GATE MATHEMATICS 2018
Previous Year Question Paper with Solution.

1.    The principal value of is

    (a) e2

    (b) e2i

    (c) e–2t

    (d) e–2

Ans.    (a)

Sol.    We have,

    

2.    Let f : C C be an entire function with f(0) = 1, f(1) = 2and = 0. If there exists M > 0 such that M for all z C, then f(2) =

    (a) 2

    (b) 5

    (c) 2 + 5i

    (d) 5 + 2i

Ans.    (b)

Sol.    We have,

        f(0) =    1, f(1) = 2, = 0

        M >    0

            M

    On integrating both sides, we get

         =    Mz + C1    ...(i)

    Again integrating, we get

    

    

    

3.    In the Laurent series expansion of valid for |z – 1| > 1, the coefficient of is

    (a) – 2

    (b) – 1

    (c) 0

    (d) 1

Ans.    (c)

Sol.    Given, where |z – 1| > 1 i.e. 1 < |z – 1| < which is shown below

    

    

    

    

    

    Clearly, coefficient of is 0 in above expansion.

4.    Let X and Y be metric spaces, and let f : X Y be a continuous map. For any subset S of X, which one of the following statements is true?

    (a)    If S is open, then f(S) is open

    (b)    If S is connected, then f(S) is connected

    (c)    If S is closed, then f(S) is closed

    (d)    If S is bounded, then f(S) is bounded

Ans.    (b)

Sol.    We know that, continuous image of a connected set is connected.

    If S is connected, then f(S) will also be connected.

6.    The general solution of the differential equation = y + for x > 0 is given by (with an arbitrary positive constant k)

    (a)

    (b)

    (c)

    (d)

Ans.    (c)

Sol.    

    

    Put,    y =    vx

    

    

    

                

    

6.    Let pn(x) be the polynomial solution of the differential equation

            

    with pn(1) = 1 for n = 1, 2, 3, ...

    

    (a) 2n

    (b) 2n + 1

    (c) 2n + 2

    (d) 2n + 3

Ans.    (d)

Sol.    Given differential equation,

    

    Which is Legendre's differential equation,

    By using Rodrigues formula,

    We have a recurrence relation,

    

    Given that,

        pn(1) =    1, for n = 1, 2, 3 ... if

    

    Comparing Eqs. (i) and (ii), we get

    

7.    In the permutation group S6, the number of elements of order 8 is

    (a) 0

    (b) 1

    (c) 2

    (d) 4

Ans.    (a)

Sol.    Given, permutation group S6, i.e. collection of all permutation on six symbols. To find the number of elements of order 8 we will partition 6 and check if any possible partition gives an LCM of 8.


    Thereforem, there is no partition that gives LCM of 8.

    Hence, there is no element of order 8 in S6.

8.    Let R be a commutative ring with 1 (unity) which is not a field. Let I R be a proper ideal such that every element of R not in I is invertible in R. Then he number of maximal ideals of R is

    (a) 1

    (b) 2

    (c) 3

    (d) infinite

Ans.    (a)

Sol.    Let R = Z2 and define the operation on R by x O y = O.

    Then, (R, +, O) is commutative ring (without identity) and I = {O} is maximal ideal, but R/I
R is not a field.

9.    Let f : R R be a twice continuosly differentiable function. The order of convergence of the secant method for finding root of the equation f(x) = 0 is

    (a)

    (b)

    (c)

    (d)

Ans.    (a)

Sol.    An iterative method is said to be of order p or has rate of convergence p if p is the largest positive real number for which a finite constant c 0 such that |Ek + 1| c |Ek|p where, Ek is the error in the kth iteration and c is asymptotic error constant.

    The order of convergence of secant method is .

10.    The Cauchy problem unx + yuy = x with u(x, 1) = 2x, when solved using its characteristic equations with an independent variable t, found to admit of a solution in the form

                   x =

    Then f(s, t) =

    (a)

    (b)

    (c)

    (d)

Ans.    (a)

Sol.    We have, uux + yuy = x with u(x, 1) = 2x.

