Question 1
Which one of the following functions is continuous at x = 3?
 A A B B C C D D
GATE CS 2013    Numerical Methods and Calculus
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Question 1 Explanation:
A function is continuous at some point c, Value of f(x) defined for x > c = Value of f(x) defined for x < c = Value of f(x) defined for x = c All values are 2 in option A
 Question 2
Function f is known at the following points:
 A 8.983 B 9.003 C 9.017 D 9.045
GATE CS 2013    Numerical Methods and Calculus
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Question 2 Explanation:
Since the intervals are uniform, apply the uniform grid formula of trapezoidal rule. This solution is contributed by Anil Saikrishna Devarasetty
 Question 3
Consider the function f(x) = sin(x) in the interval [π/4, 7π/4]. The number and location(s) of the local minima of this function are
 A One, at π/2 B One, at 3π/2 C Two, at π/2 and 3π/2 D Two, at π/4 and 3π/2
GATE CS 2012    Numerical Methods and Calculus
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Question 3 Explanation:
 Question 4
The bisection method is applied to compute a zero of the function f(x) = x4 – x3 – x2 – 4 in the interval [1,9]. The method converges to a solution after ––––– iterations
 A 1 B 3 C 5 D 7
GATE CS 2012    Numerical Methods and Calculus
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Question 4 Explanation:
In bisection method, we calculate the values at extreme points of given interval, if signs of values are opposite, then we find the middle point. Whatever sign we get at middle point, we take the corner point of opposite sign and repeat the process till we get 0. f(1) < 0 and f(9) > 0 mid = (1 + 9)/2 = 5 f(5) > 0, so zero value lies in [1, 5] mid = (1+5)/2 = 3 f(3) > 0, so zero value lies in [1, 3] mid = (1+3)/2 = 2 f(2) = 0
 Question 5
Given i=√-1, what will be the evaluation of the integral ?
 A 0 B 2 C -i D i
GATE CS 2011    Numerical Methods and Calculus
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Question 5 Explanation:
 Question 6
Newton-Raphson method is used to compute a root of the equation x2-13=0 with 3.5 as the initial value. The approximation after one iteration is
 A 3.575 B 3.676 C 3.667 D 3.607
GATE CS 2010    Numerical Methods and Calculus
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Question 6 Explanation:
In Newton-Raphson's method, We use the following formula to get the next value of f(x). f'(x) is derivative of f(x).
f(x)  =  x2-13
f'(x) =  2x

Applying the above formula, we get
Next x = 3.5 - (3.5*3.5 - 13)/2*3.5
Next x = 3.607

 Question 7
What is the value of Limn->∞(1-1/n)2n ?
 A 0 B e-2 C e-1/2 D 1
GATE CS 2010    Numerical Methods and Calculus
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Question 7 Explanation:
The value of e (mathematical constant) can be written as following   And the value of 1/e can be written as following. Limn-> ∞( 1-1/n)2n  =  (Limn-> ∞(1-1/n)n)2  =  e-2
 Question 8
Two alternative packages A and B are available for processing a database having 10k records.Package A requires 0.0001n2 time units and package B requires 10nlog10n time units to process n records.What is the smallest value of k for which package B will be preferred over A?
 A 12 B 10 C 6 D 5
GATE CS 2010    Numerical Methods and Calculus
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Question 8 Explanation:
B must be preferred on A when the time taken taken by B is more than A, i.e.,
0.0001 n2 < 10 n log10n
10-5n  < log10n
 Question 9
is equivalent to
 A 0 B 1 C ln 2 D 1/2 ln 2
GATE-CS-2009    Numerical Methods and Calculus
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Question 9 Explanation:
(1-tanx)/(1+tanx) = (cosx - sinx)/(cosx + sinx) Let cosx + sinx = t (-sinx + cosx)dx = dt (1/t)dt = ln t => ln(sinx + cosx) => ln(sin Π/4 + cos Π/4) => ln(1/√2 + 1/√2) => 1/2 ln 2
 Question 10



 A 1 B -1 C inf D -inf
Numerical Methods and Calculus    GATE CS 2008
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Question 10 Explanation:
 Question 11
The following system of equations has a unique solution. The only possible value(s) for α is/are
 A 0 B either 0 or 1 C one of 0, 1 or -1 D any real number E any real number other than 5
Numerical Methods and Calculus    GATE CS 2008
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Question 11 Explanation:
The choice E was not there in GATE paper. We have added it as the given 4 choices don't seem correct.
Augment the given matrix as

