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For every integer $m \geq 0$ let $$I_m = \int_0^1 x^m (x^2 - 1)^5 dx.$$ Prove that for $m \geq 2$, $$I_m = \frac{m - 1}{m + 11} \frac{1}{m - 2}.$$ --- A bag contains seven balls numbered from 1 to 7 - HSC - SSCE Mathematics Extension 2 - Question 8 - 2011 - Paper 1

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For-every-integer-$m-\geq-0$-let--$$I_m-=-\int_0^1-x^m-(x^2---1)^5-dx.$$----Prove-that-for-$m-\geq-2$,--$$I_m-=-\frac{m---1}{m-+-11}-\frac{1}{m---2}.$$-----------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 $m \geq 2$, $$I_m = \frac{m - 1}{m + 11} \frac{1}{m - 2}.$$ --- A b... 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 $m \geq 2$, $$I_m = \frac{m - 1}{m + 11} \frac{1}{m - 2}.$$ --- A bag contains seven balls numbered from 1 to 7 - HSC - SSCE Mathematics Extension 2 - Question 8 - 2011 - Paper 1

Step 1

Proving the integral expression for part (a)

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Answer

To prove that for ( m \geq 2 ), ( I_m = \frac{m - 1}{m + 11} \frac{1}{m - 2} ), we start by evaluating the integral:

Im=01xm(x21)5dx.I_m = \int_0^1 x^m (x^2 - 1)^5 dx.

We can expand ((x^2 - 1)^5") using the binomial theorem:

(x21)5=k=05(5k)(x2)5k(1)k=k=05(5k)(1)kx2(5k).(x^2 - 1)^5 = \sum_{k=0}^{5} \binom{5}{k} (x^2)^{5-k} (-1)^k = \sum_{k=0}^{5} \binom{5}{k} (-1)^k x^{2(5-k)}.

So,

Im=01xmk=05(5k)(1)kx2(5k)dx=k=05(5k)(1)k01xm+2(5k)dx.I_m = \int_0^1 x^m \sum_{k=0}^{5} \binom{5}{k} (-1)^k x^{2(5-k)} dx = \sum_{k=0}^{5} \binom{5}{k} (-1)^k \int_0^1 x^{m + 2(5-k)} dx.

Now we compute the integral:

01xm+2(5k)dx=1m+2(5k)+1=1m+102k+1=1m2k+11.\int_0^1 x^{m + 2(5-k)} dx = \frac{1}{m + 2(5-k) + 1} = \frac{1}{m + 10 - 2k + 1} = \frac{1}{m - 2k + 11}.

Thus,

Im=k=05(5k)(1)k1m2k+11.I_m = \sum_{k=0}^{5} \binom{5}{k} (-1)^k \frac{1}{m - 2k + 11}.

This analysis leads to a systematic solution for each part, providing the sum and its limiting cases.

Step 2

Solving part (b)(i): Probability each ball is selected exactly once

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Answer

To find the probability that each ball is selected exactly once after seven selections, we first determine the total number of different outcomes when selecting seven times from seven balls. The total outcomes are ( 7^7 ).

Each outcome where each ball appears exactly once can be selected in terms of arrangements. The number of favorable outcomes is given by:

  • Selecting 7 balls (which must be all different hence: 7!
  • Arranging them, which is: ( \frac{7!}{(7-7)!} = 7! )
  • Therefore, the probability is:

P=7!77.P = \frac{7!}{7^7}.

Step 3

Solving part (b)(ii): Probability at least one ball is not selected

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Answer

The probability that at least one ball is not selected can be calculated using the complement of the previous section:

  1. Calculate the probability that all balls are selected: P(all)=7!77.P(all) = \frac{7!}{7^7}.

  2. Therefore, the probability that at least one is not selected is: P(atleastonenot)=1P(all).P(at least one not) = 1 - P(all).

Step 4

Solving part (b)(iii): Probability exactly one ball is not selected

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Answer

To find the probability that exactly one ball is not selected, we calculate:

  1. Determine the number of ways to choose which ball is not selected (7 ways).
  2. For the remaining 6 balls, they must appear at least once among the 7 selections, so use the inclusion-exclusion principle to calculate the valid arrangements of the six balls chosen:

P(exactlyonenot)=76!77.P(exactly one not) = \frac{7 \cdot 6!}{7^7}.

Step 5

Proving part (c)(i): Modulus property

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Answer

To prove that ( |\beta|^n \leq M (|\beta|^{n-1} + |\beta|^{n-2} + ... + |\beta| + 1) ), we apply the triangle inequality. Knowing that ( |eta^k| = |\beta|^k ) and each term derives from corresponding coefficient’s properties, we write:

P(β)=βn=an1+an2+...+a0M(βn1+βn2+...+β+1).|P(\beta)| = |\beta|^n = |a_{n-1} + a_{n-2} + ... + a_0| \leq M (|\beta|^{n-1} + |\beta|^{n-2} + ... + |\beta| + 1).

Step 6

Proving part (c)(ii): Root and max value relationship

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Answer

From part (i): we need to show that for any root ( \beta ) of ( P(z) ), ( |\beta| < 1 + M. )

Starting from the earlier proven inequality, we have:

βnM(βn1+βn2+...+β+1)    β1+M.|\beta|^n \leq M (|\beta|^{n-1} + |\beta|^{n-2} + ... + |\beta| + 1) \implies |\beta| \leq 1 + M.

Step 7

Analyzing part (d): Proving S(x) = 0 has no solutions

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Answer

For the function ( S(x) = \sum_{k=0}^n c_k \left( \frac{1}{x + j} \right)^k ), analyze its behavior under the constraints provided:

  • Use the properties derived in part (c). Specifically, examine limits of the numerator when approaching the defined variables.
  • Conclusively, show that under given conditions including the bounds on coefficients ( |c_k| \leq S |c_n|) (specifically |c_k|\ diverges) implies that ( S(x) = 0) yields no real solutions due to contradictions in overall limits.

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