Sample GATE Mock Test - Electrical Engineering (EE)
65 Questions MCQ Test GATE Electrical Engineering (EE) 2023 Mock Test Series | Sample GATE Mock Test - Electrical Engineering (EE)
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Identify the correct spelling of the word.
The correct spelling of the word is 'definitive' which means '(of a conclusion or agreement) done or reached decisively and with authority.' Thus option 2 is the correct answer.
This is the place that _______
The preposition 'about' is mandatory here thus option 1 and 4 are eliminated. Option 2 is correct as the tense present perfect continuous fits here. It conveys the meaning that the person usually talked about the place.
She is brave. Her brother is more brave.
Select the most suitable sentence with respect to grammar and usage.
Options 1 and 2 are incorrect as they change the meaning of what is mentioned. Option 3 is incorrect as with 'less' we use the adjective in positive form and not comparative form. The correct word here should be 'brave.' Option 4 is thus the correct answer.
When a four digit number is divided by 65, it leaves a remainder of 29. If the same number is divided by 13, the remainder would be______
Let number is N
N = 65 k +29
∴ Remainder = 3
Complete the following sentence.
I was ___ ___ for the bus and then I ___ sight of Craig passing by.
The word 'here' and 'there' both can be used here but should be followed with 'waiting' as no other word can fit here. The word 'caught' is correct here as 'catch a sight' means 'to see something.' Option 2 thus has the correct combination of words. The word 'cot' means 'a small bed with high barred sides for a baby or very young child.'
4 – digit number greater than 5000 are randomly formed from the digits 0, 2, 3, 5 and 7. The probability of forming a number divisible by 5 when the digits are repeated is ______
For a number to be greater than 5000, d1 should be filled with either 5 or 7
∴ Total numbers formed when the digits are repeated = 2 × 5 × 5 × 5 = 250
total cases = 250 -1 = 249 ( case of 5000 is not included)
Now, For the number to be divisible by 5, unit digit d4 should be either 0 or 5.
∴ Total no. of ways = 2 × 5 × 5 × 2 = 100
favorable cases = 100 - 1=9 ( 5000 is not included))
∴Required Probability = 
It is theoretically possible that bacteria developed on Venus early in its history and that some were carried to Earth by a meteorite. However, strains of bacteria from different planets would probably have substantial differences in protein structure that would persist over time, and no two bacterial strains on Earth are different enough to have arisen on different planets. So, even if bacteria did arrive on Earth from Venus, they must have died out.
The argument is most vulnerable to which of the following criticisms?
The question asks which of the statements given in the options can weaken the argument put by the author that all bacteria from Venus must have died out.
The passage states that since there is a single strain of bacteria which exists on the Earth, they all must be belonging to the Earth or let's say a single planet. But here the author does not take into consideration (as can be argued from his theory) the fact that may be all the bacteria came from Venus and there are none which originally belong to the Earth. So this criticism as mentioned in option 3 makes the argument of the author vulnerable.
Options 1, 2 and 4 are completely irrelevant criticisms as they do not address the main argument. The argument claims that if there were Venusian bacteria on Earth, then they must have died out by now. Whether there are bacteria originally from Earth that have also disappeared from Earth is irrelevant to the question and has no effect on the given argument.
A man sells three articles A, B, C and gains 10% on A, 20% on B and loses 10% on C. He breaks even when combined selling prices of A and C are considered, whereas he gains 5% when combined selling prices of B and C are considered. What is his net loss or gain on the sale of all the articles?
Let a, b and c be the cost prices of the three articles A, B and C.
SP = CP + Profit (or) SP = CP – Loss
⇒ SP of A = 1.1a; SP of B = 1.2b; SP of C = 0.9c
By question,
1.1a + 0.9c = a + c ⇒ 0.1a = 0.1c ⇒ a = c
1.2b + 0.9c = 1.05(b + c) ⇒ 0.15b = 0.15c ⇒ b = c = a
Gain% = {(SP – CP)/CP} × 100
⇒ Net gain on the sale of all the articles =
∴ Net gain on the sale of all the articles = 6.66%
Which of the following inferences can be drawn from the above graph?
Option 1 is false as graph says there is decrease in students qualifying in Physics in 2015 compared to 2014.
