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Calculus, 2ADV C4 EO-Bank 11

  1. Differeniate \(y=\dfrac{x}{x^2+1}\)  (2 marks)
  1. Hence evaluate \(\displaystyle \int_0^1{\dfrac{1-x^2}{(x^2+1)^2}}\, dx\)   (2 marks)
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a.    \(\dfrac{dy}{dx} = \dfrac{1-x^2}{(x^2+1)^2}\)

b.    \(\dfrac{1}{2}\)

Show Worked Solution

a.    Using the quotient rule:

\(\dfrac{dy}{dx} = \dfrac{x^2+1-2x^2}{(x^2+1)^2} = \dfrac{1-x^2}{(x^2+1)^2}\)
 

b.    Using part a: 

 \(\displaystyle \int_0^1{\dfrac{1-x^2}{(x^2+1)^2}}\, dx\) \(=\left[\dfrac{x}{x^2+1}\right]_0^1\)  
  \(=\dfrac{1}{2} -0\)  
  \(=\dfrac{1}{2}\)  

 

Calculus, 2ADV C4 EO-Bank 10

Given that  `int_0^k ( 2x + 4 )\ dx = 21`, and  `k`  is a constant, find the value of  `k`.   (2 marks)

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`k = 3`

Show Worked Solution
`int_0^k ( 2x + 4 ) \ dx` `= 21`
`int_0^k ( 2x + 4 ) \ dx` `= [ x^2 + 4x ]_0^k`
  `= [(k^2 + 4k ) – 0 ]`
  `= k^2 + 4k`

 

`=> k^2 + 4k` `=21`
`k^2+4k-21` `= 0`
`(k-3)(k+7)` `= 0`
`k` `=3, -7`
`k` `=3\ text(as ) k >0`

Calculus, 2ADV C1 EO-Bank 11

A particle is moving along the `x`-axis. Its velocity `v` at time `t` is given by

`v = (t^2+4)/sqrt(3t+1)`  metres per second

Find the acceleration of the particle when  `t = 2`.

Express your answer as an exact value in its simplest form.  (3 marks)

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` (16sqrt(7))/49\ \ text(ms)^(−2)`

Show Worked Solution

`v = (t^2+4)/sqrt(3t+1)`

`alpha` `= (dv)/(dt)`

`text(Using quotient rule:)`

`u=t^2+4,`     `v=(3t+1)^(1/2)`  
`u^{′} = 2t,`     `v^{′} = 3/2 (3t+1)^(-1/2)`  
     
`alpha` `= (u^{′} v-v^{′} u)/v^2`
  `= (2t (3t+1)^(1/2)-3/2(t^2+4) (3t+1)^(-1/2))/(3t+1)`

 
`text(When)\ \ t = 2,`

`alpha` `= (4(7)^(1/2)-12(7)^(-1/2))/(7)`
  `= (4sqrt(7))/7 -12/(7sqrt(7))`
  `= (28sqrt(7))/49-(12sqrt(7))/49`
  `= (16sqrt(7))/49\ \ text(ms)^(−2)`

Calculus, 2ADV C1 EO-Bank 3

The displacement `x` metres from the origin at time `t` seconds of a particle travelling in a straight line is given by

`x = t^3-4t^2 +5t + 6`     when   `t >= 0`

  1.  Calculate the velocity when  `t = 3`.  (1 mark)

  2.  When is the particle stationary?  (2 marks)

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i.    `8\ text(ms)^(−1)`

ii.   `1, 5/3\ text(s)`

Show Worked Solution

i.   `x =t^3-4t^2 +5t + 6` 

`v = (dx)/(dt) = 3t^2-8t +5`

 
`text(When)\ t = 3,`

`v` `= 3 xx 3^2-8 · 3 +5`
  `= 8\ text(ms)^(−1)`

 

