## Pages

### TIFR-GS PREVIOUS YEAR SOLUTION PAPER ON MATHEMATICS

#### TIFR GS-2011 MATHEMATICS PAPER SOLUTION

Tata Institute of Fundamental Research organises a integrated ph.d & ph.d entrance examination each year in Mathematics. This is the solution of GS-2011.

Q. T/F: The equation 63x + 70y + 15z = 2010 has an integer solution.
Sol :- TRUE. Take x=15,y=0,z=71.
This is one of the integral solution of the given equation.

Q. T/F : The space of solutions of infinitely differentiable functions satisfying the equation y" + y' = 0 is infinite dimensional.
Sol:- FALSE   y"+y'=0 have the solution
y=c1sinx + c2cosx ,the solution is infinitely differentiable but it is not one dimensional,since it depends only on the value of x.

Q. T/F :- The polynomial x^4 +7x^3 -13x^2 +11x has exactly one real root.
Sol:-  FALSE. The given polynomial is of even degree which can have at most 4 roots of which it can either have 4 or 2 or 0 real roots,since complex roots occur in pairs.

Q. Let m≤n be natural numbers.The number of injective maps from a set of cardinality m to a set of cardinality n is
A. m!   B. n!   C.(n-m)!  D.none

Sol:- (D) f be a function s.t. f:A->B
O(A)=m,O(B)=n,then no of injections are=n!/(n-m)!

Q. T/F:- There exists a group with a proper subgroup isomorphic to itself.
Sol:- TRUE. since (Z,+) is a group.
Hence (nZ,+) be a subgroup of it.
=> there may be corresponding element in (nZ,+) to each element of (Z,+).
=> They are isomorphic also.

Q. T/F: The symmetric group S5 consisting of permutations on 5 symbols has element of order 5.
Sol:- TRUE. Since O(S5)=5!=120
Since 6|120,it is possible to for any a€S5 s.t. a^6=e.
=> any element will be of order 6 in S5.

Q. Consider the sequence {Xn} defined by Xn=[nx]/n for x belongs to R. [ ] denotes the integer part.Then {Xn} converges to
A. x    B.not in x    C.oscillates

Sol:- (A)  [nx]/n ≤ nx/n=x hence lim{Xn}=x.

Q. T/F: Any continuous function from the open unit interval (0,1) yo itself has a fixed point.
Sol:- TRUE. By fixed point property,
f(x):(0,1)->(0,1) has a point c in (0,1) s.t. f(c)=c exists.

Q.T/F: Logx is uniformly continuous on (1/2,infinity).
Sol:- FALSE. Let x,y€(1/2,infinity),k>0
| logx - logy |=|log(x/y)|<k => |x/y|< e^k
=> |x| < m.|y|
=> |x - y|=(1-m)x
=> logx is not uniformly continuous.

Q. T/F: There exists a set A is a subset of 65 elements from {1,2,3, . . . ,100} such that 65 can't be expressed as a sum of two elements in A.
Sol:- TRUE. Let x+y=65
So we can take all the elements from 65 to 100 in A. And if x is in A then 65-x will not be in A. So,we can select at least 30 elements from 1 to 64 in A to follow the condition.
Thus we can have a set A s.t. the condition x+y=65 is not being satisfied.

Q. T/F : The function f(x)= 0 if x is rational & f(x)=x if x is irrational, is not continuous anywhere on the real line.
Sol:- TRUE. By number theory,we know that there exists infinitely rational numbers between two irrational numbers & similarly there may be infinitely many irrationals in the neighbourhood of a rational. So, f(x) s not continuous anywhere on the real line.

Q.T/F: A is a 3*4 matrix of rank 3. Then the system of equations Ax=b has exactly one solution.
Sol:- FALSE.  Rank(A)=3, Rank(A|b)=3
=> the system of linear equation is consistent.
So,it will have 4 variables with 3 equation
=> it has infinite no of solutions.

Q. T/F: Let A be a 2*2 matrix with complex entries. The number of 2*2 matrices A with complex entries satisfying the equation A^3=A is infinite.
Sol: TRUE.   A(A^2 - I)=0
A^2=I(identity matrix) or A=0(zero matrix)
So there are infinitely many complex entire satisfying A^2=I.

Q. T/F: Let S be a finite subset of R^3 such that any three elements in S spans a two-dimensional subspace. Then S spans a two dimensional space.
Sol:- TRUE.  Easy to show.Try yourself.

Q.T/F:- Consider the map T from the vector space of polynomials of degree at most 5 over the reals R*R,given by sending a polynomial P to the pair (P(3),P'(3)),where P'(3) is the derivative of P.Then the dimension of the kernel is 3.
Sol:- FALSE.
dim(P(x))=6,P(x) is a at most 5 degree polynomial.
dim(R^2)=2.
By Silvester's law of nullity :
Nullity=dim(P(x)) - dim(R^2)=6-2=4.