1 1 2 | 1
1 2 3 | 2
1 4 a | 4

Apply R2 <- R2 - R1 and R3 <- R3 - R1

1 1 2 | 1
0 1 1 | 1
0 3 a-2 | 3

Apply R3 <- R3 - 3R2

1 1 2 | 1
0 1 1 | 1
0 0 a-5 | 0

So for the system of equations to have a unique solution,

a - 5 != 0
or

a != 5
or
a = R - {5} 
Thanks to Anubhav Gupta for providing above explanation. Readers can refer below MIT video lecture for linear algebra. https://www.youtube.com/watch?v=QVKj3LADCnA&index=2&list=PLE7DDD91010BC51F8
 Question 12
The minimum number of equal length subintervals needed to approximate  to an accuracy of at least  using the trapezoidal rule is
 A 1000 l B 1000 C 100 l D 100
Numerical Methods and Calculus    GATE CS 2008
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Question 12 Explanation:
Trapezoidal rule error : Maximum error = 1/3 * 10-6 (given) Therefore, |En| < 1/3 * 10-6 a = 1 and b = 2 (given) Therefore, f''(x) = xex + 2ex f''(x) is maximum at x = 2. Therefore, f''(x) = 4e2 Thus, option (A) is correct. Reference:  http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2008.html Please comment below if you find anything wrong in the above post.
 Question 13
The Newton-Raphson iteration  can be used to compute the.
 A square of R B reciprocal of R C square root of R D logarithm of R
Numerical Methods and Calculus    GATE CS 2008
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Question 13 Explanation:
According to Newton-Raphson method,
xn+1 = xn − f(xn) / f′(xn)
So we try to bring given equation in above form. Given equation is :
xn+1 = xn/2 + R/(2xn)
= xn − xn/2 +  R/(2xn)
= xn − (xn2 − R2)/(2xn) 
So clearly f(x) = x2 − R, so root of f(x) means x2 − R = 0 i.e. we are trying to find square root of R. So option (C) is correct. Source: http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2008.html
 Question 14
 A P = Q - k B P = Q + k C P = Q D P = Q +2 k
Numerical Methods and Calculus    GATE CS 2008
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Question 14 Explanation:
P is sum of odd integers from 1 to 2k

Q is sum of even integers from 1 to 2k

Let k = 5

P is sum of odd integers from 1 to 10
P = 1 + 3 + 5 + 7 + 9

Q is sum of even integers from 1 to 10
Q = 2 + 4 + 6 + 8 + 10

In general, Q can be written as
Q = (1 + 3 + 5 + 9..... ) + (1 + 1 + .....)
= P + k

 Question 15
A point on a curve is said to be an extremum if it is a local minimum or a local maximum. The number of distinct extrema for the curve 3x4 - 16x3 + 24x2 + 37
 A 0 B 1 C 2 D 3
Numerical Methods and Calculus    GATE CS 2008
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Question 15 Explanation:
 Question 16
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 and Q is false. B P is false and Qis true. C Both P and Q are true D Both P and Q are false.
Numerical Methods and Calculus    GATE-CS-2007
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Question 16 Explanation:
A function is continuous if for every value of 'x', we have a corresponding f(x). Here, for every x, we have f(x) which is actually the value of x itself, without the negative sign for x < 0.

But, the given function is not differentiable for x = 0 because for x < 0, the derivative is negative and for x > 0, the derivative is positive. So, the left hand derivative and right hand derivative do not match.

Hence, P is correct and Q is incorrect. Thus, A is the correct option.

Please comment below if you find anything wrong in the above post.
 Question 17
Consider the series Xn+1 = Xn/2 + 9/(8 Xn), X0 = 0.5 obtained from the Newton-Raphson method. The series converges to
 A 1.5 B sqrt(2) C 1.6 D 1.4
Numerical Methods and Calculus    GATE-CS-2007
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Question 17 Explanation:
As per Newton Rapson's Method,

Xn+1  = Xn − f(Xn)/f′(Xn)

Here above equation is given in the below form

Xn+1 = Xn/2 + 9/(8 Xn)

Let us try to convert in Newton Rapson's form by putting Xn as
first part.
Xn+1  = Xn - Xn/2 + 9/(8 Xn)
= Xn - (4*Xn2 - 9)/(8*Xn)

