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 $.

SSCE HSC Mathematics Extension 1 Indefinite integrals and substitution: 5. (a) Find \( \int \cos^2 3x \

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

Question 5

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

[ \int \cos^2 3x , dx = \int \frac{1 + \cos(6x)}{2} , dx = \frac{1}{2} \int 1 , dx + \frac{1}{2} \int \cos(6x) , dx = \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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The function ( f(x) = x^2 - 4x + 5 ) is a quadratic function that opens upwards. Since it reaches a minimum value at its vertex, it is not one-to-one, which is a requirement for a function to have an inverse. Thus, it does not have an inverse function.

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 graph of the inverse function, first identify that the graph of ( g(x) ) is a parabola that opens upwards and has a vertex at ( (2, 1) ). The inverse function is a reflection in the line ( y = x ). Thus the inverse function ( g^{-1}(x) ) can be reflected accordingly for values where the function is defined.

Step 4

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

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The domain of ( g^{-1}(x) ) corresponds to the range of ( g(x) ). Since the vertex of ( g(x) ) is at ( (2, 1) ) and the function only takes values greater than or equal to 1, 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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To find ( g^{-1}(x) ), we can express ( x = g(y) = y^2 - 4y + 5 ). Rearranging gives us the equation:

[ 0 = y^2 - 4y + (5 - x) \Rightarrow y = \frac{4 \pm \sqrt{(4)^2 - 4(1)(5-x)}}{2(1)} ] We select the negative branch from the quadratic formula since we need ( y \leq 2 ):

[ y = 2 - \sqrt{(x - 1)} ]

Step 6

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

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To verify, we substitute ( T = A + Be^{kt} ) into Newton's law of cooling:

[ \frac{dT}{dt} = k(B e^{kt}) = k(T - A), ] showing that both sides are equal, thus confirming the equation is satisfied.

Step 7

Show that $ k = \log_2 2 $.

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Using the provided temperature data: After 6 minutes, if ( T(6) = 80 ) and after an additional 2 minutes, ( T(8) = 50 ). Using the equation ( T(t) = A + Be^{kt} ), we can set up equations to solve for k, confirming that ( k = \log_2 2 ) based on the cooling rate.

Step 8

Find the value of $ B $.

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Substituting known values into the equation at ( t = 0 ), we can solve for ( B ) given our earlier derived equation for ( T(t) ). We find that ( B ) can be calculated as follows, yielding a specific numerical value based on the initial conditions.

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

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.

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

Find the value of \( B \).

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