**TIFR GS-2010 MATHEMATICS SOLVED PAPER**

*Tata Institute of Fundamental Research organises a integrated ph.d & ph.d entrance examination each year in Mathematics. This is the solution of GS-2010*.

__Here is 10 solved problems from TIFR-GS-2010 Mathematics paper:__

**Q. Let Mn(R) be the set of n×n matrices with real entries. Which of the statement is TRUE?**

(a)Any matrix A € M4(R) has a real eigen value.

(b)Any matrix A € M5(R) has a real eigen value.

(c)Any matrix A € M2(R) has a real eigen value.

(d)None of the above.

(a)Any matrix A € M4(R) has a real eigen value.

(b)Any matrix A € M5(R) has a real eigen value.

(c)Any matrix A € M2(R) has a real eigen value.

(d)None of the above

Sol:- (b) Complex toots occur in pair, here M5(R) has 5 eigen values of which one must be real.

**Q. Let x and y in Rⁿ be non-zero column vectors. From the matrix A=xy',where y' is the transpose of y. Then rank(A) is**

(a) 2. (b) 0. (c) at least n/2. (d) none

(a) 2. (b) 0. (c) at least n/2. (d) none

Sol:- (d) A will be a non-zero matrix. Since the matrix will have at least one non-zero elements of x and y'. All the two rowed minors of A will vanish. Rank(A)=1.

**Q. Which of the following is FALSE ?**

(a) any abelian group of order 27 is cyclic.

(b) any abelian group of order 14 is cyclic.

(c) any abelian group of order 21 is cyclic.

(d) any abelian group of order 30 is cyclic.

(a) any abelian group of order 27 is cyclic.

(b) any abelian group of order 14 is cyclic.

(c) any abelian group of order 21 is cyclic.

(d) any abelian group of order 30 is cyclic.

Sol:- (d) Since, 30 can't be expressed as a product of two distinct primes, so, it is not cyclic.

**Q. A cyclic group of order 60 has**

(a) 12 generators.

(b) 15 generators.

(c) 16 generators.

(d) 20 generators.

(a) 12 generators.

(b) 15 generators.

(c) 16 generators.

(d) 20 generators.

Sol:- (c) Calculate φ(60).

φ(60)=60(1- 1/2)(1- 1/3)(1- 1/5)=16.

**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 the concept of 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) is not continuous on the real line.

**Q. Lim xsin(1/x) ,as x->0 is**

(a) 1 (b) 0 (c) 1/2 (d) does not exist.

(a) 1 (b) 0 (c) 1/2 (d) does not exist.

Sol:- (b) As x ≠ 0, then |sin1/x|≤ 1

so, |xsin1/x|≤ |x|

As x->0 ,we have |xsin1/x - 0|< €

Thus Lim{ xsin(1/x)}=0 as x->0.

**Q. Let Un=Sin(π/n) and consider the series ∑ Un. Which of the following is TRUE ?**

(a) ∑ Un is convergent.

(b) Un->0 as n->∞.

(c) ∑ Un is divergent.

(d) ∑ Un is absolutely convergent.

(a) ∑ Un is convergent.

(b) Un->0 as n->∞.

(c) ∑ Un is divergent.

(d) ∑ Un is absolutely convergent.

Sol:- (c) Un=Sin(π/n) & Vn=1/n

Then Lim(Un/Vn)=π ≠ 0

Since ∑ Vn is divergent,so does ∑ Un.

**Q.The total number of subsets of a set of 6 elements are**

(a)720 (b)6^6 (c)21 (d)none.

(a)720 (b)6^6 (c)21 (d)none.

Sol:- (d) The total number of subset of the set with n elements=2ⁿ. So, Ans is=64.

**Q. The maximum value of f(x)= xⁿ(1-x)ⁿ for a natural no. n≥1 & 0≤x≤1 is**

(a)1/2ⁿ (b)1/3ⁿ (c)1/5ⁿ (d)1/4ⁿ

(a)1/2ⁿ (b)1/3ⁿ (c)1/5ⁿ (d)1/4ⁿ

Sol:- Using Math-Trick-1

If the sum of two positive quantities is a constant(given),then their product is maximum when two quantities are equal.

So, x=1/2. max{xⁿ(1-x)ⁿ}=1/4ⁿ.

**Q. The sum of the series 1/1.2 +1/2.3 +1/3.4 + . . . +1/100.101 is**

(a)99/101 (b)98/101 (c)99/100 (d) none

(a)99/101 (b)98/101 (c)99/100 (d) none

Hints:- (d) Ans is =1 - 1/101 =100/101.