**Unformatted text preview: **Department of Mechanical Engineering MCQ for Regulations 2017 T MA8353 Transforms
and Partial
Differential
Equations which implies T = c’ ekt ′ = k T C
O
M Therefore, u(x,t) = cc’ e(1+6k)x ekt
Now, u(x,0) = 10 e-x = cc’e(1+6k)x
Therefore, cc’ = 10 and k = -1⁄3
Therefore, u(x,t) = 10 e-x e-t/3. 2. Find the solution of
= 36
+ 10u if
(t = 0) = 3e
using the method of
separation of variables.
a)
e
e
b) 3e e
c) e e
d) 3e e
∂u ∂x ∂u −2x ∂x −3 −2x −t/3 2 x 3 −t/3 2x −t/3 −x −t/3 ∂t T. Mechanical
Engineering - Third
Semester ∂u O 2 SP Answer: a
Explanation: u(x, t) = X(x)T (t)
Substituting in the given equation,
′ ′ X T = 36T X + 10XT which implies X ′ = k LO
G UNIT I PARTIAL
DIFFERENTIAL
EQUATIONS X 17 .B TOPIC 1.1 FORMATION OF
PARTIAL DIFFERENTIAL
EQUATIONS - SINGULAR
INTEGRALS 1. Solve
= 6
+ u using the method of
separation of variables if u(x,0) = 10 e-x.
a) 10 e-x e-t/3
b) 10 ex e-t/3
c) 10 ex/3 e-t
d) 10 e-x/3 e-t
∂u ∂x ∂t C SE -R ∂u Answer: a
Explanation: u(x,t) = X(x) T(t)
Substituting in the given equation, X’T = 6
T’X + XT
′ X −X
6X
X = T
T T ′ = T ∂u
∂x (k−10) −2x (t = 0) = 3e = 1 + 6k ′ ′ = c e k−10
36 t kx = cc ke Therefore k = -2 and cc’ =
e
e
Hence, u(x, t) =
−3 −2x 2 −3
2 −t/3 . 3. Solve the partial differential equation
x
+ y
= 0 using method of
3 ∂u 2 ∂u ∂x ∂y separation of variables if u(0, y) = 10 e
a) 10e
b) 10e
c) 10e
d) 10e 5 5
y . 5 2x2 e y e x −5 5 2y2
−5
2
2y −5 e x −5
2x2 5 e y Answer: d
Explanation: u(x, t) = X(x)T (t)
3 ′ 2 ′ x X Y + y Y X = 0
X ′ X
Y
Y = ′ = k
3 x −k
2 y k which implies X = ce which implies Y = c e k ′ u(x, t) = cc e
u(0, y) = 10e y ′ 2
2x k
y k 2x2
5 = k kx = ce which implies T 36 ′ ′ X X e y ′ = cc e k
y which implies X = ce(1+6k)x Downloaded From: 1 Department of Mechanical Engineering MCQ for Regulations 2017 Therefore k = 5 and cc’ = 10
Hence, u(x, t) = 10e e .
−5 5 2
2x y the product of two functions which depend on
one variable only.
a) True
b) False ∂u ∂u ∂x ∂y separation of variables if u(0, y) = 9e
a) 9e e
b) 9e e
c) 9e
e
d) 9e
e
17
5 13
5 x −5y x −5y −17
5 x −5y x −5y −13
5 Answer: a
Explanation: If we have a function u(x,t),
then the function u depends on both x and t.
For using the variable separable method we
assume that it can be written as u(x,t) =
X(x).T(t) where X depends only on x and T
depends only on t. . T. −5y C
O
M 4. Solve the differential equation
5
+ 3
= 2u using the method of Answer: a
Explanation: u(x, t) = X(x)T (t)
′ ′ which implies X = ce
= 2 − k/3 which implies Y
Therefore u(x, t) = cc e e
′ X Y k = k
5 x 5 ′ k ′ ′ 2−k = e 3 y 2−k x 3 3 y y −5y 9e Hence cc’ = 9 and k = 17
Therefore, u(x, t) = 9 e 17
5 x LO
G u(0, y) = cc 5 2−k ′ = c e Y SP X O 5X Y + 3Y X = 2XY −5y e . 5. Solve the differential equation
x
+ y
= u using the method of
2 ∂u 2 ∂u ∂x ∂y 2 separation of variables if u(0, y) = e .
