Which of the following is not an exact differential?

- a)
- b)
- c)xdy - ydx
- d)

Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered • 5 hours ago

∴ expressions in (a), (b) and (d; are exact.

Since the expression

xdy - ydx

in (c) can not be expressed as du for some function u of x and y, therefore the expression in (c) is not exact.

Kilaparti Sekhar asked • 1 hour ago

The orthogonal trajectory to the family of circles x^{2} + y^{2} = 2cx (c arbitrary) is describe by the differential equation,

- a)(x
^{2}+ y^{2}) y' = 2xy - b)(x
^{2}- y^{2}) y' = 2xy - c)(y
^{2}- x^{2}) y' = xy - d)(y
^{2}- x^{2}) y' = 2xy

Correct answer is option 'D'. Can you explain this answer?

Which one of the following is a general solution of the differential equation p^{2} - 2p cos hx + 1 = 0?

- a)(y - e
^{x}- c) (y + e^{-x}- c) = 0 - b)(y + e
^{x}- c) (y - e^{-x}- c) = 0 - c)(y + log x - c) (y + e
^{-x}- c) = 0 - d)(y + e
^{x}- c) (y - log x - c) = 0

Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered • 5 hours ago

and do as in the previous question.

The value of the greatest term in the expansion of

- a)2871.11
- b)2871
- c)2872
- d)2873

Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered • 5 hours ago

Hence, t

The greatest coefficient in the expansion of (1 + x)^{2n + 2} is

- a)
- b)
- c)
- d)

Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered • 5 hours ago

Here 2n + 2 is even

Greatest coefficient

Greatest coefficient

What is the rank of the linear transformation T : R^{3} ---> R^{3} defined by T(x, y, z) = (y, 0, z)?

- a)3
- b)2
- c)l
- d)0

Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered • 5 hours ago

We are given that the linear transformation T : R^{3} ---> R^{3} defined by

We need to determine the rank of the linear transformation T.

Let (x, y, z) ∈ ker T

ThenT(x, y, z) = (0, 0, 0)

Using the definition of linear transformation, we get,

implies y = 0, z = 0 and x is arbitrary Therefore, ker T = {( x, y, z ) : y = 0, z = 0 and x is arbitrary}

Hence, Nullify of T= 1

Using Rank Nullity theorem, we get Rank T= dim R^{3} - Nullity T

= 3 - 1 = 2

We need to determine the rank of the linear transformation T.

Let (x, y, z) ∈ ker T

ThenT(x, y, z) = (0, 0, 0)

Using the definition of linear transformation, we get,

implies y = 0, z = 0 and x is arbitrary Therefore, ker T = {( x, y, z ) : y = 0, z = 0 and x is arbitrary}

Hence, Nullify of T= 1

Using Rank Nullity theorem, we get Rank T= dim R

= 3 - 1 = 2

If the linear transformation T(v) = Av rotates the vectors (-1, 0) and (0,1), x radians clockwise, then:

- a)A =
- b)A =
- c)A =
- d)A =

Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered • 5 hours ago

We are given that the linear transformation T(v) = Av rotates the vectors (-1,0) and (0,1), π radians clockwise.

We need to find the matrix A.

We know that if a vector (a, b) is rotated through an angle a under T. Then

Therefore, the matrix of T is

A =

We need to find the matrix A.

We know that if a vector (a, b) is rotated through an angle a under T. Then

Therefore, the matrix of T is

A =

If X = , the rank of X^{T}X, where X^{T} denotes the transpose of X, is

- a)0
- b)2
- c)3
- d)4

Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered • 5 hours ago

We are given that the matrix

X=

We need to find the rank of X^{T}X. The transpose of X,

X^{T}=

Now,

Rank (X^{T}X) = 3

X=

We need to find the rank of X

X

Now,

Rank (X

Which one of the following is true?

- a)dim Horn (R
^{3}, R^{4}) = 12 - b)dim Horn (R
^{4}, R^{3}) = 7 - c)dim Horn (R
^{3}, R^{4}) = 7 - d)None of these.

Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered • 5 hours ago

We need to find the dim Horn (R^{3}, R^{4}).

Since dim R^{3} = 3 and dim R^{4} = 4

Therefore,dim Horn (R^{3}, R^{4})

= 3 x 4 = 12

Let U and V be n and m dimensional vector space.

Then dim Hom (U, V) = nm

Since dim R

Therefore,dim Horn (R

= 3 x 4 = 12

Let U and V be n and m dimensional vector space.

Then dim Hom (U, V) = nm

If A = satisfies the matrix equation A^{2} - kA + 2I = 0, then what is the value of k?

- a)0
- b)1
- c)2
- d)3

Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered • 5 hours ago

We need to find the value o f k where the matrix

A = satisfies the equations A^{2} - kA + 2I = 0.

But A =

So,

Putting these values in the equation

A^{2} - kA + 2I= 0, we get

or

or 1 - 3k + 2 = 0 => k = 1

A = satisfies the equations A

But A =

So,

Putting these values in the equation

A

or

or 1 - 3k + 2 = 0 => k = 1

The system of equations

- a)a unique solutions
- b)finitely many solutions
- c)infinitely many solutions
- d)no solution

Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered • 16 hours ago

We are given that the system of equation,

It may be written in the matrix form as,

Here, the coefficient of matrix is,

and the augmented matrix is given by,

Reduce the system equation in echelon form using the operations

R_{1}" and

R_{3} -> R_{3} - R_{1}

These operation yield:

and also, R_{3} --> R_{3} - 2R_{2}

Here, Rank of A = Rank of aug(A) < number of unknowns.

Hence, the given system of equation has infinite many solutions.

It may be written in the matrix form as,

Here, the coefficient of matrix is,

and the augmented matrix is given by,

Reduce the system equation in echelon form using the operations

R

R

These operation yield:

and also, R

Here, Rank of A = Rank of aug(A) < number of unknowns.

Hence, the given system of equation has infinite many solutions.

The system of equation 2x + y = 5, x - 3y = -1

3x + 4y = k is consistent, when k is

3x + 4y = k is consistent, when k is

- a)1
- b)2
- c)5
- d)10

Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered • 16 hours ago

We are given that the system of equations,

2x + y = 5

x - 3y = - 1

3x + 4y = k is consistent.

We need to find the value of k. The given system of equation may be written as,

Since, this system of equation is consistent. Therefore

or 2(-3k + 4) -1(k +3) + 5(4 + 9) = 0

-7k + 70 = 0

k =10

2x + y = 5

x - 3y = - 1

3x + 4y = k is consistent.

We need to find the value of k. The given system of equation may be written as,

Since, this system of equation is consistent. Therefore

or 2(-3k + 4) -1(k +3) + 5(4 + 9) = 0

-7k + 70 = 0

k =10

Let :R^{3} --> R^{2} be the linear transformation given by

T(x, y, z) = (x, y),

with respect to standard basis of R^{3} and the basis {(1,0), (1, 1)} of R^{3}. What is the matrix representation of T?

T(x, y, z) = (x, y),

with respect to standard basis of R

- a)
- b)
- c)
- d)

Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered • 16 hours ago

Let T : R^{3}--> R^{2} be the linear transformation defined by

T(x, y, z) = (x, y)

we need to determine the matrix of linear transformation T w.r.t. standard basis of R^{3} and the basis {(1, 0), (1 ,1 } of R^{2}. since {(1 ,0 ,0) , (0,1,0) (0,0,1)} be the standard basis of R^{3}, therefore,

Thus, the matrix of linear transformation T w.r.t. the standard basis R^{3} and the basis

T(x, y, z) = (x, y)

we need to determine the matrix of linear transformation T w.r.t. standard basis of R

Thus, the matrix of linear transformation T w.r.t. the standard basis R

Consider, the linear transformation

T : R^{4 }----> R^{4} given by:

T(x, y, z, u) = (x, y, 0, 0),

Then, which one of the following is correct?

