GATE MATHEMATICS 2020
Previous Year Question Paper with Solution.
1. Suppose that and
are topologies on X produced by metrices d1 and d2, respectively, such that
. Then which of the following statements is true?
(a) If a sequence converges in (X, d2), then it converges in (X, d1)
(b) If a sequence converges in (X, d1), then it converges in (X, d2)
(c) Every open ball in (X, d1) is an open ball in (X, d2)
(d) The map x→x from (X, d1) to (X, d2) is continuous
Ans. (a)
Sol. is topology induced by metric d1, i.e. for any
> 0 and x
X.
(x, d1) = {y
X|d1, (x, y) <
} is open in
i.e. element of topology
.
IInd is topology induced by metric d2 i.e. for any
> 0 and x
X.
= {y
X|d2 (x, y) <
} is open in
i.e. element of topology
.
Let be a sequence in X, such that
for some x
X.
i.e. for any > 0.
k > 0 such that
xn N (x, d2)
r all n > 1x ...(i)
Now given that
Therefore for any > 0, there is
> 0
Such that for any x X.
...(ii)
Now, from Eqs. (i) and (ii) we get
xn (x, d) for all n > 4
Hence {xn} converges to x in (x, d2) also.
Hence, option (a) is correct.
2. Let D = [–1, 1] × [–1, 1]. If the function f : D R is defined by
f(x, y) =
then
(a) f is continuous at (0, 0)
(b) both the first order partial derivatives of f exist at (0, 0)
(c) is finite
(d) is finite
Ans. (c)
Sol.
Take put y = 0
f(x, y) is not continuous at (0, 0).
Therefore both the first order partial derivative of f does not exist.
dxdy is finite on the region D.
Hence, option (c) is correct.
3. The initial value problem = y3/5, y(0) = b has
(a) a unique solution if b = 0
(b) no solution if b = 1
(c) infinitely many solutions if b = 2
(d) a unique solution if b = 1
Ans. (d)
Sol. By picard's Existance and Uniqueness Theorem
=
(0, 1)
(i) y(a) = b, 0
This differential equatio has unique solution in the and of (a, b).
(ii) y(a) = 0, i.e. b = 0
This differential equation has infinite solution.
Now, given = y3/5, y(0) = b
Here b 0
So by Picard's theorem (result-1)
Differential equation = y3/5 has unique solution if b
0
i.e. = 1
Hence, option (d) is correct.
4. Consider the following statements:
I. log(|z|) is harmonic on C/{0}
II. log(|z|) is harmonic conjugate on C/{0}
Then,
(a) Both I and II are true
(b) I is true but II is false
(c) I is false but II is true
(d) Both I and II are false
Ans. (a)
Sol. I. log|z| has is harmonic on C.
II. log|z| has a harmonic conjugate on C.
f(x, y) = log|z|
=
[ f(x) = ln r + iq = ln z is correct answer (a) both are true]
Fxx + Fyy = 0
log|z| is harmonic
If z = |z| = r, f(z) = ln z = ln r
= U + iV
U = ln r it has harmonic conjugate if a function r such that
Ur = and Vr =
V = C+
Ur = = 0
=
= 1
V = 0
5. Let G and H be defined by
G = C/{z = x + iy C : x < 0, y = 0}
H = C/{z = x + iy C : x
Z, x < 0, y = 0}
Suppose f : G C and g : H
C are analytic functions.
Consdier the following statements:
I. is independent of paths g in G joining – i and i
II. is independent of paths
in H joining – i and i
Then
(a) Both I and II are true
(b) I is true but II is false
(c) I is false but II is true
(d) Both I and II are false
Ans. (a)
Sol.
So, that statement I and II both are true.
6.
If for a subset S of C, denotes the closure of S in C, then
(a)
(b)
(c)
(d)
Ans. (a, d)
Sol. For |z| > 0, we have
e1/z =
where z0 = 0 is an essential singularily.
According to Picard's theorem, the function exp(1/z) taken every possible value (except possibly one) in any neighbourhood of the origin. Clearly, the value 0 is never attainable, but we can easily check that any othe value is obtained. Let c 0 be any non zero complex number and let us solve for these z such that exp(1/z) = c. The multiple valuedness of the logarithmic says that there are infinitely many such z, satisfying
=
for k = 0, +1, +2, ......., whose moduli are given by
|z| =
Which can be small as desired by taking k as large necessary. Therefore in any neighbourhood of the origin, there are an infinite number of points for which the function exp (1/z) takes as value a given non zero complex number. Therefore or
satisfied above explanation.
7. Suppose that
Then, with respect to the Euclidean metric on R2.
(a) both U and V are disconnected
(b) U is disconnected but V is connected
(c) U is connected but V is disconnected
(d) both U and V are connected
Ans. (c)
Sol. U = IR2|{(x, y) IR2 : (x, y)
0}
U is connected because therefore infinite many number between two rational numbers.
