Lots of Harder Questions (1 Viewer)

[Trebla]

Hi guys, I encountered these questions while attempting some trial papers and I have no idea how to do them. It would be greatly appreciated if you could work them out and explain how to do it. (Also, if you're looking for some rather challenging questions these may be it) If I enounter more questions I can't do, I'll post them up as well.

1)
i) Find the probability that 6 throws of a fair die result in exactly 3 even scores
ii) Find the probability that 6 throws of a fair die result in exactly 3 even scores, all of which are different
iii) Find the probability that exactly 6 throws of a fair die are needed in order to obtain 3 even scores
iv) Find the probability that at least 6 throws of a fair die are needed in order to obtain 3 even scores

2)
n letters L(1), L(2), L(3),…, L(n) are to be placed at random into n addressed envelopes E(1), E(2), E(3),…, E(n), each bearing a different address, where E(i) bears the correct address for letter L(i) for i = 1, 2, 3,…., n. Let U(n) be the number of arrangements where no letter is placed in the correct envelope for n, a positive integer where n ≥ 2.
i) Show that U(2) = 1 and U(3) = 2
ii) Deduce that U(k + 1) = k(U(k) + U(k – 1)) for k = 4, 5, 6,…
iii) Use the results from (i) to calculate U(4) and U(5).
iv) Show by mathematical induction that U(n) = n! [1/2! – 1/3! + 1/4! – …… + (-1)ⁿ/n!]
v) If there are 5 letters and envelopes:
a) Explain why the probability no letter is placed in the correct envelope is U(5)/120
b) Show that the probability that exactly one letter is placed in the correct envelope is 5U(4)/120 and calculate this probability as a fraction
vi) Deduce that (from k = 2 to n)∑ nCk. U(k) = n! – 1

3)
An unbiased die is thrown six times. Find the probabilities that the six scores obtained will:
i) be 1, 2, 3, 4, 5, 6 in some order
ii) have a product which is an even number
iii) consist of exactly two 6’s and four odd numbers
iv) be such that a 6 occurs only on the last throw and exactly three of the first five throws result in odd numbers

4)
n co-planar lines are such that the number of intersection points is a maximum.
i) How many intersection points are there?
ii) If n such lines divide the plane into U(n) regions, show that U(n) = U(n – 1) + n. Hence deduce that
U(n) = 1 + n(n+ 1)/2. How many of these regions have finite area?

5)
By considering the stationary value of the function f(x) = x – ln x, show that for x > 0, ln x ≤ x – 1:
i) Deduce that if a(1), a(2), …. , a(n) are positive numbers and A = 1/n (from 1 to n)∑ a(n) then
(from 1 to a)∑ ln [a(n)/A] ≤ (from 1 to a)∑ a(n)/A – n = 0.
ii) Hence deduce that [a(1) + a(2) + a(3) + …….. + a(n)]/n ≥ [a(1).a(2).a(3)…..a(n)]^(1/n) .
iii) Hence or otherwise prove that if u, v, w are positive and u + v +w = 1 then 1/u² + 1/v² + 1/w² ≥ 27.

6)
A sequence of numbers a(1), a(2), a(3),….. is such that a(n+ 1) – a(n) = brⁿ where r =/= 0, 1. Given that a(n) can be expressed in the form p + qrⁿ, where p and q are independent of n, find the values of p and q in terms of a, b and r. Verify that the numbers 1, 4, 10, 22, …. Begin a sequence of the above type. Obtain a formula for the nth term of this sequence and find the sum of first n terms of the sequence

7)
In a triangle ABC, b, c and B are given such that two distinct triangles ABC are possible. Show that the difference between the two possible values of the third sides of the two triangles is 2√(b² - c²sin²B)

8)
i) If x ≥ 0, show that 2x/1 + x² ≤ 1
ii) Show that eª ≥ 1 + a² for a ≥ 0

9)
Prove that if n is positive integer and x > 0 then x^(n) + x^(–n) > x^(n –1) + x^(1–n) provided that x =/= 1

10)
The positive integers are bracketed as follows: (1), (2,3), (4,5,6), ….. where there are r integers in the rth bracket. Prove that the sum of integers in rth bracket is r(r² + 1)/2


Enjoy!...

 

[Affinity]

1. i) binomial theorem.. 3 unit stuff ANS: 6C3 * 2^-6
ii) One way to think about this is conditioning.. given that there are exactly 3 even scores what's the probability of them being different? This is 1*(2/3)*(1/3) = 2/9
so the answer to this part will be 2/9 * 6C3 * 2^-6
iii) That will be the probability of obtaining exactly 2 even scores out of 5 throws multiplied by the probability of an even score for the 6th throw which is:

5C2 * 2^-5 * (1/2)

iv) This is equal to the probability that one doesn't get 3 even scores with 5 throws, which is (5C0 + 5C1 + 5C2)* 2^-5


2. i) Trivial.. list all cases
ii) Well to misplace all k+1 letters.. think about where L(k+1) can go to... there are k places.. namely E(1)... E(k). Now suppose that L(k+1) is placed into E(i). Then we have the following 2 cases.. if L(i) is placed into E(k+1), then there are U(k-1) ways of arranging the remaining letters, otherwise if L(i) is not placed into E(k+1), there will be U(k) ways of arranging the letters, putting this altogether gives us:

u(k+1) = k*(u(k) + u(k-1))

iii) trivial
iv) This is quite standard, leave this to you to write
v) a] well obvious, U(5)/5!
b] obvious. (5 different letters that could be the one which is correctly placed, the U(4) for the remaining letters, divided by 5!)

