Fiabilidad del instrumento - Aspectos éticos

Articulo 7.- Disposiciones específicas para las contrataciones iguales o inferiores a ocho (08) unidades impositivas tributariaS.(p.5)

3.7 Aspectos éticos

4.1.2. Fiabilidad del instrumento

4. Equation of the line normal to function f(x) = (x ) at (0 5) is after decimal place. The value of

2 intervals, to 5 significant digits is

(A) 0.00000 (B) 1.0000

ME – 2010

6. Torque exerted on a flywheel over a cycle is listed in the table. Flywheel energy (in J per unit cycle) using impson’s rule is

Angle (degree) Torque (N-m)

0 0 subintervals each of length 1, equals (A) 1.000 the interval of integration into three equal sub intervals, the definite integral

∫ |x|dx is ____________

10. The value of ∫ ( ) _. calculated using the Trapezoidal rule with five sub intervals is _______

11. The definite integral ∫ is evaluated using Trapezoidal rule with a step size of 1. The correct answer is _______

12. The real root of the equation

5x 2cosx = 0 (up to two decimal accuracy) is _______

13. Consider an ordinary differential equation = t .If x =x at t = 0 , the increment in x calculated using Runge-Kutta fourth order multi-step method with a step size of Δt = 0.2 is

(A) 0.22 (B) 0.44

Linked Answer Question 1 and 2

Give a>0, we wish to calculate its reciprocal value 1/a by using Newton Raphson Method for f(x) = 0.

1. The Newton Raphson algorithm for the function will be

(A) x= (x ) (B) x= (x x ) (C) x= 2x ax (D) x= x x

2. For a = 7 and starting with x = 0.2 the first two iteration will be

(A) 0.11, 0.1299 (B) 0.12, 0.1392

3. A 2^nd degree polynomial f(x) has values of 1, 4 and 15 at x = 0, 1 and 2 respectively.

The integral ∫ f(x) dx is to be estimated by applying the trapezoidal rule to this data. What is the error (define as true

“value approximate value”) in the estimate?

(A) – (B) 0

4. The following equation needs to be numerically solved using the Newton-Raphson method

x³ + 4x – 9 = 0 the iterative equation for the purpose is (k indicates the iteration level)

(A) x = (B) x =

5. Given that one root of the equation x 10x + 31x – 30 = 0 is 5, the other two roots are

(A) 2 and (B) 2 and

6. Three values of x and y are to be fitted in a straight line in the form y = a + bx by the method of least squares. Given

∑x = 6, ∑y = 2 ∑x = and ∑xy = the values of a and b are respectively (A) 2 and 3

(B) 1 and 2

7. In the solution of the following set of linear equation by Gauss elimination using partial pivoting 5x + y + 2z = 34;

4y – 3z = 12; and 10x – 2y + z = 4; the pivots for elimination of x and y are (A) 10 and 4

(B) 10 and 2

CE – 2010

8. The table below given values of a function F(x) obtained for values of x at intervals of 0.25.

x 0 0.25 0.5 0.75 1.0 F(x) 1 0.9412 0.8 0.64 0.50 The value of the integral of the function between the limits 0 to using impson’s rule is

(A) 0.7854 (B) 2.3562

9. The square root of a number N is to be obtained by applying the Newton Raphson iterations to the equation x = 0. If i denotes the iteration index, the correct iteration scheme will be (A) x = (x )

(B) x = (x ) (C) x = (x ) (D) x = (x )

10. he error in

f(x)|

for a continuous function estimated with h=0.03 using the central difference formula

f(x)|

=⁽ ^{) (} ⁾ is 2 0. The values of and ( ) are 19.78 and 500.01, respectively. The corresponding error in the central difference estimate for h = 0.02 is approximately

(A) . 0 (B) .0 0

11. The estimate of ∫_.^. obtained using impson’s rule with three – point function evaluation exceeds the exact value by (A) 0.235

(B) 0.068

CE – 2013

12. Find the magnitude of the error (correct to two decimal places) in the estimation of following integral using impson’s ⁄ Rule. Take the step length as 1.___________

∫ (x 0)dx

CS – 2007

1. Consider the series =

= 0.5 obtained from the Newton-Raphson method. The series converges to (A) 1.5

(B) √2

2. The minimum number of equal length subintervals needed to approximate

2 x

xe dx



to an accuracy of at least1 10⁶ 3

 

using the trapezoidal rule is (A) 1000e

(B) 1000

3. The Newton-Raphson iteration

n 1 n

1 R

x x

2 x



 

   

 can be used to compute the

(A) square of R (B) reciprocal of R (C) square root of R (D) logarithm of R CS – 2010

4. Newton-Raphson method is used to compute a root of the equation

x 13 = 0 with 3.5 as the initial value.