V = IR2|{(x, y) IR2, x > 0, y = 1/x}
8. If D1 and D2 denote the dual problems of the linear programming problems P1 and P2 respectively, where
P1 : Minimize x1– 2x2 subject to –x1 + x2 = 10, x1, x2 > 0,
P2 : Minimize x1– 2x2 subject to –x1 + x2 = 10, x1 – x2 = 10, x1, x2 > 0.
Then
(a) Both D1 and D2 are infeasible
(b) P2 is infeasible and D2 is feasible
(c) D1 is infeasible and D2 is feasible but unbounded
(d) P1 is feasible but unbounded and D1 is feasible
Ans. (a)
Sol. According to duality theory
If the primal is unbounded, then the dual is infeasible. And if the dual is unbounded, then the primal is infeasible.
9. If (4, 0) and are critical points of the function f (x, y) = 5(
+
)x2 +
y2 + (
+ 1)y3 + x3, where
,
R, then
(a) is a point of local maxima of f
(b) is a saddle point of f
(c) = 4,
= 2
(d) is a point of local minima of f
Ans. (b)
Sol. f(x, y) = 5 – ( +
)x2 +
y2 + (
+ 1)y3 + x3 ...(i)
For critical point, fx = 0 and fy = 0
fx = 0 – ( +
)(2x) + 0 + 0 + 3x2
and fy = 0 – 0 + 2y + 3y2(
+ 1)
f2 = 0 –(
+
)(2x) + 3x2 = 0
and fy = 0 2
y + 3y2(
+ 1) = 0
at (4, 0) and at
+
= 6 ...(ii)
and 3– 4
= –3 ...(iii)
From Eqs. (ii) and (iii), we get
= 3,
= 2
Put the value of and
in in Eq. (i), we get
f(x, y) = 5 – 6x2 + 3y2 + 4y3 + x3
fx = – 12x + 3x2 and fxx = –12 + 6x
fy = 6y + 12y3 and fyy = 6 + 24y
fxy = 0
Let (a, b) =
D = fxxfyy –
= (–12 + 6x) (6 + 24) – 0
D = (–12 + 6 × 4)
= – 2 × 6 = –72 < 0
D < 0
If D(a, b) < 0 then point (a, b) is a saddle point.
Hence, D < 0, so point
is a saddle point.
10. Consider the iterative scheme
xn =
with initial point x0 > 0. Then, the sequence {xn}
(a) converges only if x0 > 1
(b) converges only if x0 < 3
(c) converges for any x0
(d) does not converges for any x0
Ans. (c)
Sol.
{xn} is a sequence of positive numbers.
Let {xn} converges to I
But {xn} is a sequence of positive numbers.
Hence, the sequece {Xn} converges for any x0 option (c) is correct.
11. Let C[0, 1] denote the space of all real-valued continuous functions on [0, 1] equipped with the supremum norm . Let T : C[0, 1]
C[0, 1] be the linear operator defined by
T(f)(x) =
Then
(a) ||T|| = 1
(b) I – T is not invertible
(c) T is surjective
(d) ||I + T|| = 1 + ||T||
Ans. (d)
Sol.
Such that (C[0, 1], d)
f = Tf
where, k(x, y) = e– y
= 1 – e–1
(b) (l – T) (l + T + T2 + ... + Tm), m N
= l – Tm + 1 ( ||T|| < 1)
(l – T) (l + T + T2 + ...) = l – 0 = l
So, l – T is invertible.
(c) T is not surjective.
(d) ||A + B||op ||A|| + ||B||op
||l + T||
||l|| + ||T||
= 1 + ||T||
= 1 + 1 – e–1
= 2 – e–1
f(x) =
(l + T) f(x) = 1 +
= 1 + (1 – e– x) = 2 – e– x
||(l + T) =
= 2 – e–1 = 1 + (1 – e–1) = 1 + ||T||.
12. Suppose that M is a 5 × 5 matrix with real entries and p(x) = det(xI – M). Then
(a) p(0) = det (M)
(b) every eigen value of M is real if p(1) + p(2) = 0 = p(2) + p(3)
(c) M–1 is necessarily a polynomial in M of degree 4 if M is invertible
(d) M is not invertible if M2 – 2M = 0
Ans. (c)
Sol. P(x) = det (xl – M)
P(0) = det (– M)
= (–1)5 det M
= (–1) det M
det M
Option (a) is false.
M =
P(1) = P(3) = 0
P(x) = det (xl – M)
= (x2 + 1)(x – 1)(x – 3)(x – 2)
x = +i, 1, 3, 2
Option (b) is false.
P(x) = x5 + a4x4 + a3x3 + a2x2 + a1x + (–1)5 det M
P(M) = 0
M5 + a4M4 + a3M3 + a2M2 + a1M – det M = 0
Multiplying by M– 1 on both side. [ det (M)
0, M– 1 exist]
M4 + a4M3 + a3M2 + a2M + a1l – (det (M)M– 1 = 0
M– 1 =
Option (c) is correct.
