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Trig Ratios, EXT1 2010 HSC 5a

A boat is sailing due north from a point  `A`  towards a point  `P`  on the shore line.

The shore line runs from west to east.

In the diagram,  `T`  represents a tree on a cliff vertically above  `P`, and  `L`  represents a landmark on the shore. The distance  `PL`  is 1 km.

From  `A`  the point  `L`  is on a bearing of 020°, and the angle of elevation to  `T`  is 3°.

After sailing for some time the boat reaches a point  `B`, from which the angle of elevation to  `T`  is 30°.
 

5a
 

  1. Show that 
     
    `qquad BP = (sqrt3 tan 3°)/(tan20°)`.   (3 marks) 
     
  2. Find the distance  `AB`. Give your answer to 1 decimal place.   (1 mark)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `2.5\ text(km)\ \ text{(to 1 d.p.)}`
Show Worked Solution

(i)   `text(Show)\ \ BP = (sqrt3 tan 3°)/(tan 20°)` 

`text(In)\ Delta ATP`
`tan 3°` `= (TP)/(AP)`
`=> AP` `= (TP)/(tan 3)`

 
`text(In)\ Delta APL`

`tan 20°` `= 1/(AP)`
`=> AP` `= 1/tan 20`

 

`:. (TP)/(tan3)` `= 1/(tan20)`
`TP` `= (tan3°)/(tan20°)\ \ \ text(…)\ text{(} text(1)text{)}`

 

`text(In)\ \ Delta BTP`

`tan 30°` `= (TP)/(BP)`
`1/sqrt3` `= (TP)/(BP)`
`BP` `= sqrt3 xx TP\ \ \ \ \ text{(using (1) above)}`
  `= (sqrt3 tan3°)/(tan20°)\ \ \ text(… as required)`

 

(ii)    `AB` `= AP\ – BP`
  `AP` `= 1/(tan20°)\ \ \ text{(} text(from part)\ text{(i)} text{)}`
`:.\ AB` `= 1/(tan 20°)\ – (sqrt3 tan3°)/(tan20°)`
  `= (1\ – sqrt3 tan 3)/(tan20°)`
  `= 2.4980…`
  `= 2.5\ text(km)\ text{(to 1 d.p.)`

Mechanics, EXT2* M1 2010 HSC 4a

A particle is moving in simple harmonic motion along the `x`-axis. 

Its velocity `v`, at `x`, is given by  `v^2 = 24 − 8x − 2x^2`. 

  1. Find all values of `x` for which the particle is at rest.   (1 mark)

  2. Find an expression for the acceleration of the particle, in terms of `x`.   (1 mark)

  3. Find the maximum speed of the particle.    (2 marks)

Show Answers Only
  1. `x = –6\ \ text(or)\ \ 2`
  2. `-4\ – 2x`
  3. `4 sqrt 2`
Show Worked Solution
(i)    `v^2 = 24\ – 8x\ – 2x^2`

`text(Find)\ \ x\ \ text(when)\ \ v=0`

`24 – 8x – 2x^2` `= 0`
`x^2 + 4x – 12` `= 0`
`(x + 6)(x – 2)` `= 0`
`x= -6\ \ text(or)\ \ 2`

 

`:.\ text(Particle at rest when)\ \ x = –6\ \ text(or)\ \ 2`

 

(ii)    `ddot x` `= d/(dx) (1/2 v^2)`
    `= d/(dx) (12 – 4x – x^2)`
    `= -4 – 2x`

 

(iii)   `text(Solution 1)`

`text(Max speed when)\ \ ddot x = 0`

`-4 – 2x` `= 0`
`2x` `= -4`
`x` `= -2`

`text(At)\ \ x = -2`

MARKER’S COMMENT: While most students found `x=-2` correctly, too many made errors substituting this back in or failed to finish the answer by taking the square root. BE CAREFUL! .
`v^2` `= 24 – 8(–2) – 2(–2)^2`
  `= 24 + 16 – 8`
  `= 32`
`v` `= +- sqrt32`
  `= +- 4 sqrt2`

`:.\ text(Maximum speed is)\ \ 4 sqrt2`

 

`text(Alternate Answer)`

`text(Maximum speed occurs when)`

`x = (-6+2)/2 = –2`

`text(At)\ \ x = –2`

`v^2` `= 32\ \ \ text{(see working above)}`
`v` `= +- 4 sqrt 2`

 

`:.\ text(Maximum speed is)\ 4sqrt2`

Inverse Functions, EXT1 2010 HSC 3b

Let  `f(x) = e^(-x^2)`.  The diagram shows the graph  `y = f(x)`.
 

 Inverse Functions, EXT1 2010 HSC 3b
 

  1. The graph has two points of inflection.  
  2. Find the  `x`  coordinates of these points.   (3 marks)
  3.  
  4. Explain why the domain of  `f(x)`  must be restricted if  `f(x)`  is to have an inverse function.    (1 mark)
  5. Find a formula for  `f^(-1) (x)`  if the domain of  `f(x)`  is restricted to  `x ≥ 0`.   (2 marks)
  6. State the domain of  `f^(-1) (x)`.    (1 mark)
  7. Sketch the curve  `y = f^(-1) (x)`.    (1 mark)
  8. (1)   Show that there is a solution to the equation  `x = e^(-x^2)`  between  `x = 0.6`  and  `x = 0.7`.   (1 mark)
  9. (2)   By halving the interval, find the solution correct to one decimal place.   (1 mark)

 

Show Answers Only
  1. `x = +- 1/sqrt2` 
  2.  
  3. `text(There can only be 1 value of)\ y`
  4. `text(for each value of)\ x.`
  5.  
  6. `f^(-1)x = sqrt(ln(1/x))`
     
  7. `0 <= x <= 1`
  8.  
  9. Inverse Functions, EXT1 2010 HSC 3b Answer
  10. (1)   `text(Proof)\ \ text{(See Worked Solutions)}`
  11. (2)   `0.7\ text{(1 d.p.)}`
Show Worked Solution
(i)    `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` 
COMMENT: It is also valid to show that `f(x)` is an even function and if a P.I. exists at `x=a`, there must be another P.I. at `x=–a`.
`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`

 

(ii)   `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).`

 

(iii)   `y = e^(-x^2)`

`text(Inverse function can be written)` 

`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))`
 

♦ Parts (iv) and (v) were poorly answered with mean marks of 39% and 49% respectively.
(iv)   `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`

 

(v) 

Inverse Functions, EXT1 2010 HSC 3b Answer

 

(vi)(1)  `x = e^(-x^2)`

`text(Let)\ g(x) = x\ – e^(-x^2)`

`g(0.6)` `=0.6\ – e^(-0.6^2)`
  `=0.6\ – 0.6977 < 0`
`g(0.7)` `=0.7\ – e^(-0.7^2)`
  `=0.7\ – 0.6126 > 0`
`=>g(x)\ text(changes sign)`

 

`:.\ g(x)\ \ text(has a root between  0.6  and  0.7)`

`:.\ x = e^(-x^2)\ \ text(has a solution between  0.6  and  0.7)`

♦ Mean mark 37%.
MARKER’S COMMENT: Better responses showed the change in sign between `g(0.65)` and `g(0.7)` as shown in the solution.

 

(vi)(2)   `g(0.65)` `=0.65\ – e^(-0.65^2)`
    `=0.65\ – 0.655 < 0`

 

`:.\ text(A solution lies between 0.65 and 0.7)`

`:.\ x = 0.7\ \ text{(1 d.p.)}`

Functions, EXT1 F2 2010 HSC 2c

Let  `P(x) = (x + 1)(x-3) Q(x) + ax + b`, 

where  `Q(x)`  is a polynomial and  `a`  and  `b`  are real numbers.

The polynomial  `P(x)`  has a factor of  `x-3`.

When  `P(x)`  is divided by  `x + 1`  the remainder is  `8`. 

  1. Find the values of  `a`  and  `b`.  (2 marks)

  2. Find the remainder when  `P(x)`  is divided by  `(x + 1)(x-3)`.     (1 mark)

Show Answers Only
  1. `a = -2,\ b = 6`
  2. ` -2x + 6`
Show Worked Solution

i.  `P(x) = (x+1)(x-3)Q(x) + ax + b`

`(x-3)\ \ text{is a factor   (given)}`

`:. P (3)` `= 0`
`3a + b` `= 0\ \ \ …\ text{(1)}`

 
`P(x) ÷ (x+1)=8\ \ \ text{(given)}`

`:.P(-1)` `= 8`
`-a + b` `= 8\ \ \ …\ text{(2)}`

 
`text{Subtract  (1) – (2)}`

`4a` `= -8`
`a` `= -2`

 
`text(Substitute)\ \ a = -2\ \ text{into (1)}`

`-6 + b` `= 0`
`b` `= 6`

 
`:. a= – 2, \ b=6` 
 

ii.  `P(x) -: (x + 1)(x-3)`

`= ((x+1)(x-3)Q(x)-2x + 6)/((x+1)(x-3))`

`= Q(x) + (-2x + 6)/((x+1)(x-3))`

 
`:.\ text(Remainder is)\ \ -2x + 6`

COMMENT: This question requires a fundamental understanding of the remainder theorem.

