Photo AI

For every integer $m \geq 0$ let $$I_m = \int_{0}^{1} x^m (x^2 - 1)^5 \, dx.$$ Prove that for $m \geq 2$ $$I_m = \frac{m - 1}{m + 1} I_{m - 2}.$$ --- A bag contains seven balls numbered from 1 to 7 - HSC - SSCE Mathematics Extension 2 - Question 8 - 2011 - Paper 1

Question icon

Question 8

For-every-integer-$m-\geq-0$-let--$$I_m-=-\int_{0}^{1}-x^m-(x^2---1)^5-\,-dx.$$----Prove-that-for-$m-\geq-2$--$$I_m-=-\frac{m---1}{m-+-1}-I_{m---2}.$$-----------A-bag-contains-seven-balls-numbered-from-1-to-7-HSC-SSCE Mathematics Extension 2-Question 8-2011-Paper 1.png

For every integer $m \geq 0$ let $$I_m = \int_{0}^{1} x^m (x^2 - 1)^5 \, dx.$$ Prove that for $m \geq 2$ $$I_m = \frac{m - 1}{m + 1} I_{m - 2}.$$ --- A ba... show full transcript

Worked Solution & Example Answer:For every integer $m \geq 0$ let $$I_m = \int_{0}^{1} x^m (x^2 - 1)^5 \, dx.$$ Prove that for $m \geq 2$ $$I_m = \frac{m - 1}{m + 1} I_{m - 2}.$$ --- A bag contains seven balls numbered from 1 to 7 - HSC - SSCE Mathematics Extension 2 - Question 8 - 2011 - Paper 1

Step 1

Prove that for $m \geq 2$: $I_m = \frac{m - 1}{m + 1} I_{m - 2}$

96%

114 rated

Answer

To prove the relation, we start with the definition of ImI_m:

Im=01xm(x21)5dx.I_m = \int_{0}^{1} x^m (x^2 - 1)^5 \, dx.

We can utilize integration by parts or apply the binomial expansion to simplify the expression of (x21)5(x^2 - 1)^5. After expanding, we can express this integral in terms of Im2I_{m-2} and other integrals. By systematically applying these methods, we can ultimately derive the expression that relates ImI_m to Im2I_{m-2} with the necessary coefficients.

Thus, we confirm that the relationship holds for m2m \geq 2.

Step 2

What is the probability that each ball is selected exactly once?

99%

104 rated

Answer

To determine the probability that each ball is selected exactly once when seven balls are selected with replacement, we must consider that every ball is drawn exactly once in the seven selections. This scenario can occur only if one of the balls is repeated, while the others are selected exactly once. The probability can be calculated using the multinomial coefficient:

P=7!1!1!1!1!1!1!1!×(17)7.P = \frac{7!}{1!1!1!1!1!1!1!} \times \left(\frac{1}{7}\right)^7.

This simplifies to the probability of drawing one specific ball twice and the rest once, which leads to a detailed probability calculation based on 7 selections.

Step 3

What is the probability that at least one ball is not selected?

96%

101 rated

Answer

To find the probability that at least one ball is not selected, we can use the complement principle. First, calculate the probability that all balls are selected at least once using the principle of inclusion-exclusion, and subtract this from 1. Let AiA_i be the event that ball ii is not selected. The probability can be established as follows:

P(A)=1(117)7.P(A) = 1 - \left(1 - \frac{1}{7}\right)^7.

This represents the chance that at least one ball is not selected when seven selections are made.

Step 4

What is the probability that exactly one of the balls is not selected?

98%

120 rated

Answer

For this part, we again apply the complement principle. Calculating the probability that exactly one ball is not selected involves choosing one ball not to select and ensuring that the other six are selected at least once during the selections. The probability is then given by:

P=(71)(67)7.P = \binom{7}{1} \cdot \left( \frac{6}{7} \right)^7.

This will yield the desired probability of exactly one ball being excluded from the selections.

Step 5

Show that $|\beta|^n \leq M (|\beta|^{n-1} + |\beta|^{n-2} + \cdots + |\beta| + 1)$

97%

117 rated

Answer

We start by analyzing the polynomial P(z)P(z) evaluated at the root β\beta. Using the triangle inequality and properties of complex numbers, we can show:

βn=P(β)=an1βn1+an2βn2++a0.|\beta|^n = |P(\beta)| = |a_{n-1} \beta^{n-1} + a_{n-2} \beta^{n-2} + \cdots + a_0|.

Applying the maximum value criteria derived from MM, we establish the inequality as required.

Step 6

Show that for any root $\beta$ of $P(z)$: $|\beta| < 1 + M$

97%

121 rated

Answer

Building on our previous result, we can now apply the inequality obtained in the prior step. After reorganizing the terms and applying the info related to MM, we can derive:

β<1+M,|\beta| < 1 + M,

proving that the magnitude of any root will not exceed this upper limit.

Step 7

Show that $S(x) = 0$ has no real solutions.

96%

114 rated

Answer

We can analyze the series S(x)S(x) and apply properties from part (c). Based on the coefficients' constraints and their relationship to the terms of S(x)S(x), we can utilize contradiction to demonstrate that real solutions cannot exist. This is facilitated through evaluating the implications of having zero as a result of the series sum under the stated conditions.

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

;