T : R

T(x, y, z, u) = (x, y, 0, 0),

Then, which one of the following is correct?

- a)Rank of T > Nullity of T
- b)Nullity of T > Rank of T
- c)Rank of T = Nullity of T = 3
- d)Rank of T = Nullity of T = 2

Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered • 16 hours ago

We are given that linear transformation

We need to determine Rank and Nullity of T.

Let (x,y, z, u) ∈ ker T

Then T(x, y, z, u) = (0, 0, 0, 0)

Using the definition of linear transformation we get

(x,y, 0, 0) = (0, 0, 0, 0)

implies x = 0, y = 0, z and u are arbitrary

Therefore,

Hence, Nullity of T = 2

Using Rank Nullity theorem, we get

We need to determine Rank and Nullity of T.

Let (x,y, z, u) ∈ ker T

Then T(x, y, z, u) = (0, 0, 0, 0)

Using the definition of linear transformation we get

(x,y, 0, 0) = (0, 0, 0, 0)

implies x = 0, y = 0, z and u are arbitrary

Therefore,

Hence, Nullity of T = 2

Using Rank Nullity theorem, we get

Let M = . Then, the rank of M is equal to

- a)3
- b)4
- c)2
- d)1

Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered • 16 hours ago

We need to find the rank of the matrix,

Reduce the matrix to echelon form using the operation.

These operations yield.

and also applying R_{4} —> 2R_{4} - R_{2} we have,

So, rank of M = 2.

Reduce the matrix to echelon form using the operation.

These operations yield.

and also applying R

So, rank of M = 2.

If T : R^{3} ---> R^{3} is given by

- a)
- b)
- c)
- d)

Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered • 16 hours ago

We are given that a linear tansformation T : R^{3}---> R^{3} defined by T(x, y, z) = (x - y , y + 3 z, x + 2y).

We need to find T^{-1}.

Let (x, y, z) ∈ ker T. Then

T(x,y, z) = (0, 0, 0)

Using the definition of T, we get

(x -y,y + 3z, x + 2y) = (0, 0, 0) Comparing the components on both sides, we get

x - y =0 , y + 3z = 0, x +2y = 0

Solving for x, y and z, we get

x = 0, y = 0 and z = 0

Hence, ker T= {(0, 0, 0)}, implies T is one-one.

Since T : R^{3} —> R^{3} is a linear tansformation and one-one. Hence, T is onto.

Therefore, T^{-1 }exist.

Let (a,b,c) be the image of (x, y, z) under T^{-1}.

Then

T^{-1}(x, y, z) = (a, b,c)

Implies (x, y, z) = T(a, b, c)

Using the definition of T, we get

(x, y, z) = (a - b, b + 3c, a +2b)

Comparing the components on both sides, we get

x =a - b, y = b + 3c, z = a + 2b

Solving for a, b and c, we get

Therefore, T^{-1}(x,y, z)

We need to find T

Let (x, y, z) ∈ ker T. Then

T(x,y, z) = (0, 0, 0)

Using the definition of T, we get

(x -y,y + 3z, x + 2y) = (0, 0, 0) Comparing the components on both sides, we get

x - y =0 , y + 3z = 0, x +2y = 0

Solving for x, y and z, we get

x = 0, y = 0 and z = 0

Hence, ker T= {(0, 0, 0)}, implies T is one-one.

Since T : R

Therefore, T

Let (a,b,c) be the image of (x, y, z) under T

Then

T

Implies (x, y, z) = T(a, b, c)

Using the definition of T, we get

(x, y, z) = (a - b, b + 3c, a +2b)

Comparing the components on both sides, we get

x =a - b, y = b + 3c, z = a + 2b

Solving for a, b and c, we get

Therefore, T

Solution for the system defined by the set of equations 4y + 3z = 8; 2x - z = 2 and 3x + 2y = 5 is

- a)x = 0; y = 1; z = 4/3
- b)x = 0; y = 1/2; z = 2
- c)x = 1 ;y= 1/2; z = 2
- d)not exist

Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered • 16 hours ago

We need to determine the solution for the system of equation

4y+3z= 8

2x - z = 2

3x + 2y= 5

This system of equation may be written in matrix form as,

Reduce this system of equation to echelon form using the operations

"R_{2} ⇔ R_{1}". This operations yields

and also applying "R_{3 }→ 2R_{3} - 3R_{1}" which yields

again, "R_{1}→ R_{3} - R_{2} " we get

Here the coefficient matrix

and augmented matrix is given by

Since, the rank of coefficient matrix ≠ the rank of augmented matrix.