a) e e
b) e e
c) e e
d) e e
y −3 2 y 2 y x 2 −3 y x 2 17 3 x .B x 3 y Answer: c
Explanation: u(x, t) = X(x)T (t)
2 ′ 2 ′ -R x X Y + y Y X = XY
−k X = ce
′ C SE Y = c e x k−1
y ′ −k u(x, t) = cc e u(0, y) = e 2
y x e 7. While solving a partial differential
equation using a variable separable method,
we equate the ratio to a constant which?
a) can be positive or negative integer or zero
b) can be positive or negative rational number
or zero
c) must be a positive integer
d) must be a negative integer
Answer: b
Explanation: The constant can be any
rational number. For example, we use a
positive rational number to solve a 1Dimensional wave equation, we use a
negative rational number to solve 1Dimensional heat equation, 0 when we have
steady state. The choice of constant depends
on the nature of the given problem.
8. When solving a 1-Dimensional wave
equation using variable separable method, we
get the solution if _____________
a) k is positive
b) k is negative
c) k is 0
d) k can be anything k−1
y ′ = cc e k−1
y k = 3 and cc’ = 1
Therefore u(x, t) = e 2 −3
x e y . 6. While solving a partial differential
equation using a variable separable method,
we assume that the function can be written as Downloaded From: Answer: b
Explanation: Since the given problem is 1Dimensional wave equation, the solution
should be periodic in nature. If k is a positive
number, then the solution comes out to be (c7
epx⁄c+e-px⁄cc8)(c7 ept+e-ptc8) and if k is
positive the solution comes out to be 2 Department of Mechanical Engineering MCQ for Regulations 2017 Answer: a
Explanation: The calculus of variations is a
type of analysis in the field of mathematics
(branch of calculus) which is used to find
maxima and minima of definite integrals. C
O
M (ccos(px/c) + c’sin(px/c))(c’’cospt + c’’’sinpt).
Now, since it should be periodic, the solution
is (ccos(px/c) + c’sin(px/c))(c’’cospt +
c’’’sinpt).
9. When solving a 1-Dimensional heat
equation using a variable separable method,
we get the solution if ______________
a) k is positive
b) k is negative
c) k is 0
d) k can be anything c’’’sinpx) (c e -c2 p2 t T. O Answer: a
Explanation: Partial derivatives are indicated
by the symbol ∂. This was first used in
mathematics by Marquis de Condorcet who
used it for partial differences. SP Answer: b
Explanation: Since this is a heat equation,
the solution must be a transient solution, that
is it should decay as time increases. This
happens only when k is negative and the
solution comes out to be (c’’cospx + 2. The symbol used for partial derivatives, ∂,
was first used in mathematics by Marquis de
Condorcet.
a) True
b) False 3. What is the order of the equation,
xy (
) + yx +
= 0?
a) Third Order
b) Second Order
c) First Order
d) Zero Order
3 ). ∂y 2 LO
G 10. While solving any partial differentiation
equation using a variable separable method
which is of order 1 or 2, we use the formula
of fourier series to find the coefficients at last.
a) True
b) False 17 .B Answer: a
Explanation: After using the boundary
conditions, when we are left with only one
constant and one boundary condition, then we
use Fourier series coefficient formula to find
the constant. -R TOPIC 1.2 SOLUTIONS OF
STANDARD TYPES OF FIRST
ORDER PARTIAL
DIFFERENTIAL EQUATIONS SE C 2 ∂x 1. First order partial differential equations
arise in the calculus of variations.
a) True
b) False Downloaded From: ∂y ∂x Answer: c
Explanation: The equation having only first
derivative, i.e.,
are said to be first order
differential equation. Since the given equation
satisfies this condition, it is of first order.
∂y ∂x 4. In the equation, y= x2+c,c is known as the
parameter and x and y are known as the main
variables.
a) True
b) False
Answer: a
Explanation: Given: y= x2+c, where c is
known as an arbitrary constant. It is also
referred to as the parameter to differentiate it
from the main variables x and y.
