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Solve \( \frac{3x}{x-2} \leq 1 \) - HSC - SSCE Mathematics Extension 1 - Question 4 - 2001 - Paper 1
Question 4
Solve \( \frac{3x}{x-2} \leq 1 \). An aircraft flying horizontally at \( V \) m s<sup>-1</sup> releases a bomb that hits the ground 4000 m away, measured horizontal... show full transcript
Worked Solution & Example Answer:Solve \( \frac{3x}{x-2} \leq 1 \) - HSC - SSCE Mathematics Extension 1 - Question 4 - 2001 - Paper 1
Step 1
Solve \( \frac{3x}{x-2} \leq 1 \)
Answer
To solve the inequality ( \frac{3x}{x-2} \leq 1 ), we first rearrange it:
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Multiply both sides by ( x - 2 ) (noting to consider the sign change based on the value of ( x )): [ 3x \leq x - 2 ]
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Simplifying gives: [ 3x - x \leq -2 ] [ 2x \leq -2 ] [ x \leq -1 ]
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We also need to find critical points:
- The fraction is undefined when ( x - 2 = 0 ) or ( x = 2 ).
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The solution must combine the inequality results with critical points, considering intervals created:
- Testing intervals around the critical points ( x = -1 ) and ( x = 2 ):
- ( x < -1 ): true
- ( -1 < x < 2 ): false
- ( x > 2 ): false
- Testing intervals around the critical points ( x = -1 ) and ( x = 2 ):
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Thus, the final solution is: [ x \in (-\infty, -1] \cup (2, \infty) ]
Step 2
Find the speed \( V \) of the aircraft.
Answer
Using the position equations:
- Horizontal motion: ( x = Vt )
- Vertical motion: ( y = -5t^2 )
Given ( x = 4000 ) m and using the 45° angle:
- At 45° angle, ( y = x ).
- Substitute for ( y ):
[ -5t^2 = -4000 ]
[ t^2 = 800 \implies t = \sqrt{800}
] [ t = 20\sqrt{2} ]
Now, substituting ( t ) back: [ 4000 = V(20\sqrt{2}) \Rightarrow V = \frac{4000}{20\sqrt{2}} \implies V = 100\sqrt{2} , m/s. ]
Step 3
Find \( x \) as a function of \( t \) if \( \frac{dx}{dt} = -4x \).
Answer
This is a differential equation of the form: [ \frac{dx}{dt} = -4x. ]
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Separating variables yields: [ \frac{dx}{x} = -4dt ]
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Integrating both sides: [ \ln |x| = -4t + C ]
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Exponentiating gives: [ |x| = e^{-4t + C} = Ae^{-4t}, \text{ where } A = e^C ]
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Finding constants through initial conditions:
- At ( t = 0 ), ( x = 3 ), so: [ 3 = Ae^0 \Rightarrow A = 3 ]
- Thus, ( x = 3e^{-4t} ).
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Using the other initial condition ( x = -6/3 ): substitute ( 3e^{-4t} = -2 ) to find any inconsistencies, noting feasible physical interpretations and domain of ( x ).