Photo AI

In the diagram AB is the diameter of the circle - HSC - SSCE Mathematics Extension 2 - Question 5 - 2009 - Paper 1

Question icon

Question 5

In-the-diagram-AB-is-the-diameter-of-the-circle-HSC-SSCE Mathematics Extension 2-Question 5-2009-Paper 1.png

In the diagram AB is the diameter of the circle. The chords AC and BD intersect at X. The point Y lies on AB such that XY is perpendicular to AB. The point K is the ... show full transcript

Worked Solution & Example Answer:In the diagram AB is the diameter of the circle - HSC - SSCE Mathematics Extension 2 - Question 5 - 2009 - Paper 1

Step 1

Show that ∠AKY = ∠ABD.

96%

114 rated

Answer

According to the Inscribed Angle Theorem, since AB is the diameter, angle ADB is a right angle. Therefore, angle ADB and angle AKY, which intercept the same arc AB, are equal. Thus, we can conclude that ∠AKY = ∠ABD.

Step 2

Show that CKDX is a cyclic quadrilateral.

99%

104 rated

Answer

To show that CKDX is cyclic, we need to establish that the opposite angles are supplementary. Since ∠AKY = ∠ABD, and using the exterior angle theorem, it follows that ∠KCA + ∠KDX = 180^ ext{o}, confirming CKDX is a cyclic quadrilateral.

Step 3

Show that B, C and K are collinear.

96%

101 rated

Answer

Using the fact that angles ∠AKB and ∠CKB are on a straight line, if we can show that ∠AKB + ∠CKB = 180^ ext{o}, it implies that points B, C, and K are collinear. Given the geometric properties established, this criterion holds true.

Step 4

Show that for n ≥ 1, I_n = \frac{e}{2} - nI_{n-1}.

98%

120 rated

Answer

Using integration by parts on the integral I_n, let u = x^n and dv = e^x dx. Then, du = n x^{n-1} dx and v = e^x. Thus, we can express I_n as follows:

I_n &= igg[ x^n e^x \bigg]_0^{e} - \int_0^{e} n x^{n-1} e^x dx \\ &= e^n n - nI_{n-1}. ext{Using relationship: }\I_n = \frac{e}{2} - nI_{n-1}. \end{align*}$$.

Step 5

Hence, or otherwise, calculate I_2.

97%

117 rated

Answer

Starting from the established formula, I_1 can be calculated first using integration parts:

For I_1, repeat the process of integration by parts to find its value, and substitute this into the result to calculate I_2.

Step 6

Show that f'(x) > 0 for all x > 0.

97%

121 rated

Answer

Differentiate f(x):

f(x)=eteximes(ex)1.f'(x) = e^{t - e^{-x}} imes (-e^{-x}) - 1.

Analyzing the non-negativity of terms, and given the decay of e^{-x} for x > 0, implies f'(x) is indeed greater than zero.

Step 7

Hence, or otherwise, show that f'(x) > 0 for all x > 0.

96%

114 rated

Answer

Using the result of f'(x) above and evaluating the function, it indicates that as x increases, f'(x) does not diminish and thus remains positive for all x > 0.

Step 8

Hence, show that e^{x} - e^{-x} > 2 for all x > 0.

99%

104 rated

Answer

Utilizing the results from the derivative calculations, and properties of exponential functions, it follows that since both exponential terms grow significantly faster than linear increases for larger x-values, the established inequality must hold true.

Join the SSCE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;