5. Which of the following is one of the
criterions for linearity of an equation?
a) The dependent variable and its derivatives
should be of second order 3 Department of Mechanical Engineering MCQ for Regulations 2017 Answer: d
Explanation: Lagrange’s linear equation
contains only the first-order partial
derivatives which appear only with first
power; hence the equation is of first-order
and first-degree. Answer: d
Explanation: The two criterions for linearity
of an equation are: 9. Find the general solution of the linear
partial differential equation, yzp+zxq=xy.
a) φ(x2-y2 – z2 )=0
b) φ(x2-y2, y2-z2 )=0
c) φ(x2-y2, y2-x2 )=0
d) φ(x2-z2, z2-x2 )=0 C
O
M b) The dependent variable and its derivatives
should not be of same order
c) Each coefficient does not depend on the
independent variable
d) Each coefficient depends only on the
independent variable SP O Answer: b
Explanation: Given: yzp+zxq=xy
Here, the subsidiary equations are,
dx
yz = dy = zx dz xy From the first two and last two terms, we get,
respectively,
=
or xdx-ydy=0, and
dx dy yz zx LO
G 6. Which of the following is a type of
Iterative method of solving non-linear
equations?
a) Graphical method
b) Interpolation method
c) Trial and Error methods
d) Direct Analytical methods T. The dependent variable y and its
derivatives are of first degree.
Each coefficient depends only on the
independent variable or ydx-zdy=0,
Integrating these we get two solutions
x2-y2=a , y2-z2=b
Hence, the general solution of the given
equation is,
φ(x2-y2, y2-z2 )=0.
dx .B Answer: b
Explanation: There are 2 types of Iterative
methods, (i) Interpolation methods (or
Bracketing methods) and (ii) Extrapolation
methods (or Open-end methods). -R 17 7. Which of the following is an example for
first order linear partial differential equation?
a) Lagrange’s Partial Differential Equation
b) Clairaut’s Partial Differential Equation
c) One-dimensional Wave Equation
d) One-dimensional Heat Equation C SE Answer: a
Explanation: Equations of the form Pp + Qq
= R , where P, Q and R are functions of x, y,
z, are known as Lagrange’s linear equation.
8. What is the nature of Lagrange’s linear
partial differential equation?
a) First-order, Third-degree
b) Second-order, First-degree
c) First-order, Second-degree
d) First-order, First-degree Downloaded From: ′ zx = dy ′ xy 10. The equation 2 – xy = y , is an
example for Bernoulli’s equation.
a) False
b) True
dy −2 dx Answer: b
Explanation: A first order, first degree
differential equation of the form,
+ P (x). y = Q(x). y , is known as
Bernoulli’s equation.
dy a dx 11. A particular solution for an equation is
derived by eliminating arbitrary constants.
a) True
b) False 4 Department of Mechanical Engineering MCQ for Regulations 2017 Answer: b
Explanation: A particular solution for an
equation is derived by substituting particular
values to the arbitrary constants in the
complete solution. 2. Obtain the general solution for the equation
xp2+px-py+1-y=0 where p= .
a) y=cx+
b) x=cy-(c+1)
c) x=cyd) y=cx+(c+1)
dy dx 1 C
O
M c+1 1 c+1 12. A partial differential equation is one in
which a dependent variable (say ‘y’) depends
on one or more independent variables (say
’x’, ’t’ etc.)
a) False
b) True ∂y ∂y ∂x ∂t ……(1) thus
(1) is in the Clairaut’s equation form
y=px+f(p),
thus general solution is y=cx+
. dy c LO
G ′ x
a 2 .B a) 2 + y a 2 = 1 17 b) y2=-4ax
c) y2=4ax
d) x2=-2ay C SE -R ′ 0 = x − a 2 c → c 2 = hence (1) becomes
y = √ y2=4ax a
x dy dx c a x + a√ x
a a
x → c = √ c+1
1 2 2c+1
c 2 2c+2 Answer: b
Explanation: X
2 Y = y → now p =
1 p = Answer: c
Explanation: Let p = y =
Given equation is of the form y = px + f(p),
whose general solution is y = cx + f(c)
thus the general solution is y = cx +
…………..(1), to find the value of c
we differentiate (1) partially w.r.t ‘c’ i.e 2c 2 2 1. Singular solution for the Clairaut’s
equation y = y x + is given by _______
2 c+1 2 2 a 1 c+1 dx TOPIC 1.3 LAGRANGE'S
LINEAR EQUATION y 1 p+1 3. Find the general solution for the equation
(px-py)(py+x)=2p by reducing into Clairaut’s
form by using the substitution X=x2, Y=y2
where p= .
a) y = x +
b) y = cx –
c) x = cy –
d) x = y +
2 ′ or y = px + O p+1 T. xp(p+1)+1 y = SP Answer: b
Explanation: A differential equation is
divided into two types, ordinary and partial
differential equations.