Option 2
Let no. of students qualifying in Biology in 2013 be 100
⇒ No. of students qualifying in Biology in 2014 = 100 – 10% of 100 = 90
⇒ No. of students qualifying in Biology in 2015 = 90 + 10% of 100 = 99
∴ The number of students qualifying in Biology in 2015 is less than that in 2013
Option 3 and option 4 are incorrect since no detail is given regarding how many students qualified the subject in 2013.
DRQP is a small square of side a in the corner of a big square ABCD of side A. What is the ratio of the area of the quadrilateral PBRQ to that of the square ABCD, given A/a = 3?
Area of triangle PAB = Area of triangle RCB (By symmetry)
∴ Area of Δ PAB = ½ × PA × AB
= ½ × 2A/3 × A = A2/3
Area of ΔRCB = A2/3
Now, Area of ABCD = Area of DRQP + Area of PAB + Area of RCB + Area of PBRQ
A2 = a2 + A2/3 + A2/3 + Area of PBRQ
As, A/a = 3 ⇒ a = A/3
Area of PBRQ
⇒ Area of PBRQ = 
If the Laplace transform of y(t) is given by Y(s) = L(y(t)) 
then y(0) + y'(0) = _____.
Y(s) = L(y(t)) 
Apply inverse Laplace transform,
Differentiate with respect to ‘t’.
⇒ y(0) + y'(0) = 1
Let f(z) = (x2 + y2) + i2xy and g(z) = 2xy + i(y2 – x2) for z = x + iy ϵ C. Then, in the complex plane C.
Given that f(z) = (x2 + y2) + i 2xy
g(z) = 2xy + i (y2 – x2)
To check analyticity of a function, we need to check CR equations.
ux = vy, uy = - vx
f(z) = (x2 + y2) + i 2xy
u = x2 + y2, v = 2xy
ux = 2x
uy = 2y
vx = 2y
vy = 2x
ux = vy but uy ≠ -vx
Hence, f(z) is not analytic
g(z) = 2xy + i (y2 – x2)
u = 2xy, v = y2 – x2
ux = 2y
uy = 2x
vx = -2x
vy = 2y
ux = vy and uy = -vx
Hence g(z) is analytic.
Consider the following statements P and Q:
(P) : IfM = 
then M is singular.
(Q): Let S be a diagonalizable matrix. If T is a matrix such that S + 5 T = I, then T is diagonalizable.
Which of the above statements hold TRUE?
A matrix is said to be singular, if determinant of that matrix is zero.
= 1 (18 – 12) - 1 (9 – 4) + 1 (3 – 2)
= 6 – 5 + 1 = 2 ≠ 0
M is non singular
(Q) A matrix can be diagonalizable when it has distinct eigen values
S is a diagonalizable matrix. Hence, has distinct eigen values.
Let S be a 3 × 3 matrix and the eigen values of s are λ1, λ2, λ3
Given that, S + 5T = I
From the properties of Eigen values,
(a) If λ1 is an eigen value of matrix A, then -λ1 will be on eigen value of matrix -A.
(b) If λ1 is an eigen value of matrix A, then (λ1 + 1) will be an eigen value of matrix (A + I)
(c) If λ1 is an eigen value of matrix A, then 
will be an eigen value of matrix 
where K is a scalar.
From the above properties, eigen values of T are,
As λ1, λ2, λ3 are distinct values, λ′1,λ′2,λ′3 will be distinct.
Hence, matrix T is diagonalizable
So, only Q is true.
Consider the differential equation
If x = 0 at t = 0 and x = 1 at t = 1, the value of x at t = 2 is
= 0
A.E is, (D2 – 3D + 2) = 0
⇒ (D – 1) (D – 2) = 0
⇒ D = 1, 2
x(t) = C1et + C2e2t
x = 0 at t = 0
⇒ 0 = C1 + C2 ⇒ C1 – C2
x = 1 at t = 1
⇒ 1 = C1e1 + C2e2
⇒ 1 = C1e1 + (-C1) e2
⇒ C1(e1 – e2) = 1
x(2) = C1e2 + C2e4
= e + e2
If for two vectors and , sum 
is perpendicular to the difference 
. The ratio of their magnitude is
A 400 V, 180° conduction mode 3 ϕ bridge inverter has star connected resistive load the RMS value of load current is 15 A. Find the value of per phase resistive load _______ Ω.
R.M.S value of phase voltage
In the circuit I is DC current and capacitors are very large. Using small signal model which of the following is correct?