The approximation after one iteration is (A) 3.575

(B) 3.677

CS – 2012

5. The bisection method is applied to compute a zero of the function

f(x) = x x x in the interval [1,9]. The method converges to a solution after ___________ iterations.

(A) 1 (B) 3

6. Function f is known at the following points:

x f(x)

0 0

0.3 0.09 0.6 0.36 0.9 0.81 1.2 1.44 1.5 2.25 1.8 3.24 2.1 4.41 2.4 5.76 2.7 7.29 3.0 9.00

he value of ∫ f(x)dx computed using the trapezpidal rule is

(A) 8.983 (B) 9.003

7. The function f(x) = x sin x satisfied the following equation:

( ) + f(x) + t cos x = 0.

The value of t is _________.

8. In the Newton-Raphson method, an initial guess of = 2 made and the sequence x x x .. is obtained for the function 0.75x 2x 2x = 0

Consider the statements (I) x = 0.

(II) The method converges to a solution in a finite number of iterations.

Which of the following is TRUE?

(A) Only I

(B) Only II (C) Both I and II (D) Neither I nor II

9. With respect to the numerical evaluation of the definite integral, = ∫ x dx where a and b are given, which of the following statements is/are TRUE?

(I) The value of K obtained using the trapezoidal rule is always greater then or equal to the exact value of the defined integral

(II) The value of K obtained using the impson’s rule is always equal to the exact value of the definite integral (A) I only

(B) II only (C) Both I and II (D) Neither I nor II ECE– 2005

1. Match the following and choose the correct combination

Group I Group II

(A) Newton-Raphson method

1. Solving non-linear equations (B) Runge-Kutta

method

2. Solving linear simultaneous equations (C) impson’s

Rule

3. Solving ordinary differential equations (D) Gauss

elimination method

4. Numerical integration 5. Interpolation 6. Calculation of

Eigen values (A) A-6, B-1, C-5, D-3

(B) A-1, B-6, C-4, D-3 (C) A-1, B-3, C-4, D-2 (D) A-5, B-3, C-4, D-1

ECE– 2007

2. The equation x³ x²+4x 4=0 is to be solved using the Newton-Raphson method. If x=2 is taken as the initial approximation of the solution, then the next approximation using this method will be

using Newton-Raphson method is (A) = e with all coefficients positive has (A) No real root discretised using Euler’s numerical integration method with a time step T > 0. What is the maximum permissible value of T to ensure stability of the solution of the corresponding discrete time equation?

step of ewton’s method estimate x of the solution will be given by

(A) 0.71828 (B) 0.36784

has to be solved using trapezoidal rule of integration with a step size h = 0.01 sec.

Function u(t) indicates a unit step function. If x(0)= 0, then value of x at t = 0.01 s will be given by

(A) 0.00099 (B) 0.00495

4. Let x 7 = 0. The iterative steps for the solution using Newton – aphson’s method is given by

(A) x= (x ) (B) x= x (C) x= x

(D) x= x (x )

EE– 2011

5. Solution of the variables and for the following equations is to be obtained by employing the Newton-Raphson iterative method

equation(i) 0x inx 0. = 0 equation(ii) 0x 0x cosx 0. = 0 Assuming the initial values = 0.0 and

= .0 the jacobian matrix is (A) * 0 0.

0 0. + (B) * 0 0

0 0+

0 0. + (D) * 0 0

0 0+

6. Roots of the algebraic equation x x x = 0 are

(A) ( ) (B) ( j j)

EE– 2013

7. When the Newton – Raphson method is applied to solve the equation f(x) = x 2x = 0 the solution at the end of the first iteration with the initial guess value as x = .2 is

(A) 0.82 (B) 0.49

8. The function ( ) = is to be solved using Newton-Raphson method. If the initial value of is taken as 1.0, then the absolute error observed at 2^nd iteration is ___________

IN– 2006

1. For k = 0 2 . the steps of Newton-Raphson method for solving a non-linear equation is given as

k 1 k K

2 5

x x x .