    Characteristic equations are

    

    

    From Eq. (ii), we get

        y =    c1et

    From Eqs. (i) and (iii), we get

    

    Now, from Eq. (i), we get

    

    

    

    

    Now, from Eq. (iv), we get

        

11.    An urn contains four balls, each ball having equal probability of being white or black. Three black balls are added to the urn. The probability that five balls in the urn are black is

    (a) 2/7

    (b) 3/8

    (c) 1/2

    (d) 5/7

Ans.    (b)

Sol.    Given, an urn contains four balls with euqal probability of being white or black.

    Three black balls are added to the urn. For five black balls, we need 2 balls to be black out of remaining four balls in urn.

    Total number of possibilities = 2 × 2 × 2 × 2 = 16

    As each ball in urn can either be white or black. For 2 balls to be black out of four, we have 4C2 possibilities.

    P(five balls in urn are black) = .

12.    For a liner programming problem, which one of the following statements is false?

    (a)    If a constraint is an equality, then the corresponding dual variable is unrestricted in sign

    (b)    Both primal and its dual can be infeasible

    (c)    If primal is unbounded, then its dual is infeasible

    (d)    Even if both primal and dual are feasible, the optimal values of the primal and the dual can differ

Ans.    (d)

Sol.    The fundamental theorem of duality also known as Basic Duality Theorem states that if the primal problem has a finite optimal basic feasible solution then there exists a finite optimal basic feasible solution of the dual problem such that the values of the objective function for primal and dual are same and vice-versa.

13.    Let A = , where a, b, c f are real numbers and f 0. The geometric multiplicity of the largest eigenvalue of A equals

Ans.    1

Sol.    

    Take, a = b = c = 0 and f = 1, then A becomes

        A =    

    For calculating eigenvalues, put |A – | = 0

    

    

    Here, the largest eigenvalue is . Since in a symmetric matrix arithmetic multiplicity equals geometric multiplicity and arithmetic multiplicity of is 1 so we get geometric multiplicity is equal to 1.

14.    Consider the subspaces

        W1 =    {(x1, x2, x3) R3 : x1 = x2 + 2x3}

        W2 =    {(x1, x2, x3) R3 : x1 = 3x2 + 2x3}

    of R3. Then the dimension of W1 + W2 equals.

Ans.    3

Sol.    Given,

        W1 =    {(x1, x2, x3) IR3 : x1 = x2 + 2x3}

    and    W2 =    {(x1, x2, x3) IR3 : x1 = 3x2 + 2x3}

    are subspaces of IR3.

    Now, subspace W1 can be represented as follows

        W1 =    {(x2 + 2x3, x2, x3) IR3}

        =    {x2 (1, 1, 0) + x3 (2, 0, 1)}

        =    span <(1, 1, 0), (2, 0, 1)>

        dim W1 =    2    [  the set {(1, 1, 0), (2, 0, 1} is linearly independent]

    Also, we have    W2 =    {(3x2 + 2x3, x2, x3) IR3}

        =    {x2 (3, 1, 0) + x3 (2, 0, 1)}

        =    span

        dim W2 =    2    [ the set {3, 1, 0), (2, 0, 1)} is linearly independent]

    Let us consider the subspace W1
W2.

          W1
W2 =    {(x1, x2, x3) IR3 : x1 = x2 = 2x3, x1 = 3x2 + 2x3}

    On solving the equations x1 = x2 + 2x3 and x1 = 3x2 + 2x3 we get x2 = 0 and x1 = 2x3.

    Therefore, the subspace W1
W2 can be represented as follows

        W1
W2 =    {(x1, x2, x3) R3 : x2 = 0, x1 = 2x3}

        =    

        =    

        =    

        dim (W1
W2) =    1

    Now, using the formula

        dim(W1 + W2) =    dim W1 + dim W2 – dim(W1
W2)

        =    2 + 2 – 1 = 3.

16.    Let V be the real vector space of all polynomials of degree less than or equal to 2 with real coefficients. Let T : V V be the linear transformation given by T(p) = 2p + for p V, where is the derivative of p. Then the number of nonzero entries in the Jordan cononical form of a matrix of T equals.

Ans.    5

Sol.    Given, V is a real vector space of all polynomials of degree less than or equal to 2 with real coefficients i.e. V can be written as

        V =    {a0 + a1x + a2x2| a0, a2, a2
IR}

    Given, T : V V defined as

        T(p) =    2p + for p V

        T{P(x)} =    2p(x) + (x) for p(x) V

    The standard basis of vector space V is given by {1, x, x2}.

    Let us find the image of basis elements under the transformation T.

