If k∫01x . f3xdx = ∫03t . f

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 Multiple Choice QuestionsMultiple Choice Questions

481.

If f(x) is defined [- 2, 2] by f(x) = 4 - 3x + 1 and g(x) = f- x - fxx2 + 3, then - 22gxdx is equal to :

  • 64

  • - 48

  • 0

  • 24


482.

The value of the integral 0π2sin100x - cos100xdx is

  • 1100

  • 100!100100

  • π100

  • 0


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483.

If k01x . f3xdx = 03t . ftdt, then the value of k is

  • 9

  • 3

  • 19

  • 13


A.

9

Given, k01x . f3xdx = 03t . ftdt    ...iLet    I = k 01xf3xdxLet 3x = t dx = dt3when x = 0, t = 0when x = 1, t = 3 I = k03t3ft . dt3       = k903t ftdtNow, form Eq. (i), we getk903t ftdt = 03t . ftdt        k9 = 1 k = 9


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484.

The value of 11 + cos8xdx is

  • tan2x8 + c

  • tan8x8 + c

  • tan4x4 + c

  • tan4x8 + c


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485.

The value of exx5 + 5x4 + 1dx is

  • ex . x5 + c

  • e. x5 + e+ c

  • ex + 1 . x5 + c

  • 5x4 . ex + c


486.

The value of x2 + 1x2 - 1dx is

  • logx - 1x + 1 + c

  • logx + 1x - 1 + c

  • x +logx - 1x + 1 + c

  • 5x4 . ex + c


487.

ex . x5dx is

  • ex[x5 + 5x4 + 20x3 + 60x2 + 120x + 120] + C

  • ex[x5 - 5x4 - 20x3 - 60x2 - 120x - 120] + C

  • ex[x5 - 5x4 + 20x3 - 60x2 + 120x - 120] + C

  • ex[x5 + 5x4 + 20x3 - 60x2 - 120x + 120] + C


488.

The value of - 22ax3 + bx + cdx depends on the

  • value of b

  • value of c

  • value of a

  • values of a and b


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489.

secxsecx + tanxdx is equal to

  • tanx - secx + C

  • log1 + secx +C

  • secx + tanx +C

  • logsinx - logcosx + C


490.

If fxdx = gx, then fxgxdx is equal to

  • 12f2x

  • 12g2x

  • 12g'x2

  • f'(x)g(x)


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