Engineering Mathematics Workbook_Final

Probability & Statistics

232. An exam paper has 150 multiple (a) 0.05, 1.87 (b) 1.90, 5.87
choice questions of 1 mark each, with (c) 0.05, 1.10 (d) 0.25, 1.40
each question having for choices.
Each incorrect answer fetches -0.25 [GATE-2007 (PI)]
marks. Suppose 1000 students choose 235. A fair coin is tossed repeatedly till

all their answer randomly with both head and tail appear at least
uniform probability. The sum total of once. The average number of tosses
the expected marks obtained by all required is ___.
the students is
[GATE-2014-EC SET3]
(a) 0 (b) 2550
236. Each of the nine words in the
(c) 7525 (d) 9375 sentence, “The Quick brownfox
jumps over the lazy dog” is written in
[GATE-2004 (CS)]
a separate piece of paper. These nine
233. The following data about the flow of pieces of paper are kept in a box. One
liquid was observed in a continuous of the piece is drawn at random from
chemical process plant. the box. The expected length of the
word drawn is _____.
Frequency
Flow rate
(litres / sec) (The answer should be rounded to
7.5 to 7.7 1 one decimal place)
7.7 to 7.9 5
7.9 to 8.1 35 [GATE-2014 (CS-SET2)]
8.1 to 8.3 17
8.3 to 8.5 12 237. Let the random variable X represent
8.5 to 8.7 10 the number of times a fair coin needs
Mean flow rate of the liquid is to be tossed till two consecutive
heads appear for the first time. The
(a) 8.00 litres / sec (b) 8.06 litres / expectation of X is _____.
sec
[GATE-2005]
(c) 8.16 litres / sec (d) 8.26 litres /
sec 238. Passengers try repeatedly to get a seat
reservation in any train running
[GATE-2004]
between two stations until they are

234. The random variable X takes on the successful. If there is 40% chance of
values 1, 2, (or) 3 with probabilities getting reservation in any attempt by
+
+
+
2 5P 1 3P 1.5 2P a passenger then the average number
, and of attempts that passengers need to
5 5 5
respectively the value of P and E(X) make to get a seat reserved is ____.
are respectively. [GATE-2017 EC SESSION-II]



199

Probability & Statistics

239. A person decides to toss a fair coin at the end of 11 steps he is one step
repeatedly until he gets a head. He away from the starting point is
will make at most 3 tosses. Let the
random variable Y denote the number (a)   6   5 (b) 462   6   5
of heads. The value of var(Y). where   25       25    

var(.) denotes tie variance, equals:
  1   5   1   5
7 49 (c) 538     (d)    
(a) (b)  25   25 
8 64
7 105 [JNU]
(c) (d)
64 64 243. In a manufacturing plant, the
probability of making a defective bolt
[GATE 2017 EE SESSION-II] is 0.1. The mean and standard
deviation of defective bolts in a total
BINOMIAL DISTRIBUTION
of 900 bolts are respectively

240. The probability of India winning a (a) 90 and 9 (b) 9 and 90
test match against England is 0.5.
Assuming independence of the result (c) 81 and 9 (d) 9 and 81
of various matches the chance that in [GATE, EE: 2000]
a 5 match series. India’s second win
rd
occurs at 3 test is ____. 244. An unbiassed coin is tossed an
infinite number of times. The
241. A die hastbur blank faces and two probability that the fourth head
faces marked 3. The chance of getting appears at the tent toss is
a total of 12 in 5 the is (a) 0.067 (b) 0.073

    (c) 0.082 (d) 0.091
4
1
2
5
(a) C    
  
    
3
3
    [GATE-2014-EC-SET 3]
    4 245. The probability that a screw
2
1
   
5
(b) C 4      manufactured by a company is
  
3
3
   
defective is 0.1. The company sells
 1  5 screws in packets containing 5 screws
5
(c) C 4       and gives a guarantee of replacement
 6  if one or more screws in the packet
(d) None of these are found to be defective. The
probability that a packet would have
242. A man takes a step forward with to be replaced is ____.
probability 0.4 and ackward with
probability 0.6. The probability that [GATE 2016-ME-SET 2]



200

Probability & Statistics

POSITION DISTRIBUTION 249. Suppose p is the number of cars per
minute passing through a certain road
246. The number of accidents occurring in junction between 5 PM and 6 PM,
a plant in a month follows Poisson and p has aPoisson distribution with
distribution with mean as 5.2. The mean 3. What is the probability of

probability of occurrence of less than observing fewer than 3 cars during
2 accidents in the plant during a any given minute in this interval?
randomly selected month is
3
3
9/ 2e
8/ 2e
(a) ( ) (b) ( )
(a) 0.029 (b) 0.034
( )
(c) 0.039 (d) 0.044 (c) ( ) (d) 26/ 2e
3
3
17/ 2e
[GATE, ME ; 2014 (SET-4)]
[GATE-2013 (CS)]
247. It is estimated that the average
number of events during a year is 250. An observer counts 240 veh/h at a
three. What is the probability of specific highway location. Assume
occurrence of not more than two that the vehicles arrival at the
events over a two year duration location is Poisson distributed, the
Assume that the number of events probability of having one vehicle

follow a Poisson distribution arriving over a 30-second time
interval is_____.
(a) 0.052 (b) 0.02
[GATE-2014 (CE-SET 2)]
(c) 0.072 (d) 0.002
251. The second moment of a Poisson-
[GATE, PI : 2011] distributed random variable is 2. The
mean of the random variable is
248. A manufacturer knows that eh _______. [GATE-2016]
condensers he makes contain on an
average 1% defectives. He packs 252. Vehicles arriving at an intersection

them, in boxes of 100, What is the from one of the approach roads
probability that a box picked up at follow the Poisson distribution. The
random will contain 3 or more faulty mean rate of arrival is 900 vehicles
condensers per hour. If a gap is defined as the
time difference between two
3 5
−
−
1
1
(a) 1− e (b) 1− e successive vehicle arrivals (with
2 2 vehicles assumed to be points), the
2 5 probability (up to four decimal
(c) 1− (d) 1− places) that the gap is greater than 8
e e
seconds is ____.
[GATE]
[GATE-2017 CE SESSION-I]



201

Probability & Statistics

253. If a random variable X has a Poisson  0.2, for x  1

 
distribution with mean 5, then the f ( ) x =  0.1, for 1 x  4
 2 
expectation E  ( X + ) 2  equals  0, otherwise
  

_____.
The probability P(0.5 < X < 5) is
[GATE-2017 PAPER-2 (CS)] _____. [GATE-2014-EC-SET 2]


CONTINUOUS RANDOM VARIABLE 257. The variance of the random variable
X with probability density function
254. A probability density function is of 1
the form ( ) x = Ke −  x , x  (− , ) , f ( ) x = x e − x is _____.
p
2
the value of k is
[GATE-2015 (CS-SET 3)]
(a) 0.5 (b) 1
258. The probability density function of
(c) 0.5a (d) a − x
random variable X is ( ) x = e for
P
x
[GATE-2006] x  0 and 0 otherwise. The expected
value of the function ( ) x = e 3 / 4 is
x
g
255. Consider the continuous random x
variable with probability density _______.
function
[GATE-2015 (IN)]

f t = + −  259. Two random variables X and Y are
( ) 1 t for 1 t £ 0 = 1 – t

for 0 t  1 distributed according to

The standard deviation of the random  (x + y ), 0   1, 0   1
y
x

)
variable is f , x y ( , x y =  0, otherwise


1 1
(
(a) (b) The probability P X + Y  ) 1 is
3 6
_____.
1 1
(c) (d) [GATE-2016-EC-SET 2]
3 6
260. Let the probability density function of
[GATE-2006- ME]
a random variable, X, be given as:
256. Let X be a random variable with
−
probability density function f ( ) x = 3 e u ( ) x + ae u ( ) x−
3x
4x
x
2
where u(x) is the unit step function.






202

Probability & Statistics

Then the value of ‘a’ and Prob (c)    1 ,2  
X   0 , respectively, are   2   (d) {1, 2}


1 1 [ESE 2017 (EE)]
(a) 2, (b) 4,
2 2
263. The graph of a function f(x) is shown
1 1 in the figure.
(c) 2, (d) 4,
4 4

[GATE-2016 EC SET-3]

261. For a random variable x having the
PDF shown in the figure given below


The mean and the variance are,
respectively

For f(x) to be a valid probability
density function, the value of h is

(a) 1/3 (b) 2/3


(c) 1 (d) 3

[GATE 2018 (CE-AFTERNOON

SESSION)]
(a) 0.5 and 0.66 (b) 2.0 and 1.33
UNIFORM PROBABILITY
(c) 1.0 and 0.66 (d) 1.0 and 1.33 DISTRIBUTION


[ESE 2017 (EC)] 264. A point is randomly selected with
uniform probability in the x-y plane
262. A random variable X has a
probability density function with in the rectangle with corners at
(0, 0), (1, 0), (1, 2) and (0, 2). If p is
 kx e x ; x  0 the length of the position vector of
n −

f ( ) x =  (n is
 0; otherwise the point, then the expected value of

2
'
an integer) with mean 3. The values ' p is _________
of {k, n} are
[GATE 2004]
  1     1  
(a)  ,1 (b)  ,2 265. X is a uniformly distributed random
  2     4   variable that takes values between 0