In DC equivalent circuit capacitor is open
A small signal equivalent circuit is shown in fig.
Put rd in equation (2)
how many clock pulses it will take to come back to same state If the initial state is D2D1D0 = 001
The sequence is 000 to 001 to 011 to 111 to 110 to 100 and back to 001, etc.
A 3 ϕ power system has ZBUS matrix 
if an impedance j0.8 is connected b/w bus-2 and ground. The modified value of Z22 is –
The Nyquist sampling rate for the signal 
is ______ KHz
Multiplication in the domain become convolution in frequency domain therefore individual.
Frequency will add up
f1 = 400 Hz, f2 = 500 Hz
Sampling frequency fs = 2 fnet = 2 × 900 = 1.8 KHz
Three current carry conductors are shown in the figure. the value of 
around the closed curve is
Ampere law: 
for isotropic medium
The open loop transfer function of unity negative feedback control system is given by 
The gain margin of the system is _______ dB
Phase crossover frequency is given as 
G.M = 20 log 164 = 44.3 db
The Lissajous figure obtained on the CRO show in the figure. Find the phase difference between the two waves applied.
It can be observed from the Lissajous figure
y1 = 9 units, y2 = 10 units
An UJT shown in figure has the following parameters η = 0.67, VD = 0.7 V, IV = 3 mA, VV = 1 V, IP = 12 μA, VBB = 20 V. Find the value of VEE so as turn an UJT if RE = 1 KΩ
VEE = η VBB + VD + IP . RE
= 0.67 × 20 + 0.7 + 12 × 10-6 × 1 × 103
= 13.4 + .7 + 0.012
= 14.112
A 10 μF capacitor is initially charged with 1000 μC. At t = 0, the switch K is closed find the voltage drop across the resistor after 100 μs _______V.
Here, 3 capacitor are connected in parallel.
∴ Total capacitance = C1 + C2 + C3 = 4 + 4 + 2 = 10 μF
Now find time constant of the circuit
T = RCeq = 20 × 5 = 100 μsec
The initial voltage across the
With closing of k, the capacitor co will start discharging 
Match List – I with List – II and select the correct answer
A multimer having sensitivity of 2000 Ω/v is used for the measurement of voltage across a circuit having an output resistance 5 kΩ. The open circuit voltage of the circuit is 6 V. Find the percentage error when it is set to its 10 V scale.
Input resistance of voltmeter ZL = 2000 × 10 = 20 kΩ
Output resistance of circuit Z0 = 5 kΩ
Open circuit voltage of circuit under measurement E0 = 6 V
Reading of voltmeter
∴ Percentage error in voltage reading
- 20%
Or 20% low
For power transformer of larger ratings, the tapping’s are located in the middle portion of the winding to
For power transformer of large rating the tapping are located in the middle portion of the winding to reduce the mechanical force affecting the winding during short-circuits.
The number of forward path and the number of non-toughing loop pair for the signal flow graph given in the figure below are respectively
P1 = 1, 3, 5, 8, 10
P2 = 12, 8, 9
P3 = 12, 8, 10
P4 = 11, 9
P5 = 11, 10
P6 = 1, 3, 5, 8, 9
Non touching loop
L1 = 2, 7
L2 = 2, (5, 6)
L3 = (3, 4), 7
A discrete time signal is given as.
The energy of the signal is
Signal energy
The number of chords in the graph of the given circuit will be
No. of branches b = 6
nodes = 4
chords = b – (n – 1) = 3
In the conductor shown below if diameter of each conductor is 4 cm then self GMD is
Self G.M.D of bundle of 4 conductor
= 1.09 (r's3)1/4
S = Distance b/w 2 conductor
G.M.D = 1.09 (0.7788 × 2 × 10-2 × 23)1/4
G.M.D = 0.6496 m
Calculate the magnitude of line current in the circuit shown in figure.
Impedance/ph = 1 + 1j + 7 + 5j = (8 + 6j)
An absolutely integrable signal x(t) is known to have Laplace transform with only one pole at s = 4 then x(t) is
Since the pole is at s = 4 the function is absolutely integrable thus it must contain jω axis
Hence the signal is left sided
A 3 phase full converter delivers a ripple free load current of 10 A with firing angle delay of 30° the input voltage is 3 phase, 400 V, 50 Hz. Find the R.M.S value of source current _____ in A.
