CUET Mathematics Previous Year Paper 2024

 Page 1


CUETUG2024MathematicsQuestionPaperwithSolutionSet-A
Question1: Thecornerpointsofthefeasibleregiondeterminedbyx+y= 8, 2x+y= 8,
x= 0, y = 0 are A(0,8), B(4,0), and C(8,0). If the objective function Z = ax+by has its
maximumvalueonthelinesegmentAB,thentherelationbetweenaandbis:
(1) 8a+4 =b
(2)a = 2b
(3)b = 2a
(4) 8b+4 =a
CorrectAnswer: (2)a = 2b
Solution:
The line segment AB has the points A(0,8) and B(4,0). The objective function Z = ax +by
willhaveamaximumvalueonAB if
a
b
=-
changeiny
changeinx
.
BetweenpointsAandB:
SlopeofAB =
0-8
4-0
=-2
Thus,theratio
a
b
= 2impliesa = 2b.
QuickTip
Matchcoef?cientsoftheobjectivefunctionwiththeslopeofthelineformaximization
onasegment.
Question2: Ift =e
2x
andy = ln(t
2
),then
d
2
y
dx
2
is:
(1) 0
(2) 4t
(3)
2t
4e
t
(4)
2t
e
2t
(4t-1)
CorrectAnswer: (1) 0
1
Page 2


CUETUG2024MathematicsQuestionPaperwithSolutionSet-A
Question1: Thecornerpointsofthefeasibleregiondeterminedbyx+y= 8, 2x+y= 8,
x= 0, y = 0 are A(0,8), B(4,0), and C(8,0). If the objective function Z = ax+by has its
maximumvalueonthelinesegmentAB,thentherelationbetweenaandbis:
(1) 8a+4 =b
(2)a = 2b
(3)b = 2a
(4) 8b+4 =a
CorrectAnswer: (2)a = 2b
Solution:
The line segment AB has the points A(0,8) and B(4,0). The objective function Z = ax +by
willhaveamaximumvalueonAB if
a
b
=-
changeiny
changeinx
.
BetweenpointsAandB:
SlopeofAB =
0-8
4-0
=-2
Thus,theratio
a
b
= 2impliesa = 2b.
QuickTip
Matchcoef?cientsoftheobjectivefunctionwiththeslopeofthelineformaximization
onasegment.
Question2: Ift =e
2x
andy = ln(t
2
),then
d
2
y
dx
2
is:
(1) 0
(2) 4t
(3)
2t
4e
t
(4)
2t
e
2t
(4t-1)
CorrectAnswer: (1) 0
1
Solution:
First,simplifyy = log
e
(t
2
)asfollows:
y = 2log
e
(t)
Sincet =e
2x
,wehave:
log
e
(t) = 2x?y = 2·2x = 4x
Now,takingthe?rstderivativewithrespecttox:
dy
dx
= 4
Then,takingthesecondderivativewithrespecttox:
d
2
y
dx
2
= 0
Thus,thevalueof
d
2
y
dx
2
is 0.
QuickTip
Usechainruleonexponentialtermsforquickresults.
Question 3: An objective function Z = ax +by is maximum at points (8,2) and (4,6). If
a= 0,b= 0,andab = 25,thenthemaximumvalueofthefunctionis:
(1) 60
(2) 50
(3) 40
(4) 80
CorrectAnswer: (2) 50
Solution:
ThegivenfunctionZ =ax+by attainsitsmaximumvalueatpoints (8,2)and (4,6). Atthese
points:
Z
1
= 8a+2b and Z
2
= 4a+6b
Sincebothpointsyieldthesamemaximumvalue:
8a+2b = 4a+6b
2
Page 3