        T(1) =    2 × 1 +

        =    2 + 0 = 2

        T(x) =    2x + = 2x + 1

        T(x2) =    2x2 + = 2x2 + 2x

    Now, we find the matrix representation of transformation T.

        T(1) =    2 = 2.1 + 0 . x + 0 . x2

        T(x) =    1 + 2x = 1.1 + 2.x + 0.x2

        T(x2) =    2x + 2x3 = 0 . 1 + 2.x + 2.x2

      The martix of T is transpose of matrix

        

    Now, we find the Jordan cononical form of this matrix. First, we find the characteristic polynomial of this matrix.

        

        (2 – ) [(2 – )2] – 1[0] + 0 =    0

        (2 – l)3 =    0

         =    2, 2, 2 are eigenvalues.

    Thus, characteristic polynomial is = (2 – )3.

    Now,    (A – 2l)2 =    

    Also, (A – 2l) 0

      Minimal polynomial is = (2 – )3.

    Hence, the Jordan canonical form of matrix of T is .

    Therefore, the number of non-zero entries in Jordan canonical form is 5.

16.    Let I = [2, 3], J be the set of all rational numbers in the interval [4, 6], K be the Cantor (ternary) set, and let L = {7 + x : x K}. Then the Lebesgue measure of the set I J L equals

Ans.    1

Sol.    Given,    I =    [2, 3].

    Let m denotes Lebesgue measure.

    Since,    [a, b] =    m(a, b) = m[a, b) = m(a, b]

    

    Also, we know countable sets have measure O and Q, the set of rationals is countable, so has measure O.

    

    Cantor ternary set is an uncountable set having measure O and measure is translation invariant i.e.

        m(A + x0) =    m ({x + x0| x A}) = m(A)

    

    Now, we know if is a sequence of disjoint measurable sets, then

    

17.    Let u(x, y, z) = x2 – 2y + 4z2 for (x, y, z) R3. Then the directional derivative of u in the direction at the point (5, 1, 0) is

Ans.    6

Sol.    We have,

        u =    x2 – 2y + 4z2

        

    

    Now, the directional derivatives of u0 at (5, 1, 0) in the direction of is

        

18.    If the Laplace transform of y(t) is given by Y(s) = L(y(t)) = then y(0) + equals.

Ans.    1

Sol.    

    

19.    The number of regular singular point of the different equation

    [(x – 1)2 sin x] + [cos x sin (x – 1)] + (x – 1) y = 0 in the interval is equal to

Ans.    2

Sol.    Given,

        [(x – 1)2 sin x] + [cos x sin (x – 1)] + (x – 1) y = 0

    

    On comparing the above equation with standard second order differential equation, + P(x) + Q(x)y = 0. we get

    

    

    So, x = 1 is a singular point.

    Similarly, and diverges, so x = 0 is a singular point.

    Now, for regular singular point we calculate the following limits.

    

    

    

    Therefore, x = 1 is a regular singular point.

    Similarly, gives a finite value.

    Thus, x = 0 is also a regular singular point.

    Also 0, 1

    Therefore, there are 2 regular singular points in .

20.    Let F be a field with 76 elements and let K be a subfield of F with 49 elements. Then the dimension of F as a vector space over K is

Ans.    3

Sol.    The dimension of a finite field F having cardinality Pn over its subfield K having cardinality Pm, where P is any prime number, m, n are natural numbers, m divides, n is .

    Here,     |F| = 76 and |K| = 49 = 72

    So, P = 7, n = 6 and m = 2

     Dimension of F over K as a vector space

    

21.    Let C ([0, 1]) be the real vector space of all continuous real valued functions on [0, 1], and let T be the linear operator on C ([0, 1]) given by

            (Tf)(x) =

    Then the dimension of the range space of T equals.

Ans.    2

Sol.    Given, C[0, 1] is space of continuous functions on [0, 1]

    

    

    

    Here, and are arbitrary constants. So, (Tf) (x) span {sin x, cos x}.

    Now, to show linear independence of {sin x, cos x}, we find Wronskian of them.

        

    Hence, {sin x, cos x} are linearly independent.

    Range space of T is spanned by {sin x, cos x} and {sin x, cos x} is linearly independent on [0, 1]. Thus, it forms a basis of range space of T.

    Hence, dimension (range space of T) = 2.

