Inequalities

Functions, EXT1 EQ-Bank 17

A cubic function is given by \(f(x)=\left(2 x^2+3 x-5\right)(x+2)\)

Find the zeros of \(f(x)\). (1 mark)

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Hence, or otherwise, solve \(f(x) \geqslant 0\), giving your answer in set notation. (2 marks)

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a. \(\text{Zeros at} \ \ x=-\dfrac{5}{2}, x=1 \ \ \text{and}\ \ x=-2\)

b. \(x \in\left[-\frac{5}{2},-2\right] \cup\ x \in[1, \infty)\)

Show Worked Solution

a. \(f(x)=\left(2 x^2+3 x-5\right)(x+2)=(2 x+5)(x-1)(x+2)\)

\(\text{Zeros at} \ \ x=-\dfrac{5}{2}, x=1 \ \ \text{and}\ \ x=-2\)

b. \(\text{Find} \ x \ \text{such that} \ \ f(x) \geqslant 0.\)

\(\text{At} \ \ x=0: \ (5)(-1)(2)<0\)

\(\therefore \text{Graph}\ (f(x)) \geqslant 0 \ \ \text{for} \ \ x \in\left[-\frac{5}{2},-2\right] \cup\ x \in[1, \infty)\)

Functions, EXT1 EQ-Bank 12

Find all values of \(x\) for which \((2 x+1)(x-3)(x-1) \leqslant 0\). (2 marks)

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\(x \leqslant-\dfrac{1}{2}\ \cup \ 1 \leqslant x \leqslant 3\)

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\((2 x+1)(x-3)(x-1) \leqslant 0\)

\(\text{Zeros at} \ \ x=-\dfrac{1}{2}, x=3 \ \ \text{and} \ \ x=1\)

\(\text{At} \ \ x=0: \ (1)(-3)(-1)>0\)

\(\therefore \text{Graph} \ \leqslant 0 \ \ \text{for} \ \ x \leqslant-\dfrac{1}{2}\ \cup \ 1 \leqslant x \leqslant 3\)

Functions, EXT1 EQ-Bank 22

Solve the inequality \(\left(2 x^2+3 x\right)(1-x) \geqslant 0\), expressing your answer in set notation. (3 marks)

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\(x \in\left(-\infty,-\dfrac{3}{2}\right]\ \cup\ x \in [0,1] \)

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\(\left(2 x^2+3 x\right)(1-x)=x(2 x+3)(1-x)\)

\(x(2 x+3)(1-x) \geqslant 0\)

\(\text{Zeros at} \ \ x=0, x=-\dfrac{3}{2} \ \ \text{and} \ \ x=1\)

\(\text{At} \ \ x=-1: \ (-1)(1)(2)<0\)

\(\therefore \text{Graph} \geqslant 0 \ \ \text{for} \ \ x \in\left(-\infty,-\dfrac{3}{2}\right]\ \cup\ x \in [0,1] \)

Functions, EXT1 EQ-Bank 11

Solve the inequality \((x-4)(x+2)(x-1) \geqslant 0\). (2 marks)

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\(-2 \leqslant x \leqslant 1\ \cup\ x \geqslant 4\)

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\((x-4)(x+2)(x-1) \geqslant 0\)

\(\text{Zeros at} \ \ x=4, x=-2 \ \ \text{and}\ \ x=1\)

\(\text{At} \ \ x=0: \ (-4)(2)(-1)>0\)

\(\therefore \text{Graph}\ \geqslant 0 \ \text {for} \ -2 \leqslant x \leqslant 1\ \cup\ x \geqslant 4\)

Functions, EXT1 EQ-Bank 6 MC

For which values of \(x\) is \((2x^2-3x-5)(x+2)<0\) ?

