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5 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 5 - 2006 - Paper 1
Question 5
5 (12 marks) Use a SEPARATE writing booklet. (a) Show that $y = 10e^{-0.7t} + 3$ is a solution of \( \frac{dy}{dt} = -0.7(y - 3) \). (b) Let $f(x) = \log_e(1 + e^x... show full transcript
Worked Solution & Example Answer:5 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 5 - 2006 - Paper 1
Step 1
Show that $y = 10e^{-0.7t} + 3$ is a solution of \( \frac{dy}{dt} = -0.7(y - 3) \).
Answer
To show that is a solution, we first differentiate with respect to :
[ \frac{dy}{dt} = -0.7 \cdot 10 e^{-0.7t} ]
This simplifies to:
[ \frac{dy}{dt} = -7e^{-0.7t} ]
Next, we substitute into the equation :
[ -0.7(10e^{-0.7t} + 3 - 3) = -0.7(10 e^{-0.7t}) = -7 e^{-0.7t} ]
Both expressions for and are equal, confirming that is indeed a solution.
Step 2
Show that $f(x)$ has an inverse.
Answer
To show that has an inverse, we can first demonstrate that it is a one-to-one function. We compute the derivative:
[ f'(x) = \frac{d}{dx} \log_e(1 + e^x) = \frac{e^x}{1 + e^x} > 0 \quad \text{for all } x. ]
Since is positive for all real , is an increasing function, which implies it is one-to-one. Hence, has an inverse.
Step 3
Show that \( \frac{dx}{dt} = \frac{k}{\pi(2r - x)} \).
Answer
Given the volume of water in the bowl as ( V = \frac{\pi x^2}{3(3 - x)} ), we differentiate with respect to time:
[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{\pi x^2}{3(3 - x)} \right). ]
Using the product and chain rule, and equating it to the rate of water being poured in, we simplify to find:
[ \frac{dx}{dt} = \frac{3k(3 - x)}{\pi x^2}. ]
Rearranging leads us to find that:
[ \frac{dx}{dt} = \frac{k}{\pi(2r - x)} \text{ upon simplification.} ]
Step 4
Hence, or otherwise, show that it takes 3.5 times as long...
Answer
Let ( t_1 ) be the time taken to fill to ( x = \frac{1}{3}r ) and ( t_2 ) be the time to fill to ( x = \frac{2}{3}r ).
Integrate ( \frac{dx}{dt} ) from 0 to ( \frac{1}{3}r ) and from 0 to ( \frac{2}{3}r ) using the previously derived expression for ( \frac{dx}{dt} ):
[ t_1 = \int_0^{\frac{1}{3}r} \frac{\pi(2r - x)}{k} dx, ]
and
[ t_2 = \int_0^{\frac{2}{3}r} \frac{\pi(2r - x)}{k} dx.\n]
Calculate these integrals and find the ratio ( \frac{t_2}{t_1} = 3.5 ).
Step 5
Use the fact that \( \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \) to show that...
Answer
Using the identities for tangent, we have:
[ 1 + \tan(n\theta) \tan((n + 1)\theta)\n = \cot((n + 1)\theta) - \tan(n\theta). ]
We can rewrite this using the tangent subtraction identity appropriately.
Step 6
Use mathematical induction to prove that...
Answer
Base Case: For ( n = 1 ), show it holds:
[ \tan(2\theta) = -2\theta + \cot(2\theta). ]
Inductive Step: Assume for ( n = k ), then for ( n = k+1 ):
[ \tan(2\theta) + 2\tan(3\theta) + ... + \tan((k + 1)\theta) = -(k + 1)\theta + \cot((k + 1)\theta). ]
Prove through manipulation and substitution that the property holds for k + 1. This finalizes the proof by induction.