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Calculus, 2ADV C4 2011 HSC 4d v1

  1. Differentiate  `y=sqrt(16 -x^2)`  with respect to  `x`.   (2 marks)

  2. Hence, or otherwise, find  `int (8x)/sqrt(16 -x^2)\ dx`.    (2 marks)

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  1. `- x/sqrt(16\ -x^2)`
  2. `-8 sqrt(16\ -x^2) + C`
Show Worked Solution
IMPORTANT: Some students might find calculations easier by rewriting the equation as `y=(16-x^2)^(1/2)`.
a.    `y` `= sqrt(16 -x^2)`
    `= (16 -x^2)^(1/2)`

 

`dy/dx` `=1/2 xx (16 -x^2)^(-1/2) xx d/dx (16 -x^2)`
  `= 1/2 xx (16 -x^2)^(-1/2) xx -2x`
  `= – x/sqrt(16 -x^2)`

 

b.    `int (8x)/sqrt(16 – x^2)\ dx` `= -8 int (-x)/sqrt(16 -x^2)\ dx`
    `= -8 (sqrt(16 -x^2)) + C`
    `= -8 sqrt(16 -x^2) + C`

Calculus, 2ADV C4 2024 HSC 5 MC v1

What is \( {\displaystyle \int x(3 x^2+1)^4 d x} \) ?

  1. \( \dfrac{1}{30}(3x^2+1)^5+C \)
  2. \( \dfrac{1}{5}(3x^2+1)^5+C \)
  3. \( \dfrac{5}{6}(3x^2+1)^5+C \)
  4. \( \dfrac{6}{5}(3x^2+1)^5+C \)
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\( A \)

Show Worked Solution
\[ \int x(3x^2+1)^4 dx\] \(=\dfrac{1}{5} \cdot \dfrac{1}{6x}x(3x^2+1)^5+C\)  
  \(=\dfrac{1}{30}(3x^2+1)^5+C\)  

 
\( \Rightarrow A \)

NOTE: Integrating by the reverse chain rule only works if some form of the derivative is already present outside of the brackets.

Calculus, 2ADV C1 2014 HSC 13c v1

The displacement of a particle moving along the  `x`-axis is given by

 `x =2t -3/sqrt(t+1)`,

where  `x`  is the displacement from the origin in metres,  `t`  is the time in seconds, and  `t >= 0`.

  1. Show that the acceleration of the particle is always negative.    (2 marks)

  2. What value does the velocity approach as  `t`  increases indefinitely?    (1 mark)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `2`
Show Worked Solution
a.    `x` `=2t -3/sqrt(t+1)`
    `=2t -3(t+1)^(-1/2)`

 

`dot x` `= 2\ -3(-1/2) (t+1)^(-3/2)`
  `= 2 + 3/(2(t+1)^(3/2))`

 

`ddot x` `= -(9/4)(t+1)^(-5/2)`
  `= – 9/(4sqrt((t+1)^5))`

 
`text(S)text(ince)\ \ t >= 0,`

`=> 1/sqrt((t+1)^5) > 0`

`=> – 9/(4sqrt((t+1)^5)) < 0`
 

`:.\ text(Acceleration is always negative.)`

 

b.    `text(Velocity)\ (dot x) = 2 + 3/(2(t+1)^(3/2))`

 
`text(As)\ t -> oo,\ 3/(2(t+1)^(3/2)) -> 0`

`:.\ text(As)\ t -> oo,\ dot x -> 2`

Calculus, 2ADV C1 2018 HSC 12d v1

The displacement of a particle moving along the `x`-axis is given by

`x = 1/4t^4 -t^3 -1/2t^2 +3t,`

where `x` is the displacement from the origin in metres and `t` is the time in seconds, for `t >= 0`.

  1. What is the initial velocity of the particle?  (1 mark)

  2. At which times is the particle stationary?  (2 marks)

  3. Find the position of the particle when the acceleration is `4/3`.  (2 marks)

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  1. `3\ text(ms)^(-1)`
  2. `t = 1 or 3\ text(seconds)`
  3. `-329/324\ text(m)`
Show Worked Solution

i.    `x = 1/4t^4 -t^3 -1/2t^2 +3t`

`v = (dx)/(dt) = t^3-3t^2-t+3`
 

`text(Find)\ \ v\ \ text(when)\ \ t = 0:`

`v` `= 0 -3(0) -0+ 3`
  `= 3\ text(ms)^(-1)`

 

ii.  `text(Particle is stationary when)\ \ v = 0`

`t^3-3t^2-t+3` `=0`  
`t^2(t-3)-1(t-3)` `=0`  
`(t-3)(t^2-1)` `=0`  
`(t-3)(t-1)(t+1)` `=0`  

 

`t = 1 or 3\ text(seconds), t >= 0`
 

iii.  `a = (dv)/(dt) = 3t^2-6t-1`
 

`text(Find)\ \ t\ \ text(when)\ \ a = 4/3`

`3t^2-6t-1` `= 4/3`
`3t^2-6t-7/3` `= 0`
`9t^2-18t-7` `=0`
`(3t-7)(3t+1)` `=0`
`t` `=7/3`, `t >= 0`
`x(7/3)` `= 1/4(7/3)^4 -(7/3)^3 -1/2(7/3)^2 +3(7/3)`
  `= -329/324\ text(m)`

Calculus, 2ADV C1 2019 HSC 14d v1

The equation of the tangent to the curve  `y = ae^(2x)+bx`  at the point where  `x = 0`  is  `y = 3x +2`.