\(x \in(-1,-2) \ \cup\ x \in\left(\dfrac{5}{2}, \infty\right]\)
\(x \in[-\infty,-2)\ \cup\ x \in\left(-1, \dfrac{5}{2}\right)\)
\(x \in(-1,-2)\ \cup\ x \in\left(\dfrac{5}{2}, \infty\right)\)
\(x \in(-\infty,-2)\ \cup\ x \in\left(-1, \dfrac{5}{2}\right)\)

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\(D\)

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\((2x^2-3x-5)(x+2)=(2x-5)(x+1)(x+2) \)

\(\text {Zeros at} \ \ x=\dfrac{5}{2}, \ x=-1,\ \ \text {and} \ \ x=-2\)

\(\text{At} \ \ x=0: \ (-5)(1)(2)<0\)

\(\therefore \ \text{Graph }<0\ \text{ for }\ x \in(-\infty,-2)\ \cup\ x \in\left(-1, \frac{5}{2}\right) \)

\(\Rightarrow D\)

Functions, EXT1 EQ-Bank 2 MC

The inequality \((2 x+4)(x-1)(3-x) \leqslant 0\) has solution

\(x \leqslant-2 \ \cup \ 1 \lt x \leqslant 3\)
\(-2 \leqslant x \leqslant 1 \ \cup \ x \geqslant 3\)
\(x \leqslant-2 \ \cup \ x \geqslant 3\)
\(-2 \leqslant x \leqslant 3\)

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\(B\)

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\(\text{Zeros at} \ \ x=-2, x=1 \ \ \text{and}\ \ x=3\)

\(\text{At} \ \ x=0: \ (4)(-1)(3)<0 \)

\(\therefore \ \text{Graph} \ \leqslant 0 \ \ \text{for} \ -2 \leqslant x \leqslant 1 \ \cup \ x \geqslant 3 \)

\(\Rightarrow B\)

Functions, EXT1 EQ-Bank 1 MC

What is the solution to the inequality

\((x-2)(x+1)(x-3)>0\)

\(x<-1 \ \cup \ 2<x<3\)
\(-1<x<2 \ \cup \ x>3\)
\(x<-1 \ \cup \ x>3\)
\(-1<x<3\)

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\(B\)

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\(\text {Zeros at}\ \ x=2, x=-1 \ \ \text {and} \ \ x=3\)

\(\text{At} \ \ x=0: \ (-2)(1)(-3)>0\)

\(\therefore \text {Graph}\ >0 \ \ \text{for}\ \ -1<x<2 \ \cup \ x>3\)

\(\Rightarrow B\)

Functions, EXT1 F1 2025 HSC 1 MC

What is the solution to \(\abs{2 x+3}<5\) ?

\(-4<x<1\)
\(x<-4\) or \(x>1\)
\(-1<x<4\)
\(x<-1\) or \(x>4\)

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\(A\)

Show Worked Solution

\(\abs{2 x+3}<5\)

\(-5 < 2x+3 <5\)

\(-8< 2x < 2\)

\(-4<x<1\)

\(\Rightarrow A\)

Functions, EXT1 F1 2024 HSC 12e

The diagram shows the graph of \(y=\dfrac{1}{\abs{x-5}}\).

For what values of \(x\) is \(\dfrac{x}{6} \geq\dfrac{1}{\abs{x-5}}\) ? (3 marks)

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\(x \in[2,3] \cup[6, \infty)\)

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\(\dfrac{x}{6} \geqslant \dfrac{1}{|x-5|}\)

\(x|x-5| \geqslant 6\)

\(\text{Case 1:}\)

\(x(x-5) \geqslant 6\)

\(x^2-5 x-6 \geqslant 0\)

\((x-6)(x+1) \geqslant 0\)

\(x \leqslant-1\ \ \text{or}\ \ x \geqslant 6\)

\(\text {By inspection of graph} \ \Rightarrow \ x \leqslant -1\ \text{is not a solution}\)

\(\Rightarrow x \geqslant 6\)

Mean mark 55%.