3
E X
and 1. The value of ( ) will be




203

Probability & Statistics

(a) 0 (b) 1/8 distribution on [0, 1]. The probability
P X + X  X  is the largest is
(c) 1/4 (d) 1/2 1 2 3
____. [GATE-2016 (CE-SET1)]
[GATE-2008-EE]
271. Assume that in a traffic junction, the
266. A random variable is uniformly cycle of the traffic signal lights is 2
distributed over the interval 2 to 10. minutes of green (vehicle does not
Its variance will be stop) and 3 minutes of red (vehicle

stops). Consider that the arrival time
(a) 16/3 (b) 6
of vehicles at the junction is
(c) 256/9 (d) 36 uniformly distributed over 5 minute
cycle. The expected waiting time (in
[GATE-2008 (IN)]
minutes) for the vehicle at the
267. The independent random variables X junction is _______
and Y are uniformly distributed in the [GATE 2017 – EE SESSION-1]
interval [-1, 1]. The probability that
max [X, Y] is less than 1/2 is EXPONENTIAL DISTRIBUTION


(a) 3/4 (b) 9/16 272. Assume that the duration in minutes
of a telephone conversation follows
(c) 1/4 (d) 2/3
the exponential distribution

[GATE-2012-EC, EE, IN] 1 − x /5
f ( ) x = e , x  0. The
5
268. Let X , X and X be independent probability that the conversation will
3
1
2
and identically distributed random exceed five minutes is
variables with the uniform
1
1
distribution on [0, 1]. The probability (a) (b) 1−
P X 1 e e
  is the largest is ______.
[GATE-2014 (EC-SET1)] (c) 1 (d) 1− 1
e 2 e 2
269. Suppose you break a stick of unit
length at a point chosen uniformly at [GATE-2007 (IN)]
random. Then the expected length of
the shorter stick is ____. 273. For a single server with Poisson
arrival and exponential service time,

[GATE-2016 (CE-SET1)] the arrival rate is 12 per hour. Which
one of the following service rates will
270. Let X , X and X be independent provide a steady state finite queue
2
1
3
and identically distributed random length?
variables with the uniform



204

Probability & Statistics

(a) 6 per hour (b) 10 per hour holders who main an average daily
balance more than Rs. 500 is _____.
(c) 12 per hour (d) 24 per hour
277. For a random variable
[GATE 2017 ME SESSION-II]
)
x
( x −   following normal
274. The arrival of customers over fixed distribution, the mean is  = 100 if
time intervals in a bank follow a the probability is P  for x  110.
=
poisson distribution with an average Then the probability of x laying

of 30 customers / hour. The between 90 and 110 i.e.,
probability that the time between P (90  x  110 ) and equal to
successive customer arrival is
between 1 and 3 m minutes is ____ (a) 1 2 (b) 1 
−
−
(correct to two decimal places).
−
(c) 1  /2 (d) 2
[GATE-2018 (ME-AFTERNOON
SESSION)] [GATE, PI : 2008]

275. Let X and X be two independent 278. Suppose X is a normal random
1 2
exponentially distributed random variable with mean 0 and variance 4.
variables with means 0.5 and 0.25, Then the mean of the absolute value
,
respectively. Then Y-min ( X X 2 ) of X is
1
is
1 2 2
(a) exponentially distributed with (a) 2 (b) 
mean 1/6

(b) exponentially distributed with 2 2 2
mean 2 (c)  (d) 

(c) normally distributed with mean ¾
[GATE 1999]
(d) normally distributed with mean
1/6 279. Which one of the following
statements is not true?
[GATE 2018 (ME-AFRTERNOON
SESSION)] (a) The measure of skewness is
dependent upon the amount of
NORMAL DISTRIBUTION
dispersion
276. A nationalized bank has found that

the daily balance available in its (b) In a symmetric distribution, the
savings accounts follows a normal values of mean, mode and median are
distribution with a mean of Rs. 500 the same
and a standard deviation of Rs. 50. (c) In a positively skewed
The percentage of savings account distribution, mean > median > mode



205

Probability & Statistics

(d) In a negatively skewed 283. Let U and V be two independent zero
distribution, mode > mean > median mean Gaussian random variables of
1 1
[GATE-2005] variances and respectively. The
4 9
280. The value of probability (3V  2U ) is
P

1    − x 2  
1
I = exp      dx is _____. (a) 4 (b)
2 0  8  9 2

[GATE 2006] 2 5
(c) (d)

281. The standard normal probability 3 9
function can be approximated as [GATE-2013 (EC)]

1 284. Let X be a zero mean unit variance
F ( X ) = ,
N
+
1 exp − ( 1.72555X N | X N | 0.12 ) Gaussian random variable. E[|X|] is
equal to ____.
where X = standard normal [GATE-2014-EC-SET 4]
N
deviate. If mean and standard
deviation of annual precipitation are 285. If f(x) is a continuous real valued
102 cm and 27 cm respectively, the random variable defined over the
probability that the annual interval (− ,+ ) and its occurance
precipitation will be between 90 cm is defined by the density function
and 102 cm is given as

 2

−
(a) 66.7% (b) 50.0% 1  1 x a     

f ( ) x = exp −      

(c) 33.3% (d) 16.7% 2 b   2   b    
where a and b are the statistical
[GATE-2008-CE]
attributes of the random variable {x}.
282. Let X be a random variable following The value of the integral
ND with mean +1 and variance 4. Let    2
−


Y be another normal variable with  a 1 exp − 1 x a         dx is



mean -1 and variance unknown. If − 2 b   2   b    
P ( X  − ) 1 = P (Y  ) 2 . The S.D. of
(a) 1 (b) 0.5
Y is _____. [GATE-2008]

(c)  (d) /2

[GATE-2014-CE-SET 2]







206

Probability & Statistics

286. A probability distribution with right (a) 0 (b) 0.5
skew is shown in the figure.
(c) 1 (d) 10


[GATE-2008 (IN)]

289. The cumulative distribution function
of a random variable x is the
probability that X takes the value


(a) less than or equal to x
The correct statement for the
probability is distribution is (b) equal to x

(a) Mean is equal to mode (c) greater than x


(b) Mean is greater than median but (d) zero [ESE 2017 (EC)]
less than mode 290. The probability density function

(c) Mean is greater than median and F ( ) x = ae − b x where x is a random
mode variable whose allowable value range


(d) Mode is greater than median is from x = − to x = +. The CDF
for this function for x  0 is
[GATE 2018 (CE-AFTERNOON
a
a
SESSION)] (a) e (b) (2 e − bx )
−
bx
b b
,
287. Let X 1 , X X and X be
4
3
2
a
independent normal random variables (c) − a e (d) ( 2 e − bx )
bx
−
+
with zero mean and unit variance. b b
The probability that X is the [ESE 2017 (EC)]
4
smallest among the four is _______.

[GATE-2018 (EC)]
CUMMULATIVE DISTRIBUTION
FUNCTION

288. Consider a Gaussian distributed
random variable with zero mean and
standard deviation  . The value of
its cumulative distribution function at
the origin will be






207

7















Numerical




















Methods

Numerical Methods

1. The equation e − x 4x = 2 0 has a root 4. If  and  are the forward and the
between 4 and 5. Fixed point iteration backward difference operators

1 respectively, then  − is equal to
with iteration function e x /2 .
2 (a) − (b) 

(a) diverges 
(c)  + (d)
(b) converges 

(c) oscillates 5. One root of the equation e − x 3x = 2 0
lies in the interval (3, 4). The least
(d) converges monotonically
number of iterations of the bisection
−
3
2. The formula method so that error  10 are
 1   
1

A f −  + A f ( ) 0 + A f  
2   
0  
 

2
 2  2   (a) 10 (b) 8
which approximates the integral (c) 6 (d) 4
 1 f ( ) x dx is exact for polynomials
− 1 6. The least squares approximation of
of degree less than or equal to 2 if first degree to the function
f x =
( ) sin x over the interval
4 2
(a) A = A = , A =  
2
0
3 1 3  −  ,  
  2 2  is
(b) A = A = A = 1  
1
0
2
24x 24x
4 2 (a) (b)
(c) A = A = , A = −  3  2
0 2 1
3 3
24x
(d) none of the above (c) (d) 24x

3. If x = is a double root of the
( )
f x =
f
equation ( ) 0 and if ''' x is 7. The order of the numerical
differentiation formula
continuous in a neighbourhood of  1
x
then the iteration scheme for '' f ( ) = 12h 2 [−  ( f x − 2h ) + ( f x + 2h ) +
0
0
0
determining  : 16  ( f x − ) h + ( f x + h ) 30 f− ( )]x 0
0
0
( )
2 f x (a) 2 (b) 3
x n+ = x − n has the order
( )
n
1
' f x n (c) 4 (d) 1
(a) 2 (b) 1
(c) less than 2 (d) less than 1


209

Numerical Methods

8. The method 3
+
(a) x n+ 1 = − 16 6x +
n
1 x n
y + = (k + 3k ),n = 0,1,.....
1 y +
n n 1 2
4
+
(b) x n+ 1 = 3 2x
n
( , y
k = hf x )
1 n n 3
(c) x =
 2h 2  n+ 1 x − 2

k = hf x + , y + k 1    is used n

n
2
n
 3 3  2
to solve the initial value problem (d) x n+ 1 = x − 2
n
)
' y = f ( , x y = − 10y , ( ) 0 = 2
1
y
12. While solving the equation
The method will produce stable x − 3x + = using the Newton-
2
1 0
results if the step size h satisfies
Raphson method with the initial


(a) 0.2 h  0.5 (b) 0 h  0.5 guess of a root as 1, the value of the
root after one iteration is