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

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`b = -1,\ \ a = 2`

Show Worked Solution
`y ` `= ae^(2x)+bx`
`(dy)/(dx)` `= 2ae^(2x)+b`

 
`text(When)\ \ x = 0,\ \ (dy)/(dx) = 3`

`2a + b` `= 3\ …\ (1)`

 
`text(The point)\ (0, 2)\ text(lies on)\ y:`

`a(1) +b(0)` `=2`
`a` `= 2\ …\ (2)`

  

`text(Substitute into)\ (1)`

`2(2)+b` `= 3`
`b` `= -1`

Calculus, 2ADV C1 2023 HSC 14 v1

Find the equation of the tangent to the curve  `y=x(3x+2)^2`  at the point `(1,25)`.   (3 marks)

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`y=55x-30`

Show Worked Solution
`y` `=x(3x+2)^2`  
`dy/dx` `=6x(3x+2) + (3x+2)^2`  
  `=(3x+2)(9x+2)`  

 
`text{At}\ x=1\ \ =>\ \ dy/dx=(3(1)+2)(9(1)+2)=55`
 

`text{Find equation of line}\ \ m=55,\ text{through}\ (1,25)`

`y-y_1` `=m(x-x_1)`  
`y-25` `=55(x-1)`  
`y` `=55x-30`  

Calculus, 2ADV C1 2013 HSC 11b v1

Evaluate  `lim_(x->1) ((x-1)(x+2)^2)/(x^2+x-2)`.   (2 marks)

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 `3`

Show Worked Solution

`lim_(x ->1) ((x-1)(x+2)^2)/(x^2+x-2)`

COMMENT: This question has been simplified as students no longer need to factorise the difference between 2 cubes (`x^3-2^3`).

`=lim_(x->1) ( (x -1)(x+2)^2)/( (x-1)(x+2)`

`=lim_(x->1) (x+2)`

`=3`

Calculus, 2ADV C1 2019 HSC 11c v1

Differentiate  `(4x + 3)/(3x-4)`.  (2 marks)

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`-25/(3x-4)^2`

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`text(Using quotient rule:)`

`u=4x+3,`     `v=3x-4`  
`u^{′} = 4,`     `v^{′} = 3`  
     
`y^{′}` `= (u^{′} v-v^{′} u)/v^2`
  `= (4(3x-4)-3(4x+3))/(3x-4)^2`
  `= (12x-16-12x-9)/(3x-4)^2`
  `= -25/(3x-4)^2`

Calculus, 2ADV C1 2015 HSC 12c v1

Find  `f^{′}(x)`, where  `f(x) = (2x^2-3x)/(2-x).`   (2 marks)

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`((x-3) (x + 1))/(x-1)^2`

Show Worked Solution

`f(x) = (2x^2-3x)/(2-x)`

`text(Using the quotient rule:)`

`u` `= 2x^2-3x` `\ \ \ \ \ \ v` `= 2-x`
`u^{′}` `= 4x-3` `\ \ \ \ \ \ v^{′}` `= -1`
`f^{′}(x)` `= (u^{′} v-uv^{′})/v^2`
  `= ((4x-3)(2-x)-(2x^2-3x) xx -1)/(2-x)^2`
  `= (-2x^2 + 8x-6)/(x-2)^2`
  `= (-2(x^2-4x+3)/(x-2)^2`
  `= (-2(x-3) (x-1))/(x-2)^2`

Calculus, 2ADV C1 2023 HSC 7 MC v1

It is given that  \(y=f(g(x))\), where  \(f(2)=5\), \(f^{′}(2)=3\), \(g(4)=2\)  and  \(g^{′}(4)=-2\).

What is the value of \(y^{′}\) at  \(x=4\)?

  1. \(-6\)
  2. \(-2\)
  3. \(3\)
  4. \(6\)
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\(A\)

Show Worked Solution
\(y\) \(=f(g(x))\)  
\(y^{′}\) \(=f^{′}(g(4)) \times g^{′}(4)\)  
  \(=f^{′}(2) \times -2\)  
  \(=3 \times -2\)  
  \(=-6\)  

 
\(\Rightarrow A\)

Algebra, STD2 A1 2015 HSC 28d v1

The formula  \(C=\dfrac{5}{9}(F-32)\)  is used to convert temperatures between degrees Fahrenheit \((F)\) and degrees Celsius \((C)\).

Convert 18°C to the equivalent temperature in Fahrenheit.  (2 marks)

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\(64.4\ \text{degrees}\ F\)

Show Worked Solution
\(C\) \(=\dfrac{5}{9}(F-32)\)
\(F-32\) \(=\dfrac{9}{5}C\)
\(F\)  \(=\dfrac{9}{5}C+32\)

 
\(\text{When}\ \ C = 18,\)

\(F\)  \(=\dfrac{9}{5}\times 18+32\)
  \(=64.4\ \text{degrees}\ F\)

Algebra, STD2 A1 2012 HSC 21 MC v1

Which of the following correctly expresses \(r\) as the subject of  \(V=\pi r^2+x\) ?