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

`3t^2-8t +5` `= 0`
`3t^2-3t-5t+5` `= 0`
`3t(t-1)-5(t-1)` `= 0`
`(t-1)(3t-5)` `= 0`
`t` `= 1, 5/3\ text(s)`

Calculus, 2ADV C1 EO-Bank 2

  1.  Find the equations of the tangents to the curve  `y = x^2-5x+6`  at the points where the curve cuts the `x`-axis.  (2 marks)

  2.  Where do the tangents intersect?  (2 marks)

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  1. `y = −x+2`
    `y = x-3`
  2. `(5/2, −1/2)`
Show Worked Solution
a.   `y` `= x^2-5x+6`
  `= (x-2)(x-3)`

 
`text(Cuts)\ xtext(-axis at)\ \ x = 2\ \ text(or)\ \ x = 3`
 

`(dy)/(dx) = 2x-5`

 
`text(At)\ \ x = 2 \ => \ (dy)/(dx) = -1`

`T_1\ text(has)\ \ m = −1,\ text{through (2, 0)}`

`y -0` `= -1(x-2)`
`y` `= -x+2`

  

`text(At)\ \ x = 3 \ => \ (dy)/(dx) = 1`

`T_2\ text(has)\ \ m = 3,\ text{through (3, 0)}`

`y -0` `= 1(x-3)`
`y` `= x -3`

 

b.   `text(Intersection occurs when:)`

`-x+2` `= x-3`
`2x` `= 5`
`x` `= 5/2`

  

`y = 5/2 – 3 = −1/2`

`:.\ text(Intersection at)\ \ (5/2, −1/2)`

Calculus, 2ADV C1 EO-Bank 1

Find the equation of the tangent to the curve  \(y=e^{x^2+3x}\)  at the point where \(x=1\).  (2 marks)

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`y = 5e^4x-4e^4`

Show Worked Solution
\(y\) \(=e^{x^2+3x}\)
`(dy)/(dx)` \(=(2x+3)e^{x^2+3x}\)

 
`text(When)\ x = 1,\ \ (dy)/(dx) = 5e^4`

`text(Equation of tangent through)\ (1, e^4)`

`y-e^4` `= 5e^4(x – 1)`
`y` `= 5e^4x-4e^4`

Calculus, 2ADV C1 EO-Bank 3 MC

At which point on the curve  \(y = x^{2}-6x + 8\)  can a normal be drawn such that it is inclined at 45\(^{\circ}\) to the positive \(x\)-axis?

  1. \((1,3)\)
  2. \((2,0)\)
  3. \(\left(\dfrac{5}{2}, -\dfrac{3}{4}\right)\)
  4. \((5,-7)\)
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\(C\)

Show Worked Solution

\(y = x^{2}-6x + 8\)

\(y^{′} = 2x-6\)

If the normal is inclined at \(45^{\circ}\) to the positive x-axis then:

\(m_{\text{normal}} = \tan 45^{\circ} = 1\)

\(\text{Since } m_{\text{tangent}} \times m_{\text{normal}} = -1,\)

\(\therefore m_{\text{tangent}} = -1.\)
 

Find \(x\) when \(y^{′} = -1:\)

\(2x-6\) \(=-1\)  
\(2x\) \(=5\)  
\(x\) \(=\dfrac{5}{2}\)  

 
Find \(y:\)

\(y\) \(= \left(\dfrac{5}{2}\right)^{2}-6\left(\dfrac{5}{2}\right) + 8\)  
  \(=\dfrac{25}{4}-15 + 8\)  
  \(= -\dfrac{3}{4}\)  

 
\(\Rightarrow C\)

Calculus, 2ADV C1 EO-Bank 14 v1

Evaluate `f^{′}(1)`, where `f(x) = x^2 / sqrt(2x + 3)`. (4 marks)

