If 2x + 2y = 2x  then the value of dydx at (1, 1)

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 Multiple Choice QuestionsMultiple Choice Questions

761.

The number of points at which the function fx = 1logex is discontinuous, is

  • 1

  • 2

  • 3

  • infinitely many


762.

The functions f, g and h satisfy the relations f'(x) = g (x + 1) and g'(x) = h(x - 1). Then, f"(2x) is equal to

  • h(2x)

  • 4h(2x)

  • h(2x - 1)

  • h(2x + 1)


763.

If f(x) = sin-11 - cos2x2sinx, then f'x is equal to

  • sinx

  • x

  • 0

  • 1


764.

If y xx + 1 + x + 1x, then d2ydx2 at x = 1 is equal to

  • 74

  • 78

  • 14

  • - 78


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765.

If 2x + 2y = 2 then the value of dydx at (1, 1) is equal to

  • - 2

  • - 1

  • 0

  • 1


B.

- 1

We have, 2x + 2y = 2x + y

On differentiating w.r.t. x, we get

2xlog2 + 2ylog2dydx = 2x + ylog21 + dydx             2x + 2ydydx = 2x + y1 + dydx                         dydx = 2x - 2x + y2x + y - 2y              dydx1, 1 = 2 - 44 - 2 = - 1


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766.

Let y = tan-1secx + tanx. Then dydx is equal to

  • 14

  • 12

  • 1secx + tanx

  • 1sec2x


767.

If s = sec-112x2 - 1 and t = 1 - x2, then dsdt at x = 12 is

  • 1

  • 2

  • - 2

  • 4


768.

If y = f(x) is continuous on [0, 6], differentiable on (0, 6), f(0) = - 2 and f(6) = 16, then at some point between x = 0 and x = 6, f'(x) must be equal to

  • - 18

  • - 3

  • 3

  • 14


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769.

If f(x) = 2x + 42x, then f'(2) is equal to

  • 0

  • - 1

  • 1

  • 2


770.

Let f(x + y) = f(x) f(y) and f(x) = 1 + sin(3x) g(x), where g is differentiable.The f'(x) is equal to

  • 3f(x)

  • g(0)

  • f(x)g(0)

  • 3g(x)


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