Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    10
    ReputationRep:
    Hello guys...
    I have many more problems here which I am trying to solve...
    let me share some here

    p.s.- feel free to help or ignore

    1. find greater of : e^\pi or \pi^e
    2 e^k = kx find largest value of k which e^k = kx has only 1 solution
    Ans : k=e I have no idea how
    3. last digit 17^{23}
    4. prove \frac{x}{y}+\frac{y}{x} \geq2 when xy>0
    5. cosx-ax-b=0 has two solutions show sinx=-a has at least one solution
    6. sketch graph y=x^3+ax+b What would be the relationship between a and b in which curve would have 3 roots, 2roots, 1 root.
    7. What are possible remainders when square number is divided by 4
    8. can number 4000003 be written in sum of 2 square number?
    9. A sector is cut from a circle and a cone is made from the remaining material by pulling the two freshly cut edges together. What should angle of the sector be in order to maximise the volume of the cone??
    10. How many 0's in 100!
    11. Four cards A 2 3 K written on them, odd number implies that there is a vowel on the other side of the card, how many cards do I need to turn in order to disprove the statement?
    12. Prove n^2(mod7)=(n+7)^2(mod7)
    13. sketch x^3-ax^2
    14. Sketch y=x^2-x^4 then y^2=x^2-x^4
    15. Prove lengths of 2 diagonals of a quadrilateral are greater than the perimeter of it...-question is not clear
    Offline

    17
    ReputationRep:
    1) You need to use logs, rearrange it, and you'll end up with e< ln(pi) / pi. Which you can solve by looking at it like lnx/x.

    2) Try looking at it graphically.

    3) 17x17x17....x17 the last digit is dependent on the multiples of 7. If you look at the pattern

    7 = 7 mod 10
    7*7 = 49 = 9 mod 10
    49*7 = 9 mod 10 * 7 = 3 mod 10
    49*7*7 = 3 mod 10 * 7 = 1 mod 10
    49*7*7*7 = 1 mod 10 * 7 = 7 mod 10
    ...

    So the pattern of the last digit is 7,9,3,1 repeated each time. So the 8th, 12th, 16th, 20th, 24th will be a 1, the 23rd will be a 3.
    Offline

    2
    ReputationRep:
    For number 4, you have to read it carefully

    First rearrange it
    $$\frac{x}{y} + \frac{y}{x} \geq 2$$
    $$x^2 + y^2 \geq 2xy$$
    $$x^2 + 2xy + y^2 \geq 0$$

    If both $x$ and $y$ are greater than zero, then all three terms in the inequality must be greater than zero. If they're both greater than zero then their product must be greater than zero so $$xy > 0$$ implies $$\frac{x}{y} + \frac{y}{x} \geq 2$$.
    Offline

    14
    ReputationRep:
    4. x/y + y/x>=2
    (x^2+y^2)/xy>=2
    x^2+y^2>=2xy
    x^2+y^2-2xy>=0
    (x-y)^2>=0
    As this final statement is true (because squaring an integer gives a positive result), the initial statement must also be true.

    7. 0^2=0=0 (mod 4)
    1^2=1=1 (mod 4)
    2^2=4=0 (mod 4)
    3^2=9=1 (mod 4)

    As all numbers are 0,1,2 or 3 (mod 4) a square number must have a remainder of 0 or 1 when divided by four.

    10. If you want to know the number of 0's at the end of 100!, use this. I'm not sure how to work out if there are any other 0's in it.

    The number of 0's in 100! will be determined by the number of 2's and 5's when 100! is written as a product of primes. More specifically, a single zero will be added for every (2 and 5) which appears. As there are clearly more 2's in the prime factorisation (i.e. there will be at least a 2 for every 5), the number of 0's is equal to the number of 5's. There are twenty multiples of five between 1 and 100, and four of these have two fives in their prime factorisation. Therefore there are 20+4=24 0's

    12. (n+7)^2 (mod 7)=n^2+14n+49=n^2 (mod 7) because 14 and 49=0 (mod 7)
    Offline

    17
    ReputationRep:
    4) Put everything into one fraction. Multiply by xy, move everything onto the LHS. Then factorise.

    5) cos(x) = ax + b. Which is obviously a straight line. If it only has two solutions, then I think the line has to tangential to cos(x) otherwise you'll actually end up with 4 solutions. So equate the gradients.

    6) You need to look at the turning points of the cubic, if the maximum occurs below the x axis, then there's obviously only 1 solution. If the maximum lies on the line x=0 then there's two, if the maximum lies above the x axis, and the minimum below, then there's 3 solutions. If both the maximum and minimum are above the x axis, there's only one.

    7) This is number theory, specifically modular arithmetic. There's a pattern with the square numbers when divided by 4, investigate them.

    8) This is modular arithmetic again, you need to find whether the possible mod combinations of a square number can add to give the same modular value of 4000003 (obviously using the same mod)

    9) I'll leave shapes to someone else :lol:
    Offline

    0
    ReputationRep:
    15. Shouldn't it be: perimeter of quadrilateral>sum of the lengths of its diagonals? Anyway, consider quadrilateral ABCD. By the triangle inequality, AB+BC>AC, BC+CD>BD, CD+DA>AC and DA+AB>DB. Summing up the following inequalities yields the desired result.
 
 
 
Poll
Who is your favourite TV detective?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.