\(\text {Case 2: }\)

\(-x(x-5) \geqslant 6\)

\(-x^2+5 x-6 \geqslant 0\)

\(x^2-5 x+6 \leqslant 0\)

\((x-3)(x-2) \leqslant 0\)

\(\Rightarrow 2 \leqslant x \leqslant 3\)

\(\therefore x \in[2,3] \cup[6, \infty)\)

Functions, EXT1 F1 2022 HSC 11f

Solve `(x)/(2-x) >= 5`. (3 marks)

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`5/3<=x<2`

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`(x)/(2-x) >= 5`

`text{Multiply b.s. by}\ \ (2-x)^2\ \ (>0):`

`x(2-x)`	`>=5(2-x)^2`
`2x-x^2`	`>=5(x^2-4x+4)`

`6x^2-22x+20`	`<=0`
`2(3x^2-11x+10)`	`<=0`
`(3x-5)(x-2)`	`<=0`

`text{Test}\ \ x=0:`

`(-5)(-2)=10>0`

`:. 5/3<=x<2\ \ (x!=2)`

Functions, EXT1 F1 2020 SPEC1 4

Solve the inequality `3-x > 1/|x-4|` for `x`, expressing your answer in interval notation. (3 marks)

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`x ∈ (-oo, (7-sqrt 5)/2)`

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`3-x > 1/|x-4|\ =>\ |x-4| (3-x) > 1`

`text(If)\ \ x-4 > 0\ =>\ x > 4:`

`(x-4) (3-x)`	`> 1`
`3x-x^2-12 + 4x`	`> 1`
`-x^2 + 7x-13`	`> 0`

`Delta = 7^2-4 ⋅ 1 ⋅ 13 = -3 < 0`

`=>\ text(No Solutions)`

`text(If)\ \ x-4 < 0\ =>\ x < 4:`

`-(x-4) (3-x)`	`> 1`
`x^2-7x + 12`	`> 1`
`x^2-7x + 11`	`> 0`

`x= (7 +- sqrt(7^2-4 ⋅ 1 ⋅ 11))/2= (7 +- sqrt 5)/2`

`text(Combining solutions:)`

`(x < (7-sqrt 5)/2 ∪ \ x > (7 + sqrt 5)/2) nn \ x < 4`

`x ∈ (-oo, (7-sqrt 5)/2)`

Functions, EXT1 F1 2019 HSC 11b

For what values of `x` is `x/(x + 1) < 2`? (3 marks)

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`x < -2 or x > -1`

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`x/(x + 1) < 2`

`text(Multiply b.s. by)\ \ (x + 1)^2`

`x(x + 1)`	`< 2(x + 1)^2`
`x^2 + x`	`< 2x^2 + 4x + 2`
`0`	`< x^2 + 3x + 2`
`0`	`< (x + 2)(x + 1)`

`text(From graph:)`

`x < -2 or x > -1`

Functions, EXT1* F1 2019 HSC 11g

The parabola `y = x^2` meets the line `y = x + 2` at the points `(-1, 1)` and `(2, 4)`. Do NOT prove this.

By first sketching the graphs of `y = x^2` and `y = x + 2`, shade the region which simultaneously satisfies the two inequalities `y >= x^2` and `y >= x + 2`. (2 marks)

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Functions, EXT1 F1 EQ-Bank 17

Solve `3/(|\ x-3\ |) < 3`. (3 marks)

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`x < 2\ ∪\ x > 4`

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`text(Solution 1)`

`3/(|\ x-3\ |) < 3`

`\|\ x-3\ \|`	`> 1`
`(x^2-6x + 9)`	`> 1^2`
`x^2-6x + 8`	`> 0`
`(x-4)(x-2)`	`> 0`

`:. {x: \ x < 2\ ∪\ x > 4}`

`text(Solution 2)`

`|\ x-3\ | > 1`

`text(If)\ \ (x-3)`	`> 0,\ text(i.e.)\ x >3`
`x-3`	`> 1`
`x`	`> 4`

`=> x > 4\ (text(satisfies both))`

`text(If)\ \ (x-3)`	`< 0,\ text(i.e.)\ x <3`
`-(x-3)`	`> 1`
`-x + 3`	`> 1`
`x`	`< 2`

`=> x < 2\ (text(satisfies both))`

`:. {x: \ x < 2\ ∪\ x > 4}`

Functions, EXT1 F1 EQ-Bank 19

A circle has centre `(5,3)` and radius 3.

