Math Chapter Wise

EnglishBihar (TRE 1.0) -2023

Q01.Let C be the circle (x – 1)2 + y2 = 1, oriented counter-clockwise. Then the value of the line integral. ∫ 3 4 C 4 – xy dx + x dy 3 is

A6π B8π C12π DMore than one of the above ENone of the above

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EnglishBihar (TRE 1.0) -2023

Q02.The equation xy – z log y + exz = 1 can be solved in a neighborhood of the point (0, 1, 1) as y = f(x, z) for some continuously differentiable function f. Then

A∇f(0, 1) = (2, 0) B∇f(0, 1) = (0, 2) C∇f(0, 1) = (0, 1) DMore than one of the above ENone of the above

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EnglishBihar (TRE 1.0) -2023

Q03.Let V be the solid region in 3 ℝ bounded by the paraboloid y = (x2 + z2) and the plane y = 4. Then the value of ∫∫∫ 2 2 V 15 x + z dV is y = (x2 + z2) continuous

A128π B64π C28π DMore than one of the above ENone of the above

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EnglishRPSC PGT 2022

Q04.If H = x î + 2y ĵ`+3z k̂ than ˆ ∫∫s H.nds is equal to (where A is the volume enclosed by S) –

A3A B6A CA D4A

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EnglishRPSC PGT 2022

Q05.If a is differentiable vector point function and u is a differentiable scalar point function, then ∇ × (u a) is equal to -

A(∇ u) × a +u(∇ × a) B(∇ × a). a +u(∇ × a) C∇ u. a +u(∇ × a) D∇ u × a +u. (∇ × a)

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EnglishRPSC PGT 2022

Q06.If f1 and f2 are differentiable scalar functions and v is differentiable vector function such that f1v = ∇ f2, then v. curl v is -

A1 f f f 2 2 1 f1 ∇ + ∇ B1 1 f f 2 1 f f 1 2 ∇ − ∇ C1 1 f f 2 2 f f 1 2 ∇ + ×∇ DZero

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EnglishRPSC PGT 2022

Q07.Value of 2 where C is the square in xy – plane with vertices (1, 0), (–1, 0), (0, 1), (0, –1) respectively is -

A–2 B4 C0 D2

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EnglishJSSC PGT 2018

Q08.Evaluate () ∫ 4 C x x + xy y ∂ ∂ ∂ ∂, where C is the triangular curve consisting of the line segments from (0, 0) to (1, 0), from (1, 0) to (0, 1) and from (0, 1) to (0, 0).

A0 B12 C1/6 D1

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EnglishJSSC PGT 2018

Q09.For f (x, y) = x2 y – y3, which one of the following gives the corresponding gradient vector field?

Ax2/i B2xyi + (x2 – 3y2)j Cxi + (x2 – y2)j D(x3 – y2)j

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EnglishJSSC PGT 2018

Q10.If f (x, y, z) = xsin yz, find the gradient of f at the point (1, 3, 0).

A(1, 0, 3) B(1, –1, 3) C(0, 0, 3) D(0, 0, 0)

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EnglishJSSC PGT 2018

Q11.Consider f (x, y) = xey. At the point (2, 0) what is the maximum rate of change?

A5 B4 C1 D0

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EnglishUP PGT 2021

Q12.A unit vector, which is normal to the surface x2 – xy + z 2 = 1 at the point (1, 1, 1) is–

Ai– j 2k 6 ∧ ∧ ∧ + Bi j– 2k 6 ∧ ∧ ∧ + Ci j 2k 6 ∧ ∧ ∧ + + Di– j– 2k 6 ∧ ∧ ∧

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EnglishUP PGT 2021

Q13.If r r, r r r ∧ =, then div r ∧ is equal to –

Ao B–1 C1/r D2/r

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EnglishUP PGT 2021

Q14.Value of S ˆ F.nds ∫∫, where 2 F 4xz i– y j yz k ∧ ∧ ∧ = + and S is the surface of the cube bounded by x = 0, x = 1, y = 0, y = 1, z = 0, z = 1 is–

A1 B3/2 C3 D5/2

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EnglishUP PGT 2021

Q15.If the vector () () () F x 3y i y – 2z j x – az k ∧ ∧ ∧ = + + + is solenoidal then a is equal to –

A1 B–1 C2 D–2

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EnglishUP PGT 2021

Q16.If a and b are irrotational vectors then div () × a b is equal to –

A1 B2 C3 D0

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EnglishUPPSC GIC 2021

Q17.If V is the volume enclosed by a closed surface S and ˆ ˆ ˆ F xi 2yj 3zk = + + then the value of s ˆ F.nds ∫ is

A3V B4V C6V D5V

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EnglishUPPSC GIC 2021 UPPSC Asharam Paddhati 2021

Q18.If F is a vector point function and S is an open surface bounded by a closed curve C, then the tangential line integral of F along C is described-

ABy Green's theorem BBy Stokes' theorem CBy Gauss's theorem DBy Leibnitz' theorem

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EnglishUPPSC GIC 2021 UPPSC Ashram Paddhati 2021

Q19.If ˆ ˆ ˆ r xi yj zk = + + and r | r | = then the value of curl () n r r, is-

A0 Bn rn-2/r Cn rn-1/r D(n + 3) rn/r

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EnglishUPPSC Ashram Paddhati 2021

Q20.If S denotes the surface of the cube bounded by the planes x = 0, x = 1, y = 0, y = 1, z = 0, z = 1 then the value of normal surface integral of ˆ ˆ ˆ 3 2 F = (x – yz)i – 2x y j+ 2k over S is

A1 B1/3 C1/5 D1/6

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