CUETUG2024MathematicsQuestionPaperwithSolutionSet-A
Question1: Thecornerpointsofthefeasibleregiondeterminedbyx+y= 8, 2x+y= 8,
x= 0, y = 0 are A(0,8), B(4,0), and C(8,0). If the objective function Z = ax+by has its
maximumvalueonthelinesegmentAB,thentherelationbetweenaandbis:
(1) 8a+4 =b
(2)a = 2b
(3)b = 2a
(4) 8b+4 =a
CorrectAnswer: (2)a = 2b
Solution:
The line segment AB has the points A(0,8) and B(4,0). The objective function Z = ax +by
willhaveamaximumvalueonAB if
a
b
=-
changeiny
changeinx
.
BetweenpointsAandB:
SlopeofAB =
0-8
4-0
=-2
Thus,theratio
a
b
= 2impliesa = 2b.
QuickTip
Matchcoef?cientsoftheobjectivefunctionwiththeslopeofthelineformaximization
onasegment.
Question2: Ift =e
2x
andy = ln(t
2
),then
d
2
y
dx
2
is:
(1) 0
(2) 4t
(3)
2t
4e
t
(4)
2t
e
2t
(4t-1)
CorrectAnswer: (1) 0
1
Solution:
First,simplifyy = log
e
(t
2
)asfollows:
y = 2log
e
(t)
Sincet =e
2x
,wehave:
log
e
(t) = 2x?y = 2·2x = 4x
Now,takingthe?rstderivativewithrespecttox:
dy
dx
= 4
Then,takingthesecondderivativewithrespecttox:
d
2
y
dx
2
= 0
Thus,thevalueof
d
2
y
dx
2
is 0.
QuickTip
Usechainruleonexponentialtermsforquickresults.
Question 3: An objective function Z = ax +by is maximum at points (8,2) and (4,6). If
a= 0,b= 0,andab = 25,thenthemaximumvalueofthefunctionis:
(1) 60
(2) 50
(3) 40
(4) 80
CorrectAnswer: (2) 50
Solution:
ThegivenfunctionZ =ax+by attainsitsmaximumvalueatpoints (8,2)and (4,6). Atthese
points:
Z
1
= 8a+2b and Z
2
= 4a+6b
Sincebothpointsyieldthesamemaximumvalue:
8a+2b = 4a+6b
2
Simplifytheequation:
8a-4a = 6b-2b
4a = 4b
a =b
Usingtheconditionab = 25:
a·b = 25 and a =b
a
2
= 25 ? a = 5 and b = 5
Substitutea = 5andb = 5intoZ =ax+by. Atpoint (8,2):
Z = 8a+2b = 8(5)+2(5) = 40+10 = 50
Thus,themaximumvalueofZ is 50.
QuickTip
Ensure all constraints are considered when solving for the maximum or minimum of
objectivefunctions.
Question 4: The area of the region bounded by the lines x + 2y = 12, x = 2, x = 6, and
thex-axisis:
(1)34squnits
(2)20squnits
(3)24squnits
(4)16squnits
CorrectAnswer: (4)16squnits
Solution:
To ?nd the area of the region bounded by x+2y = 12, x = 2, x = 6, and the x-axis, we start
byexpressingy intermsofxfromtheequationx+2y = 12:
y =
12-x
2
3
Page 4


