smc-975-60-Other

Calculus, 2ADV C4 EQ-Bank 1

Evaluate \(\displaystyle \int_{-2}^0 \sqrt{4-x^2} \, d x\). (3 marks)

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\(\pi\)

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\(\sqrt{4-x^2} \ \ \text{is a semicircle, centre}\ \ (0,0) \ \ \text {and radius 2.}\)

STRATEGY: This integral is beyond 2ADV integration techniques. An alternate strategy is required.

\(\displaystyle \int_{-2}^0 \sqrt{4-x^2} \, d x \ \)	\( \ =\ \text{Shaded area (above)}\)
	\(=\dfrac{1}{4} \times \pi+2^2\)
	\(=\pi\)

Calculus, 2ADV C3 SM-Bank 7

The graph of `f(x) = sqrt x (1 - x)` for `0<=x<=1` is shown below.

Calculate the area between the graph of `f(x)` and the `x`-axis. (2 marks)

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For `x` in the interval `(0, 1)`, show that the gradient of the tangent to the graph of `f(x)` is `(1 - 3x)/(2 sqrt x)`. (1 mark)

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The edges of the right-angled triangle `ABC` are the line segments `AC` and `BC`, which are tangent to the graph of `f(x)`, and the line segment `AB`, which is part of the horizontal axis, as shown below.

Let `theta` be the angle that `AC` makes with the positive direction of the horizontal axis.

Find the equation of the line through `B` and `C` in the form `y = mx + c`, for `theta = 45^@`. (3 marks)

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`4/15\ text(units)^2`
`text(Proof)\ \ text{(See Workes Solutions)}`
`y = -x + 1`

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i.	`text(Area)`	`= int_0^1 (sqrt x – x sqrt x)\ dx`
		`= int_0^1 (x^(1/2) – x^(3/2))\ dx`
		`= [2/3 x^(3/2) – 2/5 x^(5/2)]_0^1`
		`= (2/3 – 2/5) – (0 – 0)`
		`= 10/15 – 6/15`
		`= 4/15\ text(units)^2`

ii.	`f (x)`	`= x^(1/2) – x^(3/2)`
	`f prime (x)`	`= 1/2 x^(-1/2) – 3/2 x^(1/2)`
		`= 1/(2 sqrt x) – (3 sqrt x)/2`
		`= (1 – 3x)/(2 sqrt x)\ \ text(.. as required.)`

iii. `m_(AC) = tan 45^@=1`

♦♦♦ Mean mark (Vic) part (iii) 20%.
MARKER’S COMMENT: Most successful answers introduced a pronumeral such as `a=sqrtx` to solve.

`=> m_(BC) = -1\ \ (m_text(BC) _|_ m_(AC))`

`text(At point of tangency of)\ BC,\ f prime(x) = -1`

`(1 – 3x)/(2 sqrt x)`	`=-1`
`1-3x`	`=-2sqrtx`
`3x-2sqrt x-1`	`=0`

`text(Let)\ \ a=sqrtx,`

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

`:. sqrt x`	`=1`	`or`	`sqrt x=- 1/3\ \ text{(no solution)}`
`x`	`=1`

`f(1)=sqrt1(1-1)=0\ \ =>B(1,0)`

`text(Equation of)\ \ BC, \ m=-1, text{through (1,0):}`

`y-0`	`=-1(x-1)`
`y`	`=-x+1`

Calculus, 2ADV C4 2016 HSC 9 MC

What is the value of `int_-3^2 |\ x + 1\ |\ dx?`

`5/2`
`11/2`
`13/2`
`17/2`

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`C`

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♦♦♦ Mean mark 19%.

`int_-3^2 \|\ x + 1\ \|\ dx`	`=\ text(Area of 2 triangles)`
	`= 1/2 xx 2 xx 2 + 1/2 xx 3 xx 3`
	`= 13/2`

`=> C`

Calculus, 2ADV C4 2005 HSC 8b

The shaded region in the diagram is bounded by the circle of radius 2 centred at the origin, the parabola `y = x^2– 3x + 2`, and the `x`-axis.