Describe, with inequalities, the region that consists of the interior of the circle and more than 2 units above the `x`-axis. (2 marks)

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Sketch the region. (1 mark)

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a. `(x-5)^2 + (y-3)^2 < 9\ ∩\ y > 2`

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a. `text(Equation of circle:)`

`(x-5)^2 + (y-3)^2 = 3^2`

`:.\ text(Region is:)`

`(x-5)^2 + (y- 3)^2 < 9\ ∩\ y > 2`

COMMENT: The broken line on the graph represents an excluded boundary.

Functions, EXT1 F1 2017 HSC 11c

Solve `(2x)/(x + 1) > 1`. (3 marks)

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`:. x < −1\ \ text(or)\ \ x > 1`

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`(2x)/(x + 1) > 1`

`text(If)\ \ x + 1 > 0\ \ text(i.e.)\ \ x > −1`

`2x`	`> x + 1`
`x`	`> 1`

`=> x > 1\ \text{(satisfies both)}`

`text(If)\ \ x + 1 < 0,\ \ text(i.e.)\ \ x < −1`

`2x`	`< x + 1`
`x`	`< 1`

`=> x < −1\ \text{(satisfies both)}`

`:. x < −1\ \ text(or)\ \ x > 1`

Functions, EXT1* F1 2017 HSC 8 MC

The region enclosed by `y = 4 - x,\ \ y = x` and `y = 2x + 1` is shaded in the diagram below.

Which of the following defines the shaded region?

A.	`y <= 2x + 1, qquad`	`y <= 4-x, qquad`	`y >= x`
B.	`y >= 2x + 1, qquad`	`y <= 4-x, qquad`	`y >= x`
C.	`y <= 2x + 1, qquad`	`y >= 4-x, qquad`	`y >= x`
D.	`y >= 2x + 1, qquad`	`y >= 4-x, qquad`	`y >= x`

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`A`

Show Worked Solution

`text(Consider)\ \ y = 2x + 1,`

`text(Shading is below graph)`

`=> y <= 2x + 1`

`text(Consider)\ \ y = 4-x,`

`text(Shading is below graph)`

`=> y <= 4-x`

`=> A`

Functions, EXT1 F1 2016 HSC 11e

Solve `3/(2x + 5)-x > 0`. (3 marks)

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`x < -3,\ \-5/2 < x < 1/2`

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`3/(2x + 5)-x > 0`

`3 (2x + 5) > x (2x + 5)^2,\ \ x !=-5/2`

`3 (2x + 5)-x(2x+5)^2`	`> 0`
`(2x + 5) [3- x(2x + 5)]`	`> 0`
`(2x + 5) (-2x^2-5x + 3)`	`> 0`
`(2x + 5) (1-2x) (x + 3)`	`> 0`

`x < -3,\ \ \-5/2 < x < 1/2`

Functions, EXT1* F1 2016 HSC 11c

Solve `|\ x-2\ | <= 3.` (2 marks)

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`-1 <= x <= 5`

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`\|\ x-2\ \|`	`<= 3`
`(x-2)^2`	`<= 3^2`
`(x^2-4x + 4)`	`<= 9`
`x^2-4x-5`	`<= 0`
`(x-5) (x + 1)`	`<= 0`

`:. -1 <= x <= 5`

Functions, EXT1 F1 2007 HSC 3b

Find the vertical and horizontal asymptotes of the hyperbola `y = (x-2)/(x-4)` and hence sketch the graph of `y = (x-2)/(x-4)`. (3 marks)
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Hence, or otherwise, find the values of `x` for which
`qquad qquad (x-2)/(x-4) ≤ 3`. (2 marks)

