If y = tan-111 + x + x2 + 

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291.

The function f(x) = x - 2 +x is

  • differentiable at both x = 2 and x = 0

  • differentiable at x = 2 but not at x = 0

  • continuous at x = 2 but not at x = 0

  • continuous at both x = 2and x = 0


292.

If y = tan-1x2 - 1, then the ratio d2ydx2 : dydx is

  • xx2 - 11 - 2x2

  • 1 - 2x2xx2 - 1

  • 1 + 2x2xx2 + 1

  • xx2 + 11 - 2x2


293.

If f(x) = fx = logxx - 1, if x  1k,          if x  1 is continuous at x = 1, then the value of k is

  • 0

  • - 1

  • 1

  • e


294.

If r = aeθcotα where a and α, are real numbers, then d2r2 - 4rcot2α is

  • r

  • 1r

  • r

  • 0


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295.

The derivative of tan-1sinx1 + cosx with respect to tan-1cosx1 + sinx is

  • 2

  • - 1

  • 0

  • - 2


296.

ddxcoscot-12 +x2 - x is

  • 14

  • 12

  • - 12

  • - 34


297.

If y = loge1 + x + x2 + .. , then dydx is equal to

  • 11 +x2

  • 11 -x2

  • - 11 +x2

  • - 11 -x2


298.

Length of the subtangent at (x1, y1) on xnym = am + n, m, n > 0, is

  • nmx1

  • mnx1

  • nmy1

  • nmx1


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299.

If y = tan-111 +x+ x2 + tan-11x2 +2x+ 3 + tan-11x2 +5x+ 7 + ... n terms, then y'(0) is

  • π2

  • 2n1 + n2

  • n21 + n2

  • - n21 + n2


D.

- n21 + n2

Given y = tan-111 +x+ x2 + tan-11x2 +2x+ 3                   + tan-11x2 +5x+ 7 + ... n           ...iNow, tan-111 +x+ x2 = tan-1x +1 - x1 + xx + 1                                        = tan-1x +1 - tan-1xFrom Eq. (i), we get     y = tan-1x +1 - tan-1x + tan-1x +2 - tan-1x + 1              + tan-1x +n - tan-1n y = tan-1x +n - tan-1xOn differentiating w.r.t. 'x', we getdydx = 11 + x + n2 - 11 + x2dydx = 11 + n2 - 1      =  1 - 1 + n21 + n2     = - n21 + n2


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300.

If f(x) = x2 - a + 2x + ax - 2, x  22,                              x = 2 is continuous at x = 2, then at x = 2, then the value of a is

  • - 6

  • 0

  • 1

  • - 1


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