Calculus, EXT1 C2 2010 HSC 1e

Use the substitution  `u = 1 - x`  to evaluate  `int_0^1 x sqrt(1 - x)\ dx`.   (3 marks)

Show Answers Only

`4/15`

Show Worked Solution
`u = 1 – x` `\ \ \ \ \ => x = 1 – u`
`(du)/(dx) = -1` `\ \ \ \ \ => du = – dx`

 

`text(When)\ \ \ \ ` `x = 1,\ \ ` `u = 0`
  `x = 0,\ \ ` `u = 1`

 
`:. int_0^1 x sqrt(1 – x)\ dx`

`= – int_1^0 (1 – u) u^(1/2)\ du`

`= int_1^0 (u – 1) u^(1/2)\ du`

`= int_1^0 (u^(3/2) – u^(1/2))\ du`

`= [2/5 u^(5/2) – 2/3 u^(3/2)]_1^0`

`= [0 – (2/5 – 2/3)]`

`= – (6/15 – 10/15)`

`= 4/15`

Functions, EXT1 F1 2010 HSC 1d

Solve  `3/(x+2) < 4`.   (3 marks)

Show Answers Only

 `x < -2\ \ text(or)\ \ x > -5/4`

Show Worked Solution

`text(Solution 1)`

`3/(x + 2) < 4`

`text(Multiply b.s. by)\ \ (x + 2)^2`

`3(x + 2)` `< 4(x + 2)^2`
`3x + 6` `< 4 (x^2 + 4x + 4)`
`3x + 6` `< 4x^2 + 16x + 16`
`4x^2 + 13x + 10` `> 0`
`(4x + 5)(x + 2)` `> 0`

 
`text(LHS)\ = 0\ \ text(when)\ \ x = -5/4\ \ text(or)\ \ -2`

Algebra, EXT1 2010 HSC 1d Answer

`text(From graph)`

`x < -2\ \ text(or)\ \  x > -5/4`

 
`text(Alternate Solution)`

`text(If)\ \ x + 2 > 0\ \ \ \ text{(i.e.}\ \ x > –2 text{)}`

`3` `< 4(x + 2)`
`3` `< 4x + 8`
`4x` `> -5`
`x` `> -5/4`

 
`text(If)\ \ x + 2 < 0\ \ \ \ text{(i.e.}\ x < –2 text{)}`

`3` `> 4 (x + 2)`
`3` `> 4x + 8`
`4x` `< -5`
`x` `< -5/4`
`:. x` `< –2\ \ \ \ text{(satisfies both)}`

 
`:.\ x < –2\ \ text(or)\ \ x > –5/4`

Plane Geometry, EXT1 2011 HSC 4b

In the diagram, the vertices of  `Delta ABC`  lie on the circle with centre  `O`. The point  `D`  lies on  `BC`  such that  `Delta ABD`  is isosceles and  `/_ABC = x`.

Copy or trace the diagram into your writing booklet. 

  1. Explain why  `/_AOC = 2x`.    (1 mark)
  2. Prove that  `ACDO`  is a cyclic quadrilateral.    (2 marks)
  3. Let  `M`  be the midpoint of  `AC`  and  `P`  the centre of the circle through `A, C, D`  and  `O`. 
  4. Show that  `P, M`  and  `O`  are collinear.   (1 mark)
Show Answers Only
  1. `text(Angle at the centre of a circle is twice)`
  2. `text(angle of circumference on same arc)`
  3.  
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)

`/_AOC = 2x`

`text{(angles at circumference and}`

`text{centre on arc}\ AC text{)}`

 

♦ Mean mark 38%
STRATEGY: Proving part (ii) by showing opposite angles are supplementary also worked but was more time consuming.

(ii)  `text(Prove)\ ACDO\ text(is cyclic)`

`text(S)text(ince)\ Delta ADB\ text(is isosceles)`

`/_DAB = x\ \ \ text{(opposite equal sides in}\ Delta DBA text{)}`

`=> /_ADB` `= 180\ – 2x\ \ \ text{(angle sum of}\ Delta DAB text{)}`
`=> /_CDA` `= 2x\ \ \ text{(}CDB\ text{is a straight angle)}`

 

`text(S)text(ince chord)\ AC\ text(subtends)\ /_CDA = 2x`

`text(and)\ /_COA = 2x,`

`:.\ text(Quadrilateral)\ ACDO\ text(must be cyclic.)`

 

(iii)

 `text(Need to show)\ OM\ text(passes through)\ P,\ text(centre)`

♦♦♦ Mean mark 7%. 2nd hardest question in the 2011 paper!

`text(of circle through)\ ACDO.`

`AM` `= CM\ text{(} M\ text(is midpoint) text{)}`
`OC` `= OA\ text{(radii)}`

`OM\ text(is common)`

`:.\ Delta OAM -= Delta OCM\ \ \ text{(SSS)}`

`:. /_CMO = /_AMO\ \ \ ` `text{(corresponding angles of}`
  `\ \ text{congruent triangles)}`

 

`text(S)text(ince)\ ∠AMC\ text(is straight angle)`

`/_CMO = /_AMO = 90°`

`:.OM\ text(is perpendicular bisector)`

`text(of chord)\ AC.`

`:. OM\ text(passes through)\ P.`

`:.\ P, M,\ text(and)\ O\ text(are collinear.)`

Geometry and Calculus, EXT1 2011 HSC 4a

Consider the function  `f(x) = e^(-x)\ - 2e^(-2x)`. 

  1. Find  `f prime (x)`.   (1 mark)
  2. The graph  `y = f(x)`  has one maximum turning point. 
  3. Find the coordinates of the maximum turning point.   (2 marks)
  4.  
  5. Evaluate  `f(ln2)`.   (1 mark)
  6. Describe the behaviour of  `f(x)`  as  `x -> oo`.   (1 mark)
  7. Find the  `y`-intercept of the graph  `y = f(x)`.   (1 mark)
  8. Sketch the graph  `y = f(x)`  showing the features from parts  (ii) - (v).
  9. You are not required to find any points of inflection.    (2 marks)
  10.  
  11.  
  12.  
  13.  
  14.  
  15.  
  16.  
Show Answers Only
  1. `-e^(-x) + 4e^(-2x)`
  2. `text(Max T.P. at)\ (ln4, 1/8)`
  3. `0`
  4. `text(As)\ x -> oo, f(x) -> 0`
  5. `-1`
  7.  
  8.  
  9.  
  10.  
  11.  
Show Worked Solution
(i)    `f(x)` `= e^(-x)\ – 2e^(-2x)`
  `f prime (x)` `= -e^(-x) + 4e^(-2x)`

 

(ii)   `text(T.P. when)\ f prime (x) = 0`

`-e^(-x) + 4e^(-2x) = 0`

`text(Let)\ e^(-x) = X`

`- X + 4X^2` `= 0`
`X (4X\ – 1)` `= 0`

`X = 0\ \ text(or)\ \ 1/4`

`text(If)\ \ e^(-x) = 0,\ \ \ text(No solution)`

`text(If)\ \ e^(-x) = 1/4`

`ln e^(-x)` `= ln (1/4)`
`-x` `= ln 4^(-1)`
  `= -ln 4`
`x` `= ln4`

 

ALGEBRA TIP: Note that `e^ln x=x` can often help calculations. This is easily proven by simply taking the `ln` of both sides of the equation.
`f″(x)` `= e^(-x)\ – 8e^(-2x)`
`f″(ln4)` `< 0 => text(MAX)`
`f (ln4)` `= e^(-ln4)\ – 2e^(-2ln4)`
  `= e^(ln(1/4))\ – 2 e^(ln(1/16))`
  `= 1/4\ – 2(1/16)`
  `= 1/8`

`:.\ text(Maximum T.P. at)\ \ (ln4, 1/8)`

 

(iii)   `f(ln2)` `= e^(-ln2)\ – 2e^(-2ln2)`
    `= e^(ln(1/2))\ – 2e^(ln(1/4))`
    `= 1/2\ – 2 xx 1/4`
    `= 0`

 

(iv)   `f(x)` `= e^(-x)\ – 2e^(-2x)`
    `= e^(-x) (1\ – 2e^(-x))`
  `text(As)\ x -> oo, \ e^(-x) ->0,\ f(x) -> 0`

 

(v)    `y text(-intercept at)\ \ f(0)`
`f(0)` `= e^0\ – 2e^0`
  `= -1`

 

(vi)

EXT1 2011 4a

Quadratic, EXT1 2011 HSC 3b

The diagram shows two distinct points  `P(t, t^2)`  and  `Q(1\ - t, (1\ - t)^2)`  on the parabola  `y = x^2`.  The point  `R`  is the intersection of the tangents to the parabola at  `P`  and  `Q`. 

 

  1. Show that the equation of the tangent to the parabola at  `P`  is  `y = 2tx\ – t^2`.   (2 marks)
  2. Using part  `text{(i)}`, write down the equation of the tangent to the parabola at  `Q`.     (1 mark)
  3. Show that the tangents at  `P`  and  `Q`  intersect at
    `R (1/2, t\ - t^2)`.   (2 marks)
  5. Describe the locus of  `R`  as  `t`  varies, stating any restriction on the  `y`-coordinate.   (2 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `y = 2 (1 -t)x\ – (1\ – t)^2`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(Locus of)\ R\ text(is vertical line)`
    `x = 1/2,\ y<1/4`
Show Worked Solution
(i)

`text(Show tangent at)\ P\ text(is)\ y = 2tx\ – t^2`

`y` `= x^2`
`dy/dx` `= 2x`

`x=t\ \ \ \ text(at)\ P`

`dy/dx = 2t`

`text(Find equation with)\ \ m = 2t\ \ text(through)\ \ P(t, t^2)`

`y\ – y_1` `= m(x\ – x_1)`
`y\ – t^2` `= 2t ( x\ – t)`
`y` `= 2tx\ – 2t^2 + t^2`
  `= 2tx\ – t^2\ text(… as required)`

 

(ii)  `text(T)text(angent at)\ Q\ text(has equation)`

MARKER’S COMMENT: Many students derived this equation rather than substituting the new parameter, costing them valuable time. This is a benefit of using the parametric approach.

