smc-975-60-Other

Calculus, 2ADV C4 EQ-Bank 30

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 EQ-Bank 34

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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a. `4/15\ text(units)^2`

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

c. `y = -x + 1`

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a.	`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`

b.	`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.)`

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

♦♦♦ Mean mark (Vic) part (c) 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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a. `text{Proof (See Worked Solutions)}`

b. `(1/2, 5/12)`

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

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

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

f. `1 5/12-2ln2\ \ text(u²)`

Show Worked Solution

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^{prime prime}(x) = 0`

`f^{prime prime}(x)`	`=-1 + 2x = 0`
`2x`	`= 1`
`x`	`= 1/2`

`text(Check for change in concavity)`

`f^{prime prime}(1/4)`	`=-1/2 < 0`
`f^{prime prime}(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^{prime}+vu^{prime}`

`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`