Therefore, the solution does not exist.

4y+3z= 8

2x - z = 2

3x + 2y= 5

This system of equation may be written in matrix form as,

Reduce this system of equation to echelon form using the operations

"R

and also applying "R

again, "R

Here the coefficient matrix

and augmented matrix is given by

Since, the rank of coefficient matrix ≠ the rank of augmented matrix.

Therefore, the solution does not exist.

The eigen vectors o f the matrix are written in the form and . What is a + b ?

- a)0
- b)1/2
- c)1
- d)2

Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered • 16 hours ago

We are given that the eigen vectors o f the matrix are and . We need to find the value o f a + b. The characteristic equation o f the given matrix is, |A - λI| = 0

or (1 - λ)(2 - λ) = 0 or

λ = 1,2

Put λ = 1 in the equation [ A- λI ] X= 0

or a = 0 and also put X = 2 in the equation [A - λI] X= 0

or

or 1+2b = 0

or b = 1/2

Now, a + b = 0 + 1/2

a + b = 1/2

or (1 - λ)(2 - λ) = 0 or

λ = 1,2

Put λ = 1 in the equation [ A- λI ] X= 0

or a = 0 and also put X = 2 in the equation [A - λI] X= 0

or

or 1+2b = 0

or b = 1/2

Now, a + b = 0 + 1/2

a + b = 1/2

For n ≠ m , let T_{1 }: R^{n} —> R^{m} and T_{2} : R^{m} —> R^{n }be linear transformations such that T_{1}T_{2} is bijective. Then

- a)rank (T
_{1}) = n and rank (T_{2}) = m - b)rank (T
_{1}) = m and rank (T_{2}) = n - c)rank (T
_{1}) = n and rank (T_{2}) = n - d)rank (T
_{1}) = m and rank (T_{2}) = m

Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered • 16 hours ago

We are given that T_{1} : R^{n} —> R^{m }and T_{2} : R^{m} —> R^{n} be linear transformation such that T_{1}T_{2} is bijective, where n ≠ m

Thus, T_{1}T_{2 }---> R^{m} ---> R^{n} is a linear transformation given by

(T_{1}T_{2}) (n) = T_{1}(T_{2}(n))

Since T_{1}T_{2} is bijective. Therefore, T_{1}T_{2 }is non-singular and hence

Rank (T_{1}T_{2}) = m

But Rank (T_{1}T_{2}) < min Rank {T_{1}T_{2}} Therefore,m ≤ min Rank {T_{1}T2}

**Case I : **If m ≤ n, Then rank of both T_{1} and T_{2} will be m.

Thus, we get

Rank (T_{1}) = Rank (T_{2}) = m

**Case II :** If m ≥ n. Then since T_{1} is n x m matrix. Therefore, its rank cannot exceed n. Similarly, Rank T_{2} cannot exceed n. Thus, we get m ≤ n. Therefore m = n.

Thus, T

(T

Since T

Rank (T

But Rank (T

Thus, we get

Rank (T

Let T : R^{3} —> R^{3 }be defined by T(x_{1} x_{2}, x_{3})) = (x_{1} - x_{2}, x_{1} - x_{2},0).

