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4. (a) Solve \( \frac{3x}{x-2} \leq 1 \) - HSC - SSCE Mathematics Extension 1 - Question 4 - 2001 - Paper 1
Question 4
4. (a) Solve \( \frac{3x}{x-2} \leq 1 \). (b) An aircraft flying horizontally at \( V \, \text{m s}^{-1} \) releases a bomb that hits the ground 4000 m away, measur... show full transcript
Worked Solution & Example Answer:4. (a) 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 manipulate it into a more workable form:
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Subtract 1 from both sides: [ \frac{3x}{x-2} - 1 \leq 0 ] Simplifying gives: [ \frac{3x - (x-2)}{x-2} \leq 0 ] which simplifies to: [ \frac{2x + 2}{x-2} \leq 0 ]
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Factor the numerator: [ \frac{2(x + 1)}{x - 2} \leq 0 ]
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Identify critical points by setting the numerator and denominator to zero:
- The numerator is zero when ( x + 1 = 0 ) or ( x = -1 ).
- The denominator is zero when ( x - 2 = 0 ) or ( x = 2 ).
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Analyze the sign of the expression in the intervals: ( (-\infty, -1)), ( (-1, 2)), and ( (2, \infty) ).
- For ( x < -1 ), the expression is positive.
- For ( -1 < x < 2 ), the expression is negative (valid solution).
- For ( x > 2 ), the expression is positive.
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Therefore, the solution to the inequality is: [ -1 \leq x < 2 ]
Step 2
Find the speed \( V \) of the aircraft
Answer
To find the speed ( V ), we need to analyze the motion of the bomb.
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The horizontal distance ( x ) that the bomb travels is given by: [ x = Vt ] To find the time ( t ) it takes to hit the ground, we use the vertical motion: [ y = -5t^2 ] The bomb hits the ground at a vertical height of 0: [ 0 = -5t^2 \Rightarrow t^2 =\frac{4000}{5} \Rightarrow t = \sqrt{800} \approx 28.28 s ]
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Given that the bomb travels 4000 meters horizontally, [ 4000 = Vt ; \Rightarrow V = \frac{4000}{28.28} \approx 141.42 , \text{m s}^{-1} ] Therefore, the speed ( V ) of the aircraft is approximately ( 141.42 , \text{m s}^{-1} ).
Step 3
Find \( x \) as a function of \( t \)
Answer
To find ( x ) as a function of ( t ), we solve the differential equation:
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Start with the equation: [ \frac{dx}{dt} = -4x ] This can be solved via separation of variables: [ \frac{1}{x}dx = -4dt ] Integrating both sides gives: [ \ln |x| = -4t + C ] Exponentiating both sides results in: [ x = e^{-4t + C} = e^C e^{-4t} = Ke^{-4t} ] where ( K = e^C ).
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We apply the initial condition ( x(0) = 3 ): [ 3 = Ke^{0} \Rightarrow K = 3 ] Thus: [ x(t) = 3e^{-4t} ]
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When ( x = -6/3 ) at ( t = 0 ): This condition does not adjust the function as it only confirms that at ( t = 0, x = 3 ). Thus, the final expression remains: [ x(t) = 3e^{-4t} ]