Math Chapter Wise

EnglishHaryana PGT 2019

Q201.The differential equation of all circles passing through the origin and having their center on the X-axis is:

Ax2 = y2 + xy dy dx Bx2 = y2 + 3xy dy dx Cy2 = x2 + 2xy dy dx Dy2 = x2 – xy dy dx

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EnglishHaryana PGT 2019

Q202.Order and degree of the differential equation 3/2 2 d y dy – – 4 = 0 dx are respectively:

A2 and 6 B3 and 6 C1 and 4 D2 and 4

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EnglishDSSSB TGT 04-09-2021 Shift-I (Male)

Q203.If dy = 2x5 dx; then what is equation of the y in terms of x if curves passes through (1, 2)

Ax6 + 3y – 5 = 0 Bx6 – 3y + 5 = 0 Cx5 – y + 2 = 0 Dx5 + y – 2 = 0

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EnglishUK SSSC LT 2020

Q204.The particular integral of the following differential equation will be: (()) 2 y cosx D 1 = + (()) 2 y cosx D 1 = +

A1 sin x 2 Bx sin x 2 − Cx sin x 2 Dx cosx 2

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EnglishUK SSSC LT 2020

Q205.The integrating factor of the following differential equation is: dy ycotx 2cosx dx + =

Asin x Blog sin x Ccot x Dcos x

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EnglishUK SSSC LT 2020

Q206.Degree of following differential equation will be: 3/2 2 2 2 dy d y 1+ = k dx dx

A1 B3 C2 D0

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EnglishHaryana PGT 2020

Q207.Solution of the differential equation 2 y y -1 dy =, dx x x -1 with y(2) = 2 3 is:

Ay = sec–1 1 2x sec 2x 3 − B1 y sec sec x 6 − π = + C1 y sec sec x 6 − π = − D1 y sec sec x 3 − π = +

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EnglishHaryana PGT 2020

Q208.Particular Integral of the differential equation (D3 + 3D2 + 2D) y = x2 is:

A3 2 1 x 3x 7x 2 3 2 2 + − B3 2 1 x 3x 7x 2 3 2 2 − + C3 2 1 x 2x 7x 3 3 3 2 − + D3 2 1 x 3x 7x 3 3 2 2 − +

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EnglishHaryana PGT 2020

Q209.Solution of differential equation (1 + ex/y) dx + ex/y x 1- dy = 0 y is:

Ax + yex/y = c By + xex/y = c Cy – yex/y = c Dx – yex/y = c

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EnglishUP PCS (Pre) 1998

Q210.The substitution y = v secx transforms the differential equation 2 x 2 d y dy 2tanx 5y e secx dx − + = − + = − + = − + = to:

A2 x 2 d v 3v e dx + = B2 x 2 d v 4v e dx + = C2 x 2 d v 5v e dx + = D2 x 2 d v 6v e dx + =

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EnglishLT 2018

Q211.The slope of the tangent at the point P(x,y) on a curve is y 3 x 2 + − +. If the curve Passes through the origin, then the equation of the curve is

Axy 2y 3x 0 + + = B2 2 x y 2x 3y 0 − + − = Cxy 6x 0 + = Dxy 2y 3x 0 − + =

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EnglishLT 2018

Q212.If y(x) is a solution of the differential equation dy 2xy x,y(0) 0 dx + = + = + = + = then ((()) x limy x →∞ is

A–1/2 B–1 C1/2 D1

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EnglishLT 2018

Q213.If y=y(x) and (()) 2 sinx dy cosx, y 1 dx + = − + y(0)=1, then y 2 π is equal to

A1 B2/3 C−1/3 D1/3

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EnglishLT 2018

Q214.A solution of the differential equation 2 2 1 x dy 1 y dx 0 − + − = − + − = − + − = − + − = ((()) | x | 1, | y | 1 < < < < is

A2 2 x 1 y y 1 x c − + − = B1 1 xsin y ysin x c − − + = C2 2 x y c 1 x 1 y + = − − D2 2 x 1 x y 1 y c − + − =

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EnglishLT 2018

Q215.The solution of the differential equation (((()))) 3 dy x 2y y,y(0) 1 dx + = + = + = + = is

A3 x y y 0 + − = B3 x y y 0 − + = C3 x 2y 2y 0 − − − = D3 x 2y 2y 0 + − =

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EnglishDSSSB TGT (SECTION-B) 28 DEC 2014

Q216.If x y x y then dy dx is–

Ax/y By/x C1 log x/1 log y Dx 1 log y

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EnglishDSSSB TGT (SECTION-B) 28 DEC 2014

Q217.If ax bx f(x) be ae, then f (0) is equal to–

A0 B2ab Cab (a + b) Dab

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EnglishSECTION B PGT TIER-I 31.11.2014

Q218.The integrating factor of the differential equation dy (xlogx) y 2logx dx is–

Aex Blog x Clog (log x) Dx

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EnglishSECTION B PGT TIER-I 31.11.2014

Q219.Solution of differential equation xdy – ydx = 0 represents–

AA rectangular hyperbola BParabola whose vertex is at origin CStraight line passing through origin DA circle whose centre is at origin

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EnglishSECTION B PGT TIER-I 31.11.2014

Q220.If x y e (Acosx Bsinx), then y is a solution of

A2 d y dy 2 0 dx B2 d y dy 2 2y 0 dx C2 d y dy 2 2y 0 dx D2 d y 2y 0 dx

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