3 3



  

Starting from a suitable initial choice as k tends to , the iterate tends to

(A) 1.7099 (B) 2.2361

2. Identify the Newton-Raphson iteration scheme for finding the square root of 2.

(A) x= (x ) (B) x = (x ) (C) x=

(D) x= √2 x

3. The polynomial p(x) = x + x + 2 has (A) all real roots

(B) 3 real and 2 complex roots (C) 1 real and 4 complex roots (D) all complex roots

IN– 2008

4. It is known that two roots of the nonlinear equation x³ – 6x² +11x 6 = 0 are 1 and 3. The third root will be

(A) j (B) j

IN – 2009

5. The differential equation = with x(0) = 0 and the constant 0 is to be numerically integrated using the forward Euler method with a constant integration time step T. The maximum value of T such that the numerical solution of x converges is

(A) (B)

IN– 2010

6. The velocity v (in m/s) of a moving mass, starting from rest, is given as

= v t.

Using Euler’s forward difference method (also known as Cauchy-Euler method) with a step size of 0.1s, the velocity at 0.2s evaluates to

(A) 0.01 m/s (B) 0.1 m/s

7. The extremum (minimum or maximum) point of a function f(x) is to be determined by solving ^{( )} = 0 using the Newton-Raphson method.

Let f(x) = x x and x = 1 be the initial guess of x. The value of x after two iterations (x ) is

(A) 0.0141 (B) 1.4142

IN– 2013

8. While numerically solving the differential equation 2xy = 0 y(0) = using Euler’s predictor – corrector (improved Euler – Cauchy )method with a step size of 0.2, the value of y after the first step is (A) 1.00

(B) 1.03

9. The iteration step in order to solve for the cube roots of a given number N using the Newton- aphson’s method is

(A) x= x ( x ) (B) x= (2x ) (C) x= x ( x ) (D) x= (2x )

Answer Keys and Explanations

1. [Ans. C]

By N-R method , =x – ^{( )}

( ) x = f(x) = x x 7

 f( ) =

 x = x ⁽ ⁾

( )

 f (x) = x

 f ( ) = ,

 = 1 ^{( )}= (0.5) = .5

2. [Ans. C]

Given x y = 2 (i) .0 x 0.0 y = b (ii)

Multiply 0.99 is equation (i) and subtract from equation (ii); we get

( .0 0. )x = b (2 0. ) 0.02x = b .

0.02Δx = Δb Δx =

0.02= 50 units 3. [Ans. D]

4. [Ans. B]

Given f(x) = (x )^/ f (x) =2

(x )

Slope of tangent at point (0, 5) m =2

(0 )^/= Slope of normal = 3 (∵ roduct of slopes = 1) Slope of normal at point (0, 5) y 5 = (x 0)

y = x 5 5. [Ans. A]

h =b a

n =2 0

= y = sin(0) = 0 y = sin ( ) = 0.70 0

y = sin ( 2) = y = sin (

) = 0.70 0 y = sin( ) = 0

y = sin (5

) = 0.70 0 y = sin (

) = y = sin (7

) = 0.70 0 y = sin (

) = 0 Trapezoidal rule

∫ f(x)dx = [(y y ) 2(y y y )]

∫sinx dx = [(0 0) 2(0.70 0 0.70 0 0.70 0 0.70 0 0 = 0 6. [Ans. B]

ower = ω = Area under the curve.

[(y y ) (y y y ) 2(y y )]

= [(0 0) ( 0 0 55) 2 ( 2 2 )]

= 2.7 /unit cycle.

7. [Ans. C]

x 1 2 3

y =x

1 2

∫ xdx =h

(y y y )

= (

= . 8. [Ans. D]

By the definition only

9. [Ans. *] Range 1.10 to 1.12

∫ |x|dx is

∫ ydx =h

2[y 2(y y . ) y ]

y y y y

x 1 0.33 0.33 1

y 1 0.33 0.333 1

∫ |x|dx

= 2

2[ 2(0. 0. )]

= . 0

10. [Ans. *] Range 1.74 to 1.76 h = 2.5

5 = 0.

∫ ln (x)dx = [y 2y 2y 2y

2y y ]

=^.[ln(2.5) 2(ln2. ) 2 ln( . ) 2ln( . ) 2ln( .7) ln( )]

= .75

11. [Ans. *]Range 1.1 to 1.2

∫xdx by trapezoidal rule rapezoidal rule

∫ f(x)dx = [y y 2(y y . . y)]

h = iven in question

0 1 2

x 1 2 3

y 1 0.5 0.33

∫xdx =

2[y y 2(y )]

=2[ 0. 2 0.5]

= .

