Math Chapter Wise

EnglishUKPSC Lecturer (Mains) 2020

Q01.Which of the following methods is based on Bolzano theorem?

ABisection method BNewton-Raphson method CRegula-Falsi method DNone of these

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EnglishUKPSC Lecturer (Mains) 2020

Q02.The roots of the equation f(x) = 0 obtained by Newton-Raphson method converges if

A2 f '()f ''() 1 [f()] < x x x B2 f '()f ''() 1 [f '()] < x x x C2 f()f '() 1 [f ''()] < x x x DNone of these

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EnglishUKPSC Lecturer (Mains) 2020

Q03.Which of the following is not true?

Ax x e (E 1)e ∆ = − B()() 1 1 1 0 +∆ −∇ + = CE 1 = ∆+ D1 E 1 − ∇+ =

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EnglishBihar (TRE 2.0)-2023

Q04.The optimal value of the objective function is attained at the points

Aon X -axis Bon Y-axis Con the corner points of the feasible region DMore than one of the above ENone of the above

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EnglishBihar (TRE 2.0)-2023

Q05.The point which does not lie in the half-plane 2x + 3y -12 < 0 is

A(2, 3) B(–2, 3) C(1, 2) DMore than one of the above ENone of the above

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EnglishBihar (TRE 1.0) -2023

Q06.Which of the following is true?

AEvery linear programming problem has a feasible solution BIf a linear programming problem has an optimal solution then it is unique CExtreme points of the disk x2 + y2 ≤ 1 are the points on the circles x2 + y2 = 1 DMore than one of the above ENone of the above

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EnglishBihar (TRE 1.0) -2023

Q07.The Linear Programming Problem Maximize z = x1 + x2 subject to x1 + 2x2 ≤ 20 x1 + x2 ≤ 15 x2 ≤ 6 x1, x2 ≥ 0

AHas exactly one optimum solution BHas more than one optimum solution CHas no solution DMore than one of the above ENone of the above

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EnglishBihar (TRE 1.0) -2023

Q08.The dual of the linear programming problem: Minimize CT x subject to Ax ≥ b and x ≥ 0 is

AMaximize bT w subject to AT w ≥ c and BMaximize bT w subject to AT w ≤ c and CMaximize bT w subject to AT w ≤ c and w is unrestricted. DMore than one of the above ENone of the above

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EnglishBihar (TRE 1.0) -2023

Q09.A transportation problem is

Adual problem Bnon-linear programming problem Cspecial case of linear programming problem DMore than one of the above ENone of the above

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EnglishKVS PGT 2023

Q10.If the corner points of the LPP, Z = 13x - 15y subject to the constraints x + y ≤7, 2x - 3y ≥ 0, x ≥ 0 and y ≥ 0 are (0, 0), (7, 0), (3, 4) and (0, 2), then the value of Maximum Z+3 Minimum Z is:

A91 B181 C1 D-1

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EnglishKVS PGT 2023

Q11.Which among the following is NOT a corner point of the LPP, maximize Z = 4x + y subject to the constraints, x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, y ≥ 0?

A(0, 50) B(20, 30) C(50, 0) D(30, 0)

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EnglishRPSC PGT 2022

Q12.A necessary and sufficient condition for a basic feasible solution to a minimization linear programming problem to be an optimum is that (for all j)-

Azj – cj = 0 Bzj – cj ≤ 0 Czj – cj ≥ 0 Dzj –cj ≠ 0

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EnglishRPSC PGT 2022

Q13.Solution of minimal assignment problem whose effectiveness matrix is following – I II III IV A 2 3 4 5 B 4 5 6 7 C 7 8 9 8 D 3 5 8 4

AA → II, B → III, C → IV, D → I BA → III, B → II, C → IV, D → I CA → III, B → II, C → I, D → IV DAll above

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EnglishRPSC PGT 2022

Q14.The objective function of the dual problem of the following primal problem is – Max (z) = 3x1 – 3x2 s.t. x1≤4 x2≤6 x1+x2≤5 –x2≤–1 X1,x2≥0

AMax (Z’) = –4w1–6w2–5w3+w4 BMin (Z’) = –4w1–6w2+5w3–w4 CMin (Z’) = 4w1+6w2+5w3+w4 DMin (Z’) = 4w1–6w2–5w3+w4

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EnglishDSSSB PGT 2021

Q15.In hyperplane 2x1 + 3x2 + 4x3 – x4 = 5, the point (1, 2, 3, 4) lies in -

AOpen half space of type cx > z BOpen half space of type cx < z CClosed half space of type cx ≤ z DClosed half space of type cx ≥ z

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EnglishDSSSB PGT 2021

Q16.The number of basic solution to the system- 3x1 + 4x2 + 3x3 + 8x4 = 7 2x1 – 3x2 + 2x3 – 6x4 = 13

A12 B4 C9 D6

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EnglishDSSSB PGT 2021

Q17.If a constraint is multiplied by a non-zero constant of a L.P.P, then the solution

ADoes not change with original solution Bis increased by addition of that non-zero constant Cis increased by multiplication of that non-zero constant Dis decreased by subtraction of that non-zero constant

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EnglishDSSSB PGT 2021

Q18.For a L.P.P, Minimise z = x1 + 5x2 – 3x3, Subjected to 2x1 + x2 + x3 = 5, 3x1 – x2 + 4x3 = 5, x1, x2, x3 ≥ 0, the solution (1, 2, 1) is–

AA basic but not feasible solution BA basic feasible solution CA feasible but not basic solution DNeither basic nor feasible solution

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EnglishDSSSB PGT 2021

Q19.The solution (1, 0, 1) to be system x1 + x2 + x3 = 2, x1 – 3x2 + x3 = 2, 2x1 + 3x2 + 4x3 = 6 is

AA basic feasible solution BNot a basic solution CA basic but not feasible solution DA non-degenerate basic feasible solution

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EnglishDSSSB PGT 2021

Q20.The minimum cost of the assignment problem:

A52 B45 C60 D50

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