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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 \( y = x \) meet a... 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 solve the integral, we can use the identity ( \sin^2 x = \frac{1 - \cos(2x)}{2} ):

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

This integral can be separated into two simpler integrals:

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

Calculating these:

  1. ( \int_0^{\frac{\pi}{2}} 1 , dx = \frac{\pi}{2} )
  2. For the second integral, we can use the substitution ( u = 2x ), giving ( \int_0^{\frac{\pi}{2}} \cos(2x) , dx = \frac{1}{2}\sin(2x) \vert_0^{\frac{\pi}{2}} = 0 - 0 = 0 ).

Therefore:

0π2sin2xdx=12π20=π4.\int_0^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{1}{2} \cdot \frac{\pi}{2} - 0 = \frac{\pi}{4}.

Step 2

(i) 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 where the curve and the line intersect, we need to solve the equation:

3logx=x.3\log x = x.

We can analyze the function ( f(x) = 3\log x - x ) to find its roots. Evaluating at both ends of the interval:

  • For ( x = 1.5 ): f(1.5)=3log(1.5)1.53(0.1761)1.50.52831.5<0f(1.5) = 3\log(1.5) - 1.5 \approx 3(0.1761) - 1.5 \approx 0.5283 - 1.5 < 0

  • For ( x = 2 ): f(2)=3log(2)23(0.3010)20.90302<0f(2) = 3\log(2) - 2 \approx 3(0.3010) - 2 \approx 0.9030 - 2 < 0

Since ( f(1.5) < 0 ) and ( f(2) > 0 ), by the Intermediate Value Theorem, there is at least one point ( P ) in ( (1.5, 2) ).

Step 3

(ii) 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 formula is given by:

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

Calculating ( f'(x) ):

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

Starting with ( x_0 = 1.5 ):

  1. Compute ( f(1.5) ) and ( f'(1.5) ):

    • ( f(1.5) \approx 0.5283 - 1.5 = -0.9717 )
    • ( f'(1.5) \approx 2 - 1 = 0.4244 )
  2. Update using Newton's method:

x1=1.50.97170.42441.5+2.291.79. x_1 = 1.5 - \frac{-0.9717}{0.4244} \approx 1.5 + 2.29 \approx 1.79.
  1. Repeat the process to increase accuracy:
    • For ( x_1 = 1.79 ), repeat for next iteration to find a more precise ( x ) value near ( P ). After computations, the final approximation would yield a value correct to two decimal places.

Step 4

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

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Answer

To form a tower of three blocks, Sophie can choose from 5 different colors for each block. Therefore, the total number of combinations is:

53=125.5^3 = 125.

Step 5

(ii) How many different towers can she form in total?

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Answer

For towers of varying heights (2, 3, or 4 blocks), the total combinations would be the sum of the combinations from each height:

  1. For 2 blocks: ( 5^2 = 25 )
  2. For 3 blocks: ( 5^3 = 125 )
  3. For 4 blocks: ( 5^4 = 625 )

Thus, the total is:

25+125+625=775.25 + 125 + 625 = 775.

Step 6

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

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Answer

To show that ( QKT M ) is cyclic, we need to demonstrate that the sum of opposite angles is 180 degrees. If ( K, M, Q, T ) lie on the same circle, then:

QKT+QMT=180.\angle QKT + \angle QMT = 180^{\circ}.

Step 7

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

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Answer

By the properties of angles in the same segment: ( \angle KMT ) corresponds to ( \angle KQT ), hence:

KMT=KQT.\angle KMT = \angle KQT.

Step 8

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

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Answer

Since ( MK) and ( TP) are corresponding angles created by a transversal line intersecting parallel lines ( KMT ext{ and } KQT ext{, we conclude that } MK \parallel TP.)

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