12. [Ans. *] Range 0.53 to 0.56 Let f(x) = 5x 2 cos x f (x) = 5 2 sin x f(0) = f( ) = 2.

By intermediate value theorem roots lie be between 0 and 1.

et x = rad = 57. 2 By Newton Raphson method x= x f(x )

f (x )

x=2x sin x 2 cos x 5 2 sin x

x = 0.5 2 x = 0.5 25 x = 0.5 2 13. [Ans. D]

The variation in options are much, so it can be solved by integrating directly dx

dt = t

∫ dx= ∫ ( t )dt

Δx = t 2 t |

= 2t t|

Δx = 0.0 0. = 0.

1. [Ans. C]

To calculate using N-R method Set up the equation as x = i.e. = a

a = 0 i.e. f(x) = a = 0 Now f (x) = f(x ) = a f (x ) = For N-R method x = x ⁽₍⁾₎ x = x ⁽⁾

Simplifying which we get x= 2x ax

2. [Ans. B]

For a = 7 iteration equation Becomes x= 2x 7x with x = 0.2

x = 2x 7x = 2× 0.2 – 7(0.2) = 0.12 and x = 2x 7x = 2× 0.12 7(0. 2)

= 0.1392 3. [Ans. A]

f(x) = 1, 4, 15 at x = 0, 1 and 2 respectively

∫ f(x)dx = (f 2f f ) (3 points Trapezoidal Rule) Here h = 1

∫ f(x)dx = (1 + 2(4) + 15) = 12 Approximate value by rapezoidal ule

= 12

Since f(x) is second degree polynomial, let f(x) = a0 + a x + a x

f(0) = 1

 a 0 0 =

 a = f(1) = 4

 a a a =

 1+ a a =

 a a =

 f(2) = 15

 a 2a a = 5

 2a a = 5

 2a a =

Solving (i) and (ii) a = and a = f(x) = 1 – x + 4 x

Now exact value ∫ f(x)dx

= ∫ ( x x )dx

= *x + =

Error = exact – Approximate value

= 2 =

4. [Ans. A ]

Given f(x) = x x = 0 f (x) = x

Newton – Raphson formula is

x= x f(x ) f (x )

= x (x x ) ( x )

= x x x x ( x )

x = 2x x 5. [Ans. A]

Given

x – 10 x + 31x 30 = 0 One root = 5

Let the roots be α β and γ of equation ax + bx + cx + d = 0

α β γ =

and α β βγ γα = α βγ = 5 βγ = ^{( )} = 30

 βγ = (i) Also

αβ βγ γα = 5β βγ 5γ = =

 5 (β γ) βγ = ince βγ = from (i) 5 (β γ) =

 β γ = 5 βγ =

olving for β and γ β (5 β) =

 β 5β = 0

 β = 2 and γ = Alternative method

5 1 0 31 0

0 5 25 30

1 5 6 0

(x 5)(x 5x ) = 0 (x 5)(x 2)(x ) = 0 x = 2 5

6. [Ans. D]

Y = a + bx Given

n= ∑x = ∑y = 2 ∑x = 14 and ∑xy =

Normal equations are

∑y = na b∑x

∑xy = a∑x b∑x

Substitute the values and simply a = b = 2 Now we eliminate x using this pivot as follows : respectively 10 and 4

8. [Ans. A]

Error for h = 0.02 is approximately 2 0 (0.02) Approximate value – Exact value

= 1.1116 1.0986

The value of integral

∫ (x 0)dx = *x

5 0x+

= 5 0 = 2 . Magnitude of error

= 2 5. 2 . = 0.5 CS

1. [Ans. A]

Given x = +

, x = 0.5

when the series converges x = x = α α = +

α =

8α = 4α +9 α =

α = = 1.5

2. [Ans. A]

Here, the function being integrated is f(x) = xe

f (x) = xe + e = e (x + 1) f’ (x) = xe + e + e = e (x + 2)

Since, both are increasing functions of x, maximum value of f ( ) in interval 1 2, occurs at = 2 so max |f( )| = e (2 2) = e

Truncation Error for trapezoidal rule

= TE (bound)

