CUET PG 2026 — Mathematics PYQ
CUET PG | Mathematics | 2026Arrange the following in decreasing order:
(A)
(B)
(C)
(D)
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Arrange the following in decreasing order:
(A) 1231amp;3amp;6amp;11amp;5amp;10amp;38
(B) 673981amp;19amp;13amp;24amp;21amp;14amp;26
(C) 143amp;−3amp;−1amp;5amp;2amp;2amp;2
(D) 149amp;4amp;9amp;16amp;9amp;16amp;25
A, B, C, D
B, A, D, C
C, D, A, B
C, A, D, B
(Correct Answer)C, A, D, B
We will calculate the value of each determinant to determine the order.
Value of (A):
Observe the first two rows. Row 2 is exactly twice Row 1 (R2=2R1).
Since two rows are proportional, the value of the determinant is:
Value of (B):
We can simplify this using row operations. Perform R1→R1−3R2:
R1:[67−3(39),19−3(13),21−3(14)]=[67−117,19−39,21−42]=[−50,−20,−21]
Now perform R3→R3−2R2:
R3:[81−2(39),24−2(13),26−2(14)]=[81−78,24−26,26−28]=[3,−2,−2]
The determinant is roughly:
−50393amp;−20amp;13amp;−2amp;−21amp;14amp;−2
Expanding or further simplifying shows this value is negative. Specifically, performing C2−C3 helps, but direct calculation yields:
Value of (C):
Take 2 common from C3:
2143amp;−3amp;−1amp;5amp;1amp;1amp;1
Perform R2→R2−R1 and R3→R3−R1:
2132amp;−3amp;2amp;8amp;1amp;0amp;0=2×[1×(24−4)]=2×20
Value of (D):
Apply C2→C2−C1 and C3→C3−C2:
149amp;3amp;5amp;7amp;5amp;7amp;9
Apply C3→C3−C2 and C2→C2−C1 again (checking differences):
149amp;2amp;1amp;−2amp;2amp;2amp;2
Expanding along R1: 1(2−(−4))−2(8−18)+2(−8−9)=6+20−34
Comparison of Values:
(C) = 40
(A) = 0
(D) = −8
(B) = −43
Decreasing Order: C > A > D > B
Correct Option: (d)
We will calculate the value of each determinant to determine the order.
Value of (A):
Observe the first two rows. Row 2 is exactly twice Row 1 (R2=2R1).
Since two rows are proportional, the value of the determinant is:
Value of (B):
We can simplify this using row operations. Perform R1→R1−3R2:
R1:[67−3(39),19−3(13),21−3(14)]=[67−117,19−39,21−42]=[−50,−20,−21]
Now perform R3→R3−2R2:
R3:[81−2(39),24−2(13),26−2(14)]=[81−78,24−26,26−28]=[3,−2,−2]
The determinant is roughly:
−50393amp;−20amp;13amp;−2amp;−21amp;14amp;−2
Expanding or further simplifying shows this value is negative. Specifically, performing C2−C3 helps, but direct calculation yields:
Value of (C):
Take 2 common from C3:
2143amp;−3amp;−1amp;5amp;1amp;1amp;1
Perform R2→R2−R1 and R3→R3−R1:
2132amp;−3amp;2amp;8amp;1amp;0amp;0=2×[1×(24−4)]=2×20
Value of (D):
Apply C2→C2−C1 and C3→C3−C2:
149amp;3amp;5amp;7amp;5amp;7amp;9
Apply C3→C3−C2 and C2→C2−C1 again (checking differences):
149amp;2amp;1amp;−2amp;2amp;2amp;2
Expanding along R1: 1(2−(−4))−2(8−18)+2(−8−9)=6+20−34
Comparison of Values:
(C) = 40
(A) = 0
(D) = −8
(B) = −43
Decreasing Order: C > A > D > B
Correct Option: (d)