22.    Let a (–1, 1) be such that the quadrature rule is exact for all polynomials of degree less than or equal to 3. Then 3a2 equals

Ans.    1

Sol.    Given, a (–1, 1) and is exact for all polynomials of degree less than or equal to 3.

    Now, collection of polynomials of degree less than or equal to 3 can be represented as {a0 + a1x + a2x2 + a3x3|a0, a1, a2, a3, are constants}.

    It has basis {1, x, x2, x3}. We will calculate the value of integral on basis elements i.e. we take f(x) = 1, x, x2 and x3.

    Let     f(x) =    1,

        f(–a) =    1 = f(a)

    

    

        f(–a) =    –a, f(a) = a

        f(–a) + f(a) =    –a + a = 0 =

    

        f(–a) =    (–a)2 = a2, f(a) = a2

    For quadrature rule of be exact,

    

    

        f(–a) =    – a3, f(a) = a3

        f(–a) + f(a) =    – a3 + a3 = 0

    

23.    Let X and Y have joint probability density function given by

            fX,Y(x, y) =

    If fY denotes the marginal probability density function of Y, then fY(1/2) equals

Ans.    1

Sol.    We have,

        fX,Y(x, y) =    

    Clearly, marginal probability density function of Y = fY(y)

    

    

24.    Let the cumulative distribution function of the random variable X be given by

        FX(x) =    

    Then P(X = 1/2) equal to

Ans.    0.25

Sol.    We have,

        FX(x) =    

    

    

    

    

26.    Let {X1} be a sequence of independent Bernoulli random variables with P(Xj = 1) = 1/4 and let
Yn = Then Yn converges, in probability, to

Ans.    0.25

Sol.    Here, X1, X2, X3 ,.... are independent Bernoulli random variables with

    

    We know that, if X1, X2, .... are independent Bernoulli random variables with E|X1|k < then

    

26.    Let be the circle given by z = where varies from 0 to Then equals

    (a)

    (b)

    (c)

    (d)

Ans.    (c)

Sol.    Given, denotes the circle z =

        

    We use the Cauchy's integral formula for derivative which states that if f is analytic in a simply connected domain D, z0
D, then

    

    where is any closed contour in D containing z0.

    

    So, we have two singularities of f(z), i.e. 0 and 2 which are simple poles.

    Now, we are given the contour as |z| = 4 i.e. circle of radius 4 with centre (0, 0).


    Now, consider the contours and such that and enclose singularities 0 and 2 respectively.

    

    

    

    

27.    The image of the half plane Re(z) + Im(z) > 0 under the map w = is given by

    (a) Re(w) > 0

    (b) Im(w) > 0

    (c) |w| > 1

    (d) |w| < 1

Ans.    (d)

Sol.    Given,

    

    

    We put z = x + iy and w = u + iv in Eq. (i), where w is image of z under given map.

    

    

    

    Now, we have to find image of

        Re(z) + Im(z) >    0

    i.e.    x + y >    0    [ z = x + iy]

    On comparing both sides of Eq. (ii), we get

    

    

    

    

28.    Let D R2 denote the closed disc with centre at the origin and radius 2.

    Then equal to

    (a)

    (b)

    (c)

    (d)

Ans.    (a)

Sol.    

    Putting x2 + y2 = r2, we get

        r =    0 to 2 and

            

        

29.    Consider the polynomial p(X) = X4 + 4 in the ring Q[X] of polynomials in the variable X with coefficients in the field Q of rational numbers. Then

    (a)    the set of zeroes of p(X) in C forms a group under multiplication

    (b)    p(X) is reducible in the ring Q[X]

    (c)    the splitting field of p(X) has degree 3 over Q

    (d)    the splitting field of p(X) has degree 4 over Q

Ans.    (b)

Sol.    Given, p(X) = X4 + 4 in the ring Q[X], where Q is field of rational numbers.

    Consider, p(X) = X4 + 4

        =    X4 + 4X2 + 4 – 4X2

        =    (X2 + 2)2 – (2X)2

        =    (X2 + 2 + 2X) (X2 + 2 – 2X)

    Clearly, p(X) is reducible in the ring Q[X]

    For zeroes of p(X), put p(X) = 0

    

    

    Now, consider the set of zeroes equipped with binary operation *, where * denotes multiplication. [{1 + i, 1 – i, – 1 +    i, – 1 – i},*] = G

    Now, consider (1 + i) (1 – i) = 1 – i2 = 1 – (–1) = 2

    Here, 2 G. So closure property fails to hold. Thus, G does not form a group, we note that degree of splitting field of p(X) over Q is two.