= max |f( )|

Where is number of subintervals =

max |f( )|

= (b – a) max |f( )| 1 2

= (2 – 1) [e (2 + 2)]

= e

Now putting _{( )}= 0 =

h =

Now, No. of intervals, =

= ₍_)/ = 1000 e

3. [Ans. C]

x = (x ) At convergence x = x = α α = (α ) 2 α = α + 2α = 2α = α + R α = R α = √

So this iteration will compute the square root of R

4. [Ans. D]

y = x dy

dx= 2x f(x)= x x = .5 ^._. = . 07 5. [Ans. B]

f( ) = 5

f( ) = 5 72 = 57 7 0 f( ) 0 f( ) 0

x = (

2 ) = 5 f(x ) 0 oot lies between and x = (

2 ) = 2 f(x ) 0 After ' ' interations we get the root 6. [Ans. D]

∫ f(x)dx=h

2[f(0) f( ) 2(∑f)]

=^.[0 2(0.0 0.

0. . . 7.2 ) ]

= ^.[ 5 . ]

= 9.045

7. [Ans. – ] impson’s rule is used then value is exact Hence both statements are TRUE

ECE

By Newton Raphson method, x= x f(x ) If the above equation have complex roots, then they must be in complex conjugate pair, because it’s given all co-efficients are positive ( they are real )

So if complex roots are even no. (in pair) then real roots will also be even.

ption ( )is wrong From the equation

roduct of roots = ( 0) As no. of roots = 4,

Product of roots < 1 either one root 0 (or)

Product of three roots < 0 ption ( )is rong.

Now, take option (A), Let us take it is correct .

Roots are in complex conjugate pairs

6. [Ans. A]

sin x 2 cos x

= (x –x

) 2 ( –x

2 ) 7. [Ans. B]

8. [Ans. D]

∑n =

2 . . .

= e

as e = x

2 . . . x in EE

1. [Ans. D]

Here, = f(x y) = x

Euler’s method equation is x= x h. f(x y) x= x h ( x

) x= ( h

) x h or stability | h

| h

since h = Δ here Δ

Δ 2

o maximum permissible value of Δ is 2 . 2. [Ans. A]

Here f(x) = e f (x) = e

The Newton Raphson iterative equation is x = x ⁽₍⁾₎

f(x) = e f’(x) = e

x = x i.e. x = ^–(⁾ = ⁽⁾

Now put i = 0 x = ⁽ ⁾

Put x = as given, x = [e( 2) ]/e

= 0.71828 3. [Ans. C]

= eu(t)

x = ∫ eu(t) dt = ∫ f(t) dt At t = 0.01, x = Area of trapezoidal

= [f(0) f(0.0 )] =^.[ e^.]

= 0.0099 4. [Ans. A]

x = x ⁽₍⁾₎ = x = *x +

5. [Ans. B]

u(x x ) = 0x sin x 0. = 0 v(x x ) = 0x 0x cosx 0. = 0 The Jacobian matrix is

[ u x

u x v

x v x ]

=[ 0x cos x 0sinx 0x sinx 20x 0cosx ] The matrix at x = 0 x = is

= * 0 0 0 0+ 6. [Ans. D]

x x x = 0 (x )(x ) = 0 x = 0 x = 0 x = x = j

t x

7. [Ans. C]

x = x f(x ) f (x )

= .2 ( .2) 2( .2) ( .2) 2

= 0.705

8. [Ans. *] Range 0.05 to 0.07

Clearly, x = 0 is root of the equation f(x) = e = 0

f (x) = e and x = .0

Using ewton raphson method x = x f(x )

f (x )= (e^. ) e^. =

e and x = x f(x )

f (x )= e

(e ) e

=e e

= 0. 7 0. = 0.0 Absolute error at 2nd itteration is

|0 0.0 | = 0.0 IN

1. [Ans. A]

As k ∞ xk+1 ≈xk

xk = x x x = x x = 5

x = 5^/= 1.70 2. [Ans. A]

Assume x = √ f(x) = x = 0 x= x f(x )

f (x )=

2[x 2 x ] 3. [Ans. C]

Given p(x) = x + x + 2

There is no sign change, hence at most 0 positive root ( rom escarte’s rule of signs)

p( x) = x x + 2

There is one sign change, hence at most 1 negative root ( rom escarte’s rule of signs)

Hence, it will have atleast 5 (0+1)= 4 complex roots.