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a. `text(See Worked Solutions.)`

b. `x < 4\ text(and)\ x ≥ 5`

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a. `y = (x-2)/(x-4)`

`text(Vertical asymptote at)\ x = 4`

`lim_(x → ∞) (x-2)/(x-4)= lim_(x → ∞) (1-2/x)/(1-4/x)=1`

`ytext(–intercept)\ = 1/2`

`xtext(–intercept)\ = 2`

b. `text(Find)\ x\ text(so that)\ \ (x-2)/(x-4) ≤ 3`

`text(When)\ \ (x-2)/(x-4)`	`= 3`
`x-2`	`= 3x-12`
`2x`	`= 10`
`x`	`= 5`

`(5, 3)\ text(is the intersection of)\ \ y = 3\ \ text{and}\ \ y = (x-2)/(x-4)`

`:. (x-2)/(x-4) ≤ 3\ \ text(when)\ \ x < 4\ \ text(and)\ \ x ≥ 5.`

Functions, EXT1 F1 2004 HSC 1b

Solve `4/(x + 1) < 3.` (3 marks)

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`x < −1\ \ text(or)\ \ x > 1/3`

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`4/(x + 1) < 3`

`text(Multiply both sides by)\ (x + 1)^2`

`4(x + 1)`	`< 3(x + 1)^2`
`4x + 4`	`< 3(x^2 + 2x + 1)`
`4x + 4`	`< 3x^2 + 6x + 3`

`3x^2 + 2x-1`	`> 0`
`(3x-1)(x + 1)`	`> 0`

`text(LHS) = 0\ \ text(when)\ \ x = 1/3\ \ text(or)\ \ −1`

`text(From the graph:)`

`x < -1\ \ text(or)\ \ x > 1/3`

Functions, EXT1 F1 2004 HSC 1a

Indicate the region on the number plane satisfied by `y ≥ |\ x + 1\ |.` (2 marks)

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`text(See Worked Solution)`

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`y ≥ |\ x + 1\ |`

`text(Test)\ (0, 0):`

`0 ≥ |\ 0 + 1\ |\ \ =>\ \ 0 ≥ 1\ \ =>\ text{False (i.e. (0, 0) lies outside)}`

`:.\ text(Shaded area represents)\ y ≥ |\ x + 1\ |`

Functions, EXT1 F1 2015 HSC 11c

Solve the inequality `4/(x + 3) ≥ 1`. (3 marks)

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`−3 < x ≤ 1, \ \ x ≠ −3`

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`text(Solution 1)`

`4/(x + 3) ≥ 1`

`4(x + 3)`	`≥ (x + 3)^2`
`4x + 12`	`≥ x^2 + 6x + 9`
`x^2 + 2x-3`	`≤ 0`
`(x + 3)(x-1)`	`≤ 0`

`:.−3 < x ≤ 1, \ \ x ≠ −3`

`text(Solution 2)`

`text(If)\ (x + 3)`	`> 0:`
`x`	`> -3`

`4/(x + 3)`	`≥ 1`
`4`	`≥ x + 3`
`x`	`≤ 1`

`-3 < x ≤ 1`

`text(If)\ (x + 3)`	`< 0:`
`x`	`< -3`

`4/(x + 3)`	`≥ 1`
`4`	`≤ x + 3`
`x`	`≥ 1`

`text(No solution.)`

`:. −3 < x ≤ 1`

Functions, EXT1* F1 2015 HSC 13b

Find the domain and range for the function `f(x) = sqrt (9-x^2)`. (2 marks)

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On a number plane, shade the region where the points `(x, y)` satisfy both of the inequalities
`qquad y <= sqrt (9-x^2)` and `y >= x` . (2 marks)

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i. `text(Domain)\ -3 <= x <= 3\ \ \ \ \ \ \ \ text(Range)\ 0<= y <= 3`

ii. `text(See Worked Solutions)`

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i. `f(x) = sqrt(9-x^2)`

`text(Domain:) \ 9-x^2>= 0\ =>\ x^2<= 9\ =>\ -3 <= x <= 3`

`text(Range:) \ 0 <= y <= 3`

♦ Mean mark 34%.

ii.