If N(T) and R(T) denote the null space and the range space of T respectively, then

If N(T) and R(T) denote the null space and the range space of T respectively, then

- a)dim N(T) = 2
- b)dim R(T) = 2
- c)R(T) = N(T)
- d)N(T) ⊂ R(T)

Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered • yesterday

Let T : R^{3}---> R^{3} be def ined by T (x_{1}, x_{2}, x_{3}) = (x_{1} - x_{2}, x_{1} - x_{2}, 0).

and N(T) and R(T) denote the null space and the range space of T respectively.

Let (x_{1}, x_{2}, x_{3}) ∈ ker T.

Then

T(x_{1} x_{2}, x_{3}) = (0, 0, 0)

using the definition of T, we get

(x_{1} - x_{2}, x_{1} - x_{2}, 0) = (0, 0, 0)

Implies x_{1} = x_{2}, x_{3} is arbitrary.

Therefore,ker T= {(x_{1}, x_{2}, x_{3}) : x_{1}, x_{3} arbitrary}

Hence, nullity of T = 2

and Rank of T = 3 - 2 = 1.

and N(T) and R(T) denote the null space and the range space of T respectively.

Let (x

Then

T(x

using the definition of T, we get

(x

Implies x

Therefore,ker T= {(x

Hence, nullity of T = 2

and Rank of T = 3 - 2 = 1.

A system o f linear equations x + 2 y - z = l l , 3 x + y - 2 z = 10, x - 3y = 5 has

- a)no solution
- b)exactly one solution
- c)exactly 3 solutions
- d)infinitely many solution

Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered • yesterday

We are given that the system of equation

x + 2y - z =11

3 x + y - 2z = 10

x - 3 y = 5

The given system of equation can be written in the augmented matrix as,

Applying the operations "R^{2} ---> R^{2} - 3R_{1}" and R_{3} —> R_{3} - R_{1 }these operations yield

and also "R_{3} ---> R_{3} - R_{2} ", which gives,

Here, rank of (A) ≠ rank of aug (A).

Hence, the given system of equation has no solution.

x + 2y - z =11

3 x + y - 2z = 10

x - 3 y = 5

The given system of equation can be written in the augmented matrix as,

Applying the operations "R

and also "R

Here, rank of (A) ≠ rank of aug (A).

Hence, the given system of equation has no solution.

The characteristic polynomial of 3 x 3 matrix A |λI - A| = λ^{3} + 3λ^{2} + 4λ - 3

Let x - trace (A) and y = |A|, the determinant of A.

Then,

Let x - trace (A) and y = |A|, the determinant of A.

Then,

- a)x/y = 3/4
- b)x/y = 4/3
- c)x = y = -3
- d)x = 3 and y = -3

Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered • yesterday

We are given that the characteristic polynomial o f 3 x 3 matrix, A is

λ^{3} + 3λ^{2} + 4λ - 3 = 0

and x = trace (A)

y = det(A)

The general characteristic polynomial of 3 x 3 matrix is given by,

λ^{3} - trace (A) λ^{2} + λ + |A| = 0

Comparing this equation by the given equation, we get

trace = -3

or x = -3

and |A| = -3

or y = -3

Hence, x = y = -3

λ

and x = trace (A)

y = det(A)

The general characteristic polynomial of 3 x 3 matrix is given by,

λ

Comparing this equation by the given equation, we get

trace = -3

or x = -3

and |A| = -3

or y = -3

Hence, x = y = -3

Let T : R^{3} --> R^{3} be defined by T(x_{1}, x_{2}, x_{3}) = (x_{1} + x_{2}, x_{2} + x_{3}, x_{3} + x_{1}) Then T^{-1} is

- a)
- b)
- c)
- d)None of these

Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered • yesterday

We are given that a linear transformation T :R^{3} --> R^{3} defined by

T(x_{1}, x_{2}, x_{3}) = (x_{1} + x_{2}, x_{2} + x_{3}, x_{3} + x_{4})

We need to find T^{-1}

Let (x, y, z) ∈ ker T

then T(x, y, z) =(0, 0, 0)

Using the definition of linear transformation, we get

(x_{1} + x_{2}, x_{2} + x_{3}, x_{3} + x_{4}) = (0, 0, 0)

Comparing the components of the coordinates, we get

x_{1} + x_{2} = 0

x_{2} + x_{3} = 0

x_{3} + x_{1} = 0

Implies, x_{1} = 0, x_{2} = 0 and x_{3} = 0

Therefore, ker T = {(0,0,0)} , hence, T is one-one.

