The slope of the tangent at (x, y) to a curve passing through&nbs

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 Multiple Choice QuestionsMultiple Choice Questions

151.

A spherical iron ball of radius 10cm is coated with a layer of ice of uniform thickness that melts a rate of 50cm3/min When the thickness of the ice 5cm, then the rate at which the thickness (in cm/min) of the ice decreases, is :

  • 56π

  • 118π

  • 19π

  • 136π

     


152.

If x + y = 8, then maximum value of x2y is

  • 20489

  • 204881

  • 20483

  • 204827


153.

If f(x) = x3 - 9x + 1 and x = 3, then using Newton Raphson method, first iteration is

  • 5318

  • 539

  • 3518

  • 359


154.

The maximum value of logxx is

  • 0

  • 2

  • 1/e

  • - 1


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155.

If a particle moves such that the displacement is proportional to the square of the velocity acquired, then its acceleration is

  • proportional to s2

  • proportional to 1s2

  • proportional to 1s

  • a constant


156.

The function fx = tan-1sinx + cosx, x > 0 is always an increasing function on the interval

  • 0, π

  • 0, π2

  • 0, π4

  • 0, 3π4


157.

A ladder 10 m long rests against a vertical wall with the lower end on the horizontal ground. The lower end of the ladder is pulled along the ground away from the wall at the rate of 3 cm/s. The height of the upper end while it is descending at the rate of 4 cm/s, is

  • 43 m

  • 53 m

  • 52 m

  • 6 m


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158.

The slope of the tangent at (x, y) to a curve passing through 1, π4 is given by yx - cos2yx, then the equation of the curve is

  • y = tan-1logex

  • y = xtan-1logxe

  • y = xtan-1logex

  • None of the above


C.

y = xtan-1logex

 According to the given condition,dydx = yx - cos2yxOn putting y = vx            dydx = v + xdvdx, we get       v + xdvdx = v - cos2v   dvcos2v = - dxx sec2vdv = - 1xdxOn integrating both sides, we get       tanv  = - logx + logc tanyx = - logx + logcSince, this curve is passing through (1, π4). tanπ4 = - log1 + logc tanyx = - logx + 1          y = xtan-1logex


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159.

If  the function f(x) = 2x3 - 9ax2 + 12a2x + 1 attains its maximum and minimum at p and q respectively such that p2  = q, then a equals 

  • 0

  • 1

  • 2

  • None of these


160.

On the interval [0, 1] the function x25(1 - x)75 takes its maximum value at the point

  • 0

  • 1/4

  • 1/2

  • 1/3


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