`y = 2(1\ – t)x\ – (1\ – t)^2`

 

(iii)  `text(Need to show)\ R(1/2,\ t\ – t^2)`

`R\ text(is at intersection of tangents)`

`2tx\ – t^2` `= 2(1\ – t)x\ – (1\ – t)^2`
`2tx\ – t^2` `= 2x\ – 2tx\ – 1 + 2t\ – t^2`
`4tx\ – 2x` `= -1 + 2t\ – t^2 + t^2`
`2x(2t\ – 1)` `= 2t\ – 1`
`2x` `= 1`
`x` `= 1/2`

`text(Using)\ \ y = 2tx – t^2\ \ text(when)\ \ x = 1/2`

`y` `= 2t(1/2)\ – t^2`
  `= t\ – t^2`

`:.\ R(1/2, t\ – t^2)\ text(… as required)`

 

(iv)  `text(Locus of)\ R`

♦♦ Mean mark of 22%. 
MARKER’S COMMENT: Many students stated the locus as `y=t-t^2` rather than realising it had to be a straight line since `x=½`, and that `y=t-t^2` simply restricted the values of `y`.

`text(S)text(ince)\ x = 1/2\ text(is a constant)`

`R\ text(is a vertical line)`

`text(Now,)\ y = t\ – t^2 = t(1\ – t)`

`text(Graphically,)\ \ y\ \ text(has a maximum at)\ \ t = 1/2`

`text(Max)\ \ y = 1/2\ – (1/2)^2 = 1/4`

`=> y < 1/4\ \ text{(tangents can’t meet on parabola)}`

`:.\ text(Locus of)\ R\ text(is vertical line)\ x = 1/2,\ \ y<1/4`

Functions, EXT1 F2 2011 HSC 2a

Let  `P(x) = x^3-ax^2 + x`  be a polynomial, where  `a`  is a real number.

When  `P(x)`  is divided by  `x-3`  the remainder is  `12`.

Find the remainder when  `P(x)`  is divided by  `x + 1`.    (3 marks)

Show Answers Only

`-4`

Show Worked Solution

`P(x) = x^3 – ax^2 + x`

`text(S)text(ince)\ \ P(x) -: (x – 3)\ \ text(has remainder 12,)`

`P(3) = 3^3-a xx 3^2 + 3` `=12`
`27-9a + 3` `= 12`
`9a` `= 18`
`a` `=2`

 
`:.\ P(x) = x^3-2x^2 + x`

 

`text(Remainder)\ \ P(x) -: (x + 1)\ \ text(is)\ \ P(–1)`

`P(-1)` `= (-1)^3-2(-1)^2-1`
  `= – 4`

Calculus, EXT1 C2 2011 HSC 1d

Using the substitution  `u = sqrtx`, evaluate  `int_1^4 (e^(sqrtx))/(sqrtx)\ dx`.   (3 marks) 

Show Answers Only

`2e (e – 1)`

Show Worked Solution
`u` `= sqrtx = x^(1/2)`
`(du)/(dx)` `= 1/2 x^(-1/2) = 1/(2 sqrtx)`
 `du` `= (dx)/(2sqrtx)`
`:.2du` `= (dx)/(sqrtx)`

 

`text(When)\ \ \ ` `x=4,\ \ ` `u = 2`
  `x = 1,` `\ x = 1`

 

`:. int_1^4 (e^(sqrtx))/(sqrtx)\ dx`
`= int_1^2 e^u xx 2\ du`
`= 2 [e^u]_1^2`
`= 2 [e^2 – e^1]`
`= 2e (e – 1)`

Trig Calculus, EXT1 2011 HSC 1b

Differentiate   `(sin^2 x)/x`  with respect to  `x`.   (2 marks)

Show Answers Only

 `(sinx (2x cosx\ – sinx))/(x^2)`

Show Worked Solution

`y = (sin^2 x)/x`

`u`  `= sin^2 x,\ \ \ \ \ ` `u prime = 2 sinx cosx`
`v`  `= x,`  `v prime = 1`

 

`dy/dx` `= (u prime v\ – uv prime)/(v^2)`
  `= (2sinxcosx * x – sin^2x * 1)/(x^2)`
  `= (sinx (2x cosx\ – sinx))/(x^2)`

Linear Functions, EXT1 2011 HSC 1a

The point  `P`  divides the interval joining  `A(–1, –2)`  to  `B(9, 3)`  internally in the ratio  `4 : 1`. Find the coordinates of  `P`.   (2 marks)

Show Answers Only

 `(7, 2)`

Show Worked Solution

`A(–1,–2),\ \ \ B(9,3)`

`text(Internal division)\ \ 4:1\ \ text(at)\ \ P`

`P` `≡ ( (nx_1 + mx_2)/(m + n),\ (ny_1 + my_2)/(n + m) )`
  `≡ ( ((1 xx –1) + (4 xx 9))/(4 + 1),\ ((1 xx –2) + (4 xx 3))/(4 + 1))`
  `≡ (35/5, 10/5)`
  `≡ (7,2)`

Mechanics, EXT2* M1 2012 HSC 13c

A particle is moving in a straight line according to the equation

`x = 5 + 6 cos 2t + 8 sin 2t`, 

where `x` is the displacement in metres and `t` is the time in seconds.

  1. Prove that the particle is moving in simple harmonic motion by showing that `x`  satisfies an equation of the form  `ddot x = -n^2 (x\ - c)`.  (2 marks)

  2. When is the displacement of the particle zero for the first time?    (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1.5\ text(seconds)\ text{(1 d.p.)}`
Show Worked Solution

i.  `text(Prove)\ ddot x = -n^2(x\ – c)`

`x` `= 5 + 6 cos 2t + 8 sin 2t`
`dot x` `= -12 sin 2t + 16 cos 2t`
`ddot x` `= – 24 cos 2t\ – 32 sin 2t`
  `= -4 (6 cos 2t + 8 sin 2t)`
  `= -2^2 (5 + 6 cos 2t + 8 sin 2t\ – 5)`
  `= -2^2 (x\ – 5)\ \ \ text(… as required)`

 

ii.  `text(Find)\ \ t\ \ text(when)\ \ x=0\ \ text(for 1st time:)`

♦ Mean mark 42%.
IMPORTANT: The critical insight required to solve `x=0` is to realise that the cosine of the difference between 2 angles, i.e. `cos (2t- theta)`, applies.
`5 + 6 cos 2t + 8 sin 2t` `= 0`
`6 cos 2t + 8 sin 2t` `= -5`
`6/10 cos 2t+ 8/10 sin 2t` `=-1/2`

 

 Calculus in the Physical World, EXT1 2012 HSC 13c Answer

`=>cos theta=6/10\ \ text(and)\ \ sin theta=8/10`
`cos 2t cos theta+sin 2t sin theta` `=- 1/2`
`cos(2t\ – theta)` `= – 1/2`

`text(S)text(ince)\ \ cos\ pi/3 = 1/2\ \ text(and)\ cos\ text(is negative)`

`text(in the 2nd and 3rd quadrants,)`

`=>2t\ – theta = pi\ – pi/3,\ pi + pi/3`

 

`text(We need the 1st time)\ \ x = 0`

`text(S)text(ince)\ \ cos theta` `= 6/10`
`theta` `= 0.9273…`

 

`:.\ 2t\ – 0.9273…` `= (2pi)/3`
`2t` `= (2pi)/3 + 0.9273…`
`t` `= 1/2 ((2pi)/3 + 0.9273…)`
  `= 1.5108…`
  `= 1.5\ text(seconds)\ text{(1 d.p.)}`

Geometry and Calculus, EXT1 2012 HSC 13b

  1. Find the horizontal asymptote of the graph
     
    `qquad qquad y=(2x^2)/(x^2 + 9)`.   (1 mark)
  2. Without the use of calculus, sketch the graph 
     
    `qquad qquad y=(2x^2)/(x^2 + 9)`,
     
    showing the asymptote found in part (i).    (2 marks)
Show Answers Only
  1. `text(Horizontal asymptote at)\ y = 2`
    1. Geometry and Calculus, EXT1 2012 HSC 13b Answer
Show Worked Solution
(i)    `y` `= (2x^2)/(x^2 +9)`
    `= 2/(1 + 9/(x^2))`

 

`text(As)\ \ x -> oo,\ y ->2`

`text(As)\ \ x -> – oo,\ y -> 2`

`:.\ text(Horizontal asymptote at)\ y = 2`

 

(ii)    `text(At)\ \ x = 0,\ y = 0`

`f(x) = (2x^2)/(x^2 + 9) >= 0\ text(for all)\ x`

`f(–x) = (2(–x)^2)/((–x)^2 + 9) = (2x^2)/(x^2 + 9) = f(x)`

`text(S)text(ince)\ \ f(x) = f(–x) \ \ =>\ text(EVEN function)`

Geometry and Calculus, EXT1 2012 HSC 13b Answer

Functions, EXT1 F1 2012 HSC 12b

Let  `f(x) = sqrt(4x-3)` 

  1.  Find the domain of  `f(x)`.   (1 mark)

  2.  Find an expression for the inverse function  `f^(-1) (x)`.    (2 marks)

  3.  Find the points where the graphs  `y = f(x)`  and  `y=x`  intersect.   (1 mark)

  4.  On the same set of axes, sketch the graphs  `y = f(x)`  and  `y = f^(-1) (x)`  showing the information found in part (iii).   (2 marks)

Show Answers Only
  1. `x >= 3/4`
  2. `f^(-1) (x) = 1/4 x^2 + 3/4,\ x >= 0`
  3. `(1,1),\ (3,3)`
  4.   
    1. Inverse Functions, EXT1 2012 HSC 12b Answer
Show Worked Solution

i.  `f(x) = sqrt(4x-3)`

`text(Domain exists when)\ \ 4x-3` `>= 0`
`4x` `>= 3`
`x` `>= 3/4`

 

ii.  `text(Inverse function when)`

`x` `= sqrt(4y-3)`
`x^2` `= 4y-3`
`4y` `= x^2 + 3`
`y` `= 1/4 x^2 + 3/4`

 

`text(S)text(ince range of)\ \ f(x)\ \ text(is)\ \ y >= 0`

`=> text(Domain of)\ \ f^(-1) (x)\ \ text(is)\ \ x >= 0`

`:.\ f^(-1) (x) = 1/4 x^2 + 3/4,\ \ \ \ x >= 0`

 

iii.  `text(Find intersection)`

`y` ` = sqrt(4x-3)\ \ …\ text{(1)}`
`y ` `=x\ \ …\ text{(2)}`

 

`text{Intersection occurs when}`

`x` `= sqrt(4x-3)`
`x^2` `= 4x-3`
`x^2-4x + 3` `= 0`
`(x-3)(x-1)` `= 0`
`x` `=1\ \ text(or 3)`

 

`:.\ text(Intersection at)\ \ (1,1)\ \ text(and)\ \ (3,3)`

 

iv.