30.    Which one of the following statements is true?

    (a)    Every group of order 12 has a non-trivial proper normal subgroup

    (b)    Some group of order 12 does not have a non-trivial proper normal subgroup

    (c)    Every group of order 12 has a subgroup of order 6

    (d)    Every group of order 12 has an element of order 12

Ans.    (a)

Sol.    Consider Z2 × Z6 it is an abelian group of order 12 it is non-cyclic as it does not contian an element of order 12. So, option (d) is incorrect.

    Let us consider the group A4 i.e. the collection of even permutations on a set of four symbols.

    Order of A4, |A4| = = 12. A4 does not have a subgroup of order 6.

    Hence, option (c) is incorrect.

    There are five possible groups of order 12.

    Either abelian groups Z12 or Z2 × Z6 or non-abelian group D6 (Dihedral group of order 12),
A4 (even permutations), Semidirect product of Z3 × Z4.

    Since, each subgroup of abelian group is normal. So, Z12 and Z2 × Z6 contain a non-trivial proper normal subgroup.

    Also, the smallest non-abelian group which is simple (i.e. having only two trivial normal subgroup) is A5 (i.e. collection of even permutations on five symbols) which has order = 60. So option (b) is incorrect.

    Thus, every group of order 12 has a non-trivial proper normal subgroup.

31.    For an odd prime p, consider the ring Then the element 2 in is

    (a) a unit

    (b) a square

    (c) a prime

    (d) irreducible

Ans.    (d)

Sol.    Clearly,

    

    

    Taking conjugates on both sides, we get

        

    On multiplying the respective sides of the above equations, we get

        4 =    (a2 + pb2) (c2 + pd2)

    Both the sides of the above equation are positive integers.

    Consequently, we have the following cases

    Case I: a2 + pb2 = 1 and c2 + pd2 = 4

    Case II: a2 + pb2 + 4 and c2 + pd2 = 1

    Case III: a2 + pb2 = 2 and c2 + pd2 = 2

    It is clear that case III is not possible in Z.

    Case I is possible when a = ±1, b = 0

     = ±1, which are units in

    Similarly, case II yields that c +

    which are units in .

    Hence, Z is an irreducible elements.

32.    Consider the following two statements

    P : The matrix has infinitely many LU factorizations, where L is lower triangular with each diagonal entry I and U is upper trangular.

    Q : The matrix has no LU factorization, where L is lower triangular with each diagonal entry 1 and U is upper triangular.

    Then which one of the following options is correct?

    (a)    P is TRUE and Q is FALSE

    (b)    Both P and Q are TRUE

    (c)    P is FALSE and Q is TRUE

    (d)    Both P and Q are FALSE

Ans.    (b)

Sol.    

    Let    A =    LU

    where L = and U = a, b, c, d are constants

    

    

        ab =    0, ad + c = 7

    Now,     ad + c =    7

    

    Since, ab = 0, so 'a' can take any value. Correspondingly we have infinte value of c.

    So, there are infinite LU factorisations of A.

    

        ab =    2, ad + c = 5

    Now,     ab =    2

    

    

    Now, the relation 0 = 2 is not possible.

    Hence, there is no LU factorisation possible.

33.    If the characteristic curves of the partial differential equation xuxx + 2x2uxy = ux – 1 are µ(x, y) = c1 and v(x, y) = c2, where c1 and c2 are constants, then

    (a)    m(x, y) = x2 – y, v(x, y) = y

    (b)    m(x, y) = x2 + y, v(x, y) = y

    (c)    m(x, y) = x2 + y, v(x, y) = x2

    (d)    m(x, y) = x2 – y, v(x, y) = x2

Ans.    (a)

Sol.    Given, xuxx + 2x2uxy = ux – 1

    On comparing the equation with standard second order partial differential equation

    Ruxx + Suyy + Tuxy + Pux + Quy + Au = f(x, y)

    Here,    R =    x

        S =    2x2

        T =    0

    Characteristic equation is given as

    

    

    Now, characteristic curves are given by

    

    Put = 0 and = –2x

    Put = –2x in Eq. (i), we get

    

    On integrating both sides, we get

    

    

    Now, put = 0 in Eq. (i), we get

    

    Hence µ(x, y) = x2 – y and v(x, y) = y.

