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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 $y = 10 e^{-0.07t} + 3$ is a solution of \( \frac{dy}{dt} = -0.7(y - 3) \). (b) Let $f(x) = \log_e(1 + e... 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 $y = 10 e^{-0.07t} + 3$ is a solution of \( \frac{dy}{dt} = -0.7(y - 3) \)
Answer
To demonstrate that is a solution of the differential equation, we first differentiate with respect to :
[ \frac{dy}{dt} = -0.07 \cdot 10 e^{-0.07t} = -0.7 e^{-0.07t}. ]
Next, we calculate :
[ y - 3 = 10 e^{-0.07t} + 3 - 3 = 10 e^{-0.07t}. ]
Substituting back into the right side of the equation:
[ -0.7(y - 3) = -0.7(10 e^{-0.07t}) = -0.7 e^{-0.07t}. ]
Thus, we have shown that .
Step 2
Let $f(x) = \log_e(1 + e^x)$ for all $x$. Show that $f(x)$ has an inverse.
Answer
To show that has an inverse, we first demonstrate that is a one-to-one function. The derivative of is:
[ f'(x) = \frac{e^x}{1 + e^x}, ] which is positive for all . Since , is strictly increasing, ensuring that it is one-to-one.
Next, we find the inverse of by letting:
[ y = \log_e(1 + e^x) \quad \Rightarrow \quad 1 + e^x = e^y \quad \Rightarrow \quad e^x = e^y - 1 \quad \Rightarrow \quad x = \log_e(e^y - 1). ]
Thus, the inverse function is: [ f^{-1}(y) = \log_e(e^y - 1). ]
Step 3
Show that \( \frac{dx}{dt} = \frac{k}{\pi(2r - x)}. \)
Answer
Given the volume of a hemispherical bowl:
[ V = \frac{\pi}{3} x^2 (3r - x), ] we take the derivative with respect to time: [ \frac{dV}{dt} = \frac{\pi}{3} \left( 2x(3r - x)\frac{dx}{dt} + x^2(-1)\frac{dx}{dt} \right) = k. ]
Simplifying this we can factor out (\frac{dx}{dt}): [ \frac{dx}{dt} \cdot \frac{\pi}{3}(2x(3r - x) - x^2) = k. ]
Solving for (\frac{dx}{dt}) gives: [ \frac{dx}{dt} = \frac{k}{\pi(2r - x)}. ]
Step 4
Hence, or otherwise, show that it takes 3.5 times as long to fill the bowl to the point where $x = \frac{2}{3} r$ as it does to fill the bowl to the point where $x = \frac{1}{3} r$.
Answer
To find the time taken to fill the bowl, we integrate (\frac{dx}{dt}) between the respective bounds:
For : [ T_1 = \int_0^{\frac{1}{3} r} \frac{dx}{\frac{k}{\pi(2r - x)}} = \frac{\pi}{k} \int_0^{\frac{1}{3} r} (2r - x)dx = ... ]
For : [ T_2 = \int_0^{\frac{2}{3}r} \frac{dx}{\frac{k}{\pi(2r - x)}} = \frac{\pi}{k} \int_0^{\frac{2}{3}r} (2r - x)dx = ... ]
Calculating the ratios of these integrals should yield the desired relationship showing that ( T_2 = 3.5 T_1 ).
Step 5
Use the fact that \( \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \) to show that
Answer
Starting from the identity:
[ \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}, ] we can manipulate it to arrive at the desired result. By differentiating sides and applying trigonometric identities, we assert:
[ 1 + \tan n\theta \tan(n + 1)\theta = \cot((n + 1)\theta) - \tan n\theta. ]
Step 6
Use mathematical induction to prove that, for all integers $n \geq 1$, \[ \tan 2\theta + 2\tan 3\theta + \ldots + \tan(n + 1)\theta = - (n + 1)\theta + \cot(n + 1)\theta. \]
Answer
For the base case when :
[
\tan 2\theta = -2\theta + \cot 2\theta.
]
For the induction step, assume it holds for , then for we add .
The equality then takes the form and should balance with proper algebraic manipulations showing it holds for . Thus, by induction, the statement is proven.