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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 \( \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \) . (i) By considering \( f(x) = 3 \log x - x \), show that the curve \( y = 3 \log x \) and the line \( y = x \) mee... 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, we can use the identity: ( \sin^2 x = \frac{1 - \cos(2x)}{2} ).

Thus, our integral becomes:

0π2sin2xdx=0π21cos(2x)2dx=12(0π21dx0π2cos(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} \left( \int_0^{\frac{\pi}{2}} 1 \, dx - \int_0^{\frac{\pi}{2}} \cos(2x) \, dx \right).

Calculating the first integral, we get:

0π21dx=[x]0π2=π2\int_0^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2}

The second integral can be calculated as:

0π2cos(2x)dx=[sin(2x)2]0π2=sin(π)sin(0)2=0.\int_0^{\frac{\pi}{2}} \cos(2x) \, dx = \left[ \frac{\sin(2x)}{2} \right]_0^{\frac{\pi}{2}} = \frac{\sin(\pi) - \sin(0)}{2} = 0.

Thus,

0π2sin2xdx=12(π20)=π4.\int_0^{\frac{\pi}{2}} \sin^2 x \, dx = \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 find the intersection points of ( f(x) = 3 \log x - x ), we first evaluate this function at the endpoints.

  1. For ( x = 1.5 ): [ f(1.5) = 3 \log(1.5) - 1.5 \approx 3(0.405) - 1.5 \approx 1.215 - 1.5 = -0.285. ]

  2. For ( x = 2 ): [ f(2) = 3 \log(2) - 2 \approx 3(0.693) - 2 \approx 2.079 - 2 = 0.079. ]

Thus, since ( f(1.5) < 0 ) and ( f(2) > 0 ), by the Intermediate Value Theorem, there exists a point ( P ) in the interval ( (1.5, 2) ) where ( f(P) = 0 ).

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 is given by:

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

We first calculate the derivative of ( f(x) ):

f(x)=3x1.f'(x) = \frac{3}{x} - 1.

Now, we start at ( x_0 = 1.5 ):

  1. Calculate ( f(1.5) ) and ( f'(1.5) ): [ f(1.5) \approx -0.285, ; f'(1.5) \approx \frac{3}{1.5} - 1 \approx 1.0. ]

  2. Using Newton's formula: [ x_1 = 1.5 - \frac{-0.285}{1.0} = 1.785. ]

  3. Repeat for ( x_1 = 1.785 ): [ f(1.785) \approx 0.030, ; f'(1.785) \approx 0.678. ] [ x_2 = 1.785 - \frac{0.030}{0.678} \approx 1.782. ]

  4. Continue until convergence: The answer can be approximated at ( x \approx 1.78 ) after calculus up 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 total combinations for stacking three blocks from five different colored blocks, we can use the formula:

C(n,r)=n!(nr)!r!.C(n, r) = \frac{n!}{(n - r)! r!}.

For each arrangement, we choose 3 blocks from 5,

C(5,3)=5!(53)!3!=10.C(5, 3) = \frac{5!}{(5 - 3)! 3!} = 10.

However, since the order in which the blocks are stacked matters, there are ( 3! = 6 ) ways to arrange those blocks. Thus,

Total arrangements = ( 10 \times 6 = 60 ).

Step 5

How many different towers can she form in total?

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Answer

To calculate the total number of towers she can form with 2, 3, 4, or 5 blocks high, we calculate:

  • For two blocks: ( C(5, 2) \times 2! = 10 \times 2 = 20 )
  • For three blocks: ( C(5, 3) \times 3! = 10 \times 6 = 60 )
  • For four blocks: ( C(5, 4) \times 4! = 5 \times 24 = 120 )
  • For five blocks: ( C(5, 5) \times 5! = 1 \times 120 = 120 )

Adding these together:

Total = ( 20 + 60 + 120 + 120 = 320 ).

Step 6

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

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Answer

For quadrilateral ( QKTN ) to be cyclic, the opposite angles must sum to ( 180^\circ ). We prove:

  1. By alternate segment theorem, ( \angle QKN + \angle QTN = 180^\circ ) since they subtend the same arc.

Thus, ( QKTN ) is cyclic.

Step 7

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

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Answer

From the tangent property at point ( T ): ( \angle KMT ) is equal to the angle subtended by arc ( KT ).

Since ( KQT ) subtends the same arc, we have: ( \angle KMT = \angle KQT. )

Step 8

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

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Answer

From the previous angle relationships:

  • If ( \angle KMT = \angle KQT ), it follows that lines ( MK ) and ( TP ) are parallel by the converse of the alternate interior angles theorem.

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