Math Chapter Wise

EnglishTGT 2009

Q241.The order and degree of differential equation 2 d y dy dy 2x 1 0 dx dx dx + + + = + + + = + + + = + + + =

A(2, 1) B(2, 2) C(1, 2) DNone of these

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EnglishTGT 2004

Q242.If y2 = P (x), is a polynomial of degree 2, then 2 3 2 d d y 2 y dx dx equal to:

AP(x) d 3 d P(x) x BP(x) d 2 d P(x) x C2 d P()... 2 dP(x) x dx dx D2 d 2 dP(x) x

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EnglishTGT 2004

Q243.The slope of the tangent to the curve represented by x=t2 +3t–8 and y=2t2 –2t–5 at the point (2,–1) is

A7/6 B6/7 C8/5 D– 1 2

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EnglishTGT 2003

Q244.Solution of the differential equation 2ysinx dy dx =2sinx cosx–y2 cosx, x 2 π =, y=1 is given by

Ay2 = sin x By= sin2 x Cy2 = cos x + 1 DNone of these

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EnglishJharkhand TGT 2017

Q245.The general solution of the differential equation [ [ ] ] ] ((()) y y xsec ydx xdy ycosec xdy ydx x x + = − + = − + = − + = − is

Ay sin xyc x = By log sec x Cx cos c y = Dsin xy c =

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EnglishTGT 2001

Q246.Solution of (yz+z2)dx – xzdy + xydz = 0 is:

Az = c (y+z) Bx = c (y+z) Cxz = c (y+z) DNone of these

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EnglishTGT 2001

Q247.If xy = c2 then ax + by what will be maximum value?

A2abc B2c ab C2c ab D2c abc

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EnglishTGT 2001

Q248.(()) dy sec x y dx = + = + solution is:

Ax y y tan c 2 + + = Bx y y tan c 2 + − = Cx y y tan c 2 − − = DNone of these

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EnglishTGT 2001

Q249.(((()))) x y x y dy e e e dx − = − = − solution is:

Ax y x e e e 1 ce− = − + B2 y x log x e e ce− = + Cx y x e e e 1 ce = − + Dx y x e e e ce − + = +

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EnglishTGT 2001

Q250.The general solution of differential yp – xp2 – a = 0 is.

Ay = cx + a By = cx – a Cy = cx Dy = cx + a/c

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EnglishTGT 1999

Q251.Differential equation (x-1)dy = ydx, y(0) = -5 then equation the solution is -

Ay = 5 (x-1) By = -5 (x-1) Cy = x (5+1) Dy = -5 (x+1)

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EnglishTGT 1999

Q252.Differential equation (D3 +3D2 +3D+1)y=e2x, where d D dx = then equation the solution is-

Ay = (C1+ C 2x+ C 3x2)e2x + 27 1 ex By = (C 1+ C 2x+ C3x2 + 27 1) e2x Cy = (C 1+ C 2x+ C 3x2 + 27 x)e2x DNone of these

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EnglishTGT 1999

Q253.2 2 (x y)dx 2xy dy + = + = solution is:

Akx = (x2 – y2) Bk (x2 – y2) = x C(x2 – y2) = x DNone of these

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EnglishTGT 1999

Q254.In what form is the differential equation y = px + f (p)?

ANewtonian Form BEuler's Form CGaussian Form DClairaut's Form

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EnglishPGT 2013

Q255.2 2 = + + + = + + + = + + + = + + + Z ax by a b Fractional generated from differential is the equation-

Az = px+py Bz = px+qy+q2 +p2 Cz = qx+py+p2 +q2 Dz = px+qy+p2 +q2

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EnglishPGT 2013

Q256.What will be the solution to the differential? (())) ((()) 2 2 0 + − − = + − − = + − − = + − − = x y dy x y dx

A2 2 2 + − = y x xy c B2 2 + + = xy y x c C2 2 4 + + = x xy y c D2 2 + − = xy y x c

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EnglishPGT 2013

Q257.The solution of the differential equation 2 2 + = + + = + + = + + = + p q x y is-

A2 2 = + + + Z x y a b B() () 3 3 2 2 – 3 3 = + + + Z a x y a b C2 2 = + − − Z x y a b D() 3 2 3 = − + Z x a b

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EnglishPGT 2013

Q258.The differential equation M∂ ∂ ∂x+N∂ ∂ ∂y=0 is fact if-

A∂ ∂ = ∂ ∂ M M x y B∂ ∂ = ∂ ∂ M N x y C∂ ∂ = ∂ ∂ N N x y D∂ ∂ = ∂ ∂ M N y x

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EnglishPGT 2013

Q259.The differential equation 2 3 4 0 − − = − − = − − = − − = d y dy y dx what will be the solution -

A4 1 2 y = c − − + x x e c e B4 1 2 y = c + x x e c e C4 1 2 y = c − + x x e c e D4 1 2 y = c − + x x e c e

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EnglishPGT 2013

Q260.The differential equation 3 2 2 2 1 10 + = dy dx d y dx of order and degree respectively -

A2, 2, B2, 6 C2, 3 D1, 6

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