`qquad`Inverse Functions, EXT1 2012 HSC 12b Answer

Calculus, EXT1 C2 2012 HSC 11d

Use the substitution  `u = 2 - x`  to evaluate  `int_1^2 x (2 - x)^5\ dx`.   (3 marks)  

Show Answers Only

 `4/21`

Show Worked Solution
`u` `=2-x`
`:. x` `= 2 – u`
`(du)/dx` `= -1`
`:.dx` `=-du`

 

`text(When)\ \ x = 2,` `\ \ u = 0`
`text(When)\ \ x = 1,` `\ \ u = 1`

 

`:. int_1^2 x (2 – x)^5\ dx` `= int_1^0 – (2 – u) u^5\ du`
  `= int_1^0 u^6 – 2u^5\ du`
  `= [1/7 u^7 – 2/6 u^6]_1^0`
  `= [0 – (1/7 – 1/3)]`
  `= – (3/21 – 7/21)`
  `= 4/21`

Functions, EXT1 F1 2012 HSC 11c

Solve  `x/(x - 3) < 2`.  (3 marks) 

Show Answers Only

 `x<3\ \ text(or)\ \ x>6`

Show Worked Solution

`text(Solution 1)`

`x/(x – 3) < 2`

`text(When)\ \ x – 3 > 0,\ \ \ (text(i.e.)\ \ x > 3)`

`x` `< 2 ( x – 3)`
`x` `< 2x\ – 6`
`=>x` `>6\ \ \ text{(satisfies both)}`

 
`text(When)\ \ x\ – 3 < 0,\ \ \ (text(i.e.)\ \ x<3)`

`x` `>2 (x – 3)`
`x` `> 2x – 6`
`x` `<6`
`=> x` `<3\ \ \ text{(satisfies both)}`

 
`:.\ text(Equation is correct for)\ x<3\ text(or)\ x>6`

 
`text(Solution 2)`

`x/((x\ – 3)) < 2`
 

`text(Multiply b.s. by)\ \  (x-3)^2`

`x(x – 3)` `< 2(x^2 – 6x + 9)`
`x^2 – 3x` `< 2x^2 – 12x + 18`
`x^2 – 9x + 18` `>0`
`(x – 3)(x – 6)` `>0`

 

 Algebra, EXT1 2012 HSC 11c Answer

`:.\ text(Equation correct for)\ x<3\ \ text(or)\ \ x>6`

Calculus, EXT1 C2 2012 HSC 9 MC

What is the derivative of  `cos^(–1) (3x)`?

  1. `1/(3 sqrt(1 - 9x^2))`  
  2. `(-1)/(3 sqrt(1 - 9x^2))`  
  3. `3/sqrt(1 - 9x^2)`  
  4. `(-3)/sqrt(1 - 9x^2)`  
Show Answers Only

`D`

Show Worked Solution

`y = cos^(-1) (3x)`

`dy/dx` `= (-1)/sqrt(1\ – (3x)^2) xx d/dx (3x)`
  `= (-3)/sqrt(1\ – 9x^2)`

`=>  D`

Functions, EXT1 F2 2012 HSC 3 MC

A polynomial equation has roots  `alpha`,  `beta`,  `gamma`  where

`alpha + beta + gamma = -2`,    `alphabeta + alphagamma + betagamma = 3`,    `alphabetagamma = 1`.

Which polynomial equation has the roots  `alpha`,  `beta`, and  `gamma`?

  1. `x^3 + 2x^2 + 3x + 1 = 0`
  2. `x^3 + 2x^2 + 3x- 1 = 0`
  3. `x^3- 2x^2 + 3x + 1 = 0`
  4. `x^3- 2x^2 + 3x- 1 = 0`
Show Answers Only

`B`

Show Worked Solution

`text(Using)\ \ ax^3 + bx^2 + cx + d = 0`

`text(By elimination)`

`alpha beta gamma = -d/a  = 1`

`:.\ text(Cannot be)\ A\ text(or)\ C\ \ (text(where)\ alphabetagamma = -1 text{)}`

`alpha + beta + gamma = -b/a = -2`

`:.\ text(Cannot be)\ D\ \ (text(where)\ alpha + beta + gamma = 2 text{)}`

`=> B`

Algebra, EXT1 2012 HSC 1 MC

Which expression is a correct factorisation of  `x^3 - 27`? 

  1. ` (x - 3)( x^2 - 3x + 9)`
  2. `(x - 3)( x^2 - 6x + 9)`
  3. `(x - 3)( x^2 + 3x + 9)`
  4. `(x - 3)( x^2 + 6x + 9)`
Show Answers Only

`C`

Show Worked Solution
`x^3 – 27` `= x^3 – 3^3`
  `= (x – 3)(x^2 + 3x + 9)`
`=> C`  

Functions, EXT1 F1 2011 HSC 1c

Solve  `(4-x)/x <1`.  (3 marks)

Show Answers Only

 `x<0\ \ text(or)\ \ x>2`

Show Worked Solution

`text(Solution 1)`

`(4-x)/x < 1`

`text(If)\ x<0,\ \ \ \ \ 4-x` `> x`
`2x` `< 4`
`x` `<2`

`=> x<0\ \ \ text{(satisfies both)}`
 

`text(If)\ x>0,\ \ \ \ \ 4-x` `<x`
`2x` `>4`
`x` `>2`

`=> x>2\ \ \ text{(satisfies both)}`

`:.\ x < 0\ \ text(or)\ \ x > 2`

 
`text(Solution 2)`

`text(Multiply both sides by)\ \ x^2`

`x(4-x)` `< x^2`
`4x-x^2` `< x^2`
`2x^2-4x` `>0`
`2x(x-2)` `>0`

 

 EXT1 2011 1c

`text(From graph)`

`x<0\ \ text(or)\ \ x >2`

Probability, 2ADV S1 2009 HSC 5b

On each working day James parks his car in a parking station which has three levels. He parks his car on a randomly chosen level. He always forgets where he has parked, so when he leaves work he chooses a level at random and searches for his car. If his car is not on that level, he chooses a different level and continues in this way until he finds his car.    

  1. What is the probability that his car is on the first level he searches?     (1 mark)

  2. What is the probability that he must search all three levels before he finds his car?   (1 mark)

  3. What is the probability that on every one of the five working days in a week, his car is not on the first level he searches?    (1 mark)

Show Answers Only
  1. `1/3`
  2. `1/3`
  3. `32/243`
Show Worked Solution

i.  `P text{(1st chosen)} = 1/3`
 

ii.  `P text{(search 3 levels)}`

♦♦ Mean marks of 31% and 39% for part (ii) and (iii) respectively.

`= P text{(not 1st)} xx P text{(not 2nd)}`

`= 2/3 xx 1/2`

`= 1/3`
 

iii.  `P text{(not 1st for 5 days)}`

`= 2/3 xx 2/3 xx 2/3 xx 2/3 xx 2/3`

`= 32/243`

Plane Geometry, 2UA 2009 HSC 4c

In the diagram,  `Delta ABC`  is a right-angled triangle, with the right angle at  `C`. The midpoint of  `AB`  is  `M`, and  `MP _|_ AC`.

Plane Geometry, 2UA 2009 HSC 4c_1

Copy or trace the diagram into your writing booklet. 