4. [Ans. C]

Approach- 1

Given, x³– 6x² + 11x – 6 = 0 Or (x – 1)(x – 3)(x – 2) = 0 x= 1, 2, 3.

Approach- 2

For ax³+bx² + cx +d = 0 If the three roots are p,q,r then Sum of the roots= p+q+r= b/a Product of the roots= pqr= d/a pq+qr+rp=c/a

5. [Ans. D]

dt = x f(x, y) =

x= x h(x y) = x h ( x )

= ( h

) x ( h ) or stability | h

| h

Δ 2

6. [Ans. A]

dt = v t t v dv

dt = v t 0 0 0 0+0 0. = 0 0.1 0 0+0.1 0. = 0.0 7. [Ans. C]

f(x) = x x

f (x) = x = g(x) x = initial guess g(x) = x

x = x g (x ) g(x )

= ( )

= .5 x = x g(x )

g(x )

= .5 0.75

= . 7 8. [Ans. D]

dx= 2xy x = 0 y = h = 0.2 y = y h. f(x y )

= (0.2)f(0 ) = and y = y [f(x y ) f(x y )]

= (0. )[f(0 ) f(0.2 )]

= 0.

is the value of y after first step, using Euler’s predictor – corrector method 9. [Ans. B]

For convergence

x = x = x x = (2x x ) x = x = √

Calculus

ME – 2005

1. The line integral ∫ ⃗⃗ ⃗⃗⃗⃗ of the vector function ⃗⃗ ( ) 2xyz ̂+ x²z ̂+ k²y ̂ from the origin to the point P (1,1,1)

(A) is 1 (B) is Zero (C) is 1

(D) cannot be determined without specifying the path

2. The right circular cone of largest volume that can be enclosed by a sphere of 1 m radius has a height of

(A) (B)

x(u,v) = uv, y(u,v) = v/u is double integral, the integral f(x,y) changes to f(uv, u/v) ( ). Then, ( ) (A) 2 v/u

(B) 2 u v

4. Changing the order of the integration in the double integral I = ∫ ∫ ( ) leads to

I =∫ ∫ ( ) What is q?

(A) 4y (B) 16y²

(A) ∫ (B) 2∫

6. Stoke’ theorem connects

(A) A line integral and a surface integral (B) Surface integral and a volume

integral

ME – 2006 7. If f(

x

) =

2 2

2x 7x 3 5x 12x 9

 

  ^{, then}limf(x)^{x 3}

 will be

(A) ⁄ (B) ⁄

8. Assuming i = 1 and t is a real number,

∫ dt is:

(A) ^√ (B) ^√

9. Let x denote a real number. Find out the INCORRECT statement

(A) S * + represents the set if all real numbers greater then 3

(B) S * + represents the empty set

(D) S * + represents the set of all real umbers between a and b, where and b are real number

ME – 2007

10. ⁽

)

(A) 0

(B) ⁄

11. The area of a triangle formed by the tips of vectors a , b and c is

(A) ( ) ( ) (B) |( ) ( )|

12. If √ √ √ , then y (2) =

(A) 4 or 1 (B) 4 only

the interval [1, 5] is (A) 0

(B) 1

14. The length of the curve between x = 0 and x = 1 is (A) 0.27

(B) 0.67

unbounded?

(A) ∫ (B) ∫

16. The directional derivative of the scalar function f(x, y, z) = x 2y z² ² _{at the} point P = (1, 1, 2) in the direction of the vector ⃗ ̂ ̂ is

(A) 4 (B) 2

17. Consider the shaded triangular region P shown in the figure. What is∬ xydxdy?

(A) ⁄ (B) ⁄

18. The value of _{( )} is (A) ⁄

(B) ⁄

19. In the Taylor series expansion of e^xabout x = 2, the coefficient of (x 2)⁴is

(A) ⁄ (B) ⁄

(x y) ̂ ( ) ̂ ( ) ̂ is (A) 0

(B) 1

21. Let What is

at x=2, y=1?