(c) 0 h  1 (d) 0 h  0.2
(a) 1.5 (b) 1
6
x
9. The equation x − − =
1 0 has
(c) 0.5 (d) 0
(a) no positive real roots
13. Consider the system of equations
(b) exactly one positive real root  5 2 1   x   13 
     1    




(c) exactly two positive real roots  − 2 5 2     x  2  = − 22



 − 1 2 8     x   14 



(d) all positive real roots    3   
10. The smallest degree of the With the initial guess of the solution

0
, ,x
polynomial that interpolates the data    x x 2 0 3  0  = 1,1,1 , the
1
x -2 -1 0 1 2 3 approximate value of the solution
1
x
f(x) -58 -21 -12 -13 -6 27    x 1 , ,x 1 3  after one iteration by the
2
1
Is
Gauss-Seidel method is
(a) 3 (b) 4


−
−
−
(a) 2, 4.4,1.625 (b) 2, 4, 3
(c) 5 (d) 6


−
(c) 2,4.4,1.625 (d) 2, 4,3
11. Suppose that x is sufficiently close
0
to 3. Which of the following 14. Using Euler’s method taking step size
iterations x n+ 1 = g x n = 0.1, the approximate value of y
( ) will converge
3
to the fixed point x = ? obtained corresponding to x = 0.2

210

Numerical Methods

for the initial value problem 17. The root of the equation
dy = x + 2 y , ( ) 0 = is x log x = 4.77 with initial guess
2
10
1
y
dx x = 6, obtained by using Newton
0
(a) 1.322 (b) 1.122 Raphson method after second
iteration is
(c) 1.222 (d) 1.110
18. Let the following discrete data be
15. Consider the system of equations obtained from a curve y = y ( ) x :

 5 − 1 1     10 
x
      x: 0 0.25 0.5
  



y = 
 2 4 0   12 0.75 1
     
 1 1 5     − 1  y: 1 0.98960 95890
z
  



  



Using Jacobi’s method with the initial 90890 8415
guess  , y z =  Let S be the solid of revolution
,
x
 2,3,0 , the
0 0 0
solution after two approximations is obtained by rotating the above curve
about the x-axis between x = 0 and x

−
−
(a) 2.64, 1.70, 1.12 = 1, and let V denote its volume. The
approximate value of V, obtained
−
(b) 2.64, 1.70,1.12  1
using Simpson’s rule, is
−
(c) 2.64,1.70, 1.12  3
19. Using the Gauss-Seidel iteration

(d) 2.64,1.70,1.12 method with the initial guess,
 , y z = 
,
x
 3.5,2.25,1.625 , the
 1 2 0  0 0 0
    solution after two approximations for
16. The matrix A =  1 3 1 can be
  system of equations 2x − x = 7,
   0 1 3    − x + 2x − x = 1, x + 2x =
1
2
−
decomposed uniquely into the 1 2 3 2 3 1 is
product A = LU, where L and U are (a) [5.3125, 4.4491, 2.1563]
lower and upper triangular matrices
respectively with L has diagonal (b) [5.3125, 4.3125, 2.6563]
entries 1. The solution of the system (c) [5.3125, 4.4491, 2.6563]

t
LX = 1 2 2 is
(d) [5.4991, 4.4491, 2.1563]

t
(a) 1 1 1 (b) 1 1 0  t



t
t
(c) 0 1 1 (d) 1 0 1

211

Numerical Methods

20. The fourth order Runge-Kutta 22. If
method given by p ( ) x = − ( x + ) 1 + ( x x + ) 1 −  ( x x + 1 )(x  − )
2
)
h interpolates the points ( , x y in the
u = u + K + 2K + 2K + K 
j+
j
1
6 1 2 3 4 table
is sued to solve the initial value
du x: -1 0 1 2
problem = , u u ( ) 0 =  .
dt y: 2 1 2 -7

u
If ( ) 1 = 1 is obtained by taking the then   + =
step size h = 1, then the value of K 23. If the differential equation
4
is
dy = x + y 2 , y ( ) 1 = 2 is solved
2
21. Let the integral 0  4 f ( ) x dx , where dx
 , x 0  x  2 using the Euler’s method with step-

f ( ) x =  size h = 0.1, then y(1.2) is equal to
 4 x , 2   4 (round off to 2 places of decimal).
−
x

 R
f
Consider the following statements P 24. Let  : , a b → be any function
and Q: which is twice differentiable in (a, b)
with only one root  in (a, b). Let
P: If I is the value of the integral
2 ' f '' f
obtained by the composite trapezoidal ( ) x and ( ) x denote the first
rule with two equal sub-intervals, and second order derivatives of

then I is exact. f ( ) x with respect to x. If  is a
2
simple root and is computed by the
Q: If I is the value of the integral Newton-Raphson method, then the
3
obtained by the composite trapezoidal method converges if
rule with three equal sub-intervals,
f x
'' f
then I is exact. (a) ( ) ( ) x  ' f ( ) x 2 (b)
3
( ) ( ) x 
f x ' f '' f ( ) x
Which of the above statements hold
TRUE? 2
x
'' f
' f
(c) ( ) ( ) x  f ( ) x (d)
(a) Both P, Q
( ) ( ) x 
f x '' f ' f ( ) x
(b) Only P
25. The maximum value of the error term
(c) Only Q
of the composite Trapezoidal rule
(d) Neither P nor Q when it is used to evaluate the
definite integral





212

Numerical Methods
 0.2 (sin x − log e ) x dx with 12 sub- (d) an infinite number of solutions
1.4

intervals of equal length, is equal to 28. Solution of the variables x and x
(round off to 3 places of decimal). 1 2
for the following equations is to be
26. A piecewise linear function f(x) is obtained by employing the Newton-

plotted using thick solid lines in the Raphson interactive method
figure (the plot is drawn to scale).
=
x
equation (i) 10 sin x − 0.8 0
2 1
equation (ii)
10x − 10 cos x − 0.6 0
=
2
x
1
2
2
Assuming the initial values x = 0.0
1
and x = 1.0, the Jacobian matrix is
2
 10 − 0.8   10 0 
(a)     (b)    
  0 − 0.6    0 10 


 0 − 0.8   10 0 
(c)     (d)    
If we use the Newton-Raphson   10 − 0.6    10 − 10 

method to find the roots of f(x) = 0
−
using x , x and x respectively as 29. The differential equation dx = 1 x
0
1
2
initial guesses, the roots obtained dt 
would be is discretised using Euler’s numerical
integration method with a time step
(a) 1.3, 0.6 and 0.6 respectively  T  0. What is the maximum

(b) 0.6, 0.6 and 1.3 respectively permissible value of T to ensure
stability of the solution of the
(c) 1.3, 0.6 and 1.3 respectively corresponding discrete time equation?


(d) 1.3, 0.6 and 1.3 respectively (a) 1 (b) / 2

[CS. GATE-2003] (c)  (d) 2

27. In the interval 0, the equation x = du
30. Consider the equation = 3t + 2 1
cos x has dt
with u = 0 at t = 0. This is
(a) No solution
numerically solved by using the

(b) exactly one solution forward Euler method with a step
size, t = 2. The absolute error in
(c) exactly two solution



213

Numerical Methods

nd
the solution at the end of the first is x = 1, then the value of x after 2
0
time step is __________ iteration is ____________.
[GATE-2017-CE SESSION-I] [GATE-2015-ME-SET-3]


31. There is no value of x that can 35. In Newton Raphson iterative method,
simultaneously satisfy both the given the initial guess value ( X ) is
ini
equation. Therefore, find the ‘least considered as zero, while finding the
squares error’ solution to the two roots of the equation
equations, i.e., find the value of x that ( ) x = − + 4x + 3
2 6x −
2
minimizes the sum of squares of the f 0.5x . The

errors in the two equations 2x = 3, 4x correction x , to be added to X in
ini
= 1. the first iteration is ______________.

32. Using given data points tabulated [GATE-2015-CE-SET-2]
below, a straight line passing through
the origin is fitted using least squares 36. The quadratic equation
2
4 0 is the solved
method. The slope of the line is x − 4x + =
numerically, starting with the initial
2
x y xy x guess x = 3. The Newton-Raphson
0
1 1.5 1.5 1 method is applied once to get a new
2 2.2 4.4 4 estimate and then the Secant method
3 2.7 8.1 9 is applied once using the initial guess
 xy = 14  x = 2 14 and this new estimate. The estimated

value of the root after the application
(a) 0.9 (b) 1 of the Secant method is __________.
(c) 1.1 (d) 1.5
[GATE-2015-CE-SET-1]

33. The Newton-Rapshon method is used 37. The Newton-Raphson method is used
to solve the equation to solve the equation
3
2
f ( ) x = x − 5x + 6x − = . Taking
8 0
3
2
f ( ) x = x − 5x + 6x − = .
8 0
the initial guess as x = 5, the solution
5
obtained at the end of the first Taking the initial guess as x = , the
iteration is ___________. solution obtained at the end of the
first iterations is _______.
[GATE-2015-ME-SET-2]
[GATE-2015-EC-SET-3]
34. Newton-Rapshon method is used to
find the roots of the equation, 38. How many distinct values of x satisfy
( )
x + 2x + 3x − 1 0. If the initial guess the equation sin x = x / 2, where x
=
2
3
is in radians?



214

Numerical Methods

(a) 1 (b) 2 an accuracy of 0.001, the required
minimum number of iterations is
(c) 3 (d) 4 or more
___________.

