TIFR GS 2010 Complete Solution
Here is fully solved paper of TIFR GS 2010 with justification of each and every step, hope you'll enjoy it. For any query feel free to write at sushantkala786@gmail.com.
Q.1 Ans. (c). phi(60) = 16
Justification -: We know that in a cyclic group G of order n generated by a. The order of an element a^(k) is n/(n,k)=n iff (n,k)=1. So the number of generators of G are phi(n).
Q.2 Ans. (a).
(a) False, Consider the direct product Z_3*Z_3*Z_3.
(b) True, Any Abelian group of order 14 must-have elements a and b with respective order 2 and 7 (Cauchy Theorem). Then a*b generates the group.
(c) True, Same logic as in option (b).
(d) True, Same logic as in option (b), (c).
Q.3 Ans. (c). 6
Justification -: Calculating last digit is the same as 2^(80) mod (10). We know that 2^(5) =2 (mod 10). So 5 can be treated as 1 in power. Since 80 =3*5^(2)+5, so 2^(80) = 2^(3*5^(2))*2^(5) = 2^(3)*2^(1) = 6 (mod 10).
Q.4 Ans. (d). none of the given
Justification -: Just use the following manipulation 1/((n)(n+1)) = 1/n - 1/(n+1). So the telescopic sum will be 100/101.
Q.5 Ans. (c).
Justification -:
(a) False, Take f(x) = x/2. f is one-one but not onto on [0,1].
(b) False,
(c) True, Since f is one-one and [0,1] is uncountable. Then f([0,1]) must be uncountable. So range of must have an irrational number.
(d) False, Since (b) is false.
Q.6 Ans. (a). 1/2^n
Justification -: From f'(x) = 0, we get critical points are 0,1 and 1/2. One can check using double derivative test that max value occurs at x = 1/2.
Q.7 Ans. (a). (1+\sqrt{30})/2
Justification -: Recurrence relation which define the sequence is X_n+1 = \sqrt{7+X_n}. Show that X_n is increasing and bounded, hence convergent. By uniqueness of the limit l^2-l-7(from relation). Hence X_n converges to the positive root of l^2-l-7.
Q.9 Ans. (a). 2^6
Justification -:
Q.9 Ans. (a). 2^6
Justification -: By simple combinatorics, first calculate no. of subsets having no. of elements respectively 0, 1, 2, etc. and sum them to get total no. of subsets. Sum is given by nC0 + nC1 + ...... + nCn = 2^n.
Q.10 Ans. (b).
Justification -:
(a) False, Consider 4*4 matrix A=[0 1 0 0; -1 0 0 0; 0 0 0 1; 0 0 -1 0] row wise. One can easily calculate its eigen values are not real.
(b) True, Since A is 5*5 matrix so it's characteristic polynomial p(x) will be of 5 degree. Hence it must have a real root (p(x) is real, complex root occurs in pair).
(c) False, Same logic as in (a). Consider A = [0,1; -1 0].
Q.11 Ans. (a).
Justification -: By Leibnitz test for alternate series, given series converges and by p-test convergence, it is not absolutely summable.
Q.12 Ans. (d).
Justification -: (x-a1)(x-a2)(x-a3).....(x-an) = x^n +(-1) ^(n-1) (a1+a2+a3...+an) x^(n-1)......+ (a1a2a3..an). From this we can see sum of roots is (-1) ^(n-1) times coefficient of x^(n-1)
Q.13 Ans. (b).
Justification -: we can see this by epsilon-delta definition |x*sin(1/x) - 0| <= |x| < ε whenever |x| < ε .
Q.14 Ans. (d).
Justification -: By Picard's existence uniqueness theorem given IVP has a unique solution i.e. y=0.
Q.15 Ans. (c).
Justification -: (AB)' = B' A'.
Find two such matrices A, B for which B'A' is not the same as AB.
PART - 'A'
Justification -: We know that in a cyclic group G of order n generated by a. The order of an element a^(k) is n/(n,k)=n iff (n,k)=1. So the number of generators of G are phi(n).
Q.2 Ans. (a).
(a) False, Consider the direct product Z_3*Z_3*Z_3.
(b) True, Any Abelian group of order 14 must-have elements a and b with respective order 2 and 7 (Cauchy Theorem). Then a*b generates the group.
(c) True, Same logic as in option (b).
(d) True, Same logic as in option (b), (c).
Q.3 Ans. (c). 6
Justification -: Calculating last digit is the same as 2^(80) mod (10). We know that 2^(5) =2 (mod 10). So 5 can be treated as 1 in power. Since 80 =3*5^(2)+5, so 2^(80) = 2^(3*5^(2))*2^(5) = 2^(3)*2^(1) = 6 (mod 10).
Q.4 Ans. (d). none of the given
Justification -: Just use the following manipulation 1/((n)(n+1)) = 1/n - 1/(n+1). So the telescopic sum will be 100/101.
Q.5 Ans. (c).
Justification -:
(a) False, Take f(x) = x/2. f is one-one but not onto on [0,1].
(b) False,
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(d) False, Since (b) is false.
Q.6 Ans. (a). 1/2^n
Justification -: From f'(x) = 0, we get critical points are 0,1 and 1/2. One can check using double derivative test that max value occurs at x = 1/2.
Q.7 Ans. (a). (1+\sqrt{30})/2
Justification -: Recurrence relation which define the sequence is X_n+1 = \sqrt{7+X_n}. Show that X_n is increasing and bounded, hence convergent. By uniqueness of the limit l^2-l-7(from relation). Hence X_n converges to the positive root of l^2-l-7.
Q.9 Ans. (a). 2^6
Justification -:
Q.9 Ans. (a). 2^6
Justification -: By simple combinatorics, first calculate no. of subsets having no. of elements respectively 0, 1, 2, etc. and sum them to get total no. of subsets. Sum is given by nC0 + nC1 + ...... + nCn = 2^n.
Q.10 Ans. (b).
Justification -:
(a) False, Consider 4*4 matrix A=[0 1 0 0; -1 0 0 0; 0 0 0 1; 0 0 -1 0] row wise. One can easily calculate its eigen values are not real.
(b) True, Since A is 5*5 matrix so it's characteristic polynomial p(x) will be of 5 degree. Hence it must have a real root (p(x) is real, complex root occurs in pair).
(c) False, Same logic as in (a). Consider A = [0,1; -1 0].
Q.11 Ans. (a).
Justification -: By Leibnitz test for alternate series, given series converges and by p-test convergence, it is not absolutely summable.
Q.12 Ans. (d).
Justification -: (x-a1)(x-a2)(x-a3).....(x-an) = x^n +(-1) ^(n-1) (a1+a2+a3...+an) x^(n-1)......+ (a1a2a3..an). From this we can see sum of roots is (-1) ^(n-1) times coefficient of x^(n-1)
Q.13 Ans. (b).
Justification -: we can see this by epsilon-delta definition |x*sin(1/x) - 0| <= |x| < ε whenever |x| < ε .
Q.14 Ans. (d).
Justification -: By Picard's existence uniqueness theorem given IVP has a unique solution i.e. y=0.
Q.15 Ans. (c).
Justification -: (AB)' = B' A'.
Find two such matrices A, B for which B'A' is not the same as AB.
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