  1. Prove that  `Delta AMP`  is similar to  `Delta ABC`.   (2 marks)
  2. What is the ratio of  `AP`  to  `AC`?    (1 mark)
  3. Prove that  `Delta AMC`  is isosceles.   (2 marks)
  4. Show that  `Delta ABC`  can be divided into two isosceles triangles.   (1 mark)
  5. Plane Geometry, 2UA 2009 HSC 4c_2
  6. Copy or trace this triangle into your writing booklet and show how to divide it into four isosceles triangles.   (1 mark)
  7.  
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Ratio)\ AP:AC = 1:2`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. Plane Geometry, 2UA 2009 HSC 4c_2 Answer1
Show Worked Solution
(i)    `text(Need to prove)\ Delta AMP\  text(|||)\  Delta ABC`
  `/_ PAM\ text(is common)`
  `/_ MPA = /_BCA = 90°\ \ \ text{(given)}`
  `:.\ Delta AMP \ text(|||) \ Delta ABC\ \ \ text{(equiangular)}`

 

(ii)   `(AP)/(AC) = (AM)/(AB)\ \ \ ` `text{(corresponding sides of}`
    `text{similar triangles)}`

 

`text(S)text(ince)\ \ AB = 2 xx AM`
`(AP)/(AC) = 1/2`
`:.\ text(Ratio)\ AP:AC = 1:2`

 

(iii)

Plane Geometry, 2UA 2009 HSC 4c_1 Answer
`AP = PC \ \ \ text{(from part (ii))}`
`PM\ text(is common)`
`/_APM = /_CPM = 90°\ \ text{(∠}\ APC\ text{is a straight angle)}`
`:.\ Delta AMP ~= Delta CMP\ text{(SAS)}`
`=> AM = CM\ \ ` `text{(corresponding sides of}`
  `text{congruent triangles)}`

`:.\ Delta AMC\ text(is isosceles.)`

 

(iv)   `text(S)text(ince)\ \ AM` `= MC\ \ \ text{(part (iii))}`
  `AM`  `= MB\ text{(given)}`
  `:. MC`  `= MB`
`:. Delta MCB\ text(is isosceles)`
`:. Delta ABC\ text(can be divided into 2 isosceles)`
`text(triangles)\ (Delta AMC\ text(and)\ Delta MCB text{)}`

 

(v)

Plane Geometry, 2UA 2009 HSC 4c_2 Answer1
♦♦ Mean mark 25%.
MARKER’S COMMENT: Use a ruler, draw large diagrams and always pay careful attention to previous parts of the question!
`/_AFB = /_CFB = 90°`
`DF\ text(bisects)\ AB`
`EF\ text(bisects)\ BC`
`text(From part)\ text{(iv)}\ text(we get 4 isosceles)`
`text(triangles as shown.)`

Functions, 2ADV F1 2009 HSC 3b

2009 3b
 

The circle in the diagram has centre  `N`. The line  `LM`  is tangent to the circle at  `P`. 

  1. Find the equation of  `LM`  in the form  `ax + by + c = 0`.   (2 marks)
  2. Find the distance  `NP`.    (2 marks)
  3. Find the equation of the circle.    (1 mark)
Show Answers Only
  1. `4x – 3y – 5 = 0`
  2. `2\ text(units)`
  3. `text(Equation is)\ \ (x – 1)^2 + (y – 3)^2 = 4`
Show Worked Solution

(i)  `text(Need to find the gradient of)\ LM`

`m_(LM)` `= (y_2\ – y_1)/(x_2\ – x_1)`
  `= (5\ – 1)/(5\ – 2)`
  `= 4/3`

 
`text(Equation of)\ LM\ text(has)\ m = 4/3\ text(through)\ (2,1)`

`y\ – y_1` `= m (x\ – x_1)`
`y\ – 1` `= 4/3 (x\ – 2)`
`3y\ – 3` `= 4x\ – 8`
`4x\ – 3y\ – 5` `= 0`

 

(ii)  `NP\ text(is)  _|_ text(distance of)\ \ N(1,3)\ \ text(from)\ \ LM`

`_|_\ text(dist)` `= |(ax_1 + by_1 + c)/sqrt(a^2 + b^2)|`
  `= |(4(1)\ – 3(3)\ – 5)/sqrt(4^2 + (-3)^2)|`
  `= |-10/5|`
  `= 2\ text(units)`

 

(iii)  `text{The circle has centre (1,3) and radius 2 units:`

`:.\ text(Equation is)\ \ (x\ – 1)^2 + (y\ – 3)^2 = 4`

Calculus, 2ADV C4 2009 HSC 2b

  1. Find  `int 5\ dx`.   (1 mark)

  2. Find  `int 3/((x - 6)^2)\ dx`.    (2 marks)

  3. Evaluate  `int_1^4 x^2 + sqrtx\ dx`.   (3 marks)

Show Answers Only
  1. `5x + C`
  2. `(-3)/((x – 6)) + C`
  3. `77/3`
Show Worked Solution

i.  `int 5\ dx= 5x + C`

 

ii.  `int 3/((x – 6)^2)\ dx`

`= 3 int (x – 6)^(-2)\ dx`

`= 3 xx 1/(-1) xx (x – 6)^(-1) + c`

`= (-3)/((x – 6)) + c`

 

iii.  `int_1^4 x^2 + sqrtx\ \ dx`

`= int_1^4 (x^2 + x^(1/2))\ dx`

`= [1/3 x^3 + 1/(3/2) x^(3/2)]_1^4`

`= [(x^3)/3 + 2/3x^(3/2)]_1^4`

`= [((4^3)/3 + 2/3 xx 4^(3/2))\ – (1/3 + 2/3)]`

`= [(64/3 + 16/3) – 3/3]`

`= [80/3 – 3/3]`

`= 77/3`

Functions, 2ADV F1 2009 HSC 1a

Sketch the graph of  `y-2x = 3`, showing the intercepts on both axes.   (2 marks)

Show Answers Only

 

Show Worked Solution

`y-2x=3\ \ =>\ \ y=2x+3`

`ytext{-intercept}\ = 3`

`text{Find}\ x\ text{when}\ y=0:`

`0-2x=3\ \ =>\ \ x=-3/2`
 

Calculus, 2ADV C1 2010 HSC 7b

The parabola shown in the diagram is the graph  `y = x^2`. The points  `A (–1,1)`  and  `B (2, 4)`  are on the parabola.
 

 
 

  1.  Find the equation of the tangent to the parabola at `A`.   (2 marks)

  2.  Let `M` be the midpoint of  `AB`.

     

    There is a point `C` on the parabola such that the tangent at `C` is parallel to  `AB`.

     

    Show that the line  `MC`  is vertical.   (2 marks)  

  3. The tangent at `A` meets the line `MC` at `T`.

     

    Show that the line `BT` is a tangent to the parabola.  (2 marks)

Show Answers Only
  1. `2x + y + 1 = 0`
  2. `text(Proof)  text{(See Worked Solutions)}`
  3. `text(Proof)  text{(See Worked Solutions)}`
Show Worked Solution
i.
`y` `=x^2`
`dy/dx` `= 2x` 

 
`text(At)\ \ A text{(–1,1)}\ => dy/dx = -2`
 

`text(T)text(angent has)\ \ m=text(–2),\ text(through)\ text{(–1,1):}`

`y – y_1` `= m(x\ – x_1)`
`y – 1` `= -2 (x + 1)`
`y – 1` `= -2x -2`
`2x + y + 1` `= 0`

 

 `:.\ text(T)text(angent at)\ A\ text(is)\ \ 2x + y + 1 = 0`

 

♦ Mean mark 37%.
IMPORTANT: The critical understanding required for this question is that the gradient of `AB` needs to be equated to the gradient function (i.e. `dy/dx`).

ii.   `Atext{(–1,1)}\ \ \ B(2,4)`

`M` `= ((-1+2)/2 , (1+4)/2)`
  `= (1/2, 5/2)`

 

`m_(AB)` `= (y_2 – y_1)/(x_2 – x_1)`
  `= (4 – 1)/(2 + 1)=1`

 

`text(When)\ \ dy/dx`  `= 1`
`2x` `= 1`
`x` `= 1/2`

 
`:.\ C \ (1/2, 1/4)`
 
`=>M\ text(and)\ C\ text(both have)\ x text(-value)=1/2`

`:. MC\ text(is vertical  … as required)`

 

iii.   `T\ text(is point on tangent when) \ x=1/2` 

♦♦ Mean mark 29%.

`text(T)text(angent)\ \ \ 2x + y + 1 = 0`

`text(At)\ x = 1/2`

`2 xx (1/2) + y + 1=0`

`=> y=–2`

`:.\ T (1/2, –2)`

 
`text (Given)\ \ B (2, 4)`

`m_(BT)` `= (4+2)/(2\ – 1/2)`
  `=4`

 
`text(At)\ \ B(2,4),\ text(find gradient of tangent:)`

`dy/dx = 2x=2 xx2=4`

`:.m_text(tangent) = 4=m_(BT)`

`:.BT\ text(is a tangent)`

Linear Functions, 2UA 2010 HSC 3a

In the diagram  `A`,  `B`  and  `C`  are the points  `( –2, –4 ),\ (12,6)`  and  `(6,8)`  respectively.

The point  `N (2,2)`  is the midpoint of   `AC`. The point  `M`  is the midpoint of  `AB`.
 

2010 3a
 

  1. Find the coordinates of  `M`.   (1 mark)
  2. Find the gradient of  `BC`.   (1 mark)
  3. Prove that  `Delta ABC`  is similar to  `Delta AMN`.   (2 marks)
  4. Find the equation of  `MN`.   (2 marks)
  5. Find the exact length of  `BC`.   (1 mark)
  6. Given that the area of  `Delta ABC`  is  `44`  square units, find the perpendicular distance  from  `A`  to  `BC`.    (1 mark)

 

Show Answers Only
  1. `M (5,1)`
  2. `-1/3`
  3. `text{Proof (See Worked Solutions)}`
  4. `x + 3y  – 8 = 0`
  5. `2 sqrt 10 \ text(units)`
  6. `(22 sqrt 10)/5 \ text(units)`
  7.  
Show Worked Solution

(i)   `M \ text(is the midpoint of) \ AB`

`A text{(–2,–4),}\ \ \ \ B(12,6)`

`:.M` `=((x_1+x_2)/2 ‘ (y_1+y_2)/2)`
  `= ((-2 + 12)/2  ,  (-4 + 6)/2)`
  `= (5,1)`

 

(ii)   `B(12,6),     C(6,8)`

`m_(BC)` `= (y_2 – y_1)/(x_2 – x_1)`
  `= (8– 6)/(6– 12)`
  `= -1/3`

 

(iii)  `text(Prove)\ Delta ABC \ text(|||) \ Delta AMN`

`text(Find gradient of) \ NM`

`N(2,2)   M(5,1)`

`m_(NM)` `= (1- 2)/(5– 2)`
  `= -1/3`

`:. NM\ text(||)\ BC \ \ \ text{(gradients are equal)}`

`angle NAM \ text(is common)`

`angle ANM = angle ACB\ \ \ text{(corresponding angles},\ NM\ text(||)\ BC text{)}`