By considering the difference of two areas, find the area of the shaded region. (3 marks)

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`(pi − 5/6)\ \ text(u²)`

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`text(Shaded Area = Area in the quarter circle less)`

`text(the area below the parabola between)\ x= 0 and 1.`

`text(Area of)\ 1/4\ text(circle)`	`= 1/4 xx pir^2`
	`= 1/4 xx pi xx 2^2`
	`= pi\ \ \ text(u²)`

`text(Area below the parabola between)\ x= 0 and 1`

`=int_0^1y\ dx`

`= int_0^1x^2 − 3x + 2\ dx`

`= [x^3/3 − 3/2x^2 + 2x]_0^1`

`= [(1/3 − 3/2 + 2) − 0]`

`= 5/6`

`:.\ text(Shaded Area) = (pi − 5/6)\ \ \ text(u²)`

Calculus, 2ADV C4 2008 HSC 10a

In the diagram, the shaded region is bounded by `y = log_e (x – 2)`, the `x`-axis and the line `x = 7`.

Find the exact value of the area of the shaded region. (5 marks)

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`5 log_e 5 – 4\ \ \ text(u²)`

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`text(Shaded Area)\ text{(} A_1 text{)}`	`=\ text(Rectangle) – A_2`
`text(Area of Rectangle)\ `	`= 7 xx log_e 5`

`text(Finding the Area of)\ \ A_2`

`y`	`= log_e (x – 2)`
`x – 2`	`= e^y`
`x`	`= e^y + 2`

`:. A_2`	`= int_0^(log_e 5) x\ dy`
	`= int_0^(log_e 5) e^y + 2\ dy`
	`= [e^y + 2y]_0^(log_e 5)`
	`= [(e^(log_e 5) + 2 log_e 5) – (e^0 + 0)]`
	`= (5 + 2 log_e 5) – 1`
	`= 4 + 2 log_e 5`

`:.\ A_1`	`= 7 log_e 5 – (4 + 2 log_e 5)`
	`= 5 log_e 5 – 4\ \ \ text(u²)`

Calculus, 2ADV C3 2009 HSC 10

`text(Let)\ \ f(x) = x - (x^2)/2 + (x^3)/3`

Show that the graph of `y = f(x)` has no turning points. (2 marks)

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Find the point of inflection of `y = f(x)`. (1 mark)

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i. Show that `1 - x + x^2 - 1/(1 + x) = (x^3)/(1 + x)` for `x != -1`. (1 mark)

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ii. Let `g(x) = ln (1 + x)`.

Use the result in part c.i. to show that `f prime (x) >= g prime (x)` for all `x >= 0`. (2 marks)

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Sketch the graphs of `y = f(x)` and `y = g(x)` for `x >= 0`. (2 marks)

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Show that `d/(dx) [(1 + x) ln (1 + x) - (1 + x)] = ln (1 + x)`. (2 marks)

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Find the area enclosed by the graphs of `y = f(x)` and `y = g(x)`, and the straight line `x = 1`. (2 marks)

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`text{Proof (See Worked Solutions)}`
`(1/2, 5/12)`
i. `text{Proof (See Worked Solutions)}`

ii. `text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`1 5/12\ – 2ln2\ \ text(u²)`

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a.	`f(x) = x\ – (x^2)/2 + (x^3)/3`

♦♦ Mean mark 28% for all of Q10 (note that data for each question part is not available).

