Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

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91.

If for every integer n, nn + 1fxdx = n2, then the value of - 24f(x)dx is

  • 16

  • 14

  • 19

  • None of these


C.

19

Given, nn + 1fxdx = n2On putting n = - 2, - 1, 0, 1, 2, 3, we get   - 2- 1fxdx = 4, - 10fxdx = 1      01fxdx = 0, 34fxdx = 9 - 24fxdx = 4 + 1 + 0 + 1 + 4 + 9 = 19


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92.

The value of integral 0πxfsinxdx is

  • 0

  • π0π2fsinxdx

  • π40πfsinxdx

  • None of these


93.

Solution of Given equation is    dydx = xlogx2 + xsiny + ycosy siny + ycosydy = xlogx2 + xdxOn integrating both sides, we get  siny + ycosydy = xlogx2 + xdx - cosy + ysiny + cosy = x22logx2 is

  • ysiny = x2logx + C

  • ysiny = x2 + C

  • ysiny = x2 + logx + C

  • ysiny = xogx + C


94.

limx02xxexdxe4x2 equals

  • 0

  • 2

  • 12


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95.

If xpyq = (x + y)p + q, then dydx is equal to

  • yx

  • pyqx

  • xy

  • qypx


96.

Area bounded by the curve y2 = 16x and line y = mx is 23, then m is equal to

  • 3

  • 4

  • 1

  • 2


97.

The equation zz + az + az + b = 0, represents a circle, if

  • a2 = b

  • a2 > b

  • a2 <b

  •  None of the above


98.

Area included between curves y = x2 - 3x + 2 and y = - x2 + 3x - 2 is

  • 16 sq unit

  • 12 sq unit

  • 1 sq unit

  • 13 sq unit


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99.

The solution of dydx + ytanx = secx is

  • ysecx = tanx + C

  • ytanx = secx + C

  • tanx = ytanx + C

  • xsecx = ytany + C


100.

If y = tan-1sinx + cosxcosx - sinx, then dydx is

  • 12

  • π4

  • 0

  • 1


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