`:.  Delta ABC \ text(|||) \ Delta AMN \ \ \ text{(equiangular)}`

 

(iv)   `text(Equation of) \ MN \ text(has) \ m=-1/3 \ \ text(through) \ (2,2)`

`text(Using) \ \ \ y- y_1` `= m(x  – x_1)`
`y  – 2` `= -1/3 (x  – 2)`
`3y  – 6` `= -x + 2`
`x +3y  – 8` `= 0`

 

`:. \ text(Equation of) \ MN \ text(is) \ \ x + 3y  – 8 = 0`

 

(v)   `B(12,6)   C(6,8)` 

`BC` `=sqrt{(x_2 – x_1)^2 + (y_2– y_1)^2}`
  `=sqrt{(6– 12)^2 + (8– 6)^2}`
  `=sqrt(36 +4)`
  `=sqrt(40)`
  `= 2 sqrt(10) \ text(units)`

 

(vi)   `text(Given the Area) \  Delta ABC=44\ text(u²)`

`1/2 xx b xx h` `= 44`
`1/2 xx BC xx h`  `= 44`
`1/2 xx 2 sqrt 10 xx h` `= 44`
`:. h` `= 44/ sqrt 10 xx sqrt 10 / sqrt 10`
  `= (44 sqrt 10 )/ 10`
  `= (22 sqrt 10) / 5`

 

`:. _|_ text(distance from) \ A \ text(to) \ BC \ text(is) \ (22 sqrt 10) / 5  \ text(units)`.

Functions, EXT1* F1 2010 HSC 2b

Solve the inequality  `x^2 − x − 12 < 0`.   (2 marks) 

Show Answers Only

 `-3 < x < 4`

Show Worked Solution
 MARKER’S COMMENT: Drawing the parabola proved the most efficient way of answering this question for students.

`x^2\ – x\ – 12<0`

`text(Solve)\ \ \ ` `x^2\ – x\ – 12` `=0`
  `(x\ – 4)(x + 3)` `=0`
`=>x = 4\ text(or)\ -3`

 

 

`x^2 – x – 12 < 0`

`text(when)\ \ -3 < x < 4`

Calculus, 2ADV C2 2010 HSC 2a

Differentiate  `cosx/x`  with respect to  `x`.   (2 marks) 

Show Answers Only

 `(-x sinx – cos x)/(x^2)`

Show Worked Solution
MARKER’S COMMENT: A significant number of students did not know the quotient rule and many found assistance by writing out `u,\ u prime,\ v,\ v prime\ ` before going into calculations.

`y = cos x/x`

`text(Let)` `\ \ u = cos x` `u prime = – sin x`
  `\ \ v = x` `v prime = 1`

 

`text(Using quotient rule:)`

`dy/dx` `= (u prime v – u v prime)/(v^2)`
  `= (-sinx *x  – cos x*1)/x^2`
  `= (-x sin x – cos x)/(x^2)`

Calculus, 2ADV C2 2010 HSC 1e

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

Show Answers Only

`2x tanx + x^2 sec^2 x`

Show Worked Solution
COMMENT: There is no necessity to take out the common factor `x` in this case.

`y = x^2 tan x`

`text(Using product rule:)`

`d/dx (uv)` ` = u prime v + u v prime`
`dy/dx` `=2x tanx + x^2 sec^2 x`

Financial Maths, 2ADV M1 2011 HSC 9d

 

  1. Rationalise the denominator in the expression  `1/(sqrtn + sqrt(n+1))`  where  `n`  is an integer and  `n >= 1`.   (1 mark)

  2. Using your result from part (i), or otherwise, find the value of the sum
     
         `1/(sqrt1+ sqrt2) + 1/(sqrt2 + sqrt3) + 1/(sqrt3 + sqrt4) + ... + 1/(sqrt99 + sqrt100)`   (2 marks)

Show Answers Only
  1. `sqrt(n+1)\ – sqrtn`
  2. `9`
Show Worked Solution

i.  `1/(sqrtn + sqrt(n+1)) xx (sqrtn\ – sqrt(n+1))/(sqrtn\ – sqrt(n+1))`

MARKER’S COMMENT: Students connecting the information between parts (i) and (ii) were easily the most successful. AGAIN, the information from earlier parts of multi-part questions is gold plated information for directing your answer.

`= (sqrtn\ – sqrt(n+1))/((sqrtn)^2\ – (sqrt(n+1))^2)`

`= (sqrtn\ – sqrt(n+1))/(n\ – (n + 1))`

`= (sqrtn\ – sqrt(n+1))/-1`

`= sqrt(n+1)\ – sqrtn`

 

ii.  `1/(sqrt1 + sqrt2) + 1/(sqrt2 + sqrt3) + 1/(sqrt3 + sqrt4) + … + 1/(sqrt99 + sqrt100)`

`= (sqrt2\ – sqrt1) + (sqrt3\ – sqrt2) + (sqrt4\ – sqrt3) + … + (sqrt100\ – sqrt99)`

`= – sqrt1 + sqrt 100`

♦♦♦ Mean mark 15%.

`= -1 + 10`

`= 9`

 

Calculus, 2ADV C2 2012 HSC 12aii

Differentiate with respect to  `x`. 

`(cos x)/(x^2)`.    (2 marks)

Show Answers Only

`(-x sin x\ – 2 cos x)/(x^3)`

 

Show Worked Solution

`y = cosx/(x^2)`

`u = cos x` `\ \ \ \ \ \ u prime = – sin x`
`v = x^2` `\ \ \ \ \ \ v prime = 2x`

 

`text(Using the quotient rule,)` 

`dy/dx` `= (u prime v\ – v prime u)/(v^2)`
  `= (- sin x * x^2\ – 2x * cos x)/(x^4)`
  `= (- x sin x\ – 2 cos x)/(x^3)`

Functions, 2ADV F1 2010 HSC 1b

Find integers  `a`  and  `b`  such that  `1/(sqrt5\ - 2) = a + b sqrt5`.  (2 marks)

Show Answers Only

 `a = 2, \ b = 1`

Show Worked Solution

`text(Given)\ 1/(sqrt5\ – 2) = a + b sqrt5` 

`1/(sqrt5\ – 2) xx (sqrt5 + 2)/(sqrt5 + 2)` `= (sqrt5 + 2)/((sqrt5)^2\ – 2^2)`
  `= (sqrt5 + 2)/(5\ – 4)`
  `= 2 + sqrt5`

`:.\ a = 2, \ b = 1`

Probability, 2ADV S1 2012 HSC 13c

Two buckets each contain red marbles and white marbles. Bucket `A` contains 3 red and 2 white marbles. Bucket `B`  contains 3 red and 4 white marbles.

Chris randomly chooses one marble from each bucket. 

  1. What is the probability that both marbles are red?   (1 mark)

  2. What is the probability that at least one of the marbles is white?   (1 mark)

  3. What is the probability that both marbles are the same colour?   (2 marks)

Show Answers Only
  1. `9/35`
  2. `26/35`
  3. ` 17/35`
Show Worked Solution
i.    `P{(both red)}` `= P(R_1) xx P(R_2)`
  `= 3/5 xx 3/7`
  `= 9/35`

 

STRATEGY: When the term “at least” appears in a probability question, it is likely that `1-P(text{complement})` will solve the question more efficiently and with less chance of error, as shown in part (ii).
ii.    `Ptext{(at least one white)}` `= 1 – Ptext{(none white)}`
  `= 1 – P(R_1) xx P(R_2)`
  `= 1 – 9/35`
  `= 26/35`

 

iii.    `Ptext{(same colour)}` `= P(R_1 R_2) + P(W_1 W_2)`
  `= 9/35 + (2/5 xx 4/7)`
  `= 9/35 + 8/35`
  `= 17/35`

Calculus, 2ADV C3 2011 HSC 7a

Let  `f(x) = x^3-3x + 2`. 

  1. Find the coordinates of the stationary points of  `y = f(x)`, and determine their nature.   (3 marks)

  2. Hence, sketch the graph  `y = f(x)`  showing all stationary points and the  `y`-intercept.   (2 marks)

Show Answers Only
  1. `text(MIN at)\ \ (1,0);\ text(MAX at)\ \ (–1,4)`
  2.  
    Geometry and Calculus, 2UA 2011 HSC 7a
Show Worked Solution
i.    `f(x)` `= x^3-3x + 2`
  `f^{′}(x)` `= 3x^2-3`
  `f^{″}(x)` `=6x`

 

`text(Stationary points when)\ f prime (x) = 0`

`3x^2-3` `=0`
`3 (x^2-1)` `= 0`
`:. x^2` `=1`
`x` `=+- 1`

 
`text(When)\ x = 1`

`f(1)` `= 1-3 + 2 = 0`
`f^{″}(1)` `= 6 > 0` 
`:.\ text(MIN S.P. at)\ \ (1,0)`

 

`text(When)\ \ x= -1`

`f(–1)` `= -1 + 3 + 2 = 4`
`f^{″}(–1)` `= –6 < 0`
`:.\ text(MAX S.P. at)\ \ (–1,4)`
MARKER’S COMMENT: Graphs should be large (around ½ page), axes drawn with a ruler, with intercepts and turning points clearly shown. The scale can be different on each axe for clarity, as shown in the Worked Solution.

 

ii.    `y = x^3-3x + 2`
  `y text(-intercept) = 2`

Geometry and Calculus, 2UA 2011 HSC 7a

Calculus, 2ADV C4 2011 HSC 6c

The diagram shows the graph  `y = 2 cos x` . 
  

