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### TIFR GS Mathematics Previous Years Solution Paper,Gs-2012

#### TIFR GS-2012 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: If  H1,H2  are subgroups of a group G then H1H2={ h1h2 ∈ G : h1∈ H1,h2∈ H2} is a subgroup of G.

Hints: TRUE .

Since, H1H2 is closed under multiplication. So, the statement is true.

Q.  T/F: Every subgroup of order 74 in a group of 148 is normal.

Hints: TRUE .

Since O(G/A)=O(G)/O(A)=148/74=2.

Q.  T/F: If the equation xyz=1 holds in a group G, does it follow
that yzx=1.
Hints: TRUE.

Do yourself. Easy to show.

Q.  T/F: The automorphism group Aut(Z/2 x Z/2) is abelian.

Sol: TRUE.

Aut(Z/2 x Z/2) =G (let)

Then we define a mapping f : G->G by f(x)=f(x^-1) for all x∈G.

Then suppose that f  is an automorphism of G then for any a,b∈G

f(ab)=(ab)^-1

=b^-1 a^-1

=f(b)f(a)

=f(ba)

Now since f is one-one function therefore    f(ab)=f(ba) implying ab=ba, implying G is abelian.

Q.  T/F: The polynomial (t-1)(t-2),(t-2)(t-3),(t-3)(t-4),(t-4)(t-6)∈R[t] are linearly independent.
Sol: FALSE .

Let     A1= (t-1)(t-2),  A2=(t-2)(t-3),  A3=(t-3)(t-4),   A4=(t-4)(t-6) & ki's are chosen scalars.

Then if these polynomials are linearly independent then ki=0 be the only solution of the equation

k1 A1 +  k2 A2 +  k3 A3 +   k4 A4 = 0 .

.

.

.

.

On solving You get k3 = -4k4/15 .

So these polynomials are not LIN.

Q.  T/F: Any odd order skew-symmetric matrices have always zero determinants.

Sol: TRUE.

Q.  T/F:  If A∈ M2(C) and A is nilpotent then A² =0.
Sol: TRUE.

Consider the 2*2 matrix with its complex element i,i,i,-i .

A^k =0 then all the characteristics roots are all 0.

Q. T/F:  If  f :[0,∞]->[0,∞] is continuous and bounded then f has a fixed point.
Hints:-  TRUE.

If f is continuous & bounded function

f :[0,∞]->[0,∞] .

Then f(x)-x must be positive for some a in [0,∞]

And f(x)-x must be negative for some b in [0,∞].

Then f is continuous & f(x)-x assumes -ve to +ve values from a to b in [0,∞] & therefore by intermediate value theorem there exists a point c lying
between a & b in [0,∞] such that f(c)=c for all c in [0,∞].

So,f has a fixed point.

Q.  T/F: The rank of the matrix  ( 11  12  13  14 )    is 2 .

( 21  22  23   24 )

( 31  32  33   34 )

( 41  42  43   44 )

Sol: TRUE.

Apply the elementary row operation Ri+1 = Ri+1 - Ri, i=1(1)3.

Q.  T/F: The function f(x)= x│x│ & x │sinx│are not differentiable at x=0.
Sol: FALSE.

Both the differentiable at x=0.

ALSO SEE TIFR GS-2011 MATHEMATICS SOLVED PAPER HERE.