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`9 / (5sqrt5)`

Show Worked Solution

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

`f^{′}(x)` `= 2x(2x + 3)^(-1/2) + x^2(-1/2)(2x + 3)^(-3/2)(2)`

`= (2x)/(sqrt(2x + 3)) – (x^2)/(2x + 3)^(3/2)`

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

`= (3x^2 + 6x) / (2x + 3)^(3/2)`

`f^{′}(1)` `= (3(1)^2 + 6(1)) / (2(1) + 3)^(3/2)`

`= 9 / (5sqrt5)`

Calculus, 2ADV C1 EO-Bank 6

Use the definition of the derivative, `f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}`  to find  `f^{\prime}(x)`  if  `f(x)=x-3x^2`.   (2 marks)

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

`f^{′}(x)` `= \lim_{h->0} \frac{(x+h)-3(x+h)^2-(x-3x^2)}{h}`  
  `= \lim_{h->0} \frac{x+h-3x^2-6hx-3h^2-x+3x^2}{h}`  
  `= \lim_{h->0} \frac{h-6hx-3h^2}{h}`  
  `= \lim_{h->0} \frac{h(1-6x-3h)}{h}`  
  `= \lim_{h->0} 1-6x-3h`  
  `=1-6x`  

Show Worked Solution

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

`f^{′}(x)` `= \lim_{h->0} \frac{(x+h)-3(x+h)^2-(x-3x^2)}{h}`  
  `= \lim_{h->0} \frac{x+h-3x^2-6hx-3h^2-x+3x^2}{h}`  
  `= \lim_{h->0} \frac{h-6hx-3h^2}{h}`  
  `= \lim_{h->0} \frac{h(1-6x-3h)}{h}`  
  `= \lim_{h->0} 1-6x-3h`  
  `=1-6x`  

Calculus, 2ADV C4 EO-Bank 4 MC SJ

 Let  `f^(')(x)=(2)/(sqrt(2x-3))`. 

If  `f(6)=4`, then
 

  1. `f(x)=2sqrt(2x-3)`
  2. `f(x)=sqrt(2x-3)-2`
  3. `f(x)=2sqrt(2x-3)-2`
  4. `f(x)=sqrt(2x-3)+2`
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`=>C`

Show Worked Solution
`f^{‘}(x)` `=2/(sqrt(2x-3))`  
`f(x)` `=2 int(2x-3)^{- 1/2}`  
  `=2*1/2*2(2x-3)^{1/2}+c`  
  `=2sqrt(2x-3)+c`  

 
`text(When)\ \ x=6, \ f(x)=4:`

`4=2sqrt(12-3) + c \ => \ c=-2`

`:. f(x) = 2sqrt(2x-3)-2`

`=>C`

Calculus, 2ADV C1 EO-Bank 11 MC v1

Two functions, \(f\) and \(g\), are continuous and differentiable for all  \(x\in R\). It is given that  \(f(-1)=7,\ g(-1)=5\)  and  \(f^{′}(-1)=-4,\ g^{′}(-1)=-2\).

The gradient of the graph  \(y=\dfrac{f(x)}{g(x)}\)  at the point where  \(x=-1\)  is

  1. \(-\dfrac{6}{49}\)
  2. \(\dfrac{6}{49}\)
  3. \(\dfrac{6}{25}\)
  4. \(-\dfrac{6}{25}\)
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\(D\)

Show Worked Solution

\(\text{Using the Quotient Rule when}\ \ x=-1:\)

\(\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)\) \(=\dfrac{g(x)f^{′}(x)-f(x)g^{′}(x)}{g(x)^2}\)
  \(=\dfrac{g(-1)f^{′}(-1)-f(-1)g^{′}(-1)}{g(-1)^2}\)
  \(=\dfrac{5 \times -4-7 \times -2}{5^2}\)
  \(=-\dfrac{6}{25}\)

 
\(\Rightarrow D\)

