

`text{Substitute into equation from (iii)}`

`40\ tan\ theta − (40^2\ sec^2\ theta)/160`	`= 20`
`40\ tan\ theta − 10\ sec^2\ theta`	`= 20`
`40\ tan\ theta − 10(1 + tan^2\ theta)`	`= 20`
`40\ tan\ theta − 10 − 10\ tan^2\ theta`	`= 20`
`10\ tan^2\ theta − 40\ tan\ theta\ + 30`	`= 0`
`tan^2\ theta − 4\ tan\ theta + 3`	`= 0\ \ \ text(… as required)`

`text(Water hits the bottom of the wall when)`

`x = 40\ \ text(and)\ \ y = 0`

`40\ tan\ theta − (40^2\ sec^2\ theta)/160`	`= 0`
`40\ tan\ theta − 10\ sec^2\ theta`	`= 0`
`4\ tan\ theta − (1 + tan^2\ theta)`	`= 0`
`tan^2\ theta − 4\ tan\ theta + 1`	`= 0`

`text(Using the quadratic formula)`

`tan\ theta`	`= (+4 ± sqrt(16 − 4 · 1 · 1))/ (2 xx 1)`
	`= (4 ± sqrt12)/2`
	`= 2 ± sqrt3`
`theta`	`= 15^@\ \ text(or)\ \ 75^@`

`text(Water hits the top of the wall when)`

`x = 40\ text(and)\ \ y = 20`

`tan^2\ theta − 4\ tan\ theta + 3`	`= 0\ \ \ text{from (iv)}`
`(tan\ theta − 1)(tan\ theta − 3)`	`= 0`

`tan\ theta`	`= 1`	`text(or)`	`tan\ theta`	`= 3`
`theta`	`= 45^@`		`theta`	`= tan^(−1)\ 3`
				`= 71.565…`
				`= 71.6^@\ \ \text{(to 1 d.p.)}`

`:.\ text(Following the motion of the water as)\ \ theta\ \ text(increases,)`

`text(water hits the wall when)`

`15^@ ≤ theta ≤ 45^@`	`\ text(and)`
`71.6^@ ≤ theta ≤ 75^@`

Inverse Functions, EXT1 2004 HSC 5b

The diagram below shows a sketch of the graph of `y = f(x)`, where `f(x) = 1/(1 + x^2)` for `x ≥ 0`.

Copy or trace this diagram into your writing booklet.
On the same set of axes, sketch the graph of the inverse function, `y = f^(−1)(x)`. (1 mark)
State the domain of `f^(−1)(x)`. (1 mark)
Find an expression for `y = f^(−1)(x)` in terms of `x`. (2 marks)
The graphs of `y = f(x)` and `y = f^(−1)(x)` meet at exactly one point `P`.
Let `α` be the `x`-coordinate of `P`. Explain why `α` is a root of the equation
1. `x^3 + x − 1 = 0`. (1 mark)
Take 0.5 as a first approximation for `α`. Use one application of Newton’s method to find a second approximation for `α`. (2 marks)

Show Answers Only

`text(See Worked Solutions)`
`0 < x ≤ 1`
`y = sqrt((1 − x)/x), y > 0`
`text(See Worked Solutions)`
`0.714\ \ \ text{(to 3 d.p.)}`

Show Worked Solution

(i)

(ii) `text(Domain of)\ \ f^(−1)(x)\ text(is)`

`0 < x ≤ 1`

(iii) `f(x) = 1/(1 + x^2)`

`text(Inverse: swap)\ \ x↔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`

(iv) `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`

(v)	`f(x)`	`= x^3 + x − 1`
	`f′(x)`	`= 3x^2 + 1`
	`f(0.5)`	`= 0.5^3 + 0.5 − 1`
		`= −0.375`
	`f′(0.5)`	`= 3 xx 0.5^2 + 1`
		`= 1.75`

`α_2`	`= α_1 − (f(0.5))/(f′(0.5))`
	`= 0.5 − ((−0.375))/1.75`
	`= 0.5 − (−0.2142…)`
	`= 0.7142…`
	`= 0.714\ \ \ text{(to 3 d.p.)}`

Quadratic, EXT1 2004 HSC 4b

The two points `P(2ap, ap^2)` and `Q(2aq, aq^2)` are on the parabola `x^2 = 4ay`.

