If the volume of a sphere increases at the rate of 2π cm3/ s,

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 Multiple Choice QuestionsMultiple Choice Questions

431.

If u = sin-1x4 + y4x + y, then xux + yuy = ?

  • 3u

  • 4u

  • 3sin(u)

  • 3tan(u)


432.

If y = logexx and z = logex, then d2ydx2 + dydz = ?

  • e - z

  • 2e - z

  • ze - z

  • - e - z


433.

If fx = logexx - 2x + 234, then f'0 = ?

  • 14

  • 4

  • - 34

  • 1


434.

If xy  0, x + y  0 and xmyn = x + ym + n,where m, n not belongs to N, then dydx = ?

  • yx

  • x + yxy

  • xy

  • xy


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435.

If x2 + y2 = t + 1t, x4 + y4 = t2 + 1t2, thenx3ydydx = ?

  • - 1

  • 1

  • 0

  • t


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436.

If the volume of a sphere increases at the rate of 2π cm3/ s, then the rate of increase of its radius(in cm/s), when the volume is 288π cm3, is

  • 136

  • 172

  • 118

  • 19


B.

172

Given, dVdt = 2π cm3/s Volume of sphere, V = 43πr3On diffeentiating w.r.t. we getdVdt = 43π × 3r2drdt 2π = 4πr2drdt drdt = 12r2 = 12 × 62 = 172cm/s V = 288π = 43πr3  216 = r3  r = 6


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437.

Let f be a non-zero real valued continuous function satisfying f(x + y) = f(x) · f(y) for all x, y ∈ R. If f(2) = 9, then f(6) is equal

  • 32

  • 36

  • 34

  • 33


438.

limx0 tan3x - sin3xx5 = ?

  • 52

  • 32

  • 35

  • 25


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439.

If fx = 11 + 1x and gx = 11 + 1fx, then g'2 = ?

  • 15

  • 125

  • 5

  • 116


440.

If yx + xy = 2, then dydx = ?

  • x2 + y2x + y

  • x2 - y2x + y

  • 1

  • 2


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