∫x49tan-1x501 + x100dx = ktan-1x502 

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 Multiple Choice QuestionsMultiple Choice Questions

641.

022x - 22x - x2dx is equal to

  • 0

  • 2

  • 3

  • 4


642.

If sinxcosx1 + cosxdx = f(x) + c, then f(x) is equal to

  • log1 + cosxcosx

  • logcosx1 + cosx

  • logsinx1 + sinx

  • log1 + sinxsinx


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643.

x49tan-1x501 + x100dx = ktan-1x502 +c, then k is equal to

  • 150

  • - 150

  • 1100

  • - 1100


C.

1100

Let I = x49tan-1x501 + x100dx = ktan-1x502 +cLet x50 = t 50x49dx = dt I = 150tan-1t1 +t2dtLet tan-1t = u  11 +t2dt = du   I = 150udu = u2100 + c         = tan-1x502100 + cBut I = ktan-1x502 +c        as given ktan-1x502 + c = 1100tan-1x502 + c k = 1100


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644.

0π2200sinx + 100cosxsinx + cosxdx is equal to

  • 50π

  • 25π

  • 75π

  • 150π


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645.

0πθsinθ1 + cos2θ is equal to

  • π22

  • π23

  • π2

  • π24


646.

If sin-12x1 + x2dx = fx - log1 + x2 then f(x) is equal to

  • 2xtan-1(x)

  • - 2xtan-1(x)

  • xtan-1(x)

  • - xtan-1(x)


647.

If xa3 - x3dx = g(x) +c, then gx is equal to :

  • 23cos-1x

  • 23sin-1x3a3

  • 23sin-1x3a3

  • 23cos-1xa


648.

If dxx2 + 2x + 2 = fx + c, then f(x) is equal to

  • tan-1x + 1

  • 2tan-1x + 1

  • - tan-1x + 1

  • 3tan-1x + 1


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649.

Observe the following statement:

A : x2 - 1x2ex2 + 1x2dx = ex2 + 1x2 + cR : f'xefxdx = fx + c

Then which of the followmg is true ?

  • Both A and R are true and R is not thecorrect reason for A

  • Both A and R are true and R is the correct reason for A

  • A is true, R is false

  • A is false, R is true


650.

Dividing the interval [0, 6] into 6 equal parts and by using trapezoidal rule the value of 06x3dx is approximately

  • 330

  • 331

  • 332

  • 333


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