[GATE-2016; 1 MARK]
[GATE 2017 – EE – SESSION-1]
39. The root of the function 0.2
3
f ( ) x = x + − 1 obtained after first 43. What is the value of (1525 ) to 2
x
decimal places?
iteration on application of Newton
Raphson scheme using an initial (a) 4.33 (b) 4.36
guess of x = 1 is
0
(c) 4.38 (d) 4.30
(a) 0.682 (b) 0.686
[ESE 2018 (Common Paper)]

(c) 0.750 (d) 1.000 44. The quadratic equation

2
3 0
[GATE 2016 – ME – SET-1] 2x − 3x + = to be solved
numerically starting with an initial
40. Newton-Raphson method is to be
2
used to find root of equation guess as x = . The new estimate of
0
−
3x e + sin x = . If the initial trial x after the first iteration using
x
0
value for the root is taken as 0.333, Newton-Raphson method is _______.
the next approximation for the root
would be _________ (note: answer [GATE 2018 (CE – Afternoon Session)]
upto there decimal)
45. What is the cube root of 1468 to 3
[GATE 2016 – CE – SET-1] decimal places?

41. To solve the equation 2 sin x = x by (a) 11.340 (b) 11.353

Newton-Raphson method, the initial (c) 11.365 (d) 11.382
guess was chosen to be x = 2.0.
Consider x in radian only. The value [ESE 2018 (Common Paper)]
of x (in radian) obtained after one th
iteration will be closest to 46. The following table lists an n order
polynomial

(a) -8.101 (b) 1.901 f ( ) x = a x + a x n− 1 + ......... a x a 0
+
n
+
n−
1
n
1
(c) 2.099 (d) 12.101 and the forward differences
evaluated at equally spaced values of
[GATE 2016 (PI)] x. The order of the polynomial is
42. Only one of the real roots of
6
x
f ( ) x = x − − 1lies in the interval
1 x  2 and bisection method is

used to find its value. For achieving



215

Numerical Methods

49. Using a unit step size, the value of
integral 1  2 x ln x dx by trapezoidal
x f(x) f   2 f  3 f rule is ________.
- -
0.4 1.7648 0.2965 0.089 -0.03 [GATE 2015 – ME – SET-3]
- - -
0.3 1.4683 0.2075 0.059 0.0228 50. The integral x  x 2 x dx with
2
- - - 1
0.2 1.2608 0.1485 0.0362 0.0156 x  2 x  1 0 is evaluated analytically
- - -
0.1 1.1123 0.1123 0.0206 0.0084 as well as numerically using a single
- - application of the trapezoidal rule. If I
0 1 0.0122
0.0917 0.0012 is the exact value of the integral
- obtained analytically and J is the
0.1 0.9083 0.011 0.006
0.0795
approximate value obtained using the
0.2 0.8288 0.0685 0.017 0.0132
trapezoidal rule, which of the
following statements is correct about

(a) 1 (b) 2 their relationship?

(c) 3 (d) 4 (a) J > I


[GATE 2017 (IN)] (b) J < I


47. For step-size: x = 0.4 the value of (c) J = I
the following integral (d) Insufficient data to determine the

0.8
 (0.2 25x − 200x + 675x − 900x + 400x 5 relationship
+
2
3
4
) dx
0
using Simpson’s 1/3 rule is _______. [GATE 2015 – CE – SET-1]
51. The values of function f(x) at 5
[GATE 2015 – CE – SET-2]
discrete points are given below.
48. Simpson’s 1/3 rule is used to
integrate the function x 0 0.1 0.2 0.3 0.4
3 9 f(x) 0 10 40 90 160
f ( ) x = x + 2 between x = 0 and
5 5 Using Trapezoidal rule with step size

0.4
x = 1 using the least number of equal of 0.1 the value of  f ( ) x dx
sub intervals. The value of integral is 0
___________. _________.

[GATE 2015 – ME – SET-1] [GATE 2015 – ME – SET-2]

52. In numerical integration using
Simpson’s rule, the approximating
function in the interval is a




216

Numerical Methods

(a) constant (b) straight line 56. Values of f(x) in the interval [0, 4]
are given below:
(c) cubic B-Spline (d) parabola

[GATE 2015 (PI)] x 0 1 2 3 4

53. The velocity v (in kilometer / minute) f(x) 3 10 21 36 55
of a motorbike which starts from rest, Using Simpson’s 1/3 rule with a step
is given at fixed intervals of time t (in size of 1, the numerical

minutes) as follows: approximation (rounded off to the

1 1 1 1 1 2 second decimal place) of  4 f ( ) x dx
t 2 4 6 8 0
0 2 4 6 8 0
1 1 2 1 3 2 1 is _________.
v 5 2 0
0 8 5 9 2 0 1 [GATE 2016 (CH)]
The approximate distance (in
kilometers) rounded to two places of 57. P (0, 3), Q (0.5, 4) and R (1, 5) are
decimals covered in 20 minutes using three points on the curve defined by
Simpson’s 1/3 rule is ________. f(x). Numerical integration is carried
out using both Trapezoidal rule and
[GATE 2015 – CS – SET-3]
Simpson’s rule within limits x = 0
54. Numerical integration using and x = 1 for the curve. The
trapezoidal rule gives the best result difference between the two results
for a single variable function, which will be
is
(a) 0 (b) 0.25

(a) linear (b) parabolic
(c) 0.5 (d) 1
(c) logarithmic (d) hyperbolic
[GATE 2017 – ME – SESSION-1]

[GATE 2016 – ME – SET-2]
58. The ordinary differential equation
55. The error in numerically computing dx = − 3x + , with ( ) 0 = is to be
2
1
x
the integral 0   (sin x + cos ) x dx dt
using the trapezoidal rule with three solved using the forward Euler
intervals of equal length between 0 method. The largest time step that can
and  is __________. be used to solve the equation without
making the numerical solution
[GATE 2016 – ME – SET-2] unstable is __________.

[GATE 2016 – EC – SET-2]









217

Numerical Methods

59. Consider the first order initial value (III) Simphson’s Rule
−
problem y = + 2x x , ( ) 0 = , (IV) Runge-Kutta
2
1
y
1
y
)
x
(0    with exact solutions
Choose the correct set of
x
y ( ) x = x + 2 e . For x = 0.1, the combinations.
percentage difference between the
exact solution and the solution (a) P – II, Q – I, R – III, S - IV
obtained using a single iteration of (b) P – I, Q – II, R – IV, S – III
the second-order Runge-Kutta
method with step-size h = 0.1 is. (c) P – IV, Q – III, R – II, S – I

[GATE 2016 – EC – SET-3] (d) P – II, Q – I, R – IV, S – III

du [GATE 2017 (CH)]
60. Consider the equation = 3t + 2 1
dt 63. Variation of water depth (y) in a
with u = 0.1 at t = 0. This is gradually varied open channel flow is
numerically solved by using the given by the first order differential
forward Euler method with a step equation
size. t  = 2. The absolute error in
10
the solution at the end of the first dy 1 e 3 ln y
−
time step is __________. = −
−
dx 250 45e 3ln y
[GATE 2017 – CE – SESSION-1]
Given initial combination y (x = 0) =

61. Match the problem type in Group – 1 0.8 m. The depth (in m, up to three
with the numerical method in Group decimal places) of flow at a
– 2 downstream section at x = 1m from
one calculation step of Single Step
Group – 1 Euler Method is _______


(P) System of linear algebraic [GATE 2018 ( ME – Morning Session)]
equations
64. An explicit forward Euler method is
(Q) Non-linear algebraic equations
used to numerically integrate the

(R) Ordinary differential equations differential equation dy = y using a
dt
(S) Numerical integrations
time step of 0.1. With the initial

Group – 2 condition y(0) = 1, the value of y(1)
computed by this method is
(I) Newton-Raphson __________ (correct to two decimal

(II) Gauss-seidel places).




218

Numerical Methods

[GATE 2018 (ME-Morning Session)] (a) x converges to 2 with rate of
n
65. Consider the iteration function for convergence 1.
f ( ) x
Newton’s method ( ) x = x − (b) x converges to 2 with rate of
g
n
' f ( ) x convergence 2.
and its application to find (c) The given iteration is the fixed
(approximate) square root of 2, point iteration for ( ) x = x − 2 2.
f
2
starting with x = . Consider the
0
first and the second iterates x and (d) The given iteration is the
1
2
f
x , respectively; then Newton’s method for ( ) x = x − 2.
2
68. The following numerical integration

(a) 1.5 x  1 2 formula is exact for all polynomials
of degree less than or equal to 3
(b) 1.5  x  1 2
(a) Trapezoidal rule

(c) x  1 1.5; x  2 1.5 1
(b) Simpson’s rd rule
(d) x = 1 1.5; x  2 1 3

3
66. Let f be a continuous map from the (c) Simpson’s th rule
interval [0,1] into itself and consider 8

the iteration x n+ 1 = f x n (d) Gauss-Legendre 2 point formula
( ). Which of
the following maps will yield a fixed 1  2 

point for f? 69. The iteration x n+ =   x +   ,
1
2   n x n  
f
0
(a) ( ) x = x 2 / 4 n  0 for a given x  is an
0
instance of
(b) ( ) x = x 2 /8
f
(a) fixed point iteration for
f
(c) ( ) x = x 2 /16 f ( ) x = x − 2 2
(b) Newton’s method for
(d) ( ) x = x 2 /32 f ( ) x = x − 2 2
f

