∫1x2 + 4x2 + 9dx = Atan-1x2&nbs

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 Multiple Choice QuestionsMultiple Choice Questions

681.

01xtan-1xdx =

  • π4 + 12

  • π4 - 12

  • 12 - π4

  • - π4 - 12


682.

If 19 - 16x2dx = αsin-1βx + c, then α + 1β =

  • 1

  • 712

  • 1912

  • 912


683.

If 0π2logcosxdx = π2log12, then 0π2logsecxdx =

  • π2log12

  • 1 - π2log12

  • 1 + π2log12

  • π2log2


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684.

1x2 + 4x2 + 9dx = Atan-1x2 + Btan-1x3 + C, then A - B =

  • 16

  • 130

  • - 130

  • - 16


A.

16

Let I = 1x2 + 4x2 + 9dx       = 1x2 + 9 - x2 - 41x2 - 4 - 1x2 + 9dx       = 151x2 - 4 - 1x2 + 9dx       = 151x2 + 22dx - 1x2 +32dx       = 1512tan-1x2 - 13tan-1x3 + C       = 110tan-1x2 - 115tan-1x3 + CBut, it is given that, I  = Atan-1x2 + Btan-1x3 + C Atan-1x2 + Btan-1x3 + C        = 110tan-1x2 - 115tan-1x3 + COn comparing both sides, we get     A = 110 and B = - 115Now, A - B = 110 + 115 15 + 10150 = 25150 = 16


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685.

If x - 5x - 7dx = Ax2 - 12x +35 + logx - 6 + x2 - 12x + 35 + C, then A =

  • - 1

  • 12

  • - 12

  • 1


686.

03xdx = ..., where [x] is greatest integer function

  • 3

  • 0

  • 2

  • 1


687.

sec8xcscxdx =

  • sec8x8 + c

  • sec7x7 + c

  • sec6x6 + c

  • sec9x9 + c


688.

The tangent to the graph of a continuous function y = f(x) at the point with abscissa x = a forms with the X-axis an angle of π3 and at the point with abscissa x = b an angle of π4, then what the value of integral abf'x + f''xdx

(where f'(x) the derivative off w.r.t. xwhich is assumed to be continuous and similarly f"(x) the double derivative of f w.r.t. x)

  • eb3ea

  • eb - 3ea

  • eb - 3ea

  • - eb - 3ea


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689.

The point of inflection of the function y = 0xt2 - 3t + 2dt is

  • 32, 34

  • - 32, - 34

  • - 12, - 32

  • 12, 32


690.

The value of integral dxxx2 - a2

  • c - 1asin-1ax

  • c - 1acos-1ax

  • sin-1ax + c

  • c + 1asin-1ax


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