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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... 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 ( \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 integrals:

0π21dx=π2,and0π2cos(2x)dx=[sin(2x)2]0π2=[00]=0.\int_0^{\frac{\pi}{2}} 1 \,dx = \frac{\pi}{2}, \quad \text{and} \quad \int_0^{\frac{\pi}{2}} \cos(2x) \,dx = \left[\frac{\sin(2x)}{2}\right]_0^{\frac{\pi}{2}} = \left[0 - 0 \right] = 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

(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 show that the curve ( y = 3 \log x ) intersects the line ( y = x ), we need to analyze the function ( f(x) = 3 \log x - x ). By calculating the values:

  • At ( x = 1.5):
f(1.5)=3log(1.5)1.53(0.405)1.51.2151.50.285.(f(1.5)<0)f(1.5) = 3 \log(1.5) - 1.5 \approx 3(0.405) - 1.5 \approx 1.215 - 1.5 \approx -0.285. \quad (f(1.5) < 0)
  • At ( x = 2):
f(2)=3log(2)23(0.693)22.07920.079.(f(2)>0)f(2) = 3 \log(2) - 2 \approx 3(0.693) - 2 \approx 2.079 - 2 \approx 0.079. \quad (f(2) > 0)

Since ( f(1.5) < 0 ) and ( f(2) > 0 ), by the Intermediate Value Theorem, we conclude that there exists a point ( P ) in the interval (1.5, 2) where the curve meets the line.

Step 3

(ii) Use one application of Newton's method, starting at \( x = 1.5 \), to find an approximation to the \( x \)-coordinate of \( P \). Give your answer correct to two decimal places.

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Answer

Using Newton's method, we first need the derivative ( f'(x) ):

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

Starting at ( x_0 = 1.5 ):

  1. Calculate ( f(1.5) ) and ( f'(1.5) ): f(1.5)0.285extandf(1.5)31.511.0.f(1.5) \approx -0.285 \quad ext{and} \quad f'(1.5) \approx \frac{3}{1.5} - 1 \approx 1.0.

  2. Use the formula: xn+1=xnf(xn)f(xn).x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.

  3. Therefore: x1=1.50.2851.0=1.5+0.285=1.785.x_1 = 1.5 - \frac{-0.285}{1.0} = 1.5 + 0.285 = 1.785.

Now repeating the process for ( x_1 = 1.785 ):

  1. Calculate ( f(1.785) ): ( f(1.785) \approx 0.032. ) f(1.785)31.78510.681.f'(1.785) \approx \frac{3}{1.785} - 1 \approx 0.681.

  2. Calculate ( x_2 ): x2=1.7850.0320.6811.7850.047=1.738.x_2 = 1.785 - \frac{0.032}{0.681} \approx 1.785 - 0.047 = 1.738.

The approximation to two decimal places is ( x \approx 1.79 ).

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 that is three blocks high, we can choose any three blocks from five, allowing for repetition of colors since blocks can be stacked in any order. The number of distinct combinations with repetition is given by:

Number of towers=53=125.\text{Number of towers} = 5^3 = 125.

Step 5

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

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Answer

To calculate the total number of towers Sophie can form, we need to consider towers of heights 2, 3, and 4.

  • For 2 blocks: 52=25.5^2 = 25.
  • For 3 blocks: 53=125.5^3 = 125.
  • For 4 blocks: 54=625.5^4 = 625.

Therefore, the total number of towers is:

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

Step 6

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

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Answer

To show ( QKT ) is cyclic, we need to demonstrate that the opposite angles add up to 180 degrees. We evaluate angles at points ( K ) and ( T ). Since both points lie on the circle, the angles formed between segments are subtended by the same arc, confirming that:

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

Step 7

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

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Answer

Both angles ( KMT ) and ( KQT ) are subtended by the diameter of the circle, proving by the Inscribed Angle Theorem that:

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 ( KMT ) and ( KQT ) are equal, by the Corresponding Angles Postulate, it follows that:

MKTP.MK \parallel TP.

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