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

`f prime (x) = 1\ – x + x^2`

`x^2\ – x + 1 = 0`

`text(S)text(ince)\ \ Delta`	`= b^2\ – 4ac`
	`= (–1)^2\ – 4 xx 1 xx 1`
	`= -3 < 0 => text(No solution)`

`:.\ f(x)\ text(has no turning points)`

b.	`text(P.I. when)\ f″(x) = 0`

`f″(x)`	`= -1 + 2x = 0`
`2x`	`= 1`
`x`	`= 1/2`

`text(Check for change in concavity)`

`f″(1/4)`	`= -1/2 < 0`
`f″(3/4)`	`= 1/2 > 0`

`=>\ text(Change in concavity)`

`:.\ text(P.I. at)\ \ x = 1/2`

`f(1/2)`	`= 1/2\ – ((1/2)^2)/2 + ((1/2)^3)/3`
	`= 1/2\ – 1/8 + 1/24`
	`= 5/12`

`:.\ text(Point of Inflection at)\ (1/2, 5/12)`

c.i.

`text(Show)\ 1\ – x + x^2\ – 1/(1 + x) = (x^3)/(1 + x),\ \ \ x != -1`

`text(LHS)`	`= (1+x)/(1+x)\ – (x(1+x))/(1+x) + (x^2(1+x))/((1+x))\ – 1/(1+x)`
	`= (1 + x\ – x\ – x^2 + x^2 + x^3\ – 1)/(1+x)`
	`= (x^3)/(1+x)\ \ \ text(… as required)`

c.ii.	`text(Let)\ g(x) = ln(1+x)`
	`g prime (x) = 1/(1 + x)`

`f prime (x)\ – g prime (x)`	`= 1\ – x + x^2\ – 1/(1+x)`
	`= (x^3)/(1 + x)\ \ text{(using part (i))}`

`text(S)text(ince)\ (x^3)/(1 + x) >= 0\ text(for)\ x >= 0`

`f prime (x)\ – g prime (x) >= 0`

`f prime (x) >= g prime (x)\ text(for)\ x >= 0`

MARKER’S COMMENT: When 2 graphs are drawn on the same set of axes, you must label them.

e.	`text(Show)\ d/(dx) [(1 + x) ln (1 + x)\ – (1 + x)] = ln (1 + x)`
	`text(Using)\ d/(dx) uv=uv′+vu′`

`text(LHS)`	`= (1+x) xx 1/(1 + x) + ln(1+x)xx1 + – 1`
	`= 1+ ln(1+x)\ – 1`
	`= ln(1+x)`
	`=\ text(RHS … as required)`

f.	`text(Area)`	`= int_0^1 f(x)\ – g(x)\ dx`
		`= int_0^1 (x\ – (x^2)/2 + (x^3)/3\ – ln(x+1))\ dx`
		`= [x^2/2\ – x^3/6 + (x^4)/12\ – (1 + x) ln (1+x) + (1+x)]_0^1`
		`text{(using part (e) above)}`
		`= [(1/2 – 1/6 + 1/12 – (2)ln2 + 2) – (ln1 + 1)]`
		`= 5/12\ – 2ln2 + 2\ – 1`
		`= 1 5/12\ – 2 ln 2\ \ text(u²)`

`text{Shaded Area}`	`=A_text{sector}-A_Delta`
	`=(pi/4)/(2pi) xx pi r^2-1/2 xx b xx h`
	`=1/8xxpixx(sqrt2)^2-1/2xx1xx1`
	`=pi/4-1/2`
	`=(pi-2)/4\ text{u}^2`

`0`	`=a/(b-0)-1`
`a/b`	`=1`
`a`	`=b`

`1`	`=a/(b-1)-1`
`a/(b-1)`	`=2`
`a`	`=2(b-1)`
`a`	`=2b-2`
`b`	`=2b-2\ \ text{(using}\ a=b)`
`b`	`=2`

c.	`int_0^1 2/(2-x)-1\ dx`	`=int_0^1 -2 xx (-1)/(2-x)-1\ dx`
		`=[-2ln\|2-x\|-x]_0^1`
		`=[(-2ln1-1)-(-2ln2-0)]`
		`=-1+2ln2`

`:.\ text{Total Area}`	`=2ln2-1 + (pi-2)/4`
	`=(8ln2-4+pi-2)/4`
	`=(8ln2+pi-6)/4\ text{u}^2`