Calculus

Question 1
Consider the following expression.

The value of the above expression (rounded to 2 decimal places) is _______
A
0.25
Question 1 Explanation: 
Question 2

Backward Euler method for solving the differential equation dy/dx = f(x,y) is specified by, (choose one of the following).

A
yn+1 = yn + hf(xn, yn)
B
yn+1 = yn + hf(xn+1, yn+1)
C
yn+1 = yn-1 + 2hf(xn, yn)
D
yn+1 = (1 + h) f(xn+1, yn+1)
Question 2 Explanation: 
dy/dx = f(x,y)
With initial value y(x0) = y0. Here the function f and the initial data x0 and y0 are known. The function y depends on the real variable x and is unknown. A numerical method produces a sequence y0, y1, y2, ....... such that yn approximates y(x0 + nh) where h is called the step size.
→ The backward Euler method is helpful to compute the approximations i.e.,
yn+1 = yn + hf(x n+1, yn+1)
Question 3

The solution of differential equation y'' + 3y' + 2y = 0 is of the form

A
C1ex + C2e2x
B
C1e-x + C2e3x
C
C1e-x + C2e-2x
D
C1e-2x + C22-x
Question 3 Explanation: 
Note: Out of syllabus.
Question 4

(a) Determine the number of divisors of 600.
(b) Compute without using power series expansion

A
Theory Explanation.
Question 5

Consider the functions

    I. e-x
    II. x2-sin x
    III. √(x3+1)

Which of the above functions is/are increasing everywhere in [0,1]?

A
II and III only
B
III only
C
II only
D
I and III only
Question 5 Explanation: 
A function f(x) is said to be increasing if f'(x)>0 at each point in an interval.
I. e-x
II. f'(x) = -e-x
f'(x)<0 on the interval [0,1] so this is not an increasing function.
II. x2-sinx
f'(x) = 2x - cosx
at x=0, f'(0) = 2(0) - 1 = -1 < 0
f(x) = x2 - sinx is decreasing over some interval, increasing over some interval as cosx is periodic.
As the question is asked about increasing everywhere II is false.
III. √(x3+1) = (x3+1)1/2
f'(x) = 1/2(3x2/√(x3+1))>0
f(x) is increasing over [0,1].
Question 6

Consider the function f(x) = sin(x) in the interval x ∈ [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
Question 6 Explanation: 
f(x) = sin x
f’(x) = cos x

[just consider the given interval (π/4, 7π/4)]
f'(x) = 0 at π/2, 3π/2
To get local minima f ’’(x) > 0, f ’’(x) = - sin x
f ’’(x) at π/2, 3π/2
f ’’(x) = -1< 0 local maxima
f ’’ (3π/2) = 1 > 0 this is local minima
In the interval [π/4, π/2] the f(x) is increasing, so f(x) at π/4 is also a local minima.
So there are two local minima for f(x) at π/4, 3π/2.
Question 7

The formula used to compute an approximation for the second derivative of a function f at a point X0 is

A
f(x0+h) + f(x0-h)/2
B
f(x0+h) - f(x0-h)/2h
C
f(x0+h) + 2f(x0) + f(x0-h)/h2
D
f(x0+h) - 2f(x0) + f(x0-h)/h2
Question 7 Explanation: 
The formula which is used to compute the second derivation of a function f at point X is
f(x0+h) - 2f(x0) + f(x0-h)/h2
Question 8

What is the maximum value of the function f(x) = 2x2 - 2x + 6 in the interval [0,2]?

A
6
B
10
C
12
D
5.5
Question 8 Explanation: 
For f(x) to be maximum
f'(x) = 4x - 2 = 0
⇒ x = 1/2
So at x = 1/2, f(x) is an extremum (either maximum or minimum).
f(2) = 2(2)2 - 2(2) + 6 = 10
f(1/2) = 2 × (1/2)2 - 2 × 1/2 + 6 = 5.5
f(0) = 6
So, the maximum value is at x=2 which is 10 as there are no other extremum for the given function.
Question 9

Consider the function y = |x| in the interval [-1,1]. In this interval, the function is

A
continuous and differentiable
B
continuous but not differentiable
C
differentiable but not continuous
D
neither continuous nor differentiable
Question 9 Explanation: 
The given function y = |x| be continuous but not differential at x= 0.
→ The left side values of x=0 be negative and right side values are positive.
→ If the function is said to be differentiable then left side and right side values are to be same.
Question 10

(a) Find the points of local maxima and minima, if any, of the following function defined in 0 ≤ x ≤ 6.

  x3 - 6x + 9x - 15 

(b) Integrate

A
Theory Explanation.
Question 11

Which of the following statements is true?

A
S > T
B
S = T
C
S < T and 2S > T
D
2S ≤ T
Question 11 Explanation: 
S is continuously increasing function but T represent constant value so S>T.
Question 12

The differential equation
d2y/dx2 + dy/dx + siny = 0 is:

A
linear
B
non-linear
C
homogeneous
D
of degree two
Question 12 Explanation: 
Note: Out of syllabus.
d2y/dx2 + dy/dx + siny = 0
In this DE, degree is 1 then this represent linear equation.
Question 13

Fourier series of the periodic function (period 2π) defined by

But putting x = π, we get the sum of the series.