So    f(X)  =  (4*Xn2 - 9)
and  f'(X) =  8*Xn 
So clearly f(X) = 4X2 − 9. We know its roots are ±3/2 = ±1.5, but if we start from X0 = 0.5, according to equation, we cannot get negative value at any time, so answer is 1.5 i.e. option (A) is correct.
 Question 18
Consider the polynomial p(x) = a0 + a1x + a2x2 + a3x3 , where ai ≠ 0 ∀i. The minimum number of multiplications needed to evaluate p on an input x is:
 A 3 B 4 C 6 D 9
Numerical Methods and Calculus    GATE-CS-2006
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Question 18 Explanation:
Background Explanation : Horner's rule for polynomial division is an algorithm used to simplify the process of evaluating a polynomial f(x) at a certain value x = x0 by dividing the polynomial into monomials (polynomials of the 1st degree). Each monomial involves a maximum of one multiplication and one addition processes. The result obtained from one monomial is added to the result obtained from the next monomial and so forth in an accumulative addition fashion. To explain the above, let is re-write the polynomial in its expanded form; f(x0) = a0 + a1x0+ a2x0^2+ ... + anx0^n This can, also, be written as: f(x0) = a0 + x0(a1+ x0(a2+ x0(a3+ ... + (an-1 + anx0)....) The algorithm proposed by this rule is based on evaluating the monomials formed above starting from the one in the inner-most parenthesis and move out to evaluate the monomials in the outer parenthesis. Solution : Using Horner's Rule, we can write the polynomial as following a0 + (a1 + (a2 + a3x)x)x In the above form, we need to do only 3 multiplications

p = a3 X x    ------------ (1)

q = (a2 + p) X x  ---------(2)

r = (a1 + q) X x  ---------(3)

result = a0 + r 
Reference : http://www.geeksforgeeks.org/horners-method-polynomial-evaluation/ This solution is contributed by Nitika Bansal.
 Question 19
Consider the following system of equations:
3x + 2y = 1
4x + 7z = 1
x + y + z = 3
x – 2y + 7z = 0 
The number of solutions for this system is __________________
 A 1 B 0 C 2 D 3
Numerical Methods and Calculus    GATE-CS-2014-(Set-1)
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Question 19 Explanation:
rank(Augmented Matrix) = rank(Matrix) = no of unknowns. Hence it has a unique solution.
 Question 20
The value of the dot product of the eigenvectors corresponding to any pair of different eigenvalues of a 4-by-4 symmetric positive definite matrix is _____________________.
 A 0 B 1 C -1 D 2
Numerical Methods and Calculus    GATE-CS-2014-(Set-1)
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Question 20 Explanation:
The eigen vectors corresponding to different eigen values of a real symmetric matrix are orthogonal to each other. And dot product of orthogonal vectors is 0.
 Question 21
 A I only B II only C Both I and II D Neither I nor II
Numerical Methods and Calculus    GATE-CS-2014-(Set-1)
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 Question 22
A non-zero polynomial f(x) of degree 3 has roots at x = 1, x = 2 and x = 3. Which one of the following must be TRUE?
 A f(0)f(4) < 0 B f(0)f(4) > 0 C f(0) + f(4) < 0 D f(0) + f(4) > 0
Numerical Methods and Calculus    GATE-CS-2014-(Set-2)
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Question 22 Explanation:
The graph of a degree 3 polynomial f(x) = a0 + a1x + a2(x^2) + a3(x^3), where a3 ≠ 0 is a cubic curve, as can be seen here https://en.wikipedia.org/wiki/... Now as given, the polynomial is zero at x = 1, x = 2 and x = 3, i.e. these are the only 3 real roots of this polynomial. Hence we can write the polynomial as f(x) = K (x-1)(x-2)(x-3) where K is some constant coefficient. Now f(0) = -6K and f(4) = 6K ( by putting x = 0 and x = 4 in the above polynomial ) and f(0)*f(4) = -36(k^2), which is always negative. Hence option A. We can also get the answer by just looking at the graph. At x < 1, the cubic graph (or say f(x) ) is at one side of x-axis, and at x > 3 it should be at other side of x-axis. Hence +ve and -ve values, whose multiplication gives negative.
 Question 23
In the Newton-Raphson method, an initial guess of x0 = 2 is made and the sequence x0, x1, x2 … is obtained for the function