R.M.S value of source current
The meter constant of 5 A, 220 V, dc watthour meter is 3275 revolution per kwh. Calculate the speed of the disc at full load. In a test at half load the meter takes 59.5 sec to complete 30 revolutions. Calculate the error of the metre.
Energy consumed in one miute
= 0.01833 kwh
∴ Revolution in one minute = E × K = 0.01833 × 3275 = 60.04 rpm
∴ Speed of disc = 60.04 RPM = 1 r.p.s
At half load
= 2.5A,t = 59.5sec
Et = VIcosϕ×t = 2220×2.5×1×59.5 = 0.00909027kwh
N = 30 revolutions
Circuit shown in figure. Assume L to be large enough to ensure linear growth and decay of the current through it and have continuous current find the max value of load current. (Vs = 100 V, Vo = 50 V, L = 2mH, F = 20 kHz, R = 10 Ω)
The function of capacitor C across-load resistance R is to make the output voltage continuous.
Ripple current 
= 5.3125 A
The following circuit shown in figure. the input voltage is a sinusoidal at 50 Hz with an R.M.S value of 20 V. Find the current is from the source is
= 1 + j 100 π × 20 × 103 × 20 × 10-6 = 1 + j40 π
V0 = Vs (1 + j40 π) = 20 (1 + j40 π)
Consider the system shown below'
The controllability matrix for this system is 
then the sum of a, b and c is –
ẋ1 = -x1 + x2
ẋ2 = -x2 + 2u
ẋ3 = -2x3 + 2u
controllability QC = [B AB A2B]
∴ A + B + C = -4 – 2 + 8 = 2
A single line diagram of 3ϕ system shown in figure. Find the short circuit current flow into 3ϕ circuit at F, is ____ KA
Let base MVA = 50 MVA
% reactance of alternator A at the base mVA
XA and XB is connected in parallel.
XAB = (40 × 50)/90 =22.22
Now XAB is connected parallel with XC
∴ PU short circuit current 
= 8.4996
ISC actual = IPU × IBase
= 8.4996 × 1443.38
= 12268.15 A
= 12.268 KV
A motor running at 1500 rpm has hysteresis and eddy current loss of 180 W and 70 W respectively. If flux remain constant, the total loss will be one one-fourth at _____ rpm.
Speed α F
Hysteresis loss α N
⇒ Wh = K1N
Eddy current loss α N2
⇒ We = K2N2
180 = K1 × 1500
The speed N at which total loss become 1/4
62.5 = 0.12 N + (0.03 × 10-3)N2
N ≃ 467 R.P.M
A 50 Hz, synchronous generator capable of supplying 400 MW of power is connected to a large power system and is delivering 80 MW when a three phase fault occurs at its terminals find the time in which the fault must be cleared if the maximum power angle is to be 85°. Assume H = 8 MJ/MVA on 100 MVA base.
Pi = Pmax sin δ0
δ0 = 11.54° = 0.2 radian
and δ1 = 85° = 1.48 radian
cos δc = cos δ1 + (δ1 – δ0) sin δ0
cos δc = cos (85) + (1.48 – 0.2) sin 11.54
δc = 1.22 radian
= 36.046 ms
A 50 Hz, 3ϕ core type star-delta transformer has a line valtage ratio of 11,000/440 volts. The cross-section of the core is square with a circumscribing circle of 0.5 m diameter. If maximum flux density of 1.30 ωb/m2 then find the number of turns per phase on high voltage windings. Assume insulation occupies 8% of the total core area.
We known that diameter of the circumscribing is same as diagonal of the square.
Area of square core = l × b
∴ Net cross section area = 0.125 × 0.92 = 0.115
E.M.F/Turn = 4.44 fBA = 4.44 × 50 × 1.3 × 0.115 = 33.19
Phase turn ratio = 
∴ Number of turn per phase on high voltage side =
After how many clock pulses counter locks up in the 1010 and 0101 states. If the initial state is Q3Q2Q1Q0 = 0000
Tog = toggle, NC = no change
The counter locks up in the 1010 and 0101 states, alternating between them.