A
π2/4
B
π2/6
C
π2/8
D
π2/12
Question 13 Explanation: 
Note: Out of syllabus.
Question 14

Which of the following improper integrals is (are) convergent?

A
B
C
D
Question 15
A
1
Question 15 Explanation: 
Since the given expression is in 0/0 form, so we can apply L-Hospital rule.
Question 16

The radius of convergence of the power series

A
Out of syllabus.
Question 17

The value of the double integral is

A
1/3
Question 17 Explanation: 
Question 18

The differential equation yn + y = 0 is subjected to the boundary conditions.

    y (0) = 0        y(λ) = 0         

In order that the equation has non-trivial solution(s), the general value of λ is __________

A
Out of syllabus.
Question 19

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
Question 19 Explanation: 
Note: Out of syllabus.
Question 20

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-II, T-I
B
Q-III, R-II, S-I, T-IV
C
Q-II, R-I, S-IV, T-III
D
Q-I, R-IV, S-II, T-III
Question 20 Explanation: 
Note: Out of syllabus.
Question 21

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)
Question 21 Explanation: 
Note: Out of syllabus.
Question 22

What is the value of

A
-1
B
1
C
0
D
π
Question 22 Explanation: 

In the limits are be -π to π, one is odd and another is even product of even and odd is odd function and integrating function from the same negative value to positive value gives 0.
Question 23

The following definite integral evaluates to

A
1/2
B
π √10
C
√10
D
π
E
None of the above
Question 23 Explanation: 
Question 24

If f(x) is defined as follows, what is the minimum value of f(x) for x∊(0,2] ?

A
2
B
2(1/12)
C
2(1/6)
D
2(1/2)
Question 24 Explanation: 
If x = 3/2
f(x) = 25/8x = 25/8(3/2) = 25/12 = 2(1/12)
Question 25
The value of the following limit is _____________.
A
1/2
Question 25 Explanation: 

When 0 is substituted, we get 0/0
Apply L- Hospital rule


-1/2
Question 26

If , then the value of k is equal to ________.

A
4
B
5
C
6
D
7
Question 26 Explanation: 
The graph x.Sinx from 0 to 2π is

We have |xSinx|,

We can observe that it is positive from 0 to π and negative in π to 2π.
To get positive value from π to 2π we put ‘-‘ sign in the (π, 2π)
Question 27

With respect to the numerical evaluation of the definite integral ,  where a  and b are given, which of the following statements is/are TRUE?

    I) The value of obtained using the trapezoidal rule is always greater than or equal to the exact value of the definite integral.
    II) The value of obtained using the Simpson’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
Question 27 Explanation: 
Note: Numerical methods are out of syllabus for the GATE -CS.
Question 28

The value of the integral given below is

A
-2π
B
π
C
D
Question 28 Explanation: 
Question 29

Let f(x) = x -(1/3) and A denote the area of the region bounded bu f(x) and the X-axis, when x varies from -1 to 1. Which of the following statements is/are TRUE?

    I) f is continuous in [-1,1]
    II) f is not bounded in [-1,1]
    III) A is nonzero and finite
A
II only
B
III only
C
II and III only
D
I, II and III
Question 29 Explanation: 
Since f(0)→∞
∴ f is not bounced in [-1, 1] and hence f is not continuous in [-1, 1].

∴ Statement II & III are true.
Question 30

If g(x) = 1 - x and h(x) , then is:

A
h(x)/g(x)
B
-1/x
C
g(x)/h(x)
D
x/(1-x)2
Question 30 Explanation: 
g(x)= 1 – x, h(x)=x/x-1 -------- (2)
Replace x by h(x) in (1), replacing x by g(x) in (2),
g(h(x))=1-h(x)=1-x/x-1=-1/x-1
h(g(x))=g(x)/g(x)-1=1-x/-x
⇒ g(h(x))/h(g(x))=x/(x-1)(1-x)=(x/x-1)/1-x=h(x)/g(x)
Question 31
A
-1
B
-2
C
-3
D
-4
Question 31 Explanation: 
Question 32
A
0.99
B
1.00
C
2.00
D
3.00
Question 32 Explanation: 

= 2-1/1(2)+3-2/2(3)+4-3/3(4)+…+100-99/99(100)
= 1/1-1/2+1/2-1/3+1/3…+1/98-1/99+1/99-1/100
= 1-1/100
= 99/100
= 0.99
Question 33

Let f(x) be a polynomial and g(x) = f'(x) be its derivative. If the degree of (f(x) + f(-x)) is 10, then the degree of (g(x) - g(-x)) is __________.

A
9
B
10
C
11
D
12
Question 33 Explanation: 
If the degree of a polynomial is ‘n’ then the derivative of that function have (n – 1) degree.
It is given that f(x) + f(-x) degree is 10.
It means f(x) is a polynomial of degree 10.
Then obviously the degree of g(x) which is f’(x) will be 9.
There are 33 questions to complete.

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