CUETUG2024MathematicsQuestionPaperwithSolutionSet-A
Question1: Thecornerpointsofthefeasibleregiondeterminedbyx+y= 8, 2x+y= 8,
x= 0, y = 0 are A(0,8), B(4,0), and C(8,0). If the objective function Z = ax+by has its
maximumvalueonthelinesegmentAB,thentherelationbetweenaandbis:
(1) 8a+4 =b
(2)a = 2b
(3)b = 2a
(4) 8b+4 =a
CorrectAnswer: (2)a = 2b
Solution:
The line segment AB has the points A(0,8) and B(4,0). The objective function Z = ax +by
willhaveamaximumvalueonAB if
a
b
=-
changeiny
changeinx
.
BetweenpointsAandB:
SlopeofAB =
0-8
4-0
=-2
Thus,theratio
a
b
= 2impliesa = 2b.
QuickTip
Matchcoef?cientsoftheobjectivefunctionwiththeslopeofthelineformaximization
onasegment.
Question2: Ift =e
2x
andy = ln(t
2
),then
d
2
y
dx
2
is:
(1) 0
(2) 4t
(3)
2t
4e
t
(4)
2t
e
2t
(4t-1)
CorrectAnswer: (1) 0
1
Solution:
First,simplifyy = log
e
(t
2
)asfollows:
y = 2log
e
(t)
Sincet =e
2x
,wehave:
log
e
(t) = 2x?y = 2·2x = 4x
Now,takingthe?rstderivativewithrespecttox:
dy
dx
= 4
Then,takingthesecondderivativewithrespecttox:
d
2
y
dx
2
= 0
Thus,thevalueof
d
2
y
dx
2
is 0.
QuickTip
Usechainruleonexponentialtermsforquickresults.
Question 3: An objective function Z = ax +by is maximum at points (8,2) and (4,6). If
a= 0,b= 0,andab = 25,thenthemaximumvalueofthefunctionis:
(1) 60
(2) 50
(3) 40
(4) 80
CorrectAnswer: (2) 50
Solution:
ThegivenfunctionZ =ax+by attainsitsmaximumvalueatpoints (8,2)and (4,6). Atthese
points:
Z
1
= 8a+2b and Z
2
= 4a+6b
Sincebothpointsyieldthesamemaximumvalue:
8a+2b = 4a+6b
2
Simplifytheequation:
8a-4a = 6b-2b
4a = 4b
a =b
Usingtheconditionab = 25:
a·b = 25 and a =b
a
2
= 25 ? a = 5 and b = 5
Substitutea = 5andb = 5intoZ =ax+by. Atpoint (8,2):
Z = 8a+2b = 8(5)+2(5) = 40+10 = 50
Thus,themaximumvalueofZ is 50.
QuickTip
Ensure all constraints are considered when solving for the maximum or minimum of
objectivefunctions.
Question 4: The area of the region bounded by the lines x + 2y = 12, x = 2, x = 6, and
thex-axisis:
(1)34squnits
(2)20squnits
(3)24squnits
(4)16squnits
CorrectAnswer: (4)16squnits
Solution:
To ?nd the area of the region bounded by x+2y = 12, x = 2, x = 6, and the x-axis, we start
byexpressingy intermsofxfromtheequationx+2y = 12:
y =
12-x
2
3
Theareabetweenx = 2andx = 6undertheliney =
12-x
2
isgivenby:
Area =
Z
6
2
12-x
2
dx
Evaluatingthisintegral:
=
Z
6
2
12-x
2
dx =
1
2
Z
6
2
(12-x)dx
=
1
2

12x-
x
2
2

6
2
=
1
2

(12×6-
6
2
2
)-(12×2-
2
2
2
)

=
1
2
[(72-18)-(24-2)]
=
1
2
[54-22] =
1
2
×32 = 16
Therefore,theareaoftheregionis 16squnits.
QuickTip
Breakdowncomplexregionsintosimplershapesforeasierareacalculation.
Question5: Adieisrolledthrice. Whatistheprobabilityofgettinganumbergreaterthan4
inthe?rstandsecondthrowofthedice,andanumberlessthan4inthethirdthrow?
(1)
1
3
(2)
1
6
(3)
1
9
(4)
1
18
CorrectAnswer: (4)
1
18
Solution:
When a die is rolled, the outcomes are 1,2,3,4,5,6. The probabilities for the given events
areasfollows:
Anumbergreaterthan4includes{5,6}. Theprobabilityofthiseventis:
P(greaterthan4) =
2
6
=
1
3
4
Page 5