Since T is a linear transformation from R^{3} to R^{3} but R^{3} is a three dimensional (finite dimensional) vector space.

Hence, T is onto and therefore, T^{-1} must exist.

Let (x, y, z) be the image of (a,b,c) under T^{-1},

Therefore T^{-1}(a ,b ,c ) = (x,y,z) Implies, T{x, y, z) = (a, b , c ) Using the definition of linear transformation, we get

(x + y , y + z ,z + x) = (a , b, c) Comparing the components of the co-ordinates, we get

x + y = a

y + z = b

z + x - c

Solving for x, y and z, we get

and

Hence,T^{-1}(a, b, c)

That is

T^{-1}(x,y,z) = Hence,T^{-1}(x_{1}, x_{2}, x_{3})

... moreT(x

We need to find T

Let (x, y, z) ∈ ker T

then T(x, y, z) =(0, 0, 0)

Using the definition of linear transformation, we get

(x

Comparing the components of the coordinates, we get

x

x

x

Implies, x

Therefore, ker T = {(0,0,0)} , hence, T is one-one.

Since T is a linear transformation from R

Hence, T is onto and therefore, T

Let (x, y, z) be the image of (a,b,c) under T

Therefore T

(x + y , y + z ,z + x) = (a , b, c) Comparing the components of the co-ordinates, we get

x + y = a

y + z = b

z + x - c

Solving for x, y and z, we get

and

Hence,T

That is

T

The system of simultaneous linear equations

x + y + z = 0

x - y - z = 0 has

x + y + z = 0

x - y - z = 0 has

- a)no solution in R
^{3} - b)a unique solution in R
^{3} - c)infinitely many solution in R
^{3} - d)more than 2 but finitely many solutions in R
^{3}

Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered • yesterday

We are given that the system of simultaneous linear equations,

x + y + z = 0

x - y - z = 0

The coefficient matrix is given by,

rank of coefficient matrix = 2. Here, the rank of matrix < no. of unknowns therefore, the system of equation has infinitely many solution in R^{3}.

x + y + z = 0

x - y - z = 0

The coefficient matrix is given by,

rank of coefficient matrix = 2. Here, the rank of matrix < no. of unknowns therefore, the system of equation has infinitely many solution in R

Let V be the vector space of polynomial functions of degree three or less. Let the ordered basis for V consisting of the functions of the four functions x^{j} : j = 0, 1, 2, 3 and let D be the differentiation operator. Then the matrix of D in the above ordered basis is

- a)
- b)
- c)
- d)

Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered • yesterday

Let V be the vector space of polynomial functions of degree three or less. Let the ordered basis for V consisting of the functions

x^{j} : j = 0 ,1 ,2 , 3

and let D be the differentiation operator. We need to find the matrix of differentiation operator D in the basis x^{j}, j = 0,1, 2, 3, that is {1, x, x^{2}, x^{3}}.

Therefore, D(1) = d/dx(1) = 0

= 0.1 + 0.x + 0.x^{2} + 0.x^{3}

D(x) = d/dx (x) = 1 = 1.1 + 0 . x + 0 .x^{2} + 0.x^{3} ax

D(x^{2}) =d/dx(x^{2}) = 2x = 0.1 + 2.x + 0.x^{2} + 0.x^{3} ax

D(x^{3}) = d/dx (x^{3}) = 3x^{2} = 0.1 + 0.x + 3.x^{2} + 0.x^{4}

Thus, the matrix of differentiation operator is

x

and let D be the differentiation operator. We need to find the matrix of differentiation operator D in the basis x