67. Consider the iteration (c) fixed point iteration for

2
1   2   f ( ) x = x + 2
0
x =  x +  , n  for a 2x
n+
1
2   n x n  
0
given x  . Then (d) Newton’s method for
0
f ( ) x = x + 2 2



219

Numerical Methods

DIRECT METHOD 4x + 3
2
(d) X k+ 1 = k 2

70. In the interval 0, the equation x = 9x + 2
k
cos x has [GATE-2007-CE]

(a) No solution 73. Identify the Newton-Raphson

iteration scheme for finding the
(b) exactly one solution
square root of 2
(c) exactly two solution
1   2  
(d) an infinite number of solutions (a) x n+ 1 = 2    x + n x n   

[GATE-1995 (CS)]
1   2  
71. How many distinct values of x satisfy (b) x n+ 1 = 2   x − n x  
the equation sin(x) = x/2, where x is  n 
in radians? 2
(c) x n+ 1 =
(a) 1 (b) 2 x n


(c) 3 (d) 4 or more (d) x n+ 1 − 2 x
+
n
[GATE-2016]
[GATE-2007 (IN)]

NEWTON-RAPHSON METHOD
74. The recursion relation to solve
(TYPE-I) x e using Newton-Raphson
=
−
x
72. The following equation needs to be method is
numerically solved using the (a) x = e − n x
Newton-Raphson method n+ 1
x + 4x − = The Iterative (b) x = x − e − n x
3
9 0.
equation for this purpose is (k n+ 1 n
indicates the iteration level) e − n x
+
(c) x n+ 1 = (1 x n ) − n x
+
2x + 9 1 e
3
(a) X k+ = k
1
2
3X + 4 x − e − x n (1 x −
+
2
) 1
k
(d) x n+ 1 = n − n x n
2x + 4 x − e
2
n
(b) X = k
k+
1
3X + 9 [GATE-2008-EC]
2
k
2
(c) X k+ 1 = x − 3x + 4
k
k


220

Numerical Methods

NEWTON-RAPHSON METHOD 78. A numerical solution of the equation
3 0
x
(TYPE-II) f ( ) x = + x − = can be
obtained using Newton-Raphson
x 9
75. Consider the series x n+ = n + , method. If the starting value is x = 2
1
2 8x n for the iteration, the value of x that is
x = 0.5 obtained from the Newton- to be used in the next step is
0
Raphson method. The series
converges to (a) 0.306 (b) 0.739


(a) 1.5 (b) 2 (c) 1.694 (d) 2.306
[GATE-2011-EC]
(c) 1.6 (d) 1.4

)
0
[CS, GATE-2007] 79. What is the value of (1525 to 2
decimal places?
76. The Newton-Raphson iteration
1   R   (a) 4.33 (b) 4.36
x =  x +  can be used to
n+
1
2   n x n   (c) 4.38 (d) 4.30
compute the [ESE-2018 (COMMON PAPER)]
(a) square of R 80. What is the cube root of 1468 to 3


(b) reciprocal of R decimal places?

(c) square root of R (a) 11.340 (b) 11.353

(d) logarithm of R (c) 11.365 (d) 11.362

[GATE-2008 (CS)] [ESE 2018 (COMMON PAPER)]

NEWTON-RAPHSON METHOD NEWTON-RAPSON METHOD

(TYPE-III) (GENERAL QUESTION)

81. The Newton-Raphson method is to be
77. Given a > 0, we wish to calculate its used to find the root of the equation
reciprocal value 1/a by using Newton and f’(x) is the derivative of f. The
Raphson method for f(x) = 0, then for method converges
a = 7 and starting with x = 0.2 , the (a) always
0
first two iterations will be
(b) only if f is a polynomial
(a) 0,11, 0.1299 (b) 0.12, 0.1392
(c) only if ( ) 0f x 
0
(c) 0.12, 0.1416 (d) 0.13, 0.1428
(d) none of the above
[GATE-2005-CE]
[GATE-1999 (CS)]



221

Numerical Methods

82. A piecewise linear function f(x) is   10 − 0.8     10 0  
plotted using thick solid lines in the (a)  0 − 0.6  (b)  0 10 
figure (the plot is drawn to scale).      
 0 − 0.8   10 0 
(c)     (d)    
  10 − 0.6    10 − 10 



[GATE-2011 EE]

f
84. The function ( ) x = e − x 1 is to be
solved using Newton-Raphson
method. If the initial value of x is
taken as 1.0, then the absolute error

nd
If we use the Newton-Raphson observed at 2 iteration is _____.
method to find the roots of f(x) = 0 [GATE-2014-EE-SET2]
using x , x and x respectively as
0
2
1
initial guesses, the roots obtained 85. In the Newton-Raphson method, an
would be initial guess of x = is made and
2
0
(a) 1.3, 0.6 and 0.6 respectively the sequence
, , ,......0.75x −
2
x x x 3 2x − 2x + =
4 0
(b) 0.6, 0.6 and 1.3 respectively 0 1 2
Consider the statements
(c) 1.3, 1.3 and 0.6 respectively
0
(i) x =
(d) 1.3, 0.6 and 1.3 respectively 3
[CS, GATE-2003, 2 MARKS] (ii) The method converges to a
solution in finite number of iterations
83. Solution of the variables x and x
2
1
for the following equations is to be Which of the following is TRUE?
obtained by employing the Newton-
Raphson iterative method (a) only (i)

=
x
Equation (i) 10 sin x − 1 0.8 0 (b) only (ii)
2
(c) Both (i) and (ii)
Equation (ii)
2
10x − 10 cos x − 0.6 0 (d) neither (i) nor (ii)
=
x
1
2
2
Assuming the initial values x = 0.0 [GATE-14 (CS-SET2)]
1
and x = 1.0, the Jacobian matrix is 86. In Newton Raphson iterative method,
2
the initial guess value ( x ) is
ini
considered as zero, while finding the



222

Numerical Methods

roots of the equation 1 x  2 and bisection method is
2 6x −
3
2
f ( ) x = − + 4x + 0.5x . The used to find its value. For achieving

correction x , to be added to X in an accuracy of 0.001, the required
ini
minimum number of iterations is
the first iteration is _______.
_____
[GATE-2015-CE-SET II]
[GATE-2017-EE-SESSION I]
87. The quadratic equation TRAPEZOIDAL RULE
x − 4x + = is the solved
2
4 0
nd
numerically starting with the initial 91. A 2 degree polynomial f(x) has
guess x = 3. The Newton-Raphson values of 1, 4 and 15 at x = 0, 1 and
respectively.
2,
The
0
integral
method is applied once to get a new  2
estimate and then the Secant method f ( ) x dx is to be estimated by
0
is applied once using the initial guess applying the Trapezoidal rule to this
and this new estimate. The estimated data. What is the error (defined as
value of the root after the application true value-approximate value) in the
of the Secant method is ______. estimate?
4
2
[GATE-2015-CE-SET I] (a) − (b) −
3 3
BISECTION METHOD
2
88. The bisection method is applied to (c) 0 (d) 3
compute a zero of the function
4
3
2
f ( ) x = x − x − x − 4 in the 92. Using the trapezoidal rule, and
dividing the interval of integration
interval [1, 9]. The method converges into three equal subintervals, the
to a solution after _____ interations. + 1
definite integral  x dx is _______
(a) 1 (b) 3 − 1

(c) 5 (d) 7 [GATE-2014-ME-SET 1]


[CS, GATE-2012, 2 MARKS] SIMPSON’S RULE

89. The real root of the equation 5x- 93. Simpson’s rule for integration gives
2cosx-1=0 (up to two decimal exact result when f(x) is a polynomial
accuracy) is ________ degree


[GATE-2014-ME-SET 2] (a) 1 (b) 2

90. Only one of the real roots of (c) 3 (d) 4
6
f ( ) x = x − − 1 lies in the interval [GATE-1993 (ME)]
x



223

Numerical Methods

1.5
94. The estimate of  dx obtained using 98. The accuracy of Simpson’s rule
0.5 x quadrature for a step size h is

Simpson’s rule with three-point 2 3
O h
O h
function evaluation exceeds the exact (a) ( ) (b) ( )
value by
4
5
O h
O h
(c) ( ) (d) ( )
(a) 0.235 (b) 0.068
[GATE-2003]
(c) 0.024 (d) 0.012
99. Match the correct pairs
[GATE-2012-CE]
Numerical integration Order of Fitting
95. Find the magnitude of the error scheme Polynomial
(correct to two decimal places) in the P. Simpson’s 3/8 Rule 1. First
estimation of following integral using Q. Trapezoidal Rule 2. Second
Simpson’s 1/3 Rule. Take the step R. Simpson’s 1/3 Rule 3. Third
length as 1. ______ (a) P – 2, Q – 1, R – 3
(b) P – 3, Q – 2, R - 1
4
 4 ( x + 10 ) dx (c) P – 1, Q – 2, R – 3
0
(d) P – 3, Q – 1, R – 2
[GATE-2013-CE]
[GATE-2013-ME]

96. For step-size; x = 0.4 the value of 2 x 2
the following integral 100. The integral 1 x  x dx with
x  2 x  1 0 is evaluated analytically
 0.8 (0.2 25x − 200x + 675x − 900x + 400x 5 ) dx as well as numerically using a single
+
4
2
3
0
using Simpson’s 1/3 rule is ____. value of the integral obtained
analytically and J is the approximate
[GATE-2015-CE-SET II] value obtained using the trapezoidal
97. Simpson’s 1/3 rule is used to rule, which of the following
integrate the function statements is correct about their
3 9 relationship?
f ( ) x = x + 2 between x = 0 and
5 5 (a) J > 1

x = 1 using the least number of equal (b) J < 1
sub intervals. The value of integral is

______. (c) J = 1

[GATE-2015-ME-SET 1] (d) Insufficient data to determine the
relationship


[GATE-2015-CE-SET I]



224

Numerical Methods

101. Numerical integration using (a) h (b) h
3
2
trapezoidal rule gives the best result
5
4
for a single variable function, which (c) h (d) h
is
[GATE-2009 (PI)]
(a) linear (b) parabolic −
105. The differential equation dx = 1 x
(c) logarithmic (d) hyperbolic dt 
is discretised using Euler’s numerical
[GATE-2016-ME-SET II]
integration method with a time step
102. P(0, 3), Q (0.5, 4) and R (1, 5) are  T  0. What is the maximum

three points on the curve defined by permissible value of T to ensure
f(x). Numerical integration is carried stability of the solution of the
out using both Trapezoidal rule and corresponding discrete time equation?

