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#1 Super Moderator Join Date: Jun 2011 Will you please provide the Mathematics exam question paper of Tata Institute of Fundamental Research (TIFR)?

As you are looking for the Mathematics exam question paper of TIFR, so here i am providing list of few questions for your idea.
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)| |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.

TIFR Mathematics Exam Question Paper

Last edited by GaganD; June 29th, 2019 at 03:37 PM.

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