Now, M is not invertible if M2 – 2M 0 so option (d) is also false
13. Let C[0, 1] denote the space of all real-valued continuous functions on [0, 1] equipped with the supremum norm . Let f
C[0, 1] be such that |f(x) – f(y)|
m|x – y|, for all x, y
[0, 1] and for some M > 0.
For n N, let fn(x) =
If S = {fn: n
N}, then
(a) the closure of S is compact
(b) S is closed and bounded
(c) S is bounded but not totally bounded
(d) S is compact
Ans. (a)
Sol. For any x, y [0, 1]
kn|x – y|
for some constant,
0 x
1
0 x1/n
1
0
in 0 y
1
0
For n IV
For some > 0 let
=
Such that for any x, y [0, 1]
|x – y| < |fn(x) – fn(x)|
=
= M
[as x, y [0, 1]
and |f(x) – f(y)|
m|x – y| for any x, y
[0, 1]
m.kn|x – y|
Hence, fn C[0, 1] for any n
N
S = {fn : n
N}
C[0, 1]
Therefore closure of S in a closed subset of compact set C[0, 1] and hence it is compact.
Option (a) is the closure of S in compact.
14. Let K : R × (0, )
R be a function such that the solution of the initial value problem
, u(x, 0) = f(x), x
R, t > 0, is given by u(x, t) =
for all bounded continuous functions f. Then, the value of
is ......... .
Ans.
Sol.
u(x, 0) = f(x)
u(x, t) =
By heat equation on are infinite domain
ut = kuxx
x r, t > 0
u(x, 0) = f(x)
then the solution is
u(x, t) =
For comparing Eqs. (i) and (ii), we get k = 1
and comparing Eq. (ii) and Eq (iv), we get
15. The number of cyclic subgroups of the quaternion group.
Q8 = b|a4 = 1, a2 = b2, ba = a3
is ......... .
Ans.
Sol. Quaternion group Q8 =
Lagrange's theorem
O(G) = finite and H < G
then O(H)/O(G)
O(H) = 1, 2, 4, 8
Number of cyclic subgroup of order n in G
=
(i) Number of element of O(8) = 0
Number cyclic subgroup of O(8)
(ii) Number of cyclic subgroup of O(1) in Q8
=
=
(iii) Number of cyclic subgroup of O(2) in Q8
=
=
(iv) Number of cyclic subgroup of order 4 in Q8
=
=
Hence, number of cyclic subgroup of Q8
= 1 + 1 + 3 = 5.
16. The number of elements of order 3 in the symmetric group S6 is ......... .
Ans.
Sol. The order 4 elements are of the form
(i) (abc) (ii) (abc)(edf)
We have 6 elements ( given symmetric group S6)
17. Let F be the field with 4096 elements. The number of proper subfields of F is ......... .
Ans.
Sol. A finite field of order pn is isomerphic to GF (Pn)
O(F) = 4096 = 212
F GF (212)
The number of subfield in GF (Pn) = = Number of positive divisors of n
Number of sub field on
GF (212) =
= (2 + 1) (1 + 1)
= 3 × 2 = 6
Which are 2, 22, 23, 24, 26 and 212
A subfield is said to be proper of F
Hence, the number of proper subfield of F = 5.
18. If (x*1, x*2) is an optimal solution of the linear programming problem, minimize x1 + 2x2 subject to
4x1 – x2 8
2x1 + x2 10
–x1 + x2 7
x1, x2 0
and (l*1, l*2, l*3) is an optimal solution of its dual problem, then is equal to ......... (correct up to one decimal place)
Ans.
Sol. Let S1, S2 ≥ 0 be the surples variables, A1, A2 0 be the artificial variables and S3
0 be the slack variables added to the given constant to convert the given inequalities to the equalities.
min z = –max (–z)
max (–z) = –min z
max (–z) = –x1 – 2x2 + 0.S1 + 0.S2 + 0.S3 – m
Subject to
4x1 – x2 – S1 + A1 = 8
2x1 + x2 – S2 + A2 = 10
–x1 + x2 + S3 = 7
x1, x2, S1, S2, S3, A1, A2 0
Since
Solution in optimum with
x1 = 5, x2 = 0, min Z = 5
Now, the dual of the above LPP can be written as
max. Z =
Subject to
Same as per above method (table) we will have
Now, our required solution will be
+
= x1 + x2 +
= 5 + 0 + 0 + 0.5 + 0 = 5.5.
19. Let a, b, c R be such that the quadrature rule
is exact for all polynomials of degree less than or equal to 2. Then b is equal to ......... (rounded off to two decimal places).
Ans.