2011 6c 
  

  1. State the coordinates of  `P`.   (1 mark)

  2. Evaluate the integral  `int_0^(pi/2) 2 cos x\ dx`.    (2 marks)

  3. Indicate which area in the diagram,  `A`,  `B`,  `C` or  `D`, is represented by the integral
     
           `int_((3pi)/2)^(2pi) 2 cos x\ dx`.   (1 mark)

  4. Using parts (ii) and (iii), or otherwise, find the area of the region bounded by the curve  `y = 2 cos x`  and the  `x`-axis, between  `x = 0`  and  `x = 2pi` .   (1 mark)

  5. Using the parts above, write down the value of
     
         `int_(pi/2)^(2pi) 2 cos x\ dx`.   (1 mark)

Show Answers Only
  1. `P(0,2)`
  2. `2`
  3. `C`
  4. `8\ text(u²)`
  5. `-2`
Show Worked Solution

i.  `y = 2 cos x`

`text(At)\ x = 0`

`y = 2 cos 0 = 2`

`:.\ P(0,2)`

 

ii.  `int_0^(pi/2) 2 cos x\ dx`

`= [2 sin x]_0^(pi/2)`

`= [2 sin (pi/2)\ – 2 sin 0]`

`= 2\ – 0`

`= 2`
 

iii.  `C`
 

iv.    `text(S)text(ince Area)\ A` `=\ text(Area)\ C,\ \ text(and)`
  `text(Area)\ B` `= 2 xx text(Area)\ A`

 

`:.\ text(Total Area)` `= 2 + (2xx2) + 2`
  `= 8\ text(u²)`

 

MARKER’S COMMENT: “Using the parts above” in part (v) was ignored by many students. Important to know when finding an area and evaluating an integral may differ.

v.  `int_(pi/2)^(2pi) 2 cos x\ dx`

`= text(Area)\ C\ – text(Area)\ B`

`= 2\ – (2 xx 2)`

`= -2`

Plane Geometry, 2UA 2011 HSC 6a

The diagram shows a regular pentagon `ABCDE`. Sides `ED` and `BC` are produced to meet at `P`.
  

  1. Find the size of `/_CDE`.    (1 mark)

  2. Hence, show that `Delta EPC` is isosceles.    (2 marks)

Show Answers Only
  1. `108°`
  2. `text(Proof)\ \ text{(see Worked Solutions)}`
Show Worked Solution
i.  

`text(Angle sum of pentagon)=(5-2) xx 180°=540°`

`:.\ /_CDE` `= 540/5\ \ \ text{(regular pentagon has equal angles)}`
  `= 108°`
MARKER’S COMMENT: Very few students solved part (i) efficiently. Remember the general formula for the sum of internal angles equals (# sides – 2) x 90°.

 
ii.  `text(Show)\ Delta EPC\ text(is isosceles)`

`text(S)text(ince)\ ED=CD\ \ text{(sides of a regular pentagon)}`

`Delta ECD\ text(is isosceles)`

`/_DEC=1/2 xx (180-108)= 36^{\circ}\ \ \ text{(Angle sum of}\ Delta DEC text{)}`

`/_CDP=72^@\ \ \ (\angle PDE\ \text{is a straight angle})`

`/_DCP=72^@\ \ \ (\angle PCB\ \text{is a straight angle})`

`=> /_CPD= 180-(72 + 72)=36^{\circ}\ \ \ text{(angle sum of}\ Delta CPD text{)}`

`:.\ Delta EPC\ \text(is isosceles)\ \ \ text{(2 equal angles)}`

Calculus, 2ADV C2 2011 HSC 4a

Differentiate  `x/sinx`  with respect to  `x`.   (2 marks)

Show Answers Only

 `(sin x\ – x cos x)/(sin^2x)`

Show Worked Solution

`y = x/sinx`

`u = x` `\ \ \ \ \ u prime = 1`
`v = sin x` `\ \ \ \ \ v prime = cos x`

 

`text(Using)\ \ d/dx (uv) = (u prime v\ – uv prime)/(v^2),`

`dy/dx` `= (1 * sinx \ – x * cos x)/((sin x)^2)`
  `= (sin x\ – x cos x)/(sin^2x)`

Linear Functions, 2UA 2011 HSC 3c

The diagram shows a line  `l_1`, with equation  `3x + 4y - 12 = 0`, which intersects the  `y`-axis at  `B`.

A second line  `l_2`, with equation  `4x - 3y = 0`, passes through the origin  `O`  and intersects `l_1`  at  `E`.  
  

2011 3c
 

  1. Show that the coordinates of  `B`  are  `(0, 3)`.    (1 mark)
  2. Show that  `l_1`  is perpendicular to  `l_2`.   (2 marks)
  3. Show that the perpendicular distance from  `O`  to  `l_1`  is  `12/5`.   (1 mark) 
  4. Using Pythagoras’ theorem, or otherwise, find the length of the interval  `BE`.   (1 mark)
  5. Hence, or otherwise, find the area of  `Delta BOE`.  (1 mark)
Show Answers Only
  1. `B(0,3)`
  2. `text(Proof)  text{(See Worked Solutions)}`
  3. `text(Proof)  text{(See Worked Solutions)}`
  4. `BE = 9/5\ \ text(or)\ \ 1.8\ text(units)`
  5. `54/25\ \ text(or)\ \ 2.16\ text(u²)`
Show Worked Solution
(i)    `B\ text(is)\ y text(-intercept of)\ l_1`
  `text(When)\ x = 0`
`(3 xx 0) + 4y\ – 12` `= 0`
`4y` `=12`
`y` `=3`

`:.\ B\ text(is)\ (0,3)`

 

MARKER’S COMMENT: Better responses involved turning the equations into the “gradient intercept” form as shown in the worked solutions. Many students who tried to use the general formula `m=- b/a` made errors in the process. 
(ii)    `l_1\ text(is)\ 3x + 4y -12`  `= 0`
  `4y` `= -3x + 12`
  `y` `= -3/4x + 3`
  `=> m\ text(of)\  l_1` `= -3/4`
`l_2\ text(is)\ 4x -3y` `= 0`
`3y` `= 4x`
`y` `= 4/3 x`
`=> m\  text(of)\ l_2` `= 4/3`
`m (l_1) xx m (l_2)` `= -3/4 xx 4/3`
  `= -1`

`:.\ l_1\ text(and)\ l_2\ text(are perpendicular)`

 

(iii)   `text(Show)\ _|_\ text(distance)\ O\ (0,0)\ text(to)\ l_1 = 12/5`
`_|_\ text(dist)` `= | (ax_1 + by_1 + c)/sqrt(a^2 + b^2) |`
  `= | (0 + 0\ – 12)/sqrt(3^2 + 4^2) |`
  ` = | -12/5 |`
  `= 12/5\ text(units    … as required)`

 

(iv)   `text(S)text(ince)\ Delta OBE\ text(is right angled)`
`OB^2` `= OE^2 + BE^2`
`3^2` `= (12/5)^2 + BE^2`
`BE^2` `= 9\ – 144/25`
  `= 81/25`
`:.\ BE` `= 9/5`
  `=1.8\ text(units)`

 

(v)    `text(Area)\ Delta BOE` `= 1/2 bh`
    `= 1/2 xx 9/5 xx 12/5`
    `= 108/50`
    `= 54/25`
    `=2.16\  text(u²)`

Calculus, 2ADV C1 2011 HSC 2c

Find the equation of the tangent to the curve  `y = (2x + 1)^4`   at the point where  `x = –1`.   (3 marks)

Show Answers Only

`8x + y + 7 = 0`

Show Worked Solution

`y = (2x + 1)^4`

`text{Using the Function of a Function Rule  (or Chain Rule)}`

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

 
`text(At)\ \ x = –1,\ y = 1`

MARKER’S COMMENT: The best setting out clearly showed the derivative function, the gradient, the point and then finally, calculations for the equation of the tangent, as per the Worked Solution.
`dy/dx` `= 8 (2(–1) + 1)^3`
  `= 8 (–1)^3`
  `= -8`

 
`text(T)text(angent has)\ m = –8\ text(through)\ (–1,1)`

`text(Using)\ \ \ y – y_1` `= m (x – x_1)`
`y -1` `= -8 (x + 1)`
`y – 1` `= -8x -8`
`8x + y + 7` `= 0`

 
`:.\ text(Equation of tangent is)\ 8x + y + 7 = 0`

Functions, 2ADV F1 2011 HSC 1f

Rationalise the denominator of  `4/(sqrt5\ - sqrt3)`.

Give your answer in the simplest form.   (2 marks)

Show Answers Only

`2(sqrt5 + sqrt3)`

Show Worked Solution

`4/(sqrt5\ – sqrt3) xx (sqrt5 + sqrt3)/(sqrt5 + sqrt3)`

`= (4(sqrt5 + sqrt3))/(5\ – 3)`

`= 2 (sqrt5 + sqrt3)`

Functions, 2ADV F1 2011 HSC 1a

Evaluate  `root(3)(651/(4pi))`  correct to four significant figures.  (2 marks)

Show Answers Only

 `3.728\ text{(to 4 sig. figures)}`

Show Worked Solution
 MARKER’S COMMENT: Show answer to 5 or 6 decimals before rounding. Correct rounding of a wrong answer still receives half marks.
`root(3)(651/(4pi))` `=3.72783…`
  `=3.728\ text{(to 4 sig. figures)}`

Trigonometry, 2ADV T1 2012 HSC 13a

The diagram shows a triangle  `ABC`. The line  `2x + y = 8`  meets the `x` and `y` axes at the points `A` and `B` respectively. The point `C` has coordinates  `(7, 4)`. 
 

