Find \( \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \).
(a) (i) By considering \( f(x) = 3 \log x - x \), show that the curve \( y = 3 \log x \) and the line \( y = x \) ... show full transcript
Worked Solution & Example Answer:Find \( \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 3 - 2006 - Paper 1
Step 1
Find \( \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \)
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Answer
To find ( \int_0^{\frac{\pi}{2}} \sin^2 x , dx ), we can use the identity ( \sin^2 x = \frac{1 - \cos(2x)}{2} ):
Show that the curve \( y = 3 \log x \) and the line \( y = x \) meet at a point \( P \)
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Answer
To find the intersection of ( y = 3 \log x ) and ( y = x ), we set:
3logx=x.
We know that for ( x = 1.5 ), ( 3 \log(1.5) ) gives approximately ( 1.39 ), which is less than 1.5, and for ( x = 2 ), ( 3 \log(2) ) gives approximately ( 2.08 ), which is greater than 2. Hence, by the Intermediate Value Theorem there is a solution in the interval (1.5, 2).
Step 3
Use one application of Newton's method, starting at \( x = 1.5 \), to find an approximation to the x-coordinate of \( P \)
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Answer
Let ( f(x) = 3 \log x - x ). The derivative ( f'(x) = \frac{3}{x} - 1 ).
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Answer
To show that ( QKTN ) is cyclic, we need to show that the opposite angles sum to 180 degrees. This can often be done by showing that they subtend the same arc in the circle.
Step 7
Show that \( \angle KMT = \angle KQT \).
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Answer
Using properties of tangents and angles, since ( MN ) is tangent at ( T ), it follows that ( \angle KMT ) is equal to the angle subtended at the circumference, giving us ( \angle KQT ).
Step 8
Hence, or otherwise, show that \( MK \) is parallel to \( TP \).
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Answer
Since ( \angle KMT = \angle KQT ), by alternate angles, it follows that ( MK ) is parallel to ( TP ) as they are transversals of parallel lines.