Calculus, 2ADV C1 EO-Bank 1 MC v1

The derivative of  \((n^2-1) x^{3n-2}\)  can be expressed as

  1. \(3(n-1)(n^2-1) x^{3n-2}\)
  2. \(3(n-1)(n^2-1) x^{3(n-1)}\)
  3. \((3n-2) (n^2-1) x^{3(n-1)}\)
  4. \((3n-2) (n^2-1) x^{3n-2}\)
Show Answers Only

\(C\)

Show Worked Solution
\(y\) \(=(n^2-1) x^{3n-2}\)  
\(y^{′}\) \(=(3n-2) (n^2-1) x^{3n-2-1)}\)  
  \(=(3n-2) (n^2-1) x^{3(n-1)}\)  

 
\(\Rightarrow C\)

Calculus, 2ADV C1 EO-Bank 7

Differentiate  `2x(1-4x)^5`  with respect to `x`.   (2 marks)

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`y^{′}=2(1-4x)^4(1-24x)`

Show Worked Solution

`y=2x(1-4x)^5`

`text{Using the product and chain rules:}`

`y^{′}` `=2 xx (1-4x)^5-40x(1-4x)^4`  
  `=2(1-4x)^4(1-4x-20x)`  
  `=2(1-4x)^4(1-24x)`  

Algebra, STD2 A1 2004 HSC 11 MC v1

If  \(m = 8n^2\), what is a possible value of \(n\) when  \(m=7200\)?

  1. \(0.03\)
  2. \(30\)
  3. \(240\)
  4. \(900\)
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\(B\)

Show Worked Solution
\(m\) \(=8n^2\)
\(n^2\) \(=\dfrac{m}{8}\)
\(n\) \(=\pm\sqrt{\dfrac{m}{8}}\)

 
\(\text{When}\ m=7200:\)

\(n\) \(=\pm\sqrt{\dfrac{7200}{8}}\)
  \(=\pm 30\)

 
\(\Rightarrow B\)

Algebra, STD2 A1 EO-Bank 9

The volume of a sphere is given by  \(V=\dfrac{4}{3}\pi r^3\)  where  \(r\)  is the radius of the sphere.

If the volume of a sphere is  \(385\ \text{cm}^3\), find the radius, to 1 decimal place.  (3 marks)

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\(4.5\ \text{cm  (to 1 d.p.)}\)

Show Worked Solution
\(V\) \(=\dfrac{4}{3}\pi r^3\)
\(3V\) \(= 4\pi r^3\)
\(r^3\) \(=\dfrac{3V}{4\pi}\)

 

\(\text{When}\ \ V =385\)

\(r^3\) \(=\dfrac{3\times 385}{4\pi}\)
  \(=91.911\dots\)
\(\therefore\ r\) \(=\sqrt[3]{91.911\dots}\)
  \(=4.512\dots\ \ \text{(by calc)}\)
  \(=4.5\ \text{cm   (to 1 d.p.)}\)

Calculus, 2ADV C1 EO-Bank 3

  1.  Use differentiation by first principles to find \(y^{′}\), given  \(y = 4x^2 - 5x + 4\).   (2 marks)

  2.  Find the equation of the tangent to the curve when  \(x = 3\).   (1 mark)

Show Answers Only
  1.  `y^{′} = 8x-5`
  2.  `y = 19x-32`
Show Worked Solution
i.    `f(x)` `= 2x^2 + 5x`
  `f^{′}(x)` `= lim_(h->0) (f(x + h)-f(x))/h`
    `= lim_(h->0) ((4(x + h)^2-5(x + h) + 4)-(4x^2-5x + 4))/h`
    `= lim_(h->0)(4x^2 + 8xh + 4h^2-5x-5h + 4-4x^2+5x-4)/h`
    `= lim_(h->0)(8xh + 4h^2-5h)/h`
    `= lim_(h->0)(h(8x-5 + 4h))/h`

 
`:.\ y^{′} = 8x-5`
 

ii.   `text(When)\ \ x = 3, y = 25`

`y^{′} = 24-5 = 19`
 

`:. y-25` `= 19(x-3)`
`y` `= 19x-32`