Simpson’s rule within limits x = 0

and x = 1 for the curve. The (a) 1 (b) /2
difference between the two results (c)  (d) 2
will be
[GATE-2011]
(a) 0 (b) 0.25
106. The ordinary differential equation
(c) 0.5 (d) 1 dx = − 3x + , with x (0) = 1 is to be
2
[GATE-2017 ME SESSION-I] dt
solved using the forward Euler
FORWARD EULER METHOD method. The largest time step that can

du be used to solve the equation without
103. Consider the equation = 3t + 2 1 making the numerical solution
dt unstable is ________.

with u = 0 at t = 0. This is
numerically solved by using the [GATE-2016-EC-SET 2]
forward Euler method with a step
size. t = 2. The absolute error in 107. Variation of water depth (y) in a
the solution in the end of the first gradually varied open channel flow is
time step is _____. given by the first order differential
equation
[GATE-2017]
− 10 ln y
−
104. During the numerical solution of a dy = 1 e 3
−
first order differential equation using dx 250 45e − 3ln y
the Euler (also known as Euler Given initial condition y (x = 0) = 0.8
Cauchy) method step size h, the local m. The depth (in m, up to three
truncation error is of the order of



225

Numerical Methods

decimal places) of flow of 2 exact solution and the solution
downstream section at x = 1m from obtained using a single iteration of
one calculation step of Single Step the second-order Runge-Kutta
Euler Method is _________. method with step-size h = 0.1 is.


[GATE-2018 (CE-MORNING SESSION)] [GATE-2016-EC-SET 3]

BACKWARD EULER METHOD RUNGE-KUTTA METHOD OF

TH
108. The differential equation 4 ORDER
dy = 0.25y is to be solved using 111. Consider an ordinary differential
2
dx dx
the backward (implicit) Euler’s equation = 4t + 4 . If x = x at t =
0
method with the boundary condition dt

y = 1 at x = 0 and with a step size of 0, the increment in x calculated using
1. What would be the value of y at x Runge-Kutta fourth order multistep
method with a step size of t 
= 1? = 0.2
is
(a) 1.33 (b) 1.67
(a) 0.22 (b) 0.44
(c) 2.00 (d) 2.33
(c) 0.66 (d) 0.88
[GATE-2006]
[GATE-2014-ME-SET 4]

MODIFIED EULER METHOD
GAUSS SEIDAL AND
109. Given the differential equation
y = 1 x − y with initial condition y(0) GAUSS JACOBI METHOD

= 0. The value of 112. Gauss-Seidel method is used to solve
the following equations (as per the
y(0, 1) calculated numerically upto given order):
nd
the third place of decimal by the 2
5
order Runge Kutta method with step x + 2x + 3x = ,
3
2
1
size h = 0.1 is 2x + 3x + x = 1,
1
3
2
[GATE-1993 (ME)] 3x + 2x + x = 3
1
3
2
110. Consider the first order initial value Assuming initial guess as
−
problem y = + 2x x , y(0) = 1, x = x = x = 0 , the value of x after
2
1
y
3
1
2
3
)
x
(0    with exact solutions the first iteration is _____
x
y ( ) x = x + 2 e . For x = 0.1, the [GATE-2016]
percentage difference between the

226

Numerical Methods

FORWARD DIFFERENCE OPERATOR (a) 1 (b) 2

113. The values of a function f(x) are (c) 3 (d) 4
tabulated below
[GATE-2017 (IN)]
x 0 1 2 3
f(x) 1 2 1 10
Using Newton’s forward difference
formula, the cubic polynomial that
can be fitted to the above data, is


3
2
2
(a) 2x + 7x − 6x +
(b) 2x − 7x + 6x −
2
3
2
3
2
(c) x − 7x − 6x + 1
3
2
(d) 2x − 7x + 6x + 1
[GATE-2004]

114. The following table lists an nth order
polynomial
+
+
n
f ( ) x = a x + a x n− 1 + ..... a x a 0
n−
1
n
1
and the forward differences
evaluated at equally spaced values of
x. The order of the polynomial is
x f(x) f   2 f  3 f

-
-0.4 1.7648 0.089 -0.03
0.2965
- -
-0.3 1.4683 0.059
0.2075 0.0228
- -
-0.2 1.2608 0.0362
0.1485 0.0156
- -
-0.1 1.1123 0.0206
0.1123 0.0084
- -
0 1 0.0122
0.0917 0.0012
-
0.1 0.9083 0.011 0.006
0.0795

0.2 0.8288 0.0685 0.017 0.0132




227

8















Laplace




















Transforms

Laplace Transforms
)
2
1. L (sin t = (c) 6 (d) 6
(s + ) 2 4 (s + ) 3 4
1 2
(a) (b)
( sinhat =
( S S + ) 1 ( S S + ) 4 6. L t )
2
2
2 1 (a) 2as (b) 2as
(c) (d) 2 2 ) 2 2 2 ) 2
2
2
S 2 ( S + ) 4 ( S S + ) 4 ( s + a ( s − a

2. ( L t − 1/2 ) = (c) 2as (d) 2as
2
2
( s − a 2 ) ( s + a 2 )
1 
(a) (b) 2 )
( cosat =
S S 7. L t

  2s ( s − a 2 )
2
(c) (d) (a)
2 S s 2 2 3
( s + a )

3. L t 1/2
( ) =
2
2s ( s − 3a 2 )
(b)
 1  ( s + a 2 ) 3
2
(a) (b)
S 3/2 2 S 3/2
2
2s ( s − 3a 3 )
  (c)
(c) (d) 2 a 2 ) 3
S 3/2 S 1/2 ( s +

 t 2 2 )
 , where  0 t  k 2s ( s − 6a

f
4. If ( ) x =  k then (d) 3
 ( s + a 2 )
2
 1 whent  k

( L f t   sin2t  
( )) =
8. L  t    =
−
+
1 e ks 1 e ks  
(a) (b) − −
1
1
ks 2 ks 2 (a) Cos s (b) Cot s
k − e − ks e − ks − 1 − 1 s − 1 s
(c) (d) (c) Cot (d) Tan
s 2 s 2 2 2

3 −
5. ( L t e 2t ) =

2 6
(a) (b)
(s − ) 2 4 (s − ) 2 4





229

Laplace Transforms
  1 e t   14. The solution of y + 11 4y = 12t ,
−
9. L     = 1
9
0 y
 t  y ( ) 0 = , ( ) 0 = is
 s − 1   s + 1  (a) 3t + 3sin2t (b) 3t − 3sin2t
(a) log       (b) log      
 s   s  (c) 3t − 6sin2t (d) 3t + 3cos2t

 s   s − 1   S 
(c) log       (d) log       15. L − 1     =
2
 s − 1   s + 1    S + 4 


10. The value of   sint dt = (a) Cost (b) Sint
0 t (c) Cos2t (d) Sin2t
 
(a) (b)  1 
4 2 16. L  − 1   =

   (S + 1 )(S − ) 2  
(c) (d) 0
3 e 2t e t −
(a) (b)
−
1

11. The value of  e sint dt = 3 3
0 t e − e t − e + e t −
2t
2t
  (c) (d)
(a) (b) 3 3
4 2
1 
−

 17. L − 1    Cot s =

(c) (d) 0
3 sint 1 cost
−
−
3t
12. The value of   e t sint dt = (a) t (b) t
0
−
sin2t 1 cos2t
3 3 (c) (d)
(a) (b) t t
25 50
   
6 3 − 1   s − 2  
(c) (d) 3 18. L  log      =
152 5    s − 4   
13. The solution of y + 11 25y = 10cos5t e − e 4t e − e 2t
4t
2t
(a) (b)
1
2 y
y
0
, ( ) 0 = , ( ) 0 = is t t
2t
(a) 2cos5t t+ sin5t e + e 4t e − 4t − e − 2t
(c) (d)
+
(b) 2sin5t t cos5t t t
(c) 3cos5t t− sin5t
(d) 3cos5t t+ sin5t




230

Laplace Transforms

19. Let ‘y’ be the solution of If ( ) S = 2 then Lt f t =
( )
2
d y = 6cos2t , ( ) 0 = 3, ( ) 0 = 1 23. F S (1 S ) x→ 0
+
1
y
y
dt 2 __________ where ( L f t F ( ) S
( )) =
. Let laplace transform of y be F(S),
the value of F(1) = (a) 0 (b) 1
17 13 (c) 2 (d) 
(a) (b)


