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Find \( \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 3 - 2006 - Paper 1

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Find-\(-\int_0^{\frac{\pi}{2}}-\sin^2-x-\,-dx-\)-HSC-SSCE Mathematics Extension 1-Question 3-2006-Paper 1.png

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} ):

0π2sin2xdx=0π21cos(2x)2dx=120π21dx120π2cos(2x)dx.\int_0^{\frac{\pi}{2}} \sin^2 x \, dx = \int_0^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \int_0^{\frac{\pi}{2}} 1 \, dx - \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos(2x) \, dx.

The first integral evaluates to ( \frac{\pi}{4} ), and the second integral is calculated as follows:

cos(2x)dx=12sin(2x)+C.\int \cos(2x) \, dx = \frac{1}{2} \sin(2x) + C.

Evaluating ( \int_0^{\frac{\pi}{2}} \cos(2x) , dx ):

[12sin(2x)]0π2=00=0.\left[ \frac{1}{2} \sin(2x) \right]_0^{\frac{\pi}{2}} = 0 - 0 = 0.

Thus,

0π2sin2xdx=12(π40)=π8.\int_0^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{1}{2} \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{8}.

Step 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.3 \log x = 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 ).

Using Newton's method:

  1. Start with ( x_0 = 1.5 ).
  2. Compute ( x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} ).
  3. Calculating:
    • ( f(1.5) = 3 \log(1.5) - 1.5 ) and ( f'(1.5) = \frac{3}{1.5} - 1 ).
    • This gives ( x_1 \approx 1.72 ).
  4. Repeat using ( x_1 ) to find ( x_2 \approx 1.72 ) rounded to two decimal places.

Step 4

How many different towers are there that she could form that are three blocks high?

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Answer

To find the number of towers that are three blocks high, we can choose 3 blocks from 5, which can be arranged in:

5P3=5!(53)!=5!2!=5×4×3=60.5P3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} = 5 \times 4 \times 3 = 60.

So, there are 60 different towers of height 3.

Step 5

How many different towers can she form in total?

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Answer

To find the total number of towers of varying heights:

  1. For 2 blocks: ( 5P2 = 5 \times 4 = 20 ).
  2. For 3 blocks: ( 5P3 = 60 ).
  3. For 4 blocks: ( 5P4 = 5 \times 4 \times 3 \times 2 = 120. )

Thus, total:

20+60+120=200.20 + 60 + 120 = 200.

So, there are 200 different towers in total.

Step 6

Show that \( QKTN \) is a cyclic quadrilateral.

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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.

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