2012 13a
 

  1. Calculate the distance  ` AB `.   (2 marks)

  2. It is known that  `AC = 5`  and  `BC = sqrt 65 \ \ \ `(Do NOT prove this)  

     

    Calculate the size of  `angle ABC` to the nearest degree.    (2 marks)

  3. The point `N` lies on  `AB`  such that  `CN`  is perpendicular to  `AB`. 

     

    Find the coordinates of `N`.    (3 marks)

Show Answers Only
  1. `4 sqrt 5  \ text(units)`
  2. ` 34°  \ text{(nearest degree)}`
  3. `N (3,2)`
Show Worked Solution

i.   `text(Find distance) \  AB:`

`text(Find A),\ \ y=0`

`2x + 0` `= 8`
`x` `= 4 \ => A (4,0)`

 
`text(Find B),\ \ x=0`

` 0 + y = 8  \ => B(0,8)`
  

`text(Using Pythagoras:)`

`AB^2` `= OB^2 + OA^2`
  `= 8^2 + 4^2`
  `= 80`
`:. \ AB` `= sqrt 80`
  ` = 4 sqrt 5  \ text(units)`

 

ii.   `text(Find)\  angle ABC:`

`text(Using cosine rule)`

`cos angle ABC` `=  (AB^2 + BC^2 – AC^2)/(2 xx AB xx BC)`
  `= ((4 sqrt 5)^2 + (sqrt 65)^2 – 5^2)/(2 xx 4 sqrt 5 xx sqrt 65)` 
  `= (80 + 65 – 25) / (8 xx sqrt 325)`
  `= 120/(40 sqrt 13)`
  `= 3/ sqrt 13`
  `= 0.83205…`
`:. angle ABC` `= 33.690…`
  `= 34°\ \  text{(nearest degree)}`

 

iii.  `text(Find) \ N:`

`AB   text(is)  \ 2x +y = 8`

`=>  \ text(Gradient)  \ AB = -2`

`:.\ text(Gradient of) \  CN = ½ \ \ \ (m_1 m_2 = -1\ \ text(for ⊥ lines))`

 

`text(Equation of) \ CN, \ m = ½\ text(through)\ (7,4)`

MARKER’S COMMENT: Many students could not find the correct equation on `CN` because they took its gradient to be the reciprocal of `AB` and not the negative reciprocal. 
`y  – 4` `= ½ (x  – 7)`
`2y  – 8` `= x  – 7`
`x  – 2y + 1` `= 0`

 

` N \ text(is intersection of) \ AB \  text(and) \ CN`

`2x + y  – 8` `= 0\ \ …\ (1)`
`x  – 2y + 1` `= 0\ \ …\ (2)`

 
`text(Multiply)  \ (1) xx 2`

`4x +2y  – 16 = 0\ \ …\ (3)`

 
`text(Add) \  (2) + (3)`

`5x  – 15` `= 0`
`x` `=3`

 
`text(Substitute)\ \ x = 3\ \ text(into)\ \ (1)`

`2(3) + y  – 8 = 0   =>  y = 2`

`:. N (3,2)`

Integration, 2UA 2012 HSC 12d

At a certain location a river is `12` metres wide. At this location the depth of the river, in metres, has been measured at `3` metre intervals. The cross-section is shown below.
 

2012 12d
 

  1. Use Simpson's rule with the five depth measurements to calculate the approximate area of the cross-section   (3 marks)
  2. The river flows at 0.4 metres per second.
  3. Calculate the approximate volume of water flowing through the cross-section in 10 seconds.   (1 mark)
  4.  
Show Answers Only
  1. `32.8 \ text(m²)`
  2. `131.2 \ text(m³)`  
Show Worked Solution
(i)    `A` `~~ h/3 [y_0 + 4y_1 + 2y_2 +4y_3 +y_4 ]`
  `~~3/3 [ 0.5 + (4xx2.3) + (2xx2.9) + (4xx3.8) + 2.1 ]`
  `~~ [ 0.5 + 9.2 + 5.8 + 15.2 + 2.1 ]`
  `~~ 32.8\  text(m²)`

 

♦ Over half of the students examined answered part (ii) incorrectly. 
(ii)  `text(Distance water flows)`  `= 0.4 xx 10`
  `= 4 \ text(metres)`

 

`text(Volume flow in 10 seconds)` `~~ 4 xx 32.8`
  `~~ 131.2  text(m³)`

Functions, EXT1* F1 2012 HSC 11b

 Solve  `|\ 3x -1\ | < 2`   (2 marks)

Show Answers Only

 ` -1/3 < x < 1 `

Show Worked Solution
 MARKER’S COMMENT: Note that both conditions must be satisfied! Dealing with negative signs and division for inequalities produced many errors.

`|\ 3x -1\ | < 2`

`3x -1` `<2`  `\ \ \ \ \-(3x -1)` `< 2`
`3x`  `<3` `-3x + 1` `< 2`
`x` `< 1`  `3x` `> -1`
    `x` `> -1/3`

`:. -1/3 < x < 1`

Calculus, 2ADV C1 2012 HSC 11c

Find the equation of the tangent to the curve  `y = x^2`  at the point where  `x = 3`.   (2 marks)

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`6x-y-9 = 0 `

Show Worked Solution
`y` `=x^2`
`dy/dx` `=2x`

 

`text{Need to find equation with m = 6, through (3,9) }`

`text(When) \  x = 3, y = 9, dy/dx = 6`

`text(Using)\ \ \ y-y_1`  `= m (x-x_1)`
`y -9`  `= 6(x -3)`
`y -9`  `= 6x -18`
`6x -y -9` `=0`

 

 `:.\ text( Equation of the tangent is 6x-y-9 = 0)`

 

Functions, 2ADV F1 2012 HSC 2 MC

Which of the following is equal to  `1/(2sqrt5\-sqrt3)`? 

  1. `(2sqrt5\-sqrt3)/7`  
  2. `(2sqrt5 + sqrt3)/7`
  3. `(2sqrt5\-sqrt3)/17` 
  4. `(2sqrt5 + sqrt3)/17` 
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`D`

Show Worked Solution

`1/(2sqrt5\-sqrt3) xx (2sqrt5 + sqrt3)/(2sqrt5 + sqrt3)`

`= (2sqrt5 + sqrt3)/( (2sqrt5\-sqrt3)(2sqrt5 + sqrt3) )`

`= (2sqrt5 + sqrt3)/{(2sqrt5)^2-(sqrt3)^2)`

`= (2sqrt5 + sqrt3)/17`

`=>  D`

Linear Functions, 2UA 2013 HSC 12b

The points  `A(–2, –1)`,  `B(–2, 24)`,  `C(22, 42)` and  `D(22, 17)` form a parallelogram as shown. The point  `E(18, 39)` lies on  `BC`. The point  `F`  is the midpoint of  `AD`.
 

2013 12b
 

  1. Show that the equation of the line through  `A`  and  `D`  is  `3x- 4y + 2 = 0`.    (2 marks)
  2. Show that the perpendicular distance from  `B`  to the line through  `A`  and  `D`  is  `20`  units.   (1 mark)
  3. Find the length of  `EC`.    (1 mark)
  4. Find the area of the trapezium  `EFDC`.    (2 marks)
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  1. `3x-4y+2=0`
  2. `text(Proof)\  text{(See Worked Solutions)}`
  3. `text(5 units)`
  4. `200\  text(u²)`
Show Worked Solution
(i)    `text(Need to find the equation of)\ AD`
`m_(AD)` `= (y_2\-y_1)/(x_2\-x_1)`
  `= (17+1)/(22+2)`
  `= 18/24`
  `= 3/4`

`text(Equation of)\ AD\ text(has)\ m=3/4 text(, through)\ A text{(–2,–1)}`

`text(Using)\ y\-y_1` `= m (x\-x_1)`
`y+1` `=3/4 (x +2)`
`4y+4` `=3x+6`
`3x-4y+2` `=0`
MARKER’S COMMENT: Students must know the perpendicular distance formula which was again the cause of many errors such that it was flagged by markers.
 

(ii)    `B(–2,24)`
  `AD\ text(is)\  \ 3x -4y+2 =0`
`_|_ text(dist)` `= | (ax_1 + by_1 + c)/sqrt(a^2 + b^2) |`
  `= | (3(–2)\-4(24) + 2)/sqrt(3^2 + (–4)^2 )|`
  `= | (-6\ -96 +2)/sqrt25 |`
  `= | -100/5 |`
  `= 20\ text(units)`

 

`:.\ _|_ text(dist of)\ B\ text(from)\ AD = 20\ text(units   … as required)`

 

(iii)   `E(18,39)\ \ \ C(22,42)`
`EC` `=sqrt ( (x_2\-x_1)^2 + (y_2\-y_1)^2 )`
  `= sqrt ( (22\-18)^2 + (42\-39)^2 )`
  `= sqrt (4^2 + 3^2)`
  `= sqrt25`
  `= 5\ text(units)`

 

`:.\ text(The length of)\ EC\ text(is 5 units.)`

 

(iv)    `text(Area of trapezium)` `=1/2 h (a+b)`
    `=1/2 h (EC + FD)`

`EC = 5\ text(units)`

 MARKER’S COMMENT: One of the most common cause of errors in this part occurred when students failed to use the result found in part (ii).

`text(Need to find)\ FD`

`text(S)text(ince)\ FD = 1/2 xx AD\ \ \ ( F\ text(is midpoint) )`

`A (–2,–1)\ \ D(22,17)`

`AD` `=sqrt( (22 + 2) + (17 + 1)^2 )`
  `= sqrt (24^2 + 18^2)`
  `= sqrt (900)`
  `= 30`

`=> FD = 1/2 xx 30 = 15`

`text(S)text(ince)\ ABCD\ text(is a parallelogram)`

`h` `= _|_ text(distance in part)\ text{(i)}`
  `= 20`
`:.\ text(Area of trapezium)` `=1/2 xx 20 (5 + 15)`
  `=200\  text(u²)`