A partial differential equation is one in which
a dependent variable depends on one or more
independent variables.
Example: F (x, t, y, , , … …) = 0 Answer: a
Explanation: xp2+px-py+1-y=0
xp2+px+1=y(p+1) dY
dy dy
dx = = x 2 dy dY dX dY dX dx → y = 2√ax is the singular solution. = 2x dY and let P = x P i. e p = √ y dx
X
Y P now consider (px-py)(py+x)=2p substituting
the value of p we get
(√ X
Y P √ X– √ Y )(√ (P X−Y )
√Y X
Y (P + 1)√ X = 2√ is in the Clairaut’s form
hence general solution is y 2 a
x dx = 2y ∗ P ∗ 2x or p = 2y dX → P √ Y + √ X) = 2 X
Y P → (P X − Y ) 2 = cx – 2c
c+1 . 4. Find the general solution of the D.E 2y4xy’-log y’=0.
a) y(p) = – 1 +
b) y(p) = – 2 + logp
2c logp p 2 c 2p Downloaded From: 5 Department of Mechanical Engineering −1 d) x(p) = 1
2p c + p p TOPIC 1.4 LINEAR PARTIAL
DIFFERENTIAL EQUATIONS
OF SECOND AND HIGHER
ORDER WITH CONSTANT
COEFFICIENTS OF BOTH
HOMOGENEOUS AND NONHOMOGENEOUS TYPES 2 c + p 1/2 Answer: a
Explanation: Let y’=p and hence given
equation is in Lagrange equation form
i.e 2y=4xp+log p …..(1) differentiating both
sides of the equation
2dy=4xdp+4pdx+ and dy=pdx C
O
M c) x(p) = MCQ for Regulations 2017 dp
p dp p p dx 1 dx 2 −1 dp p dp p 2p dp 2 logp 2p −1 p get 2y = 4p(
y(p) = p 2 −1 c + 2p p logp –1 + 2 2 . c 2p 2 2c
p c 2 .B
p d) x(p) = 2p + p 2 2 p x(p) = ∫ 6p 2 ∫ p dp 2 = p dp + c → x(p) = 2p + c
2 p ….substituting in (1) we get y(p) = 2(2p + c
2 p c) lnx d) lnx xy + x 2 )p − 3p = c lny = c y lny – x = c y 2 = p + 2c
p dy y(x−ylny) = dx x(xlnx−y) –> x2 lnx dy-xy dy=xy dx – y2 lny dx
…….dividing by x2 y2 then
lny dy – 2 y d( lnx
y 1
xy 1 (lnx( 1 dy = ) + d( lny
x dx – xy 1 dy) + 2 −y lnx ∫ d( 17 -R
dp (2) is a linear D.E whose I.F=e
hence its solution is SE C 2 xlnx−ylny 2 Answer: b
Explanation: Let y’=p –> y = 2xp – 3p2 ….
(1) is in the Lagrange equation form
now differentiating we get dy=2xdp+2pdx6pdp and dy=pdx
thus pdx=2xdp+2pdx-6pdp→-pdx=2xdp6pdp –>
+
x– 6 = 0…(2)
dx b) = c xy lny
2 x dx dx) + (lny( xy 1
2 −x dx) + 1
xy ) = 0 on integrating we get 2c
p xlnx+ylny Answer: a
Explanation: ) + logp 5. Find the general solution of the D.E y =
2xy’ – 3(y’)2.
a) y(p) = p +
b) y(p) = p +
c) x(p) = −cp +
1/2 a) 2 c 2p y(x−ylny) x(xlnx−y) 2 −1 2 2c 2 dy dx SP 2 p 2 LO
G ∫ 1. Solution of the differential equation
=
is _____________ T. dp O –> 2pdx=4xdp+4pdx+ –> -2pdx=4pdx+
-2p
= 4x +
→
+
x =
(p≠0)…….this is a linear D.E for the function
x(p)
I.F is e
= e
= p and solution is
x(p) p = ∫ p ∗
dp + c
x(p) =
+
substituting back in (1) we . Sanfoundry Global Education & Learning
Series – Ordinary Differential Equations. Downloaded From: lny ) + ∫ d( y x ) = c…. where c is a constant of
integration.
