Let $ I_n = \int_0^1 (1 - x^2)^{n/2} \, dx $, where $ n \geq 0 $ is an integer. (i) Show that $ I_n = \frac{n}{n + 1} I_{n-2} $ for every integer $ n \geq 2 $. (ii) Evaluate $ I_5 $. The diagram shows the graph of a function $ f(x) $. Sketch the following curves on separate half-page diagrams. (i) $ y^2 = f(x) $ (ii) $ y = \frac{-1}{1 - f(x)} $ The points $ A, B, C $ and $ D $ lie on a circle of radius $ r $, forming a cyclic quadrilateral. The side $ AB $ is a diameter of the circle. The point $ E $ is chosen on the diagonal $ AC $ so that $ DE \perp AC $. Let $ \alpha = \angle DAC $ and $ \beta = \angle CAD $. (i) Show that $ AC = 2r \, \sin(\alpha + \beta) $. (ii) By considering $ \triangle ABD $ or otherwise, show that $ AE = 2r \cos \beta $. (iii) Hence, show that $ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha $.

Photo AI

Let $ I_n = \int_0^1 (1 - x^2)^{n/2} \, dx $, where $ n \geq 0 $ is an integer - HSC - SSCE Mathematics Extension 2 - Question 13 - 2013 - Paper 1

Question 13

$Let-$-I_n-=-\int_0^1-(1---x^2)^{n/2}-\,-dx-$,-where-$-n-\geq-0-$-is-an-integer-HSC-SSCE Mathematics Extension 2-Question 13-2013-Paper 1.png$

Let $ I_n = \int_0^1 (1 - x^2)^{n/2} \, dx $, where $ n \geq 0 $ is an integer. (i) Show that $ I_n = \frac{n}{n + 1} I_{n-2} $ for every integer \( n \geq 2 ... show full transcript

Worked Solution & Example Answer:Let $ I_n = \int_0^1 (1 - x^2)^{n/2} \, dx $, where $ n \geq 0 $ is an integer - HSC - SSCE Mathematics Extension 2 - Question 13 - 2013 - Paper 1

Step 1

Show that $ I_n = \frac{n}{n + 1} I_{n-2} $ for every integer $ n \geq 2 $

96%

114 rated

Answer

To show this, we will use integration by parts. Set ( u = (1 - x^2)^{n/2} ) and ( dv = dx ). Then, ( du = -nx(1 - x^2)^{(n/2) - 1} , dx ) and ( v = x ). Using integration by parts:

I_n = \int_0^1 u \, dv = uv \bigg|_0^1 - \int_0^1 v \, du.

Evaluating ( uv \bigg|_0^1 ) gives 0 at both limits: (\text{At } x = 1: (1 - 1^2)^{n/2} = 0 ) and at ( x = 0: u(0) \cdot v(0) = 0.)

Now consider the remaining integral:

\int_0^1 x \cdot (-nx(1 - x^2)^{(n/2) - 1}) \, dx = -n \int_0^1 x^2(1 - x^2)^{(n/2) - 1} \, dx.

Now, substituting ( x^2 = t ) (where # t varies from 0 to 1), we find: (dx = \frac{1}{2\sqrt{t}} , dt) leading to:

I_n = \frac{n}{2}\int_0^1 (1 - t)^{(n/2) - 1} t^{1/2} \, dt = \frac{n}{n + 1} I_{n - 2}.

Step 2

Evaluate $ I_5 $

99%

104 rated

Answer

To evaluate ( I_5 ), we can use the recursive relationship established in the previous part. Starting with ( n = 5 ):

I_5 = \frac{5}{5 + 1} I_{3} = \frac{5}{6} I_{3}.

Now find ( I_3 ):

I_3 = \frac{3}{3 + 1} I_{1} = \frac{3}{4} I_{1}.

For ( I_{1} ):

I_{1} = \int_0^1 (1 - x^2)^{1/2} \, dx = \frac{1}{2} \cdot \pi.

Thus,

I_3 = \frac{3}{4} \cdot \frac{\pi}{2} = \frac{3\pi}{8}.

Now substituting back:

I_5 = \frac{5}{6} \cdot \frac{3\pi}{8} = \frac{15\pi}{48}.

Step 3

Sketch the curve $ y^2 = f(x) $

96%

101 rated

Answer

To sketch the curve ( y^2 = f(x) ), observe that it transforms the original function by taking the square root, producing a positive and negative branch. Ensure that both branches are symmetric about the x-axis, as this is a reflection of ( f(x) ). Mark critical x-values where ( f(x) ) changes sign and draw the symmetrical curves accordingly.

Step 4

Sketch the curve $ y = \frac{-1}{1 - f(x)} $

98%

120 rated

Answer

This curve describes a transformation of the graph of ( f(x) ). Identify asymptotes at points where ( f(x) = 1 ) and sketch the curve, ensuring to note any vertical asymptotes and horizontal behavior based on the limits of ( f(x) ). The graph will typically exhibit a hyperbolic nature, moving toward the asymptotes from opposite sides.

Step 5

Show that $ AC = 2r \, \sin(\alpha + \beta) $

97%

117 rated

Answer

In the cyclic quadrilateral, using the extended Law of Sines, we establish that:

\frac{AC}{2r} = \sin(\alpha + \beta),

leading us to the desired result:

AC = 2r \, \sin(\alpha + \beta).

Step 6

Show that $ AE = 2r \cos \beta $

97%

121 rated

Answer

By considering triangle ( ABD ), applying the Law of Sines gives:

\frac{AE}{AB} = \frac{\sin \angle ADB}{\sin \angle ABD}.

Using known relationships and recognizing that ( AB = 2r ), we ascertain:

AE = 2r \cos \beta.

Step 7

Show that $ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha $

96%

114 rated

Answer

Using the angle sum identity for sine:

\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta.

This is a fundamental identity in trigonometry, relating angles to their sine and cosine components.

Let \( I_n = \int_0^1 (1 - x^2)^{n/2} \, dx \), where \( n \geq 0 \) is an integer - HSC - SSCE Mathematics Extension 2 - Question 13 - 2013 - Paper 1

Worked Solution & Example Answer:Let \( I_n = \int_0^1 (1 - x^2)^{n/2} \, dx \), where \( n \geq 0 \) is an integer - HSC - SSCE Mathematics Extension 2 - Question 13 - 2013 - Paper 1

Show that \( I_n = \frac{n}{n + 1} I_{n-2} \) for every integer \( n \geq 2 \)

Evaluate \( I_5 \)

Sketch the curve \( y^2 = f(x) \)

Sketch the curve \( y = \frac{-1}{1 - f(x)} \)

Show that \( AC = 2r \, \sin(\alpha + \beta) \)

Show that \( AE = 2r \cos \beta \)

Show that \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha \)

Join the SSCE students using SimpleStudy...