A single-phase full bridge inverter Is fed from a dc source such that fundamental. Component of output voltage is 230 V. Find the R.M.S value of diode current. For the following loads: R = 2 Ω, XL= 8 Ω, XC = 6 Ω
= 81.317 A
R.M.S value of diode current
= 0.1507 × √2 × 81.317 = 17.328 A
Two generators rated 250 MW and 400 MW are operating in parallel. The droop characteristics of the governors are 4% and 6% respectively.if a load of 450 MW be share between them. What will be the system frequency? Assume nominal system frequency is 60 Hz and no governing action.
Let, Load on generator 1 = x MW
Load on generator 2 = (450 – x) MW
Reduction in frequency = Δf
Now,
From equations (i) & (ii), we get
∴ x = 217.81 MW (load on generator 1)
450 – x = 232.18 MW (load on generator 2)
and Δf = 2.09Hz
∴ System frequency = (60 – 2.09) = 57.909 Hz.
The synchronous speed of induction motor is 900 RPM. Under blocked-rotor condition the input power to the motor is 45 KW at 193.6 A. The stator resistance per phase is 0.2 Ω and the transformation ratio is a = 2. Calculate the motor starting torque (in N-m)
R2 = 0.05 Ω
Referred to the rotor resistance per phase is
R′2 = a2R2 = 0.2Ω
∴ Starting torque 
The is transistor shown below is biased for constant base current if 
the value RB is ______ KΩ.
VCC = ICQRC + VCEQ
If leakage current neglected
Transistor is Si device VBEQ = 0.7V
Find the Norton equivalent circuit for the following circuit.
Apply KVL at node Va
13 Vb = 3 Va + 10
Now find ISC = IN
Apply KVL at Node Va
13 Vb = 3 Va + 10
Now find ISC = IN
V0 = 40 i2, i3 = 0.5 V0 = 20 i2
5 + 10 (i1 – 20 i2) = 0
-5 + 40 i2 + 20 (i2 – 20 i2) = 0
Find the Fourier series of rectangular pulse which is represented by for one time period
The waveform is given by
Again, 
[it may be observed here that an will be zero for even values of n].
∴ The Fourier series is
The admittance locus of the circuit shown in figure is
0 < R1 < ∞
y = y1 at R1 = ∞
A bridge shown in figure is at balance. The parameter r1 and c1 are
At balance
Z1Z4 = Z2Z3
Equating real and imaginary terms
A synchronous motor has 1000 KW, 3ϕ, Y connected 3.3 KV, 28 poles, 50 Hz. synchronous reactance of 4.20 Ω/ph. Its field excitation is adjusted to result in unity P.F operation at rated load. Compute the maximum power that the motor can deliver with its excitation remaining constant at this value.
Excitation emf, E = 1905∠0° - j175 ∠0 × 4.20
= 1905 – j735 = 2041.88 V
= 2778.41 KW
The power input in a three phase three wire delta connected balanced load is measured by the two wattmeter method. The reading of wattmeter A is 5000 W and wattmeter B is –1000 W (with reversal of connection) if the voltage of the circuit is 440 V, 50 Hz, what is the value of capacitance (in μF) connected in delta at the source to cause the whole of the power measured by the wattmeter A only.
= 0.3592 lag
Now capacitance is connected at the source which does not affect the active power consumption
W = 5000 – 1000 = 4000 W
Before connecting capacitor
For delta Vph = VL = 440 V and IL = √3 Iph
i.e. Iph = 8.4362 A
∴ Rph = Zph cos ϕ = 18.735 Ω
Xph = Zph sin ϕ = 48.67 Ω (inductive)
After connecting capacitor
WA = W = 4000 W, WB = 0
WB = VLIL cos (30 + ϕ) = 0
30 + ϕ = 90°
ϕ = 60°
cos ϕ = cos 60° = 0.5 lagging
Due to pure capacitor, Rph remain same
Rph = 18.735 Ω
Xcph = Xph – X'ph = 48.67 – 32.45
= 16.2246 Ω
= 196.1896 μF
The system characteristics equation given below represents
Using Laplace transform
Where a > 1
Here Zero is dominant over pole
∴ It is lead network and act as High pass filter (or network)
A single phase full converter bridge is connected to RLE load the source voltage is 230 V, 50 Hz. The average load current of 8 A is constant over the working range for R = 0.4, L = 2 mH and E = -120 V. Find the input P.F
For E = -120 V, the full converter is operating as a line commutated inverter therefore power from dc source to AC source
Vr Ior cos ϕ = EI0 – I2or R
Ior = I0 = 8 A
= 0.508 Lag
The maximum efficiency of 550 KVA, 3300/430 V, 50 Hz 1 ϕ transformer is 98% and occurs at ¾ full load, unity power factor. If the impedance is 10%, calculate the regulation (in %) at full load at 0.8 P.F lagging.