0.75x3 – 2x2 – 2x + 4 = 0

Consider the statements
(I) x3 = 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
Numerical Methods and Calculus    GATE-CS-2014-(Set-2)
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Question 23 Explanation:
In Newton's method, we apply below formula to get next value.
f'(x) = 2.25x2 – 4x - 2

x1 = 2 - (0.75*23 – 2*22 – 2*2 + 4)/
(2.25*22 – 4*2 - 2)
= 2 - (-2/-1)
= 0

x2 = 0 - (4/-2) = 2

x3 = 0

We get x = 0 and x = 2 repeatedly and it never converges.
 Question 24
 A 2 B 3 C 4 D 5
Numerical Methods and Calculus    GATE-CS-2014-(Set-3)
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 Question 25
 A I Only B II Only C Both I and II D Neither I or II
Numerical Methods and Calculus    GATE-CS-2014-(Set-3)
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 Question 26
 A A B B C C D D
Numerical Methods and Calculus    GATE-CS-2014-(Set-3)
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 Question 27
Let G(x) = 1/(1 - x)2 = , where | x | < 1. What is g(i) ?
 A i B i+1 C 2i D 2i
Numerical Methods and Calculus    GATE-CS-2005
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Question 27 Explanation:
B is the correct option. Let us put values

S = 1 + 2x + 3x2 + 4x3 + ..........
Sx =    x  + 2x2 + 3x3 + ..........
S - Sx = 1 + x + x2 + x3 + ....
S - Sx = 1/(1 - x) [sum of infinite GP series with ratio < 1 is a/(1-r)]
S = 1/(1 - x)2 
 Question 28
A piecewise linear function f(x) is plotted using thick solid lines in the figure below (the plot is drawn to scale). If we use the Newton-Raphson method to find the roots of f(x) = 0 using x0, x1 and x2 respectively as initial guesses, the roots obtained would be
 A 1.3, 0.6, and 0.6 respectively B 0.6, 0.6, and 1.3 respectively C 1.3, 1.3, and 0.6 respectively D 1.3, 0.6, and 1.3 respectively
Numerical Methods and Calculus    GATE-CS-2003
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Question 28 Explanation:
First of all, There is a mistake in coordinates of a given point. I have corrected that in red color. Now in Newton-Raphson method, we draw a tangent from our guess point, and our new guess would be the point where this tangent cuts x-axis. Now we choose initial guess points one by one :

x0 : Tangent at this point is line AB itself, and that would cut x-axis at
point (1.0,0.0)  (found using equation of line AB). So our next guess
would be 1.0. Point on the curve corresponding to this new guess 1.0
is shown as F. Now tangent at point F is line DE,   which cuts x-axis at
1.3, and at this point, value of function is zero, so we found  the root
as 1.3.

x1 : Tangent at this point is line BE, which cuts x-axis at 0.6, also function
value is zero here, so we find root as 0.6.

x2 : Tangent at this point is line CD, which cuts x-axis at 1.05 (again found
by finding equation of line CD). Point on the curve corresponding to this
new guess 1.05 is shown as G. Now tangent at point G is line DE, which cuts
x-axis at 1.3, and at this point, value of function is zero, so we found
the root as 1.3. 
Source: Question 60 of http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2003.html
 Question 29
The trapezoidal rule for integration give exact result when the integrand is a polynomial of degree:
 A 0 but not 1 B 1 but not 0 C 0 or 1 D 2
Numerical Methods and Calculus    GATE-CS-2002
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Question 29 Explanation:
 Question 30
The Newton-Raphson iteration Xn + 1 = (Xn/2) + 3/(2Xn) can be used to solve the equation
 A X2 = 3 B X3 = 3 C X2 = 2 D X3 = 2
Numerical Methods and Calculus    GATE-CS-2002
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Question 30 Explanation:
In Newton-Raphson's method, We use the following formula to get the next value of f(x). f'(x) is derivative of f(x). Option (A)
X2 = 3
f(x) = X2 - 3

Xn + 1 = Xn - (Xn2 - 3) / (2*Xn)
=  (Xn/2) + 3/(2xn) 
 Question 31
Which of the following statements is true?
 A S > T B S = T C S < T and 2S > T D 2S <= T
Numerical Methods and Calculus    GATE-CS-2000
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 Question 32
A polynomial p(x) satisfies the following:
p(1) = p(3) = p(5) = 1
p(2) = p(4) = -1 
The minimum degree of such a polynomial is
 A 1 B 2 C 3 D 4
Numerical Methods and Calculus    GATE-CS-2000
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Question 32 Explanation:
p(1) = p(3) = p(5) = 1
p(2) = p(4) = -1

The polynomial touches 0 at least once from 1 to 2, so there
is a root between 1 to 2

The polynomial touches 0 at least once from 2 to 3, so there
is a root between 2 to 3

Similarly, there is at least one root from 3 to 4 and 4 to 5.