Sol. Given
We consider, f(x) = 1
Now, from Eq. (ii), we have c = a again from Eq. (iii), we have
Now, from Eq. (i), we get
a + b = 2
20. Let f(x) = x4 and let p(x) be the interpolating polynomial of f at nodes 1, 2 and 3. Then, p(0) is equal to .........
Ans.
Sol. P(x) = a0 + a1x + a2x2 ....(A)
f(xi) = P(xi), where xi = 1, 2, 3
at xi = 1 1 = a0 + a1 + a2 ...(i)
at xi = 2 16 = a0 + 2a1 + 4a2 ...(ii)
at xi = 3 81 = a0 + 3a1 + 9a2 ...(iii)
Eq. (ii) – Eq. (i) a1 + 3a2 = 15 ...(iv)
Eq. (iii) – Eq. (ii) a1 + 5a2 = 65 ...(v)
Eq. (v) – Eq. (iv) 2a2 = 50
a2 = 25
From Eq. (iv), a1 + 3 × 25 = 15
a1 = –60
From Eq. (i), 1 = a0 – 60 + 25 a0 = 36
Now, from Eq. (A), we get
P(x) = a0 + a1x + a2x2
Put x = 0
P(0) = a0 + 0 + 0 P(0) = a0
Hence, P(0) = 36.
21. For n 2, define the sequence {xn} by
xn =
Then, the sequence {xn} converges to ......... (correct up to two decimal places)
Ans.
Sol.
22. Let L2[0, 10] = {f : [0, 10] R : f is Lebesgue measurable and
} equipped with the norm
and let T be the linear functional on L2[0, 10] given by
T(f) =
Then ||T|| is equal to ......... .
Ans.
Sol. We know that, if T LP(x)*, then
the unique g
Lq(x)
such that T(f) =
and =
if p = 2 then, q = 2
and x = [0, 10]
so define g(x) =
T(f) =
Now, ||g||2 =
=
= (9)1/2 = 3
and ||T|| = ||g||
Hence, ||T|| = 3.
23. If {x13, x22, x23 = 10, x31, x32, x34} is the set of basic variables of a balanced transportation problem seeking to minimize cost of transportation from origins to destinations, where the cost matrix is,
and then x32 is equal to .......
Ans.
Sol.
µ – 5 = 5
µ = 15
Demand = Availability
10 + 10 + 20 + 15 =
+ 15
+ 15 = 55
= 40
= 10
Now, 10 + x32 + 15 = 30
x32 = 30 – 10 – 15
x32 = 5.
24. Let Z225 be the ring of integers modulo 225. If x is the number of prime ideals and y is the number of nontrivial units in Z225, Then x + y is equal to ......... .
Ans.
Sol. The number of prime ideal Zm = Number of prime divisors of m.
Number of prime ideals of Z225 = 2 (3 and 5)
x = 2 [
225 = 25 × 9 = 52 × 32]
Non-zero unit element of a group are these elements of a group whose multiplicative inverse exist.
Zn* U(zm)
U(n)
U(n) = {x N|1
< n, gcd(x, n) = 1}
O[u(225)] = (225)
= 225 ×
= 225 ×
= 15 × 8
= 120
So, these are the number of units in Z225 but we want non-trivial units i.e. 1 is trivial units because it has inverse.
y = 120 – 1 = 119
Hence, x + y = 2 + 119 = 121.
25. Let u(x, t) be the solution of
= 0, u(x, 0) = f(x),
(x, 0) = 0, x
R, t > 0.
where f is a twice continuously differentiable function. If f(–2) = 4, f(0) = 0, and u(2, 2) = 8, then the value of u (1, 3) is ......... .
Ans.
Sol.
We have given that,
Also given that,
26. Let be an orthonormal basis for a separable Hilbert space H with the inner product
. Define fn = en –
for n
N.
Then
(a) the closure of the span {fn : n N} equal to H
(b) f = 0 if for all n
N
(c) is an orthogonal subset of H
(d) there does not exist non-zero f H such that
Ans. (a)
Sol. A finite or countable infinite subset {e1, e2, ...] of a Hilbert space H is orthogonal if (ek, el) = as orthonormal subset {e1, e2, ...} =
of H determine whether this subset spans all of H, a finite linear combinations of elements in {e1, e2, ...} are dense in H, then
is an orthonormal basis for H.
Suppose a pre-Hilbert space H0 with inner product (.,.) then we can find a Hilbert space H with inner product (.,.) such that (i) H0 H (ii) (f, g)0 = (f, g) whenever f, g
H0 (iii) H0 is dense in H.
A Hilbert space satisfying properties like H in the above following proposition is an completion of H0. The construction of H follows closely contor's familiar method of obtaining the real number as the completion of the rationals in terms of cauchy sequences of rationals. Now consider the collection of all cauchy sequences {fn} with fn H0,
one difines an equivalence relation in this collection by saying that {fn} is equivalent to
if fn –
converse to 0 as n
the collection of equivalence class is then taken to be H. Then if easily varifies that H inhertis the structure of vector space with an inner product (f, g) defined as
whether {fn} and {gn} are cauchy sequence in H0, the elements of f and g is H, if f
H0 then the sequence {fn} with fn = f for all n to represent f as an element of H giving H0
H. Let
be an cauchy sequence in H with each Fk represented by
then F
H represented by the sequence {fn} with fn =
where N(n) so that,
F in H.