The equation of the tangent to `x^2 = 4ay` at an arbitrary point `(2at, at^2)` on the parabola is `y = tx − at^2`. (Do not prove this.)
Show that the tangents at the points `P` and `Q` meet at `R`, where `R` is the point `(a(p + q), apq)`. (2 marks)
As `P` varies, the point `Q` is always chosen so that `∠POQ` is a right angle, where `O` is the origin.
Find the locus of `R`. (2 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`
`y = −4a`

Show Worked Solution

(i)

`text(Show)\ \ R(a(prq), apq)`

`text(T)text(angent equations)`

`y`	`= px − ap^2`	`\ \ …\ (1)`
`y`	`= qx − aq^2`	`\ \ …\ (2)`

`text(Substitute)\ \y = px − ap^2\ \text(into)\ (2)`

`px − ap^2`	`= qx − aq^2`
`px − qx`	`= ap^2 − aq^2`
`x(p − q)`	`= a(p^2 − q^2)`
	`= a(p + q)(p − q)`
`:.x`	`= a(p + q)`

`text(Substitute)\ \x = a(p + q)\ \ text(into)\ (1)`

`y`	`= p xx a(p + q) − ap^2`
	`= ap^2 + apq − ap^2`
	`= apq`

`:.R(a(p+q), apq)\ \ \ …text(as required.)`

(ii) `text(If)\ \ ∠POQ\ text(is a right angle)`

`M_(PO) xx M_(OQ) = −1`

`M_(PO)`	`= (ap^2 − 0)/(2ap − 0)`
	`= p/2`
`M_(OQ)`	`= (aq^2 − 0)/(2aq − 0)`
	`= q/2`

`:.p/2 xx q/2`	`= −1`
`pq`	`= −4`

`⇒R\ \ text(has coordinates)\ \ (a(p+q), −4a)`

`:.\ text(Locus of)\ R\ text(is)\ \ y = −4a`

Calculus, EXT1 C1 2004 HSC 3c

A ferry wharf consists of a floating pontoon linked to a jetty by a 4 metre long walkway. Let `h` metres be the difference in height between the top of the pontoon and the top of the jetty and let `x` metres be the horizontal distance between the pontoon and the jetty.

Find an expression for `x` in terms of `h`. (1 mark)

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When the top of the pontoon is 1 metre lower than the top of the jetty, the tide is rising at a rate of 0.3 metres per hour.

At what rate is the pontoon moving away from the jetty? (3 marks)

--- 10 WORK AREA LINES (style=lined) ---

Show Answers Only

`x=sqrt(16 − h^2)`
`0.077\ text(m/hr)`

Show Worked Solution

i. `text(Using Pythagoras,)`

`x^2 + h^2`	`= 4^2`
`x^2`	`= 16 − h^2`
`x`	`= sqrt(16 − h^2)`

ii. `text(Find)\ \ (dx)/(dt)\ \ text(when)\ \ h = 1`

`(dx)/(dt)`	`= (dx)/(dh)·(dh)/(dt)`
`x`	`= (16 − h^2)^(1/2)`
`(dx)/(dh)`	`= 1/2 xx (16 − h^2)^(−1/2) xx d/(dh)(16 − h^2)`
	`= 1/2 (16 − h^2)^(−1/2) xx −2h`
	`= (−h)/(sqrt(16 − h^2))`

`text(When)\ \ h = 1, (dh)/(dt)= −0.3\ text(m/hr)`

`text{(}h\ \ text{decreases when the tide is rising)}`

`(dx)/(dt)`	`= (−h)/(sqrt(16 − h^2)) xx −0.3`
	`= (−1)/sqrt(16 − 1^2) xx −0.3`
	`= 0.3/sqrt15`
	`= 0.0774…`
	`= 0.077\ \ \ text{metres per hr (to 2 d.p.)}`

`:.\ text(When)\ \ h = 1,\ text(the pontoon is moving away)`

`text(at 0.077 metres per hr.)`

Functions, EXT1 F2 2004 HSC 3b

Let `P(x) = (x + 1) (x − 3)Q(x) + a(x + 1) + b`, where `Q(x)` is a polynomial and `a` and `b` are real numbers.

When `P(x)` is divided by `(x + 1)` the remainder is `−11`.

When `P(x)` is divided by `(x − 3)` the remainder is `1`.

What is the value of `b`? (1 mark)

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What is the remainder when `P(x)` is divided by `(x + 1)(x − 3)`? (2 marks)

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Show Answers Only

`−11`
`3x − 8`

Show Worked Solution

i. `P(x)= (x + 1) (x − 3)Q(x) + a(x + 1) + b`

`P(-1)=-11`

`-11=(−1 + 1)(−1 − 3)Q(x) + a(−1 + 1) + b`

`-11=(0)(-4)Q(x)+a(0)+b`

`:.b = −11`

(ii) `P(3) = 1`

`:.(3 + 1)(3 − 3)Q(x) + a(3 + 1) −11 = 1`

`4a`	`= 12`
`a`	`= 3`

`text(When)\ \ P(x)\ \ text(is divided by)\ \ (x + 1)(x − 3)`

`R(x)`	`= a(x + 1) + b`
	`= 3(x + 1) − 11`
	`= 3x + 3 − 11`
	`= 3x − 8`