Therefore, D(1) = d/dx(1) = 0

= 0.1 + 0.x + 0.x

D(x) = d/dx (x) = 1 = 1.1 + 0 . x + 0 .x

D(x

D(x

Thus, the matrix of differentiation operator is

Let T be a linear transformation given by the matrix (occurring in a typical transportation problem)

then

then

- a)Rank T = 3 and T is unimodular
- b)Rank T = 3 and T is not unimodular
- c)Rank T= 1 and T is unimodular
- d)Rank T = 1 and T is not unimodular

Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered • yesterday

Let T be a linear transformation given by the matrix (occurring in a typical transportation problem) A linear transformation T is unimodular iff det (T) = ±1

Therefore, Rank T = 3. Also det (T) = 0. Hence, T is unimodular.

Therefore, Rank T = 3. Also det (T) = 0. Hence, T is unimodular.

Let T : C^{n }—> C^{n} be a linear operator of rank n - 2. Then,

- a)0 is not an eigen value of T
- b)0 must be an eigen value of T
- c)1 can never be an eigen value of T
- d)1 must be an eigen value of

Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered • yesterday

We are given that a linear operator T : C^{n}—> C^{n} of rank n - 2.

Since T : C^{n} —> C^{n }is a linear operator of rank n - 2.

Therefore, T is singular operator, Hence 0 must be an eigen value of T.

Since T : C

Therefore, T is singular operator, Hence 0 must be an eigen value of T.

f(x) = k, exp for all x ∈ R; can be a probability density function for

- a)k = 1
- b)k = 2n
- c)k = (2π)
^{-1/2} - d)k(2π)
^{-1}

Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered • yesterday

it becomes standard normal distribution.

Let x and y be independent random variables with binomial distribution B(10, 1 /3) and B(20, 1/3) respectively. E[x + y] is

- a)5
- b)10
- c)15
- d)30

Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered • yesterday

E (x + y) = E (x) + E(y)

Expectation of binomial distribution is np if B(n, p) is the binomial distribution.

Expectation of binomial distribution is np if B(n, p) is the binomial distribution.

Consider the two linear mapsT_{1} and T_{2} on V_{3} defined as T_{1}(x_{1}, x_{2}, x_{3}) = (0, x_{2}, x_{3}) and T_{2}(x_{1}, x_{2}, x_{3}) = (x_{1}, 0,0)

- a)T is idempotent but T
_{2}is not idempotent - b)T
_{2}is idempotent but T_{1}is not idempotent - c)Both T
_{1}and T_{2}are idempotent - d)Neither T
_{1}norT_{2 }are idempotent.

Correct answer is option 'C'. Can you explain this answer?

Uday Singh answered • yesterday

D

If X is uniformly distributed over (0, 10), the probability that 1 < X < 6 is

- a)3/10
- b)1/10
- c)5/10
- d)None of these

Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered • yesterday

Probability density function of X which is uniformly distributed in (0, 10) is

If are the means of two distributions such that is the mean of the combined distribution, then

- a)
- b)
- c)
- d)

Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered • yesterday

If n_{1 }elements has mean elements has mean

X is an exponential random variable with parameter λ with p.d.f. (x) = λe^{-7x }if x ≥ 0 = 0 if x < 0, identify the correct one

- a)P ( X > s + t) = P ( X > s ) P( X > t)
- b)P (X > s + t) = P (X > s) + P (X > t)
- c)P (X > s + t) = 1 - P ( X = s ) P ( X = t)
- d)P ( X > s + t) = λstP (X > s) P (X > t)

Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered • yesterday

The probability of throwing 6 at least one in four throws of a die is :

- a)1/6
- b)2/3
- c)625/1296
- d)671/1296

Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered • yesterday

Probability of throwing 6 at least once = 1 - probability of not throwing 6.