27. Suppose V is finite dimensional non-zero vector space over C and T : V V is linear transformation such that Range(T) = Nullspace(T). Then, which of the following statements is false?
(a) Then dimension of V is even
(b) 0 is the only eigenvalue of T
(c) Both 0 and 1 are eigenvalues of T
(d) T2 = 0
Ans. (c)
Sol. By Rank-Nullity theorem,
dim V = Rank(T) + Nullity(T)
= Rank (T) + Rank(T) [ Rank(T) = Nullity(T)]
= 2 Rank(T)
i.e. 2 is even number.
So, the dimension of V is even.
Now, T2(x) = 0, [T2 : V
V]
x V, T(x)
Range(T)
T2(x) = T(T(x))
Range(T) = Nullspace(T)
Hence, T2(x) = 0,
T2(x) = 0, means it is nilpotent operator.
Because we know that if
Tk = 0 for some K N
then this operator is nilpotent and here T2(x) = 0, so it is nilpotent and zero is the only eigenvalue of nilpotent operator.
Hence, zero (0) is the only eigenvalue of (T).
Hence, option (a) (b) and (d) are correct and (c) is false.
28. Let P Mm × n (R). Consider the following statements.
I. If XPY = 0 for all X M1 × m(R) and Y
Mn × 1(R), then P = 0.
II. If m = n, P is symmetric and P2 = 0, then P = 0.
Then
(a) both I and II are true
(b) I is true but II is false
(c) I is false but II is true
(d) both I and II are false
Ans. (a)
Sol. Given XPY = 0
ejPei = 0,
Pij = 0 i, j = (i, j)th entry of matrix P.
P = 0
Where, ej = [0, 0, ...1...0, 0]
P.ej = ith place.
Hence, statement I is true.
Now, if P is symmetric then it is diagonalizable.
i.e. if you have any symmetric matrix then it is always diagonalizable.
Given, P2 = 0 i.e. P is nilpotent.
Because Pk = 0 for some positive integer k.
P is nilpotent.
and non-zero nilpotent matrix is never diagonalizable
So, P has to be zero.
Hence statement II is also true.
Hence, both the statement I and II are true.
29. For n N, let Tn :
and
T : be the bounded linear operators defined by
Tn(x1, x2, ...) = (y1, y2, ...), where yj =
and T(x1, x2, ...) = (x1, x2, ...)
Then
(a) ||Tn|| does not converge to ||T|| as n
(b) ||Tn – T|| converges to zero as n
(c) for all x l1, ||Tn(x) – T(x)|| converges to zero as n
(d) for each non-zero x l1, there exists a continuous linear functional g on
such that g(Tn(x)) does not converge to g(T(x)) as
Ans. (c)
Sol. Here we only treat the case 1 p <
since the case p =
in both similar and easier suppose {xn} is a Cauchy sequence in lp and let
> 0 then there exists an integer N such that
|xmi – xni|p
for all m, n N, In particular, the sequence
is
Cauchy sequence for each i, let xi denote its limits. Given and p 1 then we have
that follows xn converges to x and that x
lp
for all x , ||Tn(x) – T(x)|| converges to zero as n
.
30. Let P(R) denote the power set of R, equipped with the matric
d(U, V) =
where and
denote the characteristic functions of the subsets U and V, respectively, of R.
The set {{m} : m Z} in the matric space (P(R), d) is
(a) bounded but not totally bounded
(b) totally bounded but not compact
(c) compact
(d) not bounded
Ans. (a)
Sol. To show that set
M = {{m} : m z} is bounded in (P
R, d) metric d is defined by
d(U, V) =
Curve =
for any U, V (P(R))
d(U, V) =
So, d is discrete metric on P(R)
and hence M is bounded.
Let 0 < < 1 then for any m
Z,
, neighbourhood of {m}
only contains {m}.
and m can not be covered by finitely many as m is an infinite set.
Hence, m is not totally bounded, so m is bounded but not totally bounded.
Hence, option (a) is correct.
31. Let f : R R be defined by f(x) =
where
is the characteristic function of the interval (n, n + 1). For a
R, let
= {x
R : f(x) > a}.
Then
(a) S1/2 is open
(b) is not measurable
(c) S0 is closed
(d) is measurable
Ans. (d)
Sol.
= (0, 1] which is not option. Option (a) is false.
By option (b) =
So, is measurable.
So, option (b) is false.
By option (c)
S0 = {x R; f(x) > 0}
= (0, ) which is not closed.
