The two curves x = y2, xy = a2 cut orthogonally at a point, then

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 Multiple Choice QuestionsMultiple Choice Questions

391.

If u = xy2tan-1yx, then xux + yuy is equal to

  • 2u

  • u

  • 3u

  • 13u


392.

If: R  R is defined by f (x) = x - [x], where[x] is the greatest integer not exceeding x, then the set of discontinuous of f is

  • the empty set

  • R

  • Z

  • N


393.

If :R→ R  defined by

f x = a2cos2x + b2sin2x,   x  0         = eax + b,                    x >0is continuous function,  then

 

  • b = 2loga

  • 2b = loga

  • b = log2a

  • b2 = loga


394.

Let f(x) = ex, g(x) = sin - 1x and h(x) = f(g(x)), then

h'xhx is equal to

  • sin-1x

  • 11 - x2

  • 11 - x

  • esin-1x


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395.

If  fx = ax + a2ax, then f'a is equal to

  • 0

  • - 1

  • 1

  • a


396.

If y = aex + be-x + c, where a, b, c are parameters, then x2y" + xy' is equal to 

  • 0

  • y

  • y'

  • y''


397.

If y = acos(log (x)) + bsin (log(x)), where a, b are parameters, then xy" + xy' is equal to

  • y

  • - y

  • 2y

  • - 2y


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398.

The two curves x = y2, xy = a2 cut orthogonally at a point, then a2 is equal to

  • 13

  • 12

  • 1

  • 2


B.

12

We have,               x = y2      ...i 2ydydx = 1  dydx = 12yand      xy = a2      ...iixdydx + y = 0  dydx = y- xon solving equations i and ii, we get the point of interaction a2, a        m1 = dydxa2, a = 12a'and m2 = dydxa2, a = - aa2 = - 1asince, curves cut orthogonally then,m1m2 = - 112a -1a = - 1                   a2 = 12


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399.

If f(x) = 1 + kx - 1 - kxx, for - 1  x< 02x2 + 3x + 2,               for 0  x  1is continuous at x = 0, then k is equal to

  • - 1

  • - 2

  • - 3

  • - 4


400.

If fx =x - 12x2 - 7x + 5, for x  1                 - 13, for x = 1, then f'1 is equal to :

  • - 19

  • - 29

  • - 13

  • 13


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