vi) hmm it should be for k = 1 to n not 2 to n.. Basically nCk * u(k) is the number of arrangements leaving exactly k letters misplaced in a group of n..
the right hand side is the total number of arrangements (n!) less 1 (for the 1 arrangement where all letters are where it should be)

3. i) 6/6* 5/6 * 4/6 * 3/6 * ... * 1/6 = 6!/6^6
ii) same as atleast 1 even score.. so that will be 1 - 2^-6
iii) 6C2 * (1/6)^2 * (3/6)^4
iv) [5C3 * (3/6)^3 * (2/6)^2] * 1/6 where [5C3 * (3/5)^3 * (2/5)^2] is the probability of getting 3 odd numbers and 2 even numbers which are not 6 in the first 5 throws

4. i) I think you mean maximal in this question, since the minimum number of intersections is 0, when all the lines are parallel. The maximum number of intersections is n*(n-1)/2 (each line intersects with another line once)
ii,iii) Consult the cambridge book

5. ln(x) <= x - 1 .. you can do this.
i) you have ln(a_1/A) <= a_1/A - 1, ln(a_2/A) <= a_2/A - 1 etc.. add them up.
ii) well you have from part 1, ln( [a(1)*a(2)*...*a(n)]/A^n ) <= 0, so [a(1)*a(2)*...*a(n)]/A^n<= 1, the rest is obvious. and that 0.25 should have been an (1/n).
iii)As it stands the question is wrong... eg if u=v=w = 1/3

6. a(n) = [a(n) - a(n-1)] + [a(n-1) - a(n-2)] + ... + [a(2) - a(1)] + a(1)
= b [ r^(n-1) + r^(n-2) + ... + r^(1)] + a(1)
= br/(r-1) * [r^(n-1) - 1] + a(1)
= b/(r-1) *r^n - br/(r-1) + a(1)
so p = a(1) - br/(r-1)
q = b/(r-1)

7. Well, you know that b^2 = a^2 + c^2 - 2ac*cos(B) and they want you to show that the 2 values of a which fits the equation has a certain difference... so what do you do.. solve the equation for a:

a^2 - 2ac*cos(B) + (c^2 - b^2) = 0

so a = [2c*cos(B) +/- sqrt(4(c * cos(B))^2 - 4(c^2 - b))]/2
and the difference between them is
sqrt(4(c * cos(B))^2 - 4(c^2 - b^2))]
= 2* sqrt(b^2 -c^2 *sin(B)^2

8. i) start with (x-1)^2 >= 0
ii) use calculus.

9. You can use induction for this (the usual way) or you can use calculus:
let f(a) = x^a + x^-a, f'(a) = ln(x) * x^a - ln(x) * x^-a which is obviously positive, (think why! consider the cases where x>1 and when x<1)
therefore f(n+1) > f(n) and the statement is proven

10. There are (r-1)r/2 numbers in total in the first r-1 brackets. so the r^th bracket is (r(r-1)/2 + 1, r(k-1)/2 + 2 , ... , r(r-1)/2 + r)
so their sum is (r-1)r^2 / 2 + r(r+1) / 2 = r(r^2 + 1)/2

 

[Trebla]

vi) hmm it should be for k = 1 to n not 2 to n..

k = 2 to n was how the notation was written on the exam paper

ii,iii) Consult the cambridge book

Which page/part of the cambridge book?

iii)As it stands the question is wrong... eg if u=v=w = 1/3

If u = v = w = 1/3 then the equality holds true doesn't it? How is it wrong?

You can use induction for this (the usual way) or you can use calculus:
let f(a) = x^a + x^-a, f'(a) = ln(x) * x^a - ln(x) * x^-a which is obviously positive, (think why! consider the cases where x>1 and when x<1)
therefore f(n+1) > f(n) and the statement is proven

How is f'(a) = ln(x) * x^a - ln(x) * x^-a? (i.e. where did the logarithms come from?)

Otherwise thanks...

 

[acmilan]

a is the independent variable, not x, so if f(a) = x^a then f'(a) = x^aln(x), just like if f(x) = 2^x then f'(x) = 2^xln2

 

[Affinity]

k = 2 to n was how the notation was written on the exam paper


Which page/part of the cambridge book?


If u = v = w = 1/3 then the equality holds true doesn't it? How is it wrong?


How is f'(a) = ln(x) * x^a - ln(x) * x^-a? (i.e. where did the logarithms come from?)

Otherwise thanks...

1.) oh sorry 2 to n works too.. coz U(1) = 0.. so 1 to n is same as 2 to n
2.) the harder 3 unit induction part
3.) oops Yeah made a mistake.. the question is right.. what you do is you prove that
(a(1)*a(2)*...*a(n))^(1/n) >= sqrt(n/(a(1)^-2 + a(2)^-2 + ...)) (use the AM GM), then you can conclude that

A >= sqrt(n/(a(1)^-2 + a(2)^-2 + ...)) and the result should follow easily

4.) Acmilan explained it