So, option (c) is false.
by option (d) = (0, 1] which is measurable
So, option (d) is correct.
32. For n N, let fn, gn :(0, 1)
R be functions defined by fn(x) = xn and gn(x) = xn(1 – x). Then,
(a) {fn} converges uniformly but {gn} does not converge uniformly
(b) {gn} converges uniformly but {fn} does not converge uniformly
(c) both {fn} and {gn} converge uniformly
(d) neither {fn} nor {gn} converge uniformly
Ans. (b)
Sol. fn(x) = xn, x (0, 1)
Point wise limit f(x) = 0
Mn =
Mn = 1
= 1
Mn test fail so, fn(x) does not converge uniformly.
Now, gn(x) = xn(1 – x) x (0, 1)
Put g(n) = 0
Mn =
Let y = xn(1 – x)
For maximum or minimum,
= 0
So, gn(x) converges uniformly.
33. Let u be a solution of the differential equation y' + xy = 0 and let = u be a solution of the differential equation + (x2 + 2) y = 0 satisfying
= 1 and
= 0. Then
is
(a)
(b)
(c)
(d)
Ans. (b)
Sol. u is a solution of the ODE, + xy = 0 ...(i)
Now, + xy = 0
Differentiate w.r. to x, we get
Multiplying Eq. (ii) by x,
Now, it is given that is solution of the ODE
Hence, we can write,
Differential w.r. to 'x', we get
Again, differentiate w.r. to. 'x' we get
Now, substitute all these derivatives in Eq. (v), we get
Using Eqs. (iii) and (iv), we get
Solving we have
Differentiate w.r. to. 'x', we get
Hence, the solution is
Hence, option (b) is correct.
34. For n {0}, let yn be a solution of the differential equation
+ ny = 0 satisfying yn(0) = 1. For which of the following function w(x), the integral
is equal to zero?
(a) e–x2
(b) e–x
(c)
(d) xe–x
Ans. (b)
Sol. Lagurre differential equation
Lagurre polynomial is given
Note that, Ln(0) = 0
Now, orthogonality property.
If Lm(x) and Ln(x) are lagurre polynomial, then it satisfy this relation
Now, comparing with the given problem, it is given
Thus w(x) e–x
Hence, option (b) is correct.
35.
are metric spaces with metrics induced by the Euclidean matric of R2. Let BX and BY be the open unit balls around (0, 0) in X andd Y, respectively.
Consider the following statements.
I. The closure of BX in X is compact.
II. The closure of BY in Y is compact.
(a) both I and II are true
(b) I is true but II is false
(c) I is false but II is true
(d) both I and II are false
Ans. (c)
Sol.
The subset of is connect because it is graph of the continuous function f:
R define by f(x) = sin(1/x) and (0,
) is connect.
Now, consider R2 with Euclidean metric, Then, every closed disk and every circle is compact, more generally, in the Euclidean space Rn, the sphere
Sn – 1 = {(x1, x2, ...., xn)} Rn| x12 + x22 + ... + x22 = 1| is
closure in compact space.
So, the closure of by in y is compact.
36. If f : C\{0} C is a function such that f(z) =
and its restriction to the unit circle is continuous, then
(a) f is continuous but not necessarily analytic
(b) f is analytic but not necessarily a constant function
(c) f is a constant function
(d) exists
Ans. (a)
Sol. f : C\{0} C
Cauchy lemann equation not satisfied.
37. For a subset S of a topological space, let Int(S) and denote the interior and closure of S, respectively.
Then which of the following statements is true?
(a) If S is open, then S = Int()
(b) If the boundary of S is empty; then S is open
(c) If the boundary of S empty, then S is not closed
(d) If \S is a proper subset of the boundary of S, then S is open
Ans. (b)
Sol. Let X be a topological space and S X.
The boundary Bd(S) in define as
Bd(S) =
We know that interior of S is the complement of while the boundary is contained in the latter. So the intersection is emply.
The set Bd(S) is emply if and only if and
are disjoint. Since the latter contains X – S it follows that
and X – S are disjoint. Since their union is X this means that
must contained in S, which implies that S is closed. If one reverse the roles of S and X – S in the preceding two sentences, it follows that X – S is also closed, hence S is both open and closed in X.
38. Suppose and
are the smallest topologies on R containing S1, S2 and S3 respectively, where
S1 =
S2 = {(a, b): a < b, a, b Q}
S3 = {(a, b): a < b, a, b R}.
Then,
(a)
(b)
(c)
(d)
Ans. (c)
Sol. Let and
be two topologies or R. Then
if and only if for each x
X there are neighbourhood base M1 and M2 of x for
and
respectively such that for every V1
M1 there is a V2
M2 with V2
V1 and there are neighbourhood bases N1 and N2 of x for
and
respectively such that for every W2
N2 there is a W1
N1 with W1
W2.
Two metric topologies defined by two metrices on the same space are equal if and only if they have the same collection of convergent sequences with the same limits.