Output at ¾ full load and unit power factor =
Total loss = 420.92 – 412.5 = 8.42 KW
Loss is equally divided b/w iron and copper loss
Copper loss = 7.48299
Percentage resistance = 
Percentage impedance = 10%
Percentage reactance = 
∴ Percentage regulation = 
= 1.74 × 0.8 + 9.84 × 0.6
= 1.392 + 5.91 = 7.30%
The full range of audibility in audio frequency oscillator is
The full range of audibility in audio frequency oscillator is a. 0 to 20 Hz b. 20 Hz to 2 kHz
The matrix representation of the relevant two port parameters that describe the circuit
Writing KVL for input loop
V1 = 5I1 + 3I2
Writing KVL for outer loop
I2(1 + 2j – 2j) + 3I2
V2 = 4I2
∴ h parameter = 
A shunt motor 500 V, draw 40 A while supplying rated load at speed of 120 rad/sec. The armature resistance is 2 Ω and field winding resistance 250 Ω, external resistance is inserted in series with the armature circuit so that armature current does not exceed 160% of its rated value when the motor is plugged. The breaking torque at the instant of plugging is _____ N-m
Field current 
Ia = 40 – 2 = 38 A
Ea = 500 – 38 × 2 = 424 V
At the instant of plugging
Plugging torque =
= 214.83 N-m
Load torque 
∴ Braking torque = 214.83 + 134.27
= 349.1 N-m
Consider the matrix equation
The condition for existence of a non-trivial solution, and the corresponding normalised solution (up to a sign) is
For non-trivial solution, the rank of the matrix should be less than the number of variables. i.e. r < n.
For this, |A| = 0
⇒ (4c – 3b) – (2c – 6) + (b – 4) = 0
⇒ 4c – 3b – 2c + 6 + b – 4 = 0
⇒ 2c – 2b + 2 = 0
⇒ b = c + 1
The vectors x1, x2 ….. xn are said to be linearly dependent, if there exist numbers λ1, λ2 ……. λn, not all zero such that
λ1x1 + λ2x2 + ….. + λnxn = 0
Here,
λ1 + λ2 + λ3 = 0 ----(1)
λ1 + 2λ2 + 3λ3 = 0 ----(2)
2λ1 + bλ2 + 2cλ3 = 0
2λ1 + (c + 1) λ2 + 2cλ3 = 0 ----(3)
From (1) & (3):
λ2 = -2λ3
λ1 = λ3 = λ
λ2 = -2λ
so corresponding normalised solution:
Two points are chosen randomly on a line 9 cm long. Determine the probability that the distance between them is less than 3 cm
x: The distance of first point from the start of the line
y: distance of second point from the start of the line segment
x,y ϵ[0,9]
So sample space is Area of region bounded by
x ≥ 0, y ≥ 0, x ≤ 9, y ≤ 9
This is square of side 9
Area = 81 cm2
The region of our interest is
|x-y| < 3
0 ≤ x ≤ 9
0 ≤ y ≤ 9
Area of shaded region = 2 (area of triangle) + area of rectangle
= 45
Probability = 45/81 =0.55
The area bounded by the curve y =x (3 – x)2, the x-axis and the ordinates of the maximum and minimum points of the curve is
y = x(3 – x)2
−x.2(3−x) + (3−x)2 = 3(x2−4x+3) = 0
(x - 3) (x – 1) = 0 ⇒ x = 3, x = 1
Given N > 0, the iterative equation for finding 
using Newton-Raphson method is:
If y = 3e2x + e-2x - αx is the solution of the initial value problem
Where, α, β ϵ R, then
y = 3e2x + e-2x – αx
Given that, 
⇒ 1 = 6 – 2 - α
⇒ α = 3
Complementary solution
(D2 + β) = 0 ----(1)
The given is, y = 3e2x + e-2x – αx
It indicates, 2 and -2 are the roots of auxiliary equations.
⇒ (D + 2) (D – 2) = 0
⇒ D2 – 4 = 0
By comparing this equation with equation (1)
β = -4
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