Functions, EXT1* F1 2005 HSC 1e

Find the values of `x` for which `|\ x-3\ | ≤ 1`. (2 marks)

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`2 ≤ x ≤ 4`

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`|\ x-3\ | ≤ 1`

`text(Solution 1)`

`(x-3)^2 ≤ 1`

`x^2-6x + 9 ≤ 1`

`x^2-6x +8 ≤ 0`

`(x-4)(x-2) ≤ 0`

`:. 2 ≤ x ≤ 4`

`text(Alternative Solution)`

`(x-3)`	`≤1`	`-(x-3)`	` ≤ 1`
`x`	`≤4`	`-x +3`	`≤ 1`
		`-x`	`≤-2`
		`x`	`≥ 2`

`:. 2 ≤ x ≤ 4`

Functions, EXT1* F1 2004 HSC 1f

Find the values of `x` for which `|\ x + 1\ |<= 5`. (2 marks)

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`-6 <= x <= 4`

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`text(Solution 1)`

`|\ x + 1\ |<= 5`

`-5`	`≤x+1≤5`
`:.-6`	`≤x≤4`

`text(Solution 2)`

`|\ x + 1\ |<= 5`

`(x+1)^2`	`<= 5^2`
` x^2 + 2x + 1`	`<= 25`
`x^2 + 2x-24`	`<= 0`
`(x + 6)(x-4)`	`<= 0`

`:.\ -6 <= x <= 4`

Functions, EXT1 F1 2008 HSC 3a

Sketch the graph of `y = |\ 2x-1\ |`. (1 mark)

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Hence, or otherwise, solve `|\ 2x-1\ | <= |\ x-3\ |`. (3 marks)

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b. `-2 <= x <= 4/3`

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b. `text(Solving for)\ \ |\ 2x-1\ | <= |\ x-3\ |`

`text(Graph shows the statement is TRUE between the)`

`text(points of intersection.)`

`=>\ text(Intersection occurs when)`

`(2x-1)`	`= (x-3)\ \ \ text(or)\ \ \ `	`-(2x-1)`	`= x-3`
`x`	`= -2`	`-2x + 1`	`= x-3`
		`-3x`	`= -4`
		`x`	`= 4/3`

`:.\ text(Solution is)\ \ {x: -2 <= x <= 4/3}`

Functions, EXT1 F1 2014 HSC 11e

Solve `(x^2 + 5)/x > 6`. (3 marks)

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`0<x<1\ \ text(or)\ \ x>5`

Show Worked Solution

`(x^2 + 5)/x`	`> 6`
`x^2 ((x^2 + 5)/x)`	`> 6x^2`
`x^3 + 5x`	`> 6x^2`
`x^3-6x^2 + 5x`	`> 0`
`x (x^2-6x + 5)`	`> 0`
`x (x-5)(x-1)`	`> 0`

`:.\ 0 < x < 1\ \ text(or)\ \ x > 5`

Functions, EXT1 F1 2009 HSC 1d

Solve the inequality `(x + 3)/(2x) > 1`. (3 marks)

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`0 < x < 3`

Show Worked Solution

`text(Solution 1)`

`(2x)^2 xx (x + 3)/(2x)`	`> (2x)^2`
`2x (x + 3)`	`> 4x^2`
`2x^2 + 6x`	`> 4x^2`
`2x^2-6x`	`< 0`
`2x (x-3)`	`< 0`

`:.\ {x: 0 < x < 3}`

`text(Solution 2)`

`text(If)\ \ x > 0:`

`x + 3`	`> 2x`
`x`	`< 3`

`:.\ 0 < x < 3`

`text(If)\ \ x < 0:`

`x + 3`	`< 2x`
`x`	`> 3\ \ \ =>\ text(No solution)`

`:.\ {x:\ 0 < x < 3}`

Functions, EXT1 F1 2013 HSC 10 MC

Which inequality has the same solution as `|\ x + 2\ | + |\ x- 3\ | = 5`?