So minimum degree is 4.
 Question 33
 A ∞ B 0 C 1 D Not Defined
Numerical Methods and Calculus    GATE-CS-2015 (Set 1)
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Question 33 Explanation:
[Tex]\lim_{x\to\infty} 1/x[/Tex] = 0. and [Tex]\lim_{x\to\infty} x^0[/Tex] = 1.
Alternate method : Using log lnm=lim x->infinity 1/x*lnx lim x->infinity lnx/x  (numerator = finite value,denominator = infinity and finite/infinite=0   ) ln(m)=0 m=e0=1
 Question 34
 A a B b C c D d
Numerical Methods and Calculus    GATE-CS-2015 (Set 1)
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Question 34 Explanation:
g(h(x)) = g(x/(x-1)) = 1 - x/(x-1) = -1/(x-1)

h(g(x)) = h(1-x) = (1-x)/((1-x) - 1) = -(1-x)/x

g(h(x)) / h(g(x))  = [-1/(x-1)] / [-(1-x)/x]
= -x/(x-1)2

-x/(x-1)2 is same as h(x) / g(x) 
 Question 35
 A 0.99 B 1 C 99 D 0.9
Numerical Methods and Calculus    GATE-CS-2015 (Set 1)
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Question 35 Explanation:
S  = 1/1*2 + 1/2*3 + 1/3*4 + ..... + 1/99*100
= (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + .... + (1/99 - 1/100)
= (1 + 1/2 + 1/3 .... 1/99) - (1/2 + 1/3 + 1/4 ... 1/100)
= 1 - 1/100
= 99/100
= 0.99

 Question 36
 A 0 B -1 C 1 D infinite
Numerical Methods and Calculus    GATE-CS-2015 (Set 1)
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Question 36 Explanation:
Let f(x) be the given function. We assume that $\frac{1}{x} = z$ Differentiating both sides, we get [Tex] $\frac{-1}{x^2} dx = dz$ Now, accordingly, the lower limit of the integral is $z = \frac{1}{\frac{1}{\pi}} = \pi$ and the upper limit for the integral is $z = \frac{1}{\frac{2}{\pi}} = \frac{\pi}{2}$ So, the given function now becomes $f(x)= - \int_\pi^{\frac{\pi}{2}} cos(z) dz$ $f(x)= \int_\frac{\pi}{2}^{\pi} cos(z) dz$ $f(x) = sin(z) ,$ and the upper limit is π and the lower limit is π/2 So, $f(x) = sin(\pi) - sin(\frac{\pi}{2})$ $f(x) = 0 - 1$ $f(x) = -1$ So, the required answer is -1. [/Tex]
 Question 37
Consider a function f(x) = 1 – |x| on –1 ≤ x ≤ 1. The value of x at which the function attains a maximum and the maximum value of the function are:
 A 0, –1 B –1, 0 C 0, 1 D –1, 2
Numerical Methods and Calculus    GATE-CS-2015 (Set 2)
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Question 37 Explanation:
f(x) = 1 – |x|
f(0) = 1
f(1) = f(-1) = 0
f(0.5) = f(-0.5) = 0.5