34.    Let f : X Y be a continuous map from a Hausdorff topological space X to a matric space Y. Consider the following two statements

    P : f is a closed map and the inverse image f–1(y) = {x X : f(x) = y} is compact for each y Y.

    Q : For every compact subset K Y, the inverse image f–1 (K) is a compact subset of X.

    Which one of the following is true?

    (a)    Q implies P but P does not imply Q

    (b)    P implies Q but Q does not imply P

    (c)    P and Q are equivalent

    (d)    neither P implies Q nor Q implies P

Ans.    (c)

Sol.    A function f : X Y is continuous if and only if f–1(k) is closed in X for every closed set K in Y. Let K be an closed set in Y, then f–1(K) Ai is relatively closed in Ai for i = 1, 2, ..., k, since the restriction of t to Ai is continuous for each i. But Ai is closed in X, and therefore a subset of Ai is relatively closed in Ai if and only if it is closed in X. If f : X Y be continuous function between topological spaces X and Y, and let A be a compact subset of X, then f(A) is a compact subset of Y.

36.    Let X denote R2 endowed with the usual topology. Let Y denote R endowed with the co-finite topology. If Z is the product topological space X × Y, then

    (a)    the topology of X is the same as the topology of Z

    (b)    the topology of X is strictly coarser (weaker) than that of Z

    (c)    the topology of Z is strictly coarser (weaker) than that of X

    (d)    the topology of X cannot be compared with that of Z

Ans.    (c)

Sol.    Let X1 denoted the topological space R with discrete topology and let X2 be R with usual topology then the product topology on R × R in nothing but the dictionary order topology on R2. Since the basis for the product topology on R × R is given by {{x1} × (a, b) : x1, ab R}. The product topology is finer than the usual topology on R2. Infact any basis element (a, b) × (c, d) of the usual topology can be expressed as the union U a < x < b {x} × (c, d) of open sets {x} × (c, d) on the product topology .

36.    Consider Rn with the usual topology for n = 1, 2, 3. Each of the following options gives topological spaces X and Y with repective induced topologies. In which option is X homeomorphic to Y?

    (a)    X = {(x, y, z) R3 : x2 + y2 = 1},

        Y = {(x, y, z) R3 : z = 0, x2 + y2
0}

    (b)    X = {(x, y) R2 : y = sin (1/x),

        0 < x } {(x, y) R2 : x = 0, – 1 y 1}, Y = [0, 1] R

    (c)    X = {(x, y) R2 : y = x sin (1/x), 0 < 0 1}, Y = [0, 1] R

    (d)    X = {(x, y, z} R3 : x2 + y2 = 1},

        Y = {(x, y, z} R3 : x2 + y2 = z2
0}

Ans.    (a)

Sol.    Let (X, ) and (Y, u) be topological spaces, and let f : X Y be a bijection f is said to be homomorphism if f continuous and its inverse f–1 is continuous.

    Here,    X =    {(x, y, z) R3 : x2 + y2 = 1}

        Y =    {(x, y, z) R3 : z = 0, x2 + y2
0}

    represents a cylinder which is a continuous subjective conclude that X and Y are homomorphic.

37.    Let {Xi} be a sequence of independent Poisson variables and let Wn = Then the limiting distribution of is the normal distribution with zero mean and variance given by

    (a) 1

    (b)

    (c)

    (d)

Ans.    (c)

Sol.    

    

    

    

38.    Let X1, X2,...,Xn be independent and identically distributed random variables with probability density function given by

        

    Also, let Then the maximum likelihood estimator of is

    (a)

    (b)

    (c)

    (d)

Ans.    (c)

Sol.    Given X1, X2, ..., Xn are independent and identically distributed distributed random variables with p.d.f. given by

        

    

    

    Now, the linelihood equations for estimating is

    

    

    

39.    Consider the Liner Programming Problem (LPP) Maximize ax1 + x2

    Subject to 2x1 + x2
6, – x1 + x2
1, x1 + x2
4, x1
0, x2
0

    where is a constant. If (3, 0) is the only optimal solution, then

    (a) < – 2

    (b) – 2 < < 1

    (c) 1 < < 2

    (d) > 2

Ans.    (d)

Sol.    We have,

    Maximize = x1 + x2

    Subject to constraints,

        2x1 + x2
    6

        –x1 + x2
    1

        x1 + x2
    4

        x1
0, x2
    0

    The graph of inequalities are

    

    

    Here, (3, 0) is the only optimal solution

    

40.    Let M2(R) be the vector space of all 2 × 2 real matrices over the field R. Define the linear transformation S : M2(R) M2(R) by S(X) = 2X + XT, where XT denotes the transpose of the matrix X. Then the trace of S equals.