The number of dissimilar terms in the expansion of (a + b + c)^{2n+1} – (a + b – c)^{2n+1} is

- a)(n + 1)
^{2} - b)(n – 1)
^{2} - c)4n
^{2}– 1 - d)none of these

Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered • yesterday

Given expansion

Number of dissimilar term = (2n + 1) + (2n -1) + ....(2n - 3) + .... + 5 + 3 + 1

Number of dissimilar term = (2n + 1) + (2n -1) + ....(2n - 3) + .... + 5 + 3 + 1

Coefficient of t^{24} in (1 + t^{2})^{12} (1 + t^{12}) (1 + t^{24}) is

- a)
^{12}C_{6}+3 - b)
^{12}C_{6}+1 - c)
^{12}C_{6} - d)
^{12}C_{6}+2

Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered • yesterday

coefficient of t

⇒ coefficient of t

- a)
- b)
- c)
- d)

Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered • yesterday

∴ (d) is correct

Allumariyam Mathews asked • 22 hours ago

Let

Then,

Then,

- a)f is continuous at (0, 0) and the partial derivatives f
_{x}, f_{y}exists at every point o f R^{2} - b)f is discontinuous at (0,0) and f
_{x}, f_{y}does not exists at every point o f R^{2} - c)f is discontinuous at (0, 0) a n d f
_{x}, f_{y}exists at (0,0) - d)None of the above

Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered • yesterday

Let us suppose (x, y) approaches (0, 0) along the line y = mx. Which is a line through the origin. Put y = mx and allows x —> 0, we get

which depends on m, therefore the limit of f(x, y) at (0, 0) does not exists. Hence, f(x, y) is discontinuous at origin.

Now,

since f_{y} exists at origin.

which depends on m, therefore the limit of f(x, y) at (0, 0) does not exists. Hence, f(x, y) is discontinuous at origin.

Now,

since f

If f(x) = then

- a)f is continuous but not differentiable at x = 0
- b)f is differentiable at x = 0
- c)f is continuous but not differentiable at x=1
- d)f is differentiable at x = 1

Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered • yesterday

Given that

Similarly, Lf'(0) = 0

Hence f(x) is differentiable at x = 0.

Similarly, Lf'(0) = 0

Hence f(x) is differentiable at x = 0.

The number of 6 digit numbers in which all the odd digits and only odd digits appear, is

- a)
- b)6!
- c)
- d)5!/2

Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered • yesterday

Clearly, one of the odd digits 1, 3, 5, 7, 9 will repeated. The number of selections of the sixth digit = ^{5}C_{1} = 5.

Required, number of numbers = 5 x (6!/2!)

Required, number of numbers = 5 x (6!/2!)

x dx is equal to

- a)8/15
- b)4/15
- c)1/5
- d)0

Correct answer is option 'D'. Can you explain this answer?

Abhi Soni answered • yesterday

Odd function

The voltage across a register is a random variable E uniform between 5V and 10 V. given the resistance of the register R = 1000 Ω, the probability density function of the power W = E^{2}/R dissipated in R is

- a)
- b)
- c)
- d)

Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered • 2 days ago

For p.d.f. integration over the given interval will be 1.

Vishal Suresh asked • yesterday

Let ∑u_{n} be a series of positive terms such that Then,

(i) if l > 1, the series converges;

(ii) if l < 1, the series diverges;

(iii) if l = 1, the series may either converge or diverge and therefore the test fails;

This theorem is known as

(i) if l > 1, the series converges;

(ii) if l < 1, the series diverges;

(iii) if l = 1, the series may either converge or diverge and therefore the test fails;

This theorem is known as

- a)d’Alembert’s test
- b)Cauchy test
- c)Raabe’s test
- d)Kummer’s test

Correct answer is option 'A'. Can you explain this answer?

Shikha Pradhan asked • yesterday

Let a_{1} = b_{1} = 0, and for each n ≥ 2, let a_{n} and b_{n} be real numbers given by

Then which one of the following is TRUE about the sequences {a_{n}} and {b_{n}}?

- a)Both {a
_{n}} and {b_{n}} are divergent - b){a
_{n}} is convergent and {b_{n}} is divergent - c){a
_{n}} is divergent and {b_{n}} is convergent - d)Both {a
_{n}} and {b_{n}} are convergent

Correct answer is option 'D'. Can you explain this answer?

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