Find $ \int \cos^2 3x \, dx $. The graph of $ f(x) = x^2 - 4x + 5 $ is shown in the diagram. (i) Explain why $ f(x) $ does not have an inverse function. (ii) Sketch the graph of the inverse function, $ g^{-1}(x) $, where $ g(x) = x^2 - 4x + 5, \, x \leq 2. $ (iii) State the domain of $ g^{-1}(x) $. (iv) Find an expression for $ y = g^{-1}(x) $ in terms of $ x $. Dr Kool wishes to find the temperature of a very hot substance using his thermometer, which only measures up to 100°C. Dr Kool takes a sample of the substance and places it in a room with a surrounding air temperature of 20°C, and allows it to cool. After 6 minutes the temperature of the substance is 80°C, and after a further 2 minutes it is 50°C. If $ T(t) $ is the temperature of the substance after $ t $ minutes, then Newton's law of cooling states that $ \frac{dT}{dt} = k(T - A) $, where $ k $ is a constant and $ A $ is the surrounding air temperature. (i) Verify that $ T = A + Be^{kt} $ satisfies the above equation. (ii) Determine the values of $ k $ and $ B $.

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Find $ \int \cos^2 3x \, dx $ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2003 - Paper 1

Question 5

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Find $ \int \cos^2 3x \, dx $. The graph of $ f(x) = x^2 - 4x + 5 $ is shown in the diagram. (i) Explain why $ f(x) $ does not have an inverse function. (ii... show full transcript

Worked Solution & Example Answer:Find $ \int \cos^2 3x \, dx $ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2003 - Paper 1

Step 1

Find $ \int \cos^2 3x \, dx $

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Answer

Using the identity ( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} ), we have:

\int \cos^2 3x \, dx = \int \frac{1 + \cos(6x)}{2} \, dx = \frac{1}{2} \int (1 + \cos(6x)) \, dx

This simplifies to:

\frac{1}{2} \left( x + \frac{1}{6} \sin(6x) \right) + C = \frac{x}{2} + \frac{1}{12} \sin(6x) + C

Step 2

Explain why $ f(x) $ does not have an inverse function.

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Answer

The function ( f(x) = x^2 - 4x + 5 ) is a quadratic function that opens upwards. Its vertex is at ( (2, 1) ), and it does not cross the x-axis, which means it is always positive. This creates a situation where it is not one-to-one, as it fails the horizontal line test. Therefore, it does not have an inverse.

Step 3

Sketch the graph of the inverse function, $ g^{-1}(x) $, where $ g(x) = x^2 - 4x + 5, \, x \leq 2. $

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To sketch the inverse, reflect the graph of ( g(x) ) over the line ( y = x ). The inverse will have a restricted domain for ( x \leq 2 ). The inverse function will be a half parabola opening to the right.

Step 4

State the domain of $ g^{-1}(x) $.

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Answer

The domain of ( g^{-1}(x) ) corresponds to the range of ( g(x) ). Since the minimum value of ( g(x) ) is 1 at ( x = 2 ) and it approaches infinity as ( x ) moves away from 2, the domain of ( g^{-1}(x) ) is ( [1, \infty) ).

Step 5

Find an expression for $ y = g^{-1}(x) $ in terms of $ x $.

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Answer

Starting from ( g(x) = x^2 - 4x + 5 ), to find ( y = g^{-1}(x) ), we set ( y = x^2 - 4x + 5 ) and solve for ( x ):

Rearranging gives ( x^2 - 4x + (5 - y) = 0 ).
Applying the quadratic formula results in:

\begin{align*} x &= \frac{4 \pm \sqrt{(-4)^2 - 4(1)(5 - y)}}{2(1)} \\ &= \frac{4 \pm \sqrt{16 - 20 + 4y}}{2} \\ &= \frac{4 \pm \sqrt{4y - 4}}{2} \\ &= \frac{4 \pm 2\sqrt{y - 1}}{2} \\ &= 2 \pm \sqrt{y - 1} \end{align*}

Given ( x \leq 2 ), we take the negative sign:

\therefore y = g^{-1}(x) = 2 - \sqrt{x - 1}.

Step 6

Verify that $ T = A + Be^{kt} $ satisfies the above equation.

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Starting with the equation ( \frac{dT}{dt} = k(T - A) ), we substitute ( T = A + Be^{kt} ):

Differentiate ( T ) to get: $\frac{dT}{dt} = kBe^{kt}.$
Substitute into the original equation: $kBe^{kt} = k((A + Be^{kt}) - A) = kBe^{kt}.$ This verifies that our expression satisfies the equation.

Step 7

Determine the values of $ k $ and $ B $.

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Using the initial conditions:

Substituting ( t = 0 ), we find: $T(0) = A + B = 80. \quad (1)$
After 2 minutes, substituting ( t = 2 ): $T(2) = A + Be^{2k} = 50. \quad (2)$

From equations (1) and (2), we can solve for ( k ) and ( B ) using logarithmic transformations. Solving these equations simultaneously will yield the values of ( k ) and ( B ).

Find \( \int \cos^2 3x \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 5 - 2003 - Paper 1

Worked Solution & Example Answer:Find \( \int \cos^2 3x \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 5 - 2003 - Paper 1

Find \( \int \cos^2 3x \, dx \)

Explain why \( f(x) \) does not have an inverse function.

Sketch the graph of the inverse function, \( g^{-1}(x) \), where \( g(x) = x^2 - 4x + 5, \, x \leq 2. \)

State the domain of \( g^{-1}(x) \).

Find an expression for \( y = g^{-1}(x) \) in terms of \( x \).

Verify that \( T = A + Be^{kt} \) satisfies the above equation.

Determine the values of \( k \) and \( B \).

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