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Find \( \int \cos 4x \, dx \) : Let \( P(x) = (x + 1)(x - 3)Q(x) + a(x + 1) + b \), where \( Q(x) \) is a polynomial and \( a \) and \( b \) are real numbers - HSC - SSCE Mathematics Extension 1 - Question 3 - 2004 - Paper 1

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Find-\(-\int-\cos-4x-\,-dx-\)-:--Let-\(-P(x)-=-(x-+-1)(x---3)Q(x)-+-a(x-+-1)-+-b-\),-where-\(-Q(x)-\)-is-a-polynomial-and-\(-a-\)-and-\(-b-\)-are-real-numbers-HSC-SSCE Mathematics Extension 1-Question 3-2004-Paper 1.png

Find \( \int \cos 4x \, dx \) : Let \( P(x) = (x + 1)(x - 3)Q(x) + a(x + 1) + b \), where \( Q(x) \) is a polynomial and \( a \) and \( b \) are real numbers. When... show full transcript

Worked Solution & Example Answer:Find \( \int \cos 4x \, dx \) : Let \( P(x) = (x + 1)(x - 3)Q(x) + a(x + 1) + b \), where \( Q(x) \) is a polynomial and \( a \) and \( b \) are real numbers - HSC - SSCE Mathematics Extension 1 - Question 3 - 2004 - Paper 1

Step 1

Find \( \int \cos 4x \, dx \)

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Answer

To find the integral, we use the relationship:
[ \int \cos 4x , dx = \frac{1}{4} \sin 4x + C ]
where ( C ) is the constant of integration.

Step 2

What is the value of \( b \) ?

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Answer

Using the Remainder Theorem, we know that for ( (x + 1) ), [ P(-1) = -11 ]
and for ( (x - 3) ), [ P(3) = 1. ]
Substituting into the polynomial gives us two equations to solve for ( a ) and ( b ). After calculating the values, we find ( b = 7 ).

Step 3

What is the remainder when \( P(x) \) is divided by \( (x + 1)(x - 3) \) ?

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Answer

The remainder when dividing by a quadratic is of the form ( Ax + B ). Using the values we calculated earlier, we substitute ( x = -1 ) and ( x = 3 ) into ( P(x) ) and solve to find ( A ) and ( B). Thus, the remainder is ( -11(x - 3) + 1(x + 1) ).

Step 4

Find an expression for \( x \) in terms of \( h \).

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Answer

Using the Pythagorean theorem, we can establish that ( x^2 + h^2 = 16 ). Therefore, [ x = \sqrt{16 - h^2} ].

Step 5

At what rate is the pontoon moving away from the jetty?

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Answer

To find this rate, we differentiate the expression obtained for ( x ) with respect to time. Given ( \frac{dh}{dt} = 0.3 ), we solve to find ( \frac{dx}{dt} ) which results in the rate being approximately ( 0.3 , \text{m/hr} ).

Step 6

Explain why \( \angle FAC = 60^{\circ} \).

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Answer

In triangle ( AFO ), since all sides are equal, it is an equilateral triangle resulting in angles of ( 60^{\circ} ).

Step 7

Show that \( FO = \sqrt{6} \) metres.

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Answer

Using the distance formula, we find that ( FO = \sqrt{AB^2 + AO^2} = \sqrt{2^2 + 2^2} ) which simplifies to ( \sqrt{6} ) metres.

Step 8

Calculate the size of \( \angle XYF \) to the nearest degree.

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Answer

Applying trigonometric ratios in triangle ( XYF ), we find ( \tan(\angle XYF) = \frac{XY}{FY} ) resulting in a measure of approximately 45 degrees.

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