39. Let M = . Consider the following statements.
I. There exists a lower triangular matrix L such that M = LLt, where Lt denotes transpose of L.
II. Gauss-Seidel method for Mx = b(b R3) converges for any initial choice x0
R3.
Then
(a) I is not true when = 3
(b) II is not true when = – 1
(c) II is not true when a = 4,
(d) I is true when
Ans. (d)
Sol. The matrix is given by
M =
M = LLT, we must have |M| 0
Now, det (M) = |M| =
Thus, if statement I is true then
Now, let us check the
Now it is given = 4.5
Thus, when , the statement I is true.
Hence option (a) is false.
Option (b) By Gauss-seidel method.
Thus, statement II is true.
Hence, option (b) is false.
Hence, statement II is true.
Thus, option (c) is false.
Thus, statement I is true.
Therefore, option (d) is true.
40. Let I and J be the ideals generated by and
in the ring
=
a, b
Z}, respectively. Then
(a) both I and J are maximal ideals
(b) I is a maximal ideal but J is not a prime ideal
(c) I is not a maximal ideal but J is a prime ideal
(d) neither I nor J is a maximal ideal
Ans. (b)
Sol. Maximal : R is commutative ring. An ideal A R is said to be a maximal ideal of R if
an ideal B such that A
R then either A = B or B = R
Hence, option (b) is correct.
41. Suppose V is finite dimensional vector space over R. If W1, W2 and W3 are subspaces of V, then which of the following statements is true?
(a) If W1 + W2 + W3 = V, then span (W1 W2)
span (W2
W3)
span (W3
W1) = V
(b) If W1 W2 = {0} and W1
W3 = {0}, then W1
(W2 + W3) = {0}
(c) If W1 + W2 = W1 + W3, then W2 = W3
(d) If W1 V, then span (V\W1) = V
Ans. (d)
Sol. Let V = R3
then w1 = {1, 0, 0}, w2 {(0, 1, 0)}
w3 = {(0, 0, 1)}
w1 + w2 + w3 = V
span (w1
w2) = w1 + w2
and span (w2 w3) = w2 + w3
and span (w3 w1) = w3 + w1
Given, span (w0 w2)
span (w2
w3)
span (w3
w1) ...(i)
= (w1 + w2) (w2 + w3)
(w3 + w1)
= {(a, b, 0)} {(0, b, c)}
{(0, 0, c)}
(1, 1, 1)
V but Doesn't belongs to Eq. (i) so, option (a) is false.
Now, Let V = R3 and
w1 = {(a, b, c); a, b R}
w2 = {(0, b, b); b R}
w3 = {0, 0, c}; c R}
w1 + w2 = w1 + w3 w2 = w3
Hence, option (c) is false.
Now, Let V = R2
w1 = {(1, 1)}, w2 = {(1, 0)}, w3 = {(0, 1)}
w1 w2 = {0}, w1
w2 = {0}
w1 (w2 + w3)
0
Option (b) is false.
Option (a), (b) and (c) are false.
Hence, option (d) is correct.
42. Let 0.
x1 – 2x2 + ax3 = 8
x1 – x2 + x4 =
x1, x2, x3, x4 a
has no basic feasible solution, if
(a)
(b)
(c)
(d)
Ans. (d)
Sol. Suppose we have m equation and n variables then a solution obtain by setting (n – m) variables equal to zero and solving remaining (n – m) equation in (n – m) variable is called a basic solution.
If the solution satisfies the non-negativity condition then it is called the basic solution.
Now, in this question, given that
m =2 equation and n = 4 varibles
Thus possible cases are, we may set
Here, Feasibility of basic solution depends on sign of and
. So option (d) is satisfied.
43. Let 0 < p < 1 and left
Then,
(a)
(b)
(c)
(d) if fn converges to f pointwise on R, then
Ans. (c)
Sol.
f and g are similar kind of function
Hence, option (a) and (b) are false.
Hence, option (c) is correct.
Option (d) is also false.
44. Suppose that and 0
are linearly independent solutions of the differential equation
– (x +
+ (x2 – 2)y = 0, and
(0) = 0. Then, the smallest positive integer n such that
is ......... .
Ans.
Sol. Given differential equation is
…(i)
Which is Bessel's equation in the form of + (x2 – n2)y = 0 where n is a non-negative real number.
Assume its solution, in series as
Substituting these in Eq. (i), we get
This should be an identify in x equating to zero the coefficient of the lowest degree term in x, we get
2a0 m(m – 1) – a0m – 2a0 = 0
Similarly, equating to zero the coefficial of xm + r in (ii) and get a2, a3, ...... and the get and
, using
and
solve
The smallest positive integer n = 3.
45. Suppose that f(z) = and
If
then the value of
is equal to ..........
Ans.
Sol.
By argument theorem
46. If [0, 2] and
then
is equal to .......... (correct up to one decimal place)
Ans.