`5/(3-x) >= 1`
`1/(x-3)\-1/(x + 2) <= 0`
`x^2-x-6 <= 0`
`|\ 2x-1\ | >= 5`

Show Answers Only

`C`

Show Worked Solution

♦♦ Mean mark 39%
COMMENT: The quick elimination of `A, B` and `D` is sufficient without proving `C`.

`text(In)\ A\ text(and)\ B, \ x ≠ 3\ text(but when)\ x=3,`

`|\ 3 + 2\ | + |\ 3-3\ | = 5\ \ text(is correct.)`

`:.\ text(Not)\ A\ text(or)\ B.`

`text(Consider)\ D`

`x -> oo\ text(satisfies)\ |\ 2x-1\ | >= 5,\ \ text(but)`

`text(obviously not)\ |\ x + 2\ | + |\ x-3\ | = 5.`

`text(Consider)\ C`

`x^2-x-6`	`<= 0`
`(x-3)(x + 2)`	`<= 0`

`text(True when)\ \ -2 <= x <= 3.`

`text(In this range:)`

`(x + 2) >= 0\ \ text(and)\ \ (x-3)<= 0`

`:.\ text(We can write)`

`\|\ x + 2\ \| + \|\ x-3\ \|`	`= (x + 2)\-(x-3)`
	`= x + 2-x + 3`
	`= 5`

`:. C\ text(has the same solution)`

`=> C\ text(is correct.)`

Functions, EXT1 F1 2010 HSC 1d

Solve `3/(x+2) < 4`. (3 marks)

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`x < -2\ \ text(or)\ \ x > -5/4`

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`text(Solution 1)`

`3/(x + 2) < 4`

`text(Multiply b.s. by)\ \ (x + 2)^2`

`3(x + 2)`	`< 4(x + 2)^2`
`3x + 6`	`< 4 (x^2 + 4x + 4)`
`3x + 6`	`< 4x^2 + 16x + 16`

`4x^2 + 13x + 10`	`> 0`
`(4x + 5)(x + 2)`	`> 0`

`text(LHS)\ = 0\ \ text(when)\ \ x = -5/4\ \ text(or)\ \ -2`

`text(From graph:)`

`x < -2\ \ text(or)\ \ x > -5/4`

`text(Alternate Solution)`

`text(If)\ \ x + 2 > 0\ \ text{(i.e.}\ \ x > –2 text{)}`

`3`	`< 4(x + 2)`
`3`	`< 4x + 8`
`4x`	`> -5`
`x`	`> -5/4`

`text(If)\ \ x + 2 < 0\ \ text{(i.e.}\ x < –2 text{)}`

`3`	`> 4 (x + 2)`
`3`	`> 4x + 8`
`4x`	`< -5`
`x`	`< -5/4`
`:. x`	`< –2\ \ \ \ text{(satisfies both)}`

`:.\ x < –2\ \ text(or)\ \ x > –5/4`

Functions, EXT1 F1 2012 HSC 11c

Solve `x/(x-3) < 2`. (3 marks)

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`x<3\ \ text(or)\ \ x>6`

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`text(Solution 1)`

`x/(x-3) < 2`

`text(When)\ \ x-3 > 0,\ \ (text(i.e.)\ \ x > 3)`

`x`	`< 2 ( x-3)`
`x`	`< 2x\-6`
`=>x`	`>6\ \ \ text{(satisfies both)}`

`text(When)\ \ x-3 < 0,\ \ (text(i.e.)\ \ x<3)`

`x`	`>2 (x-3)`
`x`	`> 2x-6`
`x`	`<6`
`=> x`	`<3\ \ \ text{(satisfies both)}`

`:.\ text(Equation is correct for)\ x<3\ text(or)\ x>6`

`text(Solution 2)`

`x/((x-3)) < 2`

`text(Multiply b.s. by)\ (x-3)^2`

`x(x-3)`	`< 2(x^2-6x + 9)`
`x^2-3x`	`< 2x^2-12x + 18`
`x^2-9x + 18`	`>0`
`(x-3)(x-6)`	`>0`