(A) 0 (B) ln2

22. The area enclosed between the curves y24x and x²4yis

(A) ⁄ (B) 8

23. The distance between the origin and the point nearest to it on the surface

z 1 xy2  is (A) 1 (B) √ ⁄

24. A path AB in the form of one quarter of a circle of unit radius is shown in the figure.

Integration of



^{x y}^



² on path AB traversed in a counter-clockwise sense is

(A) (B)

25. The divergence of the vector field ̂ ̂ ̂ at a point (1,1,1) is equal to

(A) 7 (B) 4

X Y

0 1

P x

2 y

x+2y=2

ME – 2010

26. Velocity vector of a flow field is given as

⃗⃗ ̂ ̂. The vorticity vector at (1, 1, 1) is

(A) 4 ̂ ̂ (B) 4 ̂ ̂

(A) o o ∀ R ∀ R (B) o o ∀ R

∀ R except at x = 3/2

(D) o o ∀ R except at x = 3 ∀ R

28. The value of the integral ∫ is (A) –π

(B) –π

29. The parabolic arc √ is revolved around the x-axis. The volume of the solid of revolution is

(A) π (B) π

30. If f(x) is an even function and is a positive real number, then ∫ ( ) dx equals (A) 0

(B)

(C)

(D) ∫ ( )

31. What is equal to?

(A) (B)

32. A series expansion for the function is (A)

(B) (C) (D)

ME – 2012

33. Consider the function ( ) in the interval . At the point x = 0, f(x) is

(A) Continuous and differentiable.

(B) Non – continuous and differentiable.

(D) Neither continuous nor differentiable.

34. ./ is (A) 1/4

(B) 1/2

35. At x = 0, the function f(x) = has (A) A maximum value

(B) A minimum value (C) A singularity (D) A point of inflection 36. For the spherical surface

the unit outward normal vector at the point

.√ √ / (A) √ ̂

√ ̂ (B) √ ̂

√ ̂ (C) ̂

(D) √ ̂

√ ̂

37. The area enclosed between the straight line y = x and the parabola y = in the

x – y plane is (A) 1/6 (B) 1/4

38. The following surface integral is to be evaluated over a sphere for the given steady velocity vector field

defined with respect to a cartesian coordinate system having i, j and k as unit base vectors.

∫ ∫ ( )

Where S is the sphere, and n is the outward unit normal vector to the sphere. The value of the surface integral is

(A) π (B) π

∫ √ ( ) is (A) √

(B) √

ME – 2014

40.

is (A) 0

(B) 1

(D)Not defined 41. Which one of the following describes the

relationship among the three vectors ̂ ̂ ̂ ̂+ ̂+ ̂ ̂ ̂ ̂ (A) The vectors are mutually

perpendicular

(B) The vectors are linearly dependent (C) The vectors are linearly independent (D) The vectors are unit vectors

42. ._{( )}/ is equal to (A) 0

(B) 0.5

⃗ ̂ ̂ ̂ (A) ( ) ̂ ̂ ̂ (B) ( ) ̂ ̂ ̂ (C) ̂ ̂ ̂

(D) ̂ ̂ ̂

44. The best approximation of the minimum value attained by (100x) for ≥ is _______

45. If a function is continuous at a point, (A) the limit of the function may not exist

at the point

(B) the function must be derivable at the point

(D) the limit must exist at the point and the value of limit should be same as the value of the function at that point 46. Divergence of the vector field

̂ ̂ ̂ ( ) is (A) 0

(B) 3

∫ _{( )}^{( )} ( )^{( )} (A) 3

(B) 0

48. The value of the integral ∫ ∫ is

(A) ( ) (B) ( )

CE – 2005

1. Value of the integral ∮ ( ). Where, c is the square cut from the first quadrant by the lines x = 1 and y = 1 will ( G ’ h o o h h line integral into double integral) (A) ⁄

(B) 1

2. A rail engine accelerates from its stationary position for 8 seconds and travels a distance of 280 m. According to the Mean Value theorem, the speedometer at a certain time during acceleration must read exactly.

(A) 0 kmph (B) 8 kmph

CE – 2006

3. What is the area common to the circles o

(A) 0.524 a² (B) 0.614 a²

f(x, y, z) = 2 + 3 + at the point P (2, 1, 3) in the direction of the vector a = k is

(A) (B)

5. Potential function  is given as

 = . When will be the stream function () with the condition

 = 0 at x = y = 0?

