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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 \( \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \). (a) By considering \( f(x) = 3 \log x - x \), show that the curve \( y = 3 \log x \) and the line \( y = x \) meet... 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 the integral ( \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=120π2(1cos(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 - \cos(2x)) \, dx.

Calculating the integral gives:

=12[x12sin(2x)]0π2=12(π20)=π4.= \frac{1}{2} \left[ x - \frac{1}{2} \sin(2x) \right]_0^{\frac{\pi}{2}} = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}.

Step 2

By considering \( f(x) = 3 \log x - x \), show that the curve \( y = 3 \log x \) and the line \( y = x \) meet at a point \( P \) whose x-coordinate is between 1.5 and 2.

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Answer

To show that the curve and line intersect, we must find roots of the equation:

3logxx=0.3 \log x - x = 0.

We can evaluate this at both ends of the interval:

  • For ( x = 1.5 ): 3log(1.5)1.53(0.405)1.51.2151.5<0.3 \log(1.5) - 1.5 \approx 3(0.405) - 1.5 \approx 1.215 - 1.5 < 0.
  • For ( x = 2 ): 3log(2)23(0.693)22.0792>0. 3 \log(2) - 2 \approx 3(0.693) - 2 \approx 2.079 - 2 > 0.

Since the function changes signs between 1.5 and 2, by the Intermediate Value Theorem, there exists at least one root in that interval.

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

Newton's method uses the formula:

xn+1=xnf(xn)f(xn). x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.

We compute:

  • ( f'(x) = \frac{3}{x} - 1 ).

Starting with ( x_0 = 1.5 ):

  • Calculate ( f(1.5) ) and ( f'(1.5) ).
  • After one iteration, we find ( x_1 \approx 1.75 ). Thus, the x-coordinate of ( P ) is approximately ( 1.75 ).

Step 4

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

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Answer

For a tower that is three blocks high using five different colored blocks: The number of different permutations for selecting and arranging 3 blocks from 5 is:

5P3=5!(53)!=5!2!=5×4×31=60.5P3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3}{1} = 60.

Step 5

How many different towers can she form in total?

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Answer

To find the total number of towers of height 2, 3, 4, or 5:

  • For height 2: ( 5P2 = 20 ).
  • For height 3: ( 5P3 = 60 ).
  • For height 4: ( 5P4 = 120 ).
  • For height 5: ( 5P5 = 120 ). Adding these gives:\n20+60+120+120=320. 20 + 60 + 120 + 120 = 320. So, the total number of different towers Sophie can create is 320.

Step 6

Show that \( QKT M \) is a cyclic quadrilateral.

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Answer

For ( QKT M ) to be cyclic, we need to show that the sum of opposite angles equal 180 degrees. Angles ( \angle QKM ) and ( \angle QTM ) subtend the same arc ( QM ). Thus, ( \angle QKM + \angle QTM = 180^{\circ} ).

Step 7

Show that \( \angle KMT = \angle KQT. \)

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Answer

Since ( KM ) is a tangent to the circle at ( T ), we have that ( \angle KMT ) is equal to the angle subtended by the arc ( QT ), which is ( \angle KQT ) (alternate segment theorem). Thus, ( \angle KMT = \angle KQT. )

Step 8

Hence, or otherwise, show that \( MK \) is parallel to \( TP \).

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Answer

From the previous part, we found that ( \angle KMT = \angle KQT ). If two lines are cut by a transversal (which is ( TP )), and the corresponding angles are equal, then by the converse of the Alternate Interior Angles Theorem, the lines ( MK ) and ( TP ) are parallel.

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