Ans.    10

Sol.    Given,

    M2(IR) = vector space of all 2 × 2 real matrices over IR

    

    A basis (standard) of M2(IR) is given as follows

            

    A linear transformation is defined as below

        S : M2(IR)     M2(IR)

        S(X) =    2X + XT, XT is transpose of X.

    Let us find the image of basis elements under the map S.

    

    

    Now, we express image of each basis elements in linear combination of standard basis elements.

    

    

    Thus, matrix representation of linear transformation S is transpose of matrix

    

    Therefore, trace (S) = 3 + 2 + 2 + 3 = 10.

41.    Consider R3 with the usual inner product. If d is the distance from (1, 1, 1) to the subspace span {(1, 1, 0), (0, 1, 1,) of R3, then 3d2 equal to

Ans.    1

Sol.    Let v be any vector belongs to linear span of {(1, 1, 0), (0, 1, 1)} i.e. v = (1, 1, 0) + (0, 1, 1) = for some R.

    Now, the square of the distance from (1, 1, 1) to any vector v is given by

    

    Now, to minimise D2 differentiate it w.r.t. and , and setting it equal to zero.

    

                …(i)

    

                    …(ii)

    On solving Eqs. (i) and (ii), we get

    

    

42.    Consider the matrix A = I9 – 2uTu with u = [1, 1, 1, 1, 1, 1, 1, 1, 1], where I9 is the 9 × 9 identity matrix and uT is the transpose of u. If and µ are two distinct eigenvalues of A, then equals

Ans.    2

Sol.    Given, A = I9 – 2uT u, I9 is 9 × 9 identity matrix with u = [1, 1, 1, 1, 1, 1, 1, 1, 1]1 × 9

    Since, u is a row vector. So, uT is a column vector as follows

            

    

    

    If we have a square matrix of order n with all entries as 1, it has characteristic polynomial as xn – 1(x – n). So, it has eigenvalues.

    'n' of multiplicity 1 and eigenvalues '0' of multiplicity (n – 1). Here, we have a square matrix of order 9 so it has two distinct eigenvelues 9 and 0 with multiplicity 1 and 8 respectively.

    Now, uTu has eigenvalues

    i.e. 1 and 0 [ if is an eigenvalue of matrix A, then is an eigenvalue of matrix where is constant]

    Now, eigenvalues of I9 are 1 with multiplicity 9.

    Hence, eigenvalues of A = I9 – 2uTu are 1 – 2 × 1 and 1 – 2 × 0 i.e. – 1 and 1.

    [ for any matrix A, having eigenvalue the matrix I + A has eigenvalue 1 + ]

    

43.    Let f(z) = for z C and let be the circle z = where varies from 0 to Then equals

Ans.    6

Sol.    

    

    

    

    

    

44.    Let S be the surface of the solid V = {(x, y, z) : 0 x 1, 0 y 2, 0 z 3}

    Let denote the unit outward normal to S and let (x, y, z) =

    Then the surface integral equals

Ans.    18

Sol.    We have,

        V = {(x, y, z) : 0 x 1, 0 y 2, 0 z 3}

         (x, y, z) =

    

    

    

    

46.    Let A be a 3 × 3 matrix with real entries. If three solutions of the linear system of differential equations x(t) = Ax(t) are given by

    

    then the sum of the diagonal entries of A is equal to

Ans.    2

Sol.    Given, x(t) = A(x (t)

        Let A =    

    matrix with real entries i.e. a, b, c, d, e, f, g, h, i IR

    Now, is a solution of given system of differential equation.

    Therefore, it satisfies Eq. (i).

    Here, denotes the derivative of x(t).

