SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Functions, 2ADV F1 EQ-Bank 19

\(R\left(r, r^2\right), S\left(s, s^2\right)\) and \(T\left(t, t^2\right)\) are points on the parabola  \(y=x^2\).

Given \(RT\) is parallel to \(SO\), show  \(r+t=s\)   (2 marks)
 

Show Answers Only

\(R\left(r, r^2\right), S\left(s, s^2\right), T\left(t, t^2\right)\)

\(m_{S O}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{s^2-0}{s-0}=s\)

\(m_{R T}=\dfrac{t^2-r^2}{t-r}=\dfrac{(t-r)(t+r)}{(t-r)}=t+r\)

\(\text{Given}\ R T \ \| \  SO \ \Rightarrow \ m_{SO}=m_{R T}\)

\(\therefore s=r+t\ \ …\ \text{as required} \)

Show Worked Solution

\(R\left(r, r^2\right), S\left(s, s^2\right), T\left(t, t^2\right)\)

\(m_{S O}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{s^2-0}{s-0}=s\)

\(m_{R T}=\dfrac{t^2-r^2}{t-r}=\dfrac{(t-r)(t+r)}{(t-r)}=t+r\)

\(\text{Given}\ R T \ \| \  SO \ \Rightarrow \ m_{SO}=m_{R T}\)

\(\therefore s=r+t\ \ …\ \text{as required} \)

Functions, 2ADV F1 EQ-Bank 17

The tangent to the parabola  \(y=x^2+2 x-4\)  is  \(y=px-5\)  where  \(p>0\).

Find the value of \(p\).   (2 marks)

Show Answers Only

\(p=4\)

Show Worked Solution

\(\text{Intersection occurs when:}\)

\(x^2+2x-4\) \(=px-5\)  
\(x^2+(2-p)x+1\) \(=0\)  

 
\(\text{Tangent touches once}\ \Rightarrow\ \text{Discriminant}\ \Delta=0\)

\((2-p)^2-4 \times 1 \times 1\) \(=0\)  
\(4-4p+p^2-4\) \(=0\)  
\(p(p-4)\) \(=0\)  
\(p\) \(=4\ \ \ (p\gt 0)\)  
COMMENT: Key is to recognise this is a discriminant question, not a calculus application.

Functions, 2ADV F1 SM-Bank 25

Show that the parabola  \(2x^2-kx+k-2\)  has at least one real root.  (3 marks)

Show Answers Only

\(2x^2-kx+k-2=0\)

\(\Delta=b^2-4ac=(-k)^2-4 \times 2(k-2) = k^{2}-8k+16\)

\(\text{Real roots:}\ \ \Delta \geqslant 0\)

\(k^2-8k+16\) \(\geqslant 0\)  
\((x-4)^2\) \(\geqslant 0\)  

 
\(\therefore\ \text{At least one root exists for all}\ k\)

Show Worked Solution

\(2x^2-kx+k-2=0\)

\(\Delta=b^2-4ac=(-k)^2-4 \times 2(k-2) = k^{2}-8k+16\)

\(\text{Real roots:}\ \ \Delta \geqslant 0\)

\(k^2-8k+16\) \(\geqslant 0\)  
\((x-4)^2\) \(\geqslant 0\)  

 
\(\therefore\ \text{At least one root exists for all}\ k\)

Functions, 2ADV F1 2023 HSC 10 MC

The graph  \(y = x^2\)  meets the line  \(y = k\)  (where \(k>0\)) at points \(P\) and \(Q\) as shown in the diagram. The length of the interval \(PQ\) is \(L\).
 

Let \(a\) be a positive number. The graph  \(y=\dfrac{x^2}{a^2}\)  meets the line  \(y=k\)  at points \(S\) and \(T\).

What is the length of \(ST\)?

  1. \(\dfrac{L}{a}\)
  2. \(\dfrac{L}{a^2}\)
  3. \(aL\)
  4. \(a^2L\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Intersection of}\ \ y=x^2\ \ \text{and}\ \ y=k:\)

\(x^2=k\ \ \Rightarrow\ \ x=\pm \sqrt k\)

\(\therefore L=2\sqrt k\)

\(\text{Intersection of}\ \ y=\dfrac{x^2}{a^2}\ \ \text{and}\ \ y=k:\)

\(\dfrac{x^2}{a^2} \) \(=k\)  
\(x^2\) \(=a^2k\)  
\(x\) \(=\pm a\sqrt k\)  

\(\therefore ST=a \times 2\sqrt k = aL \)

\(\Rightarrow C\)

♦♦♦ Mean mark 24%.

Functions, 2ADV F1 SM-Bank 23

Find the values of `k` for which the expression  `x^2-3x + (4-2k)`  is always positive.  (3 marks)

Show Answers Only

`k < 7/8`

Show Worked Solution

`x^2-3x + (4-2k) > 0`

`x^2-3x + (4-2k) = 0\ \ text(is a concave up parabola)`

`=>\ text{Always positive (no roots) if}\ \ Delta < 0`
 

`b^2-4ac < 0`

`(−3)^2-4 · 1 · (4-2k)` `< 0`
`9-16 + 8k` `< 0`
`8k` `< 7`
`k` `< 7/8`

Functions, 2ADV F1 2016 HSC 11e

Find the points of intersection of  `y=-5-4x`  and  `y=3-2x-x^2.`  (3 marks)

Show Answers Only

`(4, – 21) and (– 2, 3)`

Show Worked Solution

`y = 3 – 2x – x^2`

`text(Substitute)\ \ y = -5 – 4x\ \ text(into equation)`

`-5 – 4x` `= 3 – 2x – x^2`
`x^2 – 2x – 8` `= 0`
`(x – 4) (x + 2)` `= 0`

  
`:. x = 4 or -2`
 

`text(When)\ \ x = 4,\ \ y = -5 – 4(4) = -21`

`text(When)\ \ x = -2,\ \ y = -5 – 4 (-2) = 3`  
 

`:.\ text(Intersection at)\ \ (4, – 21) and (– 2, 3)`

Functions, 2ADV F1 2013 HSC 1 MC

What are the solutions of   `2x^2-5x-1 = 0`? 

  1. `x = (-5 +-sqrt17)/4` 
  2. `x = (5 +-sqrt17)/4`
  3. `x = (-5 +-sqrt33)/4`
  4. `x = (5 +-sqrt33)/4`
Show Answers Only

`D`

Show Worked Solution

`2x^2-5x-1 = 0`

`text(Using)\ x = (-b +- sqrt( b^2-4ac) )/(2a)`

`x` `= (5 +- sqrt{\ \ (-5)^2-4 xx 2 xx(-1) })/ (2 xx 2)`
  `= (5 +- sqrt(25 + 8) )/4`
  `= (5 +- sqrt(33) )/4`

 
`=>  D`