(A) 2xy (B) +

6. Evaluate ∫ (A) π

(B) π ⁄

̅ = 5xy + 2 y² + 3yz²⃗ . The divergence of this velocity vector at (1 1 1) is (A) 9

(B) 10

CE – 2008

8. The equation

= 0 can be transformed to

= 0 by substituting (A)

(B)

9. The inner (dot) product of two vectors ⃗⃗

and ⃗⃗ is zero. The angle (degrees) between in two vectors is

(A) 0 (B) 30

10. The value of ∫ ∫ ( ) is (A) 13.5

(B) 27.0

11. For a scalar function

f(x, y, z) = + 3 + 2 the gradient at the point P (1, 2, 1) is

(A) 2 + 6 + 4 ⃗ (B) 2 + 12 – 4 ⃗

12. For a scalar function

f(x, y, z) = the directional derivative at the point P(1, 2, 1) in the direction of a vector ⃗ is (A)

(B) √

13. A parabolic cable is held between two supports at the same level. The horizontal span between the supports is L. The sag at the mid-span is h. The equation of the parabola is y = 4h where x is the horizontal coordinate and y is the vertical coordinate with the origin at the centre of the cable. The expression for the total length of the cable is

(A) ∫ √ (B) 2∫√ (C) ∫√ (D) ∫√

14. The _^.^/ is (A) 2/3

(B) 1

( ) The optimal value of f(x, y)

(A) Is a minimum equal to 10/3 (B) Is a maximum equal to 10/3 (C) Is a minimum equal to 8/3 (D) Is a maximum equal to 8/3 CE – 2011

16. ∫ _{√ √}^√ ? (A) 0

(B) a/2

17. Wh ho h o λ such that the function defined below is continuous π ?

f(x)={

π π (A) 0

(B) π

18. If ⃗ and ⃗ are two arbitrary vectors with magnitudes a and b respectively. | ⃗ ⃗ | will be equal to

(A) – ( ⃗ ⃗ ) (B) ab ⃗ ⃗

19. For the parallelogram OPQR shown in the sketch, ̅̅̅̅ ̂ ̂ and R̅̅̅̅ ̂ ̂.

The area of the parallelogram is

(A) ad –bc (B) ac+bd

o o o (A) sec

(B)

CE – 2013

21. The value of ∫ o (A) 0

(B) 1/15

CE – 2014

22. ./ o (A)

(B)

23. With reference to the conventional Cartesian (x, y) coordinate system, the vertices of a triangle have the following coordinates:

( ) ( ) ( ) ( ) ( ) ( ). The area of the triangle is equal to

(A) ⁄ (B) ⁄

parametric equation are : and z = 2 sin (5t), where x, y and z show variations of the distance covered by the particle (in cm) with time t (in s). The magnitude of the acceleration of the particle (in cm ) at t = 0 is ___________

25. If {x} is a continuous, real valued random variable defined over the interval

( ) and its occurrence is defined by the density function given as:

( )

√

wh ‘ ’ ‘ ’ the statistical attributes of the random variable {x}. The value of the integral

∫ √

dx is

(A) 1 (B) 0.5

26. The expression o (A) log x

(B) 0

CS – 2005

2. Consider the following two statements about the function f(x) =|x|:

P: f(x) is continuous for all real values of x Q: f(x) is differentiable for all real values of x

Which of the following is true?

(A) P is true Q is false

where k is a positive integer. Then (A)

(B)

it is a local minimum or a local maximum.

The number of distinct extrema for the curve 3x 16x 24x 37⁴ ³ ² _is evaluation of the definite integral

∫

9. Consider the function f(x)= sin(x) in the interval x ,π π⁄ ⁄ -. The number and location(s) of the local minima of this function are

10. Which one the following function is continuous at x =3? derivation of f with respect to . Which of the following statements is/are TRUE?

(I) There exists

. / h h ( )

(II) There exists

. / h h ( ) (A) I only

(B) II only (C) Both I and II (D) Neither I nor II

12. A function f (x) is continuous in the interval [0, 2]. It is known that

f(0) = f(2) = 1 and f(x) = 1. Which one of the following statements must be true?

(A) There exists a y in the interval (0,1) such that f(y) = f(y + 1)

(B) For every y in the interval (0, 1), f(y) = f(2 y)

(D) There exists a y in the interval (0, 1) such that f(y) ( )

13. If and are 4 – dimensional subspace of a 6 – dimensional vector space V, then the smallest possible dimension of is ____________.

14. If ∫ dx = π, then the value of k is equal to_______.

15. The value of the integral given below is

In document FACULTAD DE CIENCIAS EMPRESARIALES ESCUELA PROFESIONAL DE CONTABILIDAD (página 42-78)