5 5  sin , t 0 t 

f
24. If ( ) t =  then
26 11  0,   t  2 
(c) (d)
( )) =
5 5 ( L f t
20. The solution of initial value problem 1
1
y + 2y + 10y = 6 ( ) t , ( ) 0 = , (a) 1 e −  s ( s + 2 ) 1
11
0
y
−
f
0
y 1 ( ) 0 = where ( ) t is direct delta 1
function is (b) − 
+
1 e s ( s + 2 ) 1
t
t
e
e
(a) 2 sin3t (b) 6 sin3t
+
1 e −  s
(c) 2e t − sin3t (d) 6e t − sin3t (c) s + 1
2
−
21. Consider the initial value problems (d) 1 e −  s
2
=
1
y + 2y + y  ( ) t with s + 1
11
dy LAPLACE TRANSFORM
y ( ) t = − 2and = 0. The
t=
0
dt t= 0 USING MAIN DEFINITION
dy  1
numerical value of is  , where  0 t  k

dt t= 0 + 25. If ( ) t =  k then
f


(a) -2 (b) -1  1 when t  k
( L f t
( )) =
(c) 0 (d) 1
−
1 e − ks
5s + 2 23s + 6 (a)
22. If ( ) f = L F ( ) S = ks 2
( s s + 2 2s + ) 2 − ks
+
( )
then Lt f t = (b) 1 e
x→ 0 ks 2
(a) 3 (b) 5 k − e − ks
(c) 2
17 s
(c) (d) 0 −
2 1 + (k − ) 1 e ks
(d)
ks


231

Laplace Transforms

2
−
−
+
26. Laplace transform of (a bt ) where a b e s (a b )
(a) (b)
‘a’ and ‘b’ are constants is given by: S S
−
2
(a) (a bs+ ) (c) e − as − e − bs (d) e ( s a b )
S S
1
(b) [GATE-2015-EC-SET 2]
(a bs+ ) 2

29. If x(t) is as shown in the figure, its
a 2 2ab 2b 2 Laplace transform is
(c) + +
s s 2 s 3

a 2 2ab b 2
(d) + +
s s 2 s 3

27. The Laplace Transform of the

following function is
 sint for 0 t 



f ( ) t = 
 0 for t  (a) 2e + 5s + 2e − 5s


s 2
1
(a) for alls > 0 + 5s − 5s
+
+
1 s 2 (b) 2e − 4 2e
s 2
1
(b) for alls <  + −
+
+
1 s 2 2e 5s − 2 2e 5s
(c)
s 2
+
1 e − ks
(c) for alls > 0 + 5s − 5s
+
+
1 s 2 2e + 4 2e
(d)
s 2
e − ks
(d) for alls > 0
+
1 s 2 [EEE-2018 (EC)]
[GATE-2002]
28. The bilateral Laplace transform of a
 1if a t b



function ( ) t =  is
f
 0 otherwise







232

Laplace Transforms

PROPERTIES OF LAPLACE 34. If F(s) is the Laplace transform of the
TRANSFORM function f(t) then Laplace transform

)
2
30 L (sin t = of 0  t f ( )d  is
1
( )
1 2 (a) F s
(a) (b) s
( s s + ) 1 ( s s + ) 4
2
2
1
( )
−
2 4 s 2 (b) F s − f ( ) 0
(c) (d) s
2
2
s 2 ( s + ) 4 2s ( s + ) 4
( )
(c) sF s − f ( ) 0
 sin2t 
31. L       =
 t  (d)  F ( ) s ds [GATE-2007 (ME)]
−
−
1
1
(a) cos s (b) cot s 35. The unilateral Laplace transform of
(c) cot − 1 s (d) tan − 1 s f(t) is 1 . The unilateral
2 2 s + 2 s + 1
 1 e t  Laplace transform of t f(t) is
−
32. L       = s
 t  (a) −
( s + 2 s + ) 1 2
 s − 1   s  
+
(a) log       (b) log       2s + 1
 s   s  (b) − 2
 s   s − 1  ( s + 2 s + ) 1
(c) log       (d) log      
 s − 1   s + 1  s
(c) 2
33. The Laplace transform of e t  cos t  ( s + 2 s + ) 1
is equal to … 2s + 1
(d)
−
s  ( s + 2 s + ) 1 2
(a)
(s  − ) + 2  2
[GATE-2012 (EC, EE, IN)]
+
s 
(b) 36. The Laplace transform of
(s  − ) + 2  2
f ( ) 2t = / t  is s − 3/ 2 . The Laplace
1
g
(c) transform of ( ) t = 1/ t  is
−
(s  ) 2
(a) 3s − 5/ 2 /2 (b) s − 1/ 2
(d) None (c) s 1/ 2 (d) s 3/ 2

[GATE-1997-EC] [GATE-2014-EE-SET 2]




233

Laplace Transforms

5 i t
37. The Laplace transform of e where 41. Evaluate 0   sint dt
i = − 1 is t

s − 5i s + 5i 
(a) (b) (a)  (b)
s − 25 s + 25 2
2
2
s + 5i s − 5i  
(c) (d) (c) (d)
s − 25 s + 25 4 8
2
2
[GATE 2015 – ME-SET 2] [GATE-2007]


38. F(S) is the Laplace transform of the 42. The value of the integral
f t =
t −
2
function ( ) 2t e  sin2 t  
2  −       dt is equal to
F(1) is _______ (correct to two  t  
decimal places). (a) 0 (b) 0.5


[GATE 2018 (ME-MORNING (c) 1 (d) 2
SESSION)] [GATE-2016 (EE-SET 2)]

APPLICATION OF LAPLACE INVERSE LAPLACE TRANSFORM
TRANSFORM IN REAL INTEGRALS  


39. The value of the integral 43. L  − 1 1   =
  (s + 1 )(s − ) 2  
(4 t 
 − 12cos (2 t  ) sin 4 t  ) dt is ____ e 2t e t −


(a)
3 (b) 3
[2015 EC-SET 2] e − e t − e + e t −
2t
2t
(c) (d)
t −

40. The value of  e sint dt = 3 3
0 t
44. The inverse Laplace transform of
 
(a) (b) s + 9
4 2 s + 2 6s + 13 is

(c) (d) 0
3 (a) cos2t + 9sin2t

(b) e − 3t cos2t − 3e − 3t sin2t

(c) e − 3t sin2t + 3e − 3t cos2t

(d) e − 3t cos2t + 3e − 3t sin2t

[GATE-1995]




234

Laplace Transforms

45. The inverse Laplace transform of the 2 4
s + 5 (a) s + 1 (b) s + 1
function is ….
(s + 1 )(s + ) 3
4 2
(c) (d)
4
2
t −
t −
(a) 2e − e − 3t (b) 2e + e − 3t s + 1 s + 1
t −
t −
(c) e − 2e − 3t (d) e + 2e − 3t [GATE-2013-ME]
LAPLACE TRANSFORM OF
[GATE-1996-EC]
PERIODIC FUNCTIONS
46. The Laplace transform of a function

1 49. The Laplace Transform of the
f(t) is . The function f(t) is periodic function f(t) described by the
s 2 (s + ) 1 curve below


+
+
t −
t −
(a) t − 1 e (b) t + 1 e  sin , t if (2n − ) 1  t 2n   (n = 1,2,3,... )

t −

−
+
t
(c) 1 e (d) 2t + e f ( ) t =  0 otherwise

APPLICATION OF LAPLACE
TRANSFORM IN DIFFERENTIAL
EQUATION

47. Solve the initial value problem [GATE-1993 (ME)]
2
d y − 4 dy + 3y = 0 with y = 3 and
dx 2 dx INITIAL & FINAL VALUE THEOREM
dy = 7 at x = 0 using the Laplace 5s + 2 23s + 6

dx 50. If L ( ) f = F ( ) s = 2
transform technique. ( s s + 2s + ) 2
( )
then lim f t = _____ .
[GATE-1997-ME] t→ 


48. The function f(t) satisfies the 2
If F ( ) s = then
2
d f s (1 s+ )
differential equation + f = 0
dt 2 lim f ( ) t = _____ where L(f(t)) =
and the auxiliary conditions, f(0) = 0, t→ 
df ( ) 0 = . The Laplace transform of F(s).
4
dt If  f c ( 2 s + ) 1
L
f(t) is given by ( ) = s + 2 2s + 1 then f(0*)
f
and ( )  given by ….







235

Laplace Transforms

(a) 0, 2 respectively s + 1
2
(a)
(b) 2, 0 respectively s + 3

(c) 0, 1 respectively 1
(b)
(d) 2/5, 0 respectively s + 3

[GATE-1995-EC] (c) s + 2 1 + s + 2
2
 (s + 3 )(s + ) 2 s + 1
( ) =
L
51. If  f t then the value
s +  2 (d) None of these
2
of lim f ( ) .............t =
t→  [GATE-2000-EC]

(a) can not be determined 54. A delayed unit step function is
 0, for t  a

−
(b) zero defined as ( u t a = )  1, for t  a .