    

    On comparing the coefficients of like terms, we get

        a – b + c =    1

        –a + b + c =    –2

        d – e + f =    –1

        –d + e + f =    2

        g – h + i =    1

        –g + h + i =    2

    

    Satisfy the system of differential equations given by Eq. (i).

    Solving in a similar manner, we get

        a =    0, d = –1, g = 1

    

    Hence, sum of diagonal entries of A = 0 + = 2.

46.    If y1(x) = is a solution of the differential equation for some real numbers and then equal to

Ans.    4

Sol.    We have,

    

    

    Multiplying be x, we get

    

    Comparing with

    

47.    Let L2 ([0, 1]) be the Hilbert space of all real valued square integrable functions on [0, 1] with the usual inner product. Let be the linear functional on L2 ([0, 1]) defined by

        

    Where µ denotes the Lebesgue measure on [0, 1].

    Then equal to

Ans.    3

Sol.    Use the difinition of norm on the space L2[a, b] given by
= for 0 < c < 1 and for each n with 0 < c – define xn
c [0, 1] xn(t) = 1, if c 1

    

    

48.    Let U be an orthonormal set in a Hilbert space H and let x H be such that ||x|| = 2. Consider the set E = .

    Then the maximum possible number of elements in E is

Ans.    64

Sol.    We know that, if U H is any orthonormal subset of a Hilbert space H and if x belongs to H, then maximum possible number of elements in is given by n2 ||x||2.

    Maximum possible number of elements in E is

        42||x||2 =    16 × 22 = 16 × 4 = 64

49.    If p(x) = 2 – (x + 1) + x(x + 1) – x(x + 1) (x – ) interpolates the points (x, y) is the table


    then equals.

Ans.    3

Sol.    Given,

    

    

    

    

50.    

Ans.    2

Sol.    

    We know fourier series of f(x) defined on [a, b] is given by

    

    

    According to given series expansion, we get

    

    

    

    

    

    Hence, the required value is 2.

51.    For n = 1, 2, ..., lef fn(x) = [0, 1]. Then equals.

Ans.    1

Sol.    Use the following result,

    If < fn > is sequence of continuous functions defined on an interval [a, b] which converges uniformly to a function f on [a, b]. Then f is continuous and

        

52.    Let X1, X2, X3, X4 be independent exponential random variables with mean 1, 1/2, 1/3, 1/4 respectively. Then Y = min (X1, X2, X3, X4) has exponential distribution with mean equal to

Ans.    0.1

Sol.    Given, X1, X2, X3, X4 follow exponential distribution with mean respectively.

    X1 ~ exp(1), X2 ~ exp(2), X3 ~ exp(3) and X4 ~ exp(4)


    Now, as we know that, if X1, X2,....., Xn are independent random variables Xi having an exponential distribution with parameter λi, i = 1, 2... n; then Y = min (X1, X2, ... Xn) has exponential distribution

    

    

53.    Let X be the number of heads in 4 tosses of a fair coin by Person 1 and let Y be the number of heads in 4 tosses of a fair coin by Person 2. Assume that all the tosses are independent. Then the value of P (X = Y) correct up to three decimal places is.

Ans.    0.27

Sol.    The probability distribution of X and Y is given below

    

    Now, required probability,

    

    

    

54.    Let X1 and X2 be independent geometric random variables with the same probability mass function given by P (X = k) = p (1 – p)k – 1, k = 1, 2,... . Then the value of P (X1 = 2| X1 + X2 = 4) correct up to the decimal place is

Ans.    0.333

Sol.    Given X1, X2 ~ Geo (p)

    

    

    

56.    A certain commdity is produced by the manufacturing plants P1 and P2 whose capacities are 6 and 5 units, respectively. The commodity is shipped to markets M1, M2, M3 and M4 whose requirements are 1, 2, 3, and 5 units, respectively. The transportation cost per unit form plant Pi to market Mj is as follows


    Then the optimal cost of transportation is

Ans.    57

Sol.    Given,


    We may note that given transportation problem is balanced because total quantity manufactured equals total requirement i.e, 1 + 2 + 3 + 5 = 6 + 5 = 11.

    We apply North-West corner rule to find the initial basic feasible solution as follows


    Transportation cost

        =    1 × 1 + 2 × 3 + 3 × 5 + 5 × 7

        =    1 + 6 + 15 + 35 = 7 + 50 = 57.