The maximum is attained at x = 0, and the maximum value is 1.
 Question 38
Let f(x) = x –(1/3) and A denote the area of the region bounded by f(x) and the X-axis, when x varies from –1 to 1. Which of the following statements is/are True?
1. f is continuous in [–1, 1]
2. f is not bounded in [–1, 1]
3. A is nonzero and finite 
 A 2 only B 3 only C 2 and 3 only D 1, 2 and 3
Numerical Methods and Calculus    GATE-CS-2015 (Set 2)
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Question 38 Explanation:
1 is false: function is not a Continuous function. As a change of 1 in x leads to ∞ change in f(x). For example when x is changed from -1 to 0. At x = 0, f(x) is ∞ and at x = 1, f(x) is finite. 2 is True: f(x) is not a bounded function as it becomes ∞ at x = 0. 3 is true: A denote the area of the region bounded by f(x) and the X-axis. This area is bounded, we can calculate it by doing integrating the function [See this]
 Question 39
 A 0 B 1/2 C 1 D ∞
Numerical Methods and Calculus    GATE-CS-2015 (Set 3)
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Question 39 Explanation:
This can be solved using L'Hôpital's rule that uses derivatives to help evaluate limits involving indeterminate forms. Since [Tex] \lim_{x \to c}f(x)=\lim_{x \to c}g(x)=\infty, and \lim_{x\to c}\frac{f'(x)}{g'(x)} exists[/Tex] We get [Tex] \lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}. [/Tex] [Tex] \lim_{x\to \infty}\frac{1 + x^2}{e^x} = \lim_{x\to \infty}\frac{2x}{e^x} = \lim_{x\to \infty}\frac{2}{e^x} = 0 [/Tex]
 Question 40
 A A B B C C D D
Numerical Methods and Calculus    GATE-CS-2015 (Set 3)
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 Question 41
The velocity v (in kilometer/minute) of a motorbike which starts from rest, is given at fixed intervals of time t(in minutes) as follows:
t   2  4  6  8  10  12  14  16 18 20
v  10  18 25 29 32  20  11  5  2  0 
The approximate distance (in kilometers) rounded to two places of decimals covered in 20 minutes using Simpson’s 1/3rd rule is _________.
 A 309.33 B 105.33 C 110 D 405.6
Numerical Methods and Calculus    GATE-CS-2015 (Set 3)
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Question 41 Explanation:
 Question 42
Let A be an Let A be an n × n matrix of the following form. What is the value of the determinant of A?
 A A B B C C D D
Numerical Methods and Calculus    GATE-IT-2004
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Question 42 Explanation:
The first thing you need to get by seeing these type of questions is:-
Go for substitution method.
For n=2, the values will be
A) 16
B) 26
C) 7
D) 8
As all the values are unique for a small value of n, it does not take much time.
The given matrix will be A = [3 1]
[1 3]
So, det(A) = 3*3-1*1 = 8

   Alternative method:-
You can frame the relations in between det(An+1), det(An), det(An-1)
i.e. d(An+1) = 3*d(An) - d(An-1)
X   = 3*X^0 - X^-1
X^2 = 3*x - 1
Solution for this equation is (3+sqrt(5))/2, (3-sqrt(5))/2
The only option which has roots of type (3+sqrt(5)) is D.
From this, you can match the options easily.

This explanation has been provided by Anil Saikrishna.
 Question 43
Let X and Y be two exponentially distributed and independent random variables with mean α and β, respectively. If Z = min(X,Y), then the mean of Z is given by
 A 1/α+β B min(α ,β) C α β/α + β D α + β
Numerical Methods and Calculus    GATE-IT-2004
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 Question 44
If f(1) = 2,f(2) = 4 and f(4) = 16,what is the value of f(3)using Lagrange’s interpolation formula?
 A 8 B 8 1/3 C 8 2/3 D 9
Numerical Methods and Calculus    GATE-IT-2004
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Question 44 Explanation:

Using Lagrange’s interpolation formula :
f(x) = ((x - x2)(x - x4)/(x1 - x2)(x1 - x4)) * f1 + ((x - x1)(x - x4)/(x2 - x1)(x2 - x4)) * f2 + ((x - x1)(x - x2)/(x4 - x1)(x4 - x2)) * f4
f(3) = ((3 - 2)(3 - 4)/(1 - 2)(1 - 4)) * 2 + ((3 - 1)(3 - 4)/(2 - 1)(2 - 4)) * 4 + ((3 - 1)(3 - 2)/(4 - 1)(4 - 2)) * 16 f(3) = -2/3 + 4 + 16/3 f(3) = 8 (2/3)

Thus, option (C) is correct.