lnx lny + y x 2. Solution of the differential equation
= e
+ x e
is ______
=
+
+ c
a)
dy 3x−2y 2 −2y dx e 2y e 3 b)
c)
d) e 3x 2 x 3 3y (e 2x 2 3 +x ) + c 6
e 2y (e 3x 3 +x ) + c 6
e 2y 2 = e 3x 3 + 3 Answer: d
Explanation:
dy
dx −2y = e 3 dy
dx 3x (e x + c 3x−2y = e 2 −2y + x e 2 + x ) separating the variable 6 d Department of Mechanical Engineering MCQ for Regulations 2017 e 2 3x 1 3 3 1 3 Answer: c
Explanation: sec2 x tany dx + sec2 y tanx
dy=0
dividing throughout by tan y.tan x we get
dx +
dy = 0……separating the
variable
now integrating we get
sec y
tany sec x
tanx dx + ∫ sec y
tany dy t dt + ∫ p y b) .B √2
1 ( 4x+2y+1 d) √2 dy 1 dt 2 dx dt
dx 2 dy = dt
dx x
→ dy
dx = 1
2 2 2 dx + ∫ 1 dx dy– log x– x = c dy
dx 2 2 x +y = 3xy 2 dy = dx 1+ 2 x +y
3xy = 3 2
y we can x2
y
x dx dv = v + x dx 2 = 1+v
3v separating the variables and integrating –2 = t = 2t dy y = vx → dx dx 1+y x clearly see that it is an homogeneous equation
hence substituting Answer: a
Explanation:
= (4x + 2y + 1)
here we use substitution for
4x + 2y + 1 = t → 4 + 2 1 Explanation: ) = c cot-1(4x+2y+1)=x+c SE C √2 1 dy = ∫ 1+y x Answer: b (4x + 2y + 1) = x + c −1 tan x 6. Solve the differential equation
is _______
a) xp=(x2+2y2)-3
b) x2 p=(x2-2y2)3
c) x4 p=(x2-2y2)-3
d) x6 p=(x2+2y2)3 √2 −1 cot -R c) 1 1+x y – log(1+y) – log x – x = c
(y-x) – log(x(1+y)) = c is the solution. 4x+2y+1 −1 (1+y)−1 ∫ 1 dy– ∫ 17 1 = 1 + x + y + xy dx x dy = ∫ 1+y ∫ 2 2√ 2 1+x dy = y ∫ 4. Solution of the differential equation
= (4x + 2y + 1) is ______
a)
tan
(
) = x + c
dy dy dx = (1 + x) + y(1 + x) = (1 + x)(1 + y) dx 1+y dy = c dp = c is the solution. separating the variables & hence integrating log t + log p = c –>log(tan x)+log(tan y) = c =
log k….since it is an unknown constant
log(tan x .tan y) = log k
(tan x tan y) = k is the solution. dx dy LO
G → ∫ 1 ) = x + c √2 dx xy substituting tan x = t & tan y=p→sec2 x
dx=dt & sec2 y dy=dp
1 ( Answer: a
Explanation: xy 2 2 ∫ tan 4x+2y+1 SP 2 tanx = x + c √2 5. Solution of the differential equation
xy
= 1 + x + y + xy is ______
a) (y-x)-log(x(1+y))=c
b) log(x(1+y))=c
c) (y+x)-log(x)=c
d) (y-x)-log(y(1+x))=c 2 sec x tan −1 2√ 2 3. Solution of the differential equation sec2 x
tany dx + sec2 y tanx dy=0 is _______
a) (sec x. sec y)=k
b) (sec x .tany)=k
c) (tan x. tany)=k
d) (sec x .tan x)+(sec y .tan y)=k t −1 2√ 2 x dt = ∫ dx T. 2y 2 2t +4 O e 1 ∫ C
O
M e2y dy = (e3x+x2)dx…..integrating
∫ e2y dy =∫ (e3x+x2)dx
=
+
+ c. + 4 separating the variable and integrating Downloaded From: dv dx
dt ∫
dx 2 = –3v
2
2 1−2v 2 1+v –v = 3v dv = ∫ 1
x 1−2v dx 3v …….substituting 1- 2v2=t→-4v dv=dt we get
−3
4 log t = log x + log c →
2 −3log( −3
4 x −2y
2 x 2 log(1 − 2v ) = 2 2 ) = 4log cx → log( 2 x −2y
2 x −3 ) = 7 Department of Mechanical Engineering 6 x 4 2 2 = kx 3 (x −2y ) → x 2 p = (x 2 2 3 − 2y ) MCQ for Regulations 2017 2 is the (v +1)x solution where p is constant. 2 → y dx x
y x x +y
x+y = c → k(x 2 x
y x
y 2 x
y UNIT II FOURIER SERIES x dx y y x x = v + x dv = v + tan v dx 1. How many dirichlet’s conditions are there?