CUETUG2024MathematicsQuestionPaperwithSolutionSet-A
Question1: Thecornerpointsofthefeasibleregiondeterminedbyx+y= 8, 2x+y= 8,
x= 0, y = 0 are A(0,8), B(4,0), and C(8,0). If the objective function Z = ax+by has its
maximumvalueonthelinesegmentAB,thentherelationbetweenaandbis:
(1) 8a+4 =b
(2)a = 2b
(3)b = 2a
(4) 8b+4 =a
CorrectAnswer: (2)a = 2b
Solution:
The line segment AB has the points A(0,8) and B(4,0). The objective function Z = ax +by
willhaveamaximumvalueonAB if
a
b
=-
changeiny
changeinx
.
BetweenpointsAandB:
SlopeofAB =
0-8
4-0
=-2
Thus,theratio
a
b
= 2impliesa = 2b.
QuickTip
Matchcoef?cientsoftheobjectivefunctionwiththeslopeofthelineformaximization
onasegment.
Question2: Ift =e
2x
andy = ln(t
2
),then
d
2
y
dx
2
is:
(1) 0
(2) 4t
(3)
2t
4e
t
(4)
2t
e
2t
(4t-1)
CorrectAnswer: (1) 0
1
Solution:
First,simplifyy = log
e
(t
2
)asfollows:
y = 2log
e
(t)
Sincet =e
2x
,wehave:
log
e
(t) = 2x?y = 2·2x = 4x
Now,takingthe?rstderivativewithrespecttox:
dy
dx
= 4
Then,takingthesecondderivativewithrespecttox:
d
2
y
dx
2
= 0
Thus,thevalueof
d
2
y
dx
2
is 0.
QuickTip
Usechainruleonexponentialtermsforquickresults.
Question 3: An objective function Z = ax +by is maximum at points (8,2) and (4,6). If
a= 0,b= 0,andab = 25,thenthemaximumvalueofthefunctionis:
(1) 60
(2) 50
(3) 40
(4) 80
CorrectAnswer: (2) 50
Solution:
ThegivenfunctionZ =ax+by attainsitsmaximumvalueatpoints (8,2)and (4,6). Atthese
points:
Z
1
= 8a+2b and Z
2
= 4a+6b
Sincebothpointsyieldthesamemaximumvalue:
8a+2b = 4a+6b
2
Simplifytheequation:
8a-4a = 6b-2b
4a = 4b
a =b
Usingtheconditionab = 25:
a·b = 25 and a =b
a
2
= 25 ? a = 5 and b = 5
Substitutea = 5andb = 5intoZ =ax+by. Atpoint (8,2):
Z = 8a+2b = 8(5)+2(5) = 40+10 = 50
Thus,themaximumvalueofZ is 50.
QuickTip
Ensure all constraints are considered when solving for the maximum or minimum of
objectivefunctions.
Question 4: The area of the region bounded by the lines x + 2y = 12, x = 2, x = 6, and
thex-axisis:
(1)34squnits
(2)20squnits
(3)24squnits
(4)16squnits
CorrectAnswer: (4)16squnits
Solution:
To ?nd the area of the region bounded by x+2y = 12, x = 2, x = 6, and the x-axis, we start
byexpressingy intermsofxfromtheequationx+2y = 12:
y =
12-x
2
3
Theareabetweenx = 2andx = 6undertheliney =
12-x
2
isgivenby:
Area =
Z
6
2
12-x
2
dx
Evaluatingthisintegral:
=
Z
6
2
12-x
2
dx =
1
2
Z
6
2
(12-x)dx
=
1
2

12x-
x
2
2

6
2
=
1
2

(12×6-
6
2
2
)-(12×2-
2
2
2
)

=
1
2
[(72-18)-(24-2)]
=
1
2
[54-22] =
1
2
×32 = 16
Therefore,theareaoftheregionis 16squnits.
QuickTip
Breakdowncomplexregionsintosimplershapesforeasierareacalculation.
Question5: Adieisrolledthrice. Whatistheprobabilityofgettinganumbergreaterthan4
inthe?rstandsecondthrowofthedice,andanumberlessthan4inthethirdthrow?
(1)
1
3
(2)
1
6
(3)
1
9
(4)
1
18
CorrectAnswer: (4)
1
18
Solution:
When a die is rolled, the outcomes are 1,2,3,4,5,6. The probabilities for the given events
areasfollows:
Anumbergreaterthan4includes{5,6}. Theprobabilityofthiseventis:
P(greaterthan4) =
2
6
=
1
3
4
Anumberlessthan4includes{1,2,3}. Theprobabilityofthiseventis:
P(lessthan4) =
3
6
=
1
2
The probability of the required outcome (a number greater than 4 on the ?rst and second
throws,andanumberlessthan4onthethirdthrow)istheproductoftheprobabilities:
P(requiredoutcome) =P(greaterthan4)·P(greaterthan4)·P(lessthan4)
P(requiredoutcome) =
1
3
·
1
3
·
1
2
P(requiredoutcome) =
1
18
Thus,theprobabilityoftherequiredoutcomeis
1
18
.
QuickTip
Theprobabilitiesofindependenteventsaremultipliedto?ndthecombinedprobability.
Question6: Evaluatetheintegral
R
p
x
n+1
-x
dx:
Options:
(1)
p
n
log
e


x
n
-1
x
n


+C
(2) log
e


x
n
+1
x
n
-1


+C
(3)
p
n
log
e


x
n
+1
x
n


+C
(4)plog
e


x
n
x
n
-1


+C
CorrectAnswer: (1)
p
n
log
e


x
n
-1
x
n


+C
Solution:
Beginwiththeintegral:
Z
p
x
n+1
-x
dx
Factorthedenominator:
x
n+1
-x =x·(x
n
-1)
Thus,theintegralbecomes:
Z
p
x·(x
n
-1)
dx
5