Please comment below if you find anything wrong in the above post.
 Question 45
Consider the following iterative root finding methods and convergence properties: Iterative root finding Convergence properties methods (Q) False Position                        (I) Order of convergence = 1.62 (R) Newton Raphson                 (II) Order of convergence = 2 (S) Secant                                         (III) Order of convergence = 1 with guarantee of convergence (T) Successive Approximation (IV) Order of convergence = 1 with no guarantee of convergence
 A Q-II R-IV S-lIlT-I B Q-IIIR-II S-I T-IV C Q-IIR-I S-IVT-III D Q-I R-IV S-Il T-III
Numerical Methods and Calculus    GATE-IT-2004
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Question 45 Explanation:
These type of  questions are standard type. You can answer these if you have strong command on the subjects. a) False Position - Order of convergence = 1 with guarantee of convergence b) Newton Raphson - Order of convergence = 2 c) Secant  -  Order of convergence = 1.62 d) Successive Approximation - Order of convergence = 1 with no guarantee of convergence   This solution is contributed by Anil Saikrishna Devarasetty.
 Question 46
Let f(n), g(n) and h(n) be functions defined for positive inter such that f(n) = O(g(n)), g(n) ≠ O(f(n)), g(n) = O(h(n)), and h(n) = O(g(n)). Which one of the following statements is FALSE?
 A f(n) + g(n) = O(h(n)) + h(n)) B f(n) = O(h(n)) C fh(n) ≠ O(f(n)) D f(n)h(n) ≠ O(g(n)h(n))
Numerical Methods and Calculus    GATE-IT-2004
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Question 46 Explanation:
f(n), g(n), h(n) are three functions defined over n Given f(n) = O(g(n)) but g(n) != O(f(n))   g(n) = O(h(n)) and h(n) = O(g(n)) So, f(n)*h(n) = O(g(n))*h(n) using above given relations But it is stated that f(n)*h(n)!=O(g(n))*h(n) which is false So, answer is option (D).
This solution is contributed by Anil Saikrishna Devarasetty .
 Question 47
If the trapezoidal method is used to evaluate the integral obtained 01x2dx ,then the value obtained
 A is always > (1/3) B is always < (1/3) C is always = (1/3) D may be greater or lesser than (1/3)
Numerical Methods and Calculus    Gate IT 2005
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 Question 48
The determinant of the matrix given below is
 A -1 B 0 C 1 D 2
Numerical Methods and Calculus    Gate IT 2005
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Question 48 Explanation:
Matrices don’t have value associated with them, but determinant have value associated with them. Determinant of a matrix can be find out by taking any one row or one column, in this row or column, multiplying each element with its cofactor and summing the value up. This solution is contributed by Sandeep Pandey.
 Question 49
 A 0 B 1 C 2 D 3
Numerical Methods and Calculus    GATE-CS-2016 (Set 1)
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Question 49 Explanation:
Put y = x - 4. So, the problem becomes limy->0 (sin y) / y = 1. (Property of Limits on sin)   Thus, B is the correct choice.
 Question 50
The trapezoidal method is used to evaluate the numerical value of . Consider the following values for the step size h.
i. 10-2
ii. 10-3
iii. 10-4
iv. 10-5
For which of these values of the step size h, is the computed value guaranteed to be correct to seven decimal places. Assume that there are no round-off errors in the computation.
 A (iv) only B (iii) and (iv) only C (ii), (iii) and (iv) only D (i), (ii), (iii) and (iv)
Numerical Methods and Calculus    Gate IT 2007
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 Question 51
Find the Integral value of f(x) = x * sinx within the limits 0, π.
 A π B 2π C π/2 D 0
Numerical Methods and Calculus    GATE 2017 Mock
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Question 51 Explanation:
Let I = ∫x*sinx ----------------------------------------------------------> 1
I = ∫( π-x)*sin(π-x) = ∫(( π-x)*sin(x) ---------------- 2
=> I + I = ∫ π*sinx = π*[-cosx] = - π*[cos π – cos0]
=> 2I = - π*-2
=> I = π
The value of Integral is π
 Question 52
The value of the constant 'C' using Lagrange's mean value theorem for f(x) = 8x - x2 in [0,8] is:
 A 4 B 8 C 0 D None of these
Numerical Methods and Calculus    GATE 2017 Mock
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Question 52 Explanation:
f(x) = 8x - x^2 in [0,8]
C = (0 + 8) / 2 = 4
Since, the value of ‘C’ for px^2 + qx + r (irrespective of p,q,r) defined in [a,b] using mean value theorem is mid point of the interval, i.e. C = (a + b) / 2
There are 52 questions to complete.

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