5. (a) Find $ \int \cos^2 3x \, dx $. (b) 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 $. (c) 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) Show that $ k = \log_2 2 $. (iii) Find the value of $ B $.

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

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

Worked Solution & Example Answer:5. (a) 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

To solve ( \int \cos^2 3x , dx ), we use the identity ( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} ). Thus:

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

This can be separated into two integrals:

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

The first integral is ( \frac{x}{2} ) and for the second integral, we have:

\int \cos(6x) \, dx = \frac{1}{6} \sin(6x)

Combining these, we get:

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

Step 2

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

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Answer

To find the inverse function ( g^{-1}(x) ), we first express ( y = g(x) ) as:

y = x^2 - 4x + 5

Next, we re-arrange to solve for ( x ):

y = (x-2)^2 + 1 \\ (x-2)^2 = y - 1 \\ x - 2 = \pm\sqrt{y - 1} \\ x = 2 \pm\sqrt{y - 1}

Since we are interested in the domain where ( x \leq 2 ), we take the negative:

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

This is the equation for the inverse function.

Step 3

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

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The domain of ( g^{-1}(x) ) is determined by the range of the original function ( g(x) = x^2 - 4x + 5 ). The minimum value occurs at ( x = 2 ) and is equal to 1; thus, the range is ( [1, \infty) ). Therefore, the domain of the inverse function is:

x \in [1, \infty)

Step 4

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

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Answer

As found previously, the inverse function is given by:

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

Step 5

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

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Answer

To verify this, we substitute ( T = A + Be^{kt} ) into the differential equation ( \frac{dT}{dt} = k(T - A) ):

Calculate ( \frac{dT}{dt} ): $\frac{dT}{dt} = kBe^{kt}$
Substitute into the equation: $kBe^{kt} = k((A + Be^{kt}) - A)$ Simplifying, we confirm: $kBe^{kt} = kBe^{kt}$

Thus, it satisfies the equation.

Step 6

Show that $ k = \log_2 2 $.

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Answer

Using the cooling equation, we know after 6 minutes, ( T(6) = 80 ) and after 8 minutes, ( T(8) = 50 ). We substitute these values:

From ( T(6) = A + Be^{6k} ): ( 80 = 20 + Be^{6k} \implies Be^{6k} = 60 )
From ( T(8) = A + Be^{8k} ): ( 50 = 20 + Be^{8k} \implies Be^{8k} = 30 )

Now, divide the two:

$\frac{Be^{8k}}{Be^{6k}} = \frac{30}{60} \implies e^{2k} = \frac{1}{2}$

Taking the natural log of both sides gives:

$2k = \log{\frac{1}{2}} \implies k = \frac{1}{2} \log{\frac{1}{2}} = \log_2 2$ .

Step 7

Find the value of $ B $.

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Answer

Using the earlier equation ( Be^{6k} = 60 ) and knowing ( k = \log_2 2 ), we substitute:

Be^{6 \log_2 2} = 60 \\ Be^{\log_2 64} = 60 \\ B \cdot 64 = 60 \\ B = \frac{60}{64} = \frac{15}{16}

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

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

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

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.

Show that \( k = \log_2 2 \).

Find the value of \( B \).

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