(c) unity Its Laplace transform is

(d) infinite [GATE-1998-EC] (a) a e − as (b) e − as

 3s + 1  s
52. Given L  − 1   . If e as e as
  s + 4s + (K − ) 3 s  (c) s (d) a
3
2
lim f ( ) 1t = , then the value of K is   
t→    
55. The integral    t −    6sin ( ) t dt
−  6 
(a) 1 (b) 2
evaluates to

(c) 3 (d) 4 (a) 6 (b) 3

[GATE-2010-EE] (c) 1.5 (d) 0

s + 2 [GATE-10 (IN)]
( ) =
L
53. If  f t ,
s + 1 56. Given two continuous time signal
2
t −
( ) e=
x ( ) t = e and y t − 2t which
s + 1
2
( ) =
L g t exists for t > 0 then the convolution
(s + 3 )(s + ) 2
z(t) = x(t) * y(t) is
( ) ( g t T dT−
h ( ) t =  t f T ) then t − − 2t − 2t
0 (a) e − e (b) e
L{h(t)} is ….. t − t − −
(c) e (d) e + e 3t

[GATE-2011]






236

Laplace Transforms

57. The solution of the differential
equation, for t > 0,

'' y ( ) 2 't + y ( ) t + y ( ) 0t = with

initial conditions y(0) = 0 and y’(0) =
1, is? (u(t) denotes the unit step

function),

( )
t −
(a) te u t
t −
u
(b) (e − te t − ) ( ) t

−
u
(c) ( e + t − te t − ) ( ) t
( )
t −
(d) e u t [GATE-2016]

58. The Laplace transform of

n −
t 
f ( ) t = t e u ( ) t is

(n + ) 1 ! ! n
(a) n+ 1 (b) n
+
+
(s  ) (s  )

(n − ) 1 ! ! n
(c) n+ (d) n+ 1
(s  ) 1 (s  )
+
+

[ESE-2018 (EE)]





























237

9















Fourier




















Series

Fourier Series


1. A function with a period  is shown (d)  4 (1 sin n

)
+
below. The Fourier series for the n= 1 2 n 2
function is given by
[GATE-2003]
3. The period of the signal
  
x t = ( ) 8sin 0.8 t  +       is
 4 


(a) 0.4 s (b) 0.8 s

(c) 1.25 s (d) 2.5 s

[GATE-2010]
1  2   n  
(a) ( ) x = +  sin     cosnx
f
2 n= 1 n  2  4. The Fourier series of the function,

f x = −   x  0
( ) 0 ,

(b) ( ) x =  2 sin    n      cosnx =  x , 0 x  −  
f
n= 1 n  2 

1  2   n   in the interval  − ,  is
(c) ( ) x = +  sin     sin nx
f
2 n= 1 n  2   2 cos x cos3x 


f ( ) x = 4   +   1 2 + 3 2 + .... +



(d) ( ) x =  2 sin    n      sin nx  
f
n= 1 n  2   sin x sin 2x sin3x 

  + + + ......

[GATE-2000 (CE)]   1 2 3  
2. The Fourier series expansion of a The convergence of the above
symmetric and even function, ( ) x Fourier series at x = 0 gives
f
where  1  2
(a)  =
 1 2 / , −   0 n= 1 n 2 6

+

x
x

f ( ) x = 
 1 2 / , 0    − n− 1  2

−
x 
x

(b)  ( ) 1 2 = 12
n

)
+
(a)  4 2 (1 cosn n= 1
2
n= 1 n  1  2
(c)  (2n − ) 1 = 8

(b)  4 (1 cosn− ) n= 1
n= 1 2 n 2  ( ) 1− n+ 1 

(c)  4 2 (1 sin n− ) (d)  (2n − ) 1 = 4
n=
1
2
n= 1 n
[GATE-2016-CE-SET-2; 1 MARK]

239

Fourier Series

) 0, be a function
5. Let  : 0, → ) Select the correct answer using the
g
defined by ( ) x = x −   x , where [x] codes given below:
g
represents the integer part of x. (That (a) 1, 2 and 3 (b) 1 and 3 only
is, it is the largest integer which is

less than or equal to x). The value of (c) 1 and 2 only (d) 2 and 3 only
the constant term in the Fourier series [ESE-2017 (EE)]
expansion of g(x) is ___________
8. Given the Fourier series in ( − , )
[GATE-2014-EE-SET-1]
for ( ) x = x cos x, the value of a will
f
0
6. A periodic signal x(t) has a be
trigonometric Fourier series
expansion (a) − 2  (b) 0
2
3
0 
x ( ) t = a +  (a n cosn t b n sin n + 0 ) t 2

−
0
n= 1 ( ) 1 2n
(c) 2 (d) n −
2
If ( ) t = x ( ) t − = − ( x t − / 0 ) . We 1
x
can conclude that [ESE-2017 (EE)]
(a) a are zero for all n and b are 9. The Fourier series expansion of the
n
n
f
x
zero for n even saw-toothed waveform ( ) x = in
( − , ) of period 2 gives the series,
(b) a are zero for all n and b are 1 1 1
n
n
=
zero for n odd 1− + − + ....... ?
3 5
4
(c) a are zero for n even and b are   2
n
n
zero for n odd (a) 2 (b) 4
(d) a are zero for n odd and b are  2 
n
n
zero for n even (c) 16 (d) 4

[GATE-2017 EC SESSION-1] [ESE-2017 (EE)]

7. Fourier series of any periodic signal 10. For the function
x(t) can be obtained if
 − 2, −   0

x

f ( ) x = 

T

I.  x ( ) t dt    2, 0  x 
0
The value of a in the Fourier series
II. Finite number of discontinuities n
within finite time interval t expansion of f(x) is

III. Infinite number of discontinuities


240

Fourier Series

(a) 2 (b) 4 The same function x(t) can also be
considered as a periodic function with
(c) 0 (d) -2
period T = 40. Let b be the Fourier
1
k
[ECE 2017 (COMMON PAPER)] series coefficient when period is
1
 −  ,if −   x  0 taken as T . If   a = 16, then
k=−
k

11. Let ( ) x =  
f
  , if 0  x    k=− b is equal to

k
be a periodic function of period 2 . (a) 256 (b) 64
The coefficient of sin 5x in the (c) 16 (d) 4
Fourier series expansion of f(x) in the
interval  − ,  is [GATE-2018 (EC)]


4 5 14. The Fourier series of the function,
(a) (b)
( ) 0,   
5 4 f x = − x 0
4 3 =  x , 0 x  −   in the interval
(c) (d)
3 4
 − ,  is
[ESE-2018 (COMMON PAPER)]

 2 cos x cos3x  

12. The Fourier cosine series for an even f ( ) x = 4 +     1 2 + 3 2 + ........   
function f(x) is given by  sin x sin2x sin3x 

+   + + + ...... .


f ( ) x = a + 0  a n cosn ( ) x   1 2 3  
n= 1 The convergence of the above Fourier

The value of the coefficient a for the series at x = 0 gives
2

function ( ) x = cos 2 ( ) x in 0, is  1  2
f
(a)  2 =
(a) -0.5 (b) 0.0 n= 1n 6


(c) 0.5 (d) 1.0  ( ) 1− n− 1  2
(b)  2 =
[GATE-2018 (ME-Afternoon Session)] n= 1 n 12


13. Let x(t) be a periodic function with (c)  1 =  2
period T = 10. The Fourier series n= 1 (2n − ) 1 8
coefficients for this series are denoted
by a , that is  ( ) 1− n− 1 
k
(d)  1 = 4
1 2n −
n=

x ( ) t =  a e jk 2 t [GATE-2016-CE-SET 2]
T
k
k=−


241

Fourier Series

) 0, be a function
15. Let g  : 0, → ) 18. The Fourier cosine series for an even
defined by ( ) x = x −   x , where [x] function f(x) is given by
g

represents the integer part of x. (That f ( ) x = a + 0  a n cosn ( ) x . The
is, it is the largest integer which is n= 1
2
less than or equal to x). The value of value of the coefficient a for the
( )

f x =
2
the constant term in the Fourier series function ( ) cos x in 0, is
expansion of g(x). is _______
(a) -0.5 (b) 0.0
[GATE-2014-EE-SET 1]
(c) 0.5 (d) 1.0
16. Fourier series of any periodic signa
x(t) can be obtained if [GATE-2018 (ME-AFTERNOON
SESSION)]
T
1.  x ( ) t dt  
0 19. In the Fourier series expansion of
2
f ( ) x = x in  − ,  the sum of
2. Finite number of
discontinuities within finite time absolute values of the Fourier
interval t coefficients of f is _____.


3. Infinite number of  2  2
discontinuities (a) (b)
6 3
Select the correct answer using the
codes given below: 2 2
2
(c) (d) 
(a) 1, 2 and 3 (b) 1 and 3 only 3

(c) 1 and 2 only (d) 2 and 3 only 20. The Fourier series of the periodic
function
[ESE 2017 (EE)]

−
f ( ) x = x , 1 x  1, ( f x + ) 2 = f ( ) x
17. The Fourier series expansion of the 1  cos (2n − ) 1  x
( )
x
saw-toothed waveform f x = in is given by 2 −  4 2 2 .
( − ,  ) of period 2 gives the n= 1 (2n − ) 1 
1 1 1 Using the above, the sum of
series, 1− + − + ...... ? 1 1
=
3 5 4 1+ + + ..... is ______
3 2 5 2
  2
(a) (b)  2 3 2
2 4 (a) (b)
4 8
 2 
(c) (d) 2 2
2 4 (c)  (d) 
8 2
[ESE 2017 (EE)]



242

Fourier Series

21. The values of the fourier coefficient
A of the series
0


 0  A n cosnx B n sinnx of a
A +
+
n= 1
function ( ) x = x with period 2
2
f
defined over an interval 0 x  2

is _______.

4 2 2 2
(a) (b)
3 3


 2  2
(c) (d)
3 6


























































243

ENGINEERING MATHEMTICS WORKBBOK