a) One
b) Two
c) Three
d) Four 1 1 dv = ∫ tanv x SP separating the variables and integrating we
get
∫ dx log(sin v) = log x + log c
where c is constant. y
x ) = xc is the solution LO
G sin v = xc → sin( TOPIC 2.1 DIRICHLET'S
CONDITIONS O dy y = vx → dy T. Answer: d
Explanation:
=
+ tan
we can
clearly see that it is an homogeneous equation
substituting
dx 8. Particular solution of the differential
=
equation
given y=-1 at x=1.
a) y=x
b) y+x=2
c) y=-x
d) y-x=2
2 2 2 2 y −2xy−x y +2xy−x .B dy dx 17 Answer: c Explanation: dy dx = 2 2 2 2 y −2xy−x y +2xy−x 2
y = -R ……. is a homogeneous equation
thus put y = vx → = v + x – 2
x
2
y
x2 + 2y
x
2y
x –1 −1 2 dv v −2v−1 = dx 2 v +2v−1 separating the variables and integrating we
get SE C dy dx x dv = dx 3 2 v −2v−1
2 v +2v−1 –v = − 2 v +2v−1 ∫ ∫ dv = ∫ 2 (v +1)(v+1) 2 (v +1)(v+1)
2v
2 v +1 − x 2 (v +v +v+1)
2 v +2v−1 2 = − 1
v+1 Answer: c
Explanation: There are three dirichlet’s
conditions. These conditions are certain
conditions that a signal must possess for its
fourier series to converge at all points where
the signal is continuous.
2. What are the Dirichlet’s conditions?
a) Conditions required for fourier series to
diverge
b) Conditions required for fourier series to
converge if continuous
c) Conditions required for fourier series to
converge
d) Conditions required for fourier series to
diverge if continuous
Answer: b
Explanation: Dirichlet’s conditions are
Conditions required for fourier series to
converge. That is there are certain conditions
that a signal must posses for its fourier series
to converge at all points where the signal is
continuous. (v +1)(v+1)
2 v +2v−1 dx 2 2v(v+1)−(v +1) ∫( −1 2 + y ) = (x + y) C
O
M y 2
where k=1/c
at x=1, y=-1 substituting we get 2k=0→k=0
thus the particular solution is y=-x. 7. The solution of differential equation
=
+ tan
is ______
a) cot( ) = xc
b) cos( ) = xc
c) sec ( ) = xc
d) sin( ) = xc
dy = c (v+1) dv = log c– log x ) dv = log c– log x log(v2+1) – log(v+1) + log x = log c –> Downloaded From: 8 Department of Mechanical Engineering MCQ for Regulations 2017 3. What is the first Dirichlet’s condition?
a) Over any period, signal x(t) must be
integrable
b) Multiplication of the signals must be
continuous
c) x(t) should be continuous only
d) A signal can be integrable except break
points d) In majority finite interval, x(t) is of
unbounded variation Answer: a
Explanation: In the case of Dirichlet’s
conditions, the first property leads to the
integration of signal. It states that over any
period, signal x(t) must be integrable.
That is ∫|x(t)|dt<∞. 7. There are maxima and minima not possible
in dirichlet’s conditions.
a) True
b) False C
O
M T. O LO
G Answer: b
Explanation: Dirichlet’s conditions is not
possible in case of discrete signals. That is
these are certain conditions that a signal must
posses for its fourier series to converge at all
points where the signal is continuous only. Answer: b
Explanation: Maxima and minima are
possible if they are infinite number as stated
by the second dirichlet’s condition. In any
finite interval, x(t) is of bounded variation.
That is there are no more than a finite number
of maxima and minima during a single period
of the signal. SP 4. Is dirichlet’s condition possible in case of
discrete signals...

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