Sol. …(i)
From Eq. (i) and (ii), we get
47. Let K = where
is a primitive cube root of unity. Then, the degree of extension of K over Q is ......... .
Ans.
Sol. The degree of extension over the field is equal to degree of irreducible polynomial over the field.
Irreducible polynomial of degree 2.
Hence, degree of irreducible polynomial of k over
Q = 2 × 2 = 4.
48. Let R. If (3, 0, 0,
is an optimal solution of the linear programming problem minimize
x1 + x2 + x3 – ax4 subject to
2x1 – x2 + x3 = 6
–x1 + x2 + x4 = 3
x1, x2, x3, x4 0, then the maximum value of
is ......... .
Ans.
Sol. Min. Z = x1 + x2 + x3 + x4
Subject to 2x1 – x2 + x3 = 6
– x1 + x2 + x4 = 3
Point (3, 0, 0,
–3 + 0 + x4 = 3
x4 = 3 + 3 + 6 = 4
then = 6 – 3 = 3
49. Suppose that T : R4 R[x] is a linear transformation over R satisfying T(–1, 1, 1, 1) = x2 + 2x4, T(1, 2, 3, 4) = 1 – x2 T(2, –1, –1, 0) = x3 – x4.
Then, the coefficient of x4 in T (–3, 5, 6, 6) is .........
Ans.
Sol. Given T (–1, 1, 1, 1) = x2 + 2x4
T(1, 2, 3, 4) = 1 – x2
T(2, –1, – 1, 0) = x3 – x4
Now, (–3, 5, 6, 6) = C1(–1, 1, 1, 1) + C2(1, 2, 3, 4) + C3(2, –1, –1, 0)
Comparing on both sides
– c1 + c2 + 2c3 = – 3 ...(i)
c1 + 2c2 – c3 = 5 ...(ii)
c1 + 3c2 – c3 = 6 ...(iii)
c1 + 4c2 = 6 ...(iv)
Now, Eq. (iii) – Eq (ii), gives, c2 = 1
Thus, (–3, 5, 6, 6) = 2 (–1, 1, 1, 1) + 1(1, 2, 3, 4) – 1(2, –1, – 1, 0)
T(–3, 5, 6, 6) = 2T (–1, 1, 1, 1) + 1.T(1, 2, 3, 4) – 1.T(2, –1, 1, 0)
= 2(x2 + 2x4) + (1 – x2) – (x3 – x4)
T(–3, 5, 6, 6) = 2x2 + 4x4 + 1 – x2 – x3 + x4
= 1 + x2 – x3 + 5x4
Therefore the coefficient of x4 in T(–3, 5, 6, 6) is 5.
50. Let F(x, y, z) = (2x – 2y cos x) + (2y – y2 sin x)
and let S be the surface of the tetrahedron bounded by the planes x = 0, y = 0, z = 0 and x + y + z = 1.
If is the unit outward normal to the tetrahedron, then the value of is ......... (rounded off to two decimal places).
Ans.
Sol.
= 2 + 2y sin x + 2 – 2y sin x + 4
= 8
51. Let F = (x + 2y) and let S be the surface x2 + y2 + z = 1, z
0. If is a unit normal to S and
Then,
is equal to .................
Ans.
Sol. x2 + y2 + z = 1
an XY plane z = 0
x2 + y2 = 1
…(i)
and given
…(ii)
On comparing Eqs. (i) and (ii), we get
52. Let G be a non-cyclic group of order 57. Then, the number of elements of order 3 in G is .........
Ans.
Sol. o(G) = pq with p < q and p|q – 1 then a non-cyclic group also.
If G be non-abelian group of order pq and p < q then the number of element of order q = and number of element of order p =
o(q) = q – 1, o(p) = q(p – 1)
o(G) = 57 = 3 × 19, 3 < 19
o(3) =
= 19.(3 – 1)
= 19 × 2 = 38
Hence, the number of elements of order 3 in G is 38.
53. The coefficient of (x – 1)5 in the Taylor expansion about x = 1 of the function
F(x) =
is ......... (correct up to two decimal places)
Ans.
Sol.
by Leibnitz rule,
From Eqs. (i) and (ii), we get
Hence, the coefficient of (x – 1)5 is 0.04.
54. Let u(x, y) be the solution of the initial value problem = 0, u(x, 0) = 1 + x2.
Then, the value of u(0, 1) is ......... (rounded off to three decimal places)
Ans.
Sol. …(i)
u(x, 0) = 1 + x2
u(0, 1) = ?
Pux + Quy = R ...(ii)
[From Eqs. (i) and (ii), P = 1, Q = R = 0]
From III, du = 0 u = C1 ...(iii)
From (i) and (ii), we get
…(iv)
…(v)
…(vi)
Hence, u (0, 1) is 1.618.
55. The value of dx is ......... (rounded off to three decimal places).
Ans.
Sol.