  1. \(r=\pm\sqrt{\dfrac{V}{\pi}}-x\)
  2. \(r=\pm\sqrt{\dfrac{V}{\pi}-x}\)
  3. \(r=\pm\sqrt{\dfrac{V-x}{\pi}}\)
  4. \(r=\pm\dfrac{\sqrt{V-x}}{\pi}\)
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\(C\)

Show Worked Solution
\(V\) \(=\pi r^2+x\)
\(\pi r^2\) \(=V-x\)
\(r^2\) \(=\dfrac{V-x}{\pi}\)
\(\therefore\ r\) \(=\pm\sqrt{\dfrac{V-x}{\pi}}\)

\(\Rightarrow C\)

Algebra, STD2 A1 2011 HSC 18 MC v1

Which of the following correctly expresses  \(b\)  as the subject of  \(y= ax+\dfrac{1}{4}bx^2\)?

  1. \(b=\dfrac{4y-ax}{x^2}\)
  2. \(b=\dfrac{4(y-ax)}{x^2}\)
  3. \(b=\dfrac{\dfrac{1}{4}y-ax}{x^2}\)
  4. \(b=\dfrac{\dfrac{1}{4}(y-ax)}{x^2}\)
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\(B\)

Show Worked Solution
\(y\) \(= ax+\dfrac{1}{4}bx^2\)
\(\dfrac{1}{4}bx^2\) \(=y-ax\)
\(bx^2\)  \(=4(y-ax)\)
\(b\) \(=\dfrac{4(y-ax)}{x^2}\)

 
\(\Rightarrow B\)

Algebra, STD2 A1 2010 HSC 18 MC v1

Which of the following correctly express  \(h\)  as the subject of  \(A=\dfrac{bh}{2}\) ?

  1. \(h=\dfrac{A-2}{b}\)
  2. \(h=2A-b\)
  3. \(h=\dfrac{2A}{b}\)
  4. \(h=\dfrac{Ab}{2}\)
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\(C\)

Show Worked Solution
\(A\) \(=\dfrac{bh}{2}\)
\(bh\) \(=2A\)
\(\therefore\ h\) \(=\dfrac{2A}{b}\)

 
\(\Rightarrow C\)

Algebra, STD2 A1 2006 HSC 18 MC v1

What is the formula for \(g\) as the subject of \(7d=8e+5g^2\)?

  1. \(g =\pm\sqrt{\dfrac{8e-7d}{5}}\)
  2. \(g =\pm\sqrt{\dfrac{7d-8e}{5}}\)
  3. \(g =\pm\dfrac{\sqrt{7d+8e}}{5}\)
  4. \(g =\pm\dfrac{\sqrt{8e-7d}}{5}\)
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\(B\)

Show Worked Solution
\(7d\) \(=8e+5g^2\)
\(5g^2\) \(=7d-8e\)
\(g^2\) \(=\dfrac{7d-8e}{5}\)
\(g\) \(=\pm\sqrt{\dfrac{7d-8e}{5}}\)

\(\Rightarrow B\)

Algebra, STD2 A1 2007 HSC 19 MC v1

Which of the following correctly expresses  \(X\)  as the subject of  \(Y=4\pi\Bigg(\dfrac{X}{4}+L\Bigg)\)?

  1. \(X=\dfrac{Y}{\pi}-L\)
  2. \(X=\dfrac{Y}{\pi}-4L\)
  3. \(X=4L-\dfrac{Y}{2\pi}\)
  4. \(X=\dfrac{Y}{8\pi}-\dfrac{L}{4}\)
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\(B\)

Show Worked Solution
\(Y\) \(=4\pi\Bigg(\dfrac{X}{4}+L\Bigg)\)
\(\dfrac{Y}{4\pi}\) \(=\dfrac{X}{4}+L\)
\(\dfrac{X}{4}\) \(=\dfrac{Y}{4\pi}-L\)
\(X\) \(=4\Bigg(\dfrac{Y}{4\pi}-L\Bigg)\)
\(X\) \(=\dfrac{Y}{\pi}-4L\)

 
\(\Rightarrow B\)

Algebra, STD1 A1 2019 HSC 34 v1

Given the formula  \(D=\dfrac{B(x+1)}{18}\), calculate the value of  \(x\)  when  \(D=90\)  and  \(B=400\).  (3 marks)

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\(3.05\)

Show Worked Solution

\(\text{Make}\ x\ \text{the subject:}\)

\(D\) \(=\dfrac{B(x+1)}{18}\)
\(18D\) \(=B(x+1)\)
\(x+1\) \(=\dfrac{18D}{B}\)
\(x\) \(=\dfrac{18D}{B}-1\)
\(\text{When }\) \(D=90, B=400\)
\(\therefore\ x\) \(=\dfrac{18\times 90}{400}-1=3.05\)