`:.\ text(Equation correct for)\ x<3\ \ text(or)\ \ x>6`

Functions, EXT1 F1 2011 HSC 1c

Solve `(4-x)/x <1`. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

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`x<0\ \ text(or)\ \ x>2`

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`text(Solution 1)`

`(4-x)/x < 1`

`text(If)\ x<0,\ \ \ \ \ 4-x`	`> x`
`2x`	`< 4`
`x`	`<2`

`=> x<0\ \ \ text{(satisfies both)}`

`text(If)\ x>0,\ \ \ \ \ 4-x`	`<x`
`2x`	`>4`
`x`	`>2`

`=> x>2\ \ \ text{(satisfies both)}`

`:.\ x < 0\ \ text(or)\ \ x > 2`

`text(Solution 2)`

`text(Multiply both sides by)\ \ x^2`

`x(4-x)`	`< x^2`
`4x-x^2`	`< x^2`
`2x^2-4x`	`>0`
`2x(x-2)`	`>0`

`text(From graph)`

`x<0\ \ text(or)\ \ x >2`

Functions, EXT1* F1 2009 HSC 3c

Shade the region in the plane defined by `y >= 0` and `y <= 4-x^2`. (2 marks)

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`text(Shaded area is region where)`

`y >= 0\ text(and)\ y >= 4-x^2`

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COMMENT: This past “Advanced” HSC question now fits into the Ext1 (new) syllabus.

`text(Shaded area is region where)`

`y >= 0\ \ text(and)\ \ y >= 4-x^2`

Functions, EXT1* F1 2011 HSC 4e

The diagram shows the graphs `y = |\ x\ |-2` and `y = 4-x^2`.

Write down inequalities that together describe the shaded region. (2 marks)

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`y <= 4-x^2`

`y >= |\ x\ |-2`

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♦ Mean mark 46%.

`text(Inequalities are)`

`y <= 4-x^2`

`y >= |\ x\ |-2`

Functions, EXT1* F1 2012 HSC 11b

Solve `|\ 3x -1\ | < 2` (2 marks)

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` -1/3 < x < 1 `

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MARKER’S COMMENT: Note that both conditions must be satisfied! Dealing with negative signs and division for inequalities produced many errors.

`|\ 3x -1\ | < 2`

`3x -1`	`<2`	`\ \ \ \ \-(3x -1)`	`< 2`
`3x`	`<3`	`-3x + 1`	`< 2`
`x`	`< 1`	`3x`	`> -1`
		`x`	`> -1/3`

`:. -1/3 < x < 1`

Functions, EXT1* F1 2012 HSC 8 MC

The diagram shows the region enclosed by `y = x-2` and `y^2 = 4-x`.

Which of the following pairs of inequalities describes the shaded region in the diagram?

`y^2 <= 4-x\ \ and\ \ y <= x-2`
`y^2 <= 4-x\ \ and\ \ y >= x-2`
`y^2 >= 4-x\ \ and\ \ y<= x-2`
`y^2 >= 4-x\ \ and\ \ y >= x-2`

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`A`

Show Worked Solution

♦ Mean mark 44%.

`text(Using information from diagram:)`

`(3,0)\ text(is in the shaded region)`

`text{Substituting (3,0) into}\ \ y^2<=4-x\ =>\ 0 <= 4-3\ \text{(true)}`

`:.\ text(Cannot be)\ C\ text(or)\ D`

`text(Similarly:)`

`(3,0)\ text(must satisfy other inequality)`

`text(i.e.)\ \ y <= x-2\ \ text(becomes)\ \ 0<= 3-2 \ =>\ text(true)`

`=> A`

Functions, EXT1* F1 2013 HSC 11g

Sketch the region defined by `(x-2)^2 + ( y-3)^2 >= 4`. (3 marks)

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`text(The region is the exterior of a circle,)`

COMMENT: This past “Advanced” HSC question now fits into the Ext1 (new) syllabus.

`text(centre)\ text{(2,3)}\ text(and radius 2.)`