B.SC. MATH ENTRANCE 2012

**ANSWER FIVE 6 MARK QUESTIONS AND SEVEN OUT 10 MARK QUESTIONS.**

**6 mark questions**

- Find the number of real solutions of \(x = 99 \sin (\pi ) x \)
- Find \({\displaystyle\lim_{xto\infty}\dfrac{x^{100} \ln(x)}{e^x \tan^{-1}(\frac{\pi}{3} + \sin x)}}\)
- (part A)Suppose there are k students and n identical chocolates. The chocolates are to be distributed one by one to the students (with each student having equal probability of receiving each chocolate). Find the probability of a particular student getting at least one chocolate.

(part B) Suppose the number of ways of distributing the k chocolates to n students be \((\dbinom{n+k-1}{k})\). Find the probability of a particular student getting at least one chocolate. - Show that \((\dfrac{\ln 12}{\ln 18})\) is an irrational number.
- Give an example of a polynomial with real coefficients such that \((P(\sqrt{2} + i)=0)\). Further given an example of a polynomial with rational coefficients such that \((P(\sqrt{2} + i)=0)\).
- Say f(1) = 2; f(2) = 3, f(3) = 1; then show that f'(x) = 0 for some x (given that f is a continuously differentiable function defined on all real numbers).

**10 mark questions**

- (part A) Suppose a plane has 2n points; n red points and n blue points. One blue point and one red point is joined by a line segment. Like this n line segments are drawn by pairing a red and a blue point. Prove that each such scheme of pairing segments will have two segments which do not intersect each other.

(part B) Suppose the position of the n red points are given. Prove that we can put n blue points in such a way that there are two segments (produced in the manner described in part A) which do not intersect each other. - (part A) Let ABCD be any quadrilateral. E, F, G and H be the mid points of the sides AB, BC, CD and DA respectively. Prove that EFGH is a parallelogram whose area is half of the quadrilateral ABCD.

(part B) Suppose the coordinates of E, F, G, H are given: E (0,0) , F(0, -1), G (1, -1) , H (1, 0). Find all points A in the first quadrant such that E, F, G, H be the midpoints of quadrilateral ABCD. - Let f be a function whose domain and codomain be non negative natural numbers such that f(f(f(n))) < f(n+1). Prove that:

(a) If f(n) = 0 then n = 0.

(b) f(n) < n+1

(c) If f(x) < n then x<n

Using the above prove that f is an identity function, that is f(n) = n. - Consider a sequence \((c_{n+2} = a c_{n+1} + b c_n) for (n \ge 0)\) where \((c_0 = 0)\). If gcd(b, k) = 1 then show that k divides n for infinitely many n.
- Find out the value of \((x^{2012} + \dfrac{1}{x^{2012}})\) when \((x + \dfrac{1}{x} = \dfrac{\sqrt{5} + 1}{2})\).

Hint

(a) Show that \((|{r +\dfrac{1}{r}}|\ge 2)\) for all real r.

(b) Prove that \((\sin \dfrac{\pi}{5} < \cos \dfrac{2\pi}{5} < \sin \dfrac{3\pi}{5})\). - A polynomial P(x) takes values \((\prime^{positive number})\) for every positive integer n,then show that p(x) is a constant polynomial.

If such a polynomial exist then show that there also exists a polynomial g(x)= \((\prime^l)\) where l is a fixed number. - Consider a set A = {1, 2, … , n}. Suppose \((A_1 , A_2 , … , A_k ) \) be subsets of set A such that any two of them consists at least one common element. Show that the greatest value of k is \((2^{n-1})\). Further, show that if they any two of them have a common element but intersection of all of them is a null set then the greatest value of k is \((2^{n-1})\).
- Suppose \((\displaystyle x = \sum_{i=1}^{10} \dfrac{1}{10 \sqrt 3} . \dfrac{1}{1+ (\dfrac{i}{10 \sqrt 3})^2}) and (\displaystyle y = \sum_{i=0}^9 \dfrac{1}{10 \sqrt 3} . \dfrac{1}{1+ (\dfrac{i}{10 \sqrt 3})^2})\)
- Show that \((x < \dfrac{\pi}{6} < y ) \)
- \((\dfrac{x+y}{2} < \dfrac{\pi}{6} ) \)

[b]OTHERS PLEASE CONTRIBUTE THE REST OF THE QUESTIONS (AND SOLUTIONS). WE ARE TRYING ON OUR END TO DO THE SAME[/b]

Can you provide the solutions to these CMI questions ??Thanks

solution of 5- x+1/x=2cos36=e^i(pi)/5+e^-i(pi)/5 /so (x-e^i(pi)/5)(1-1/xe^i(pi)/5)=0 /x=either e^i(pi)/5 or e^-i(pi)/5 /put the value x^2012+1/x^2012=2cos2012(pi)/5=2cos2(pi)/5=2cos72=2sin18=square root of 5-2