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 Multiple Choice QuestionsMultiple Choice Questions

181.

The maximum value of fx = logxxx  0, x  1 is

  • e

  • 1e

  • e2

  • 1e2


182.

If the volume of spherical ball is increasing at the rate of 4π cm3/s, then the rate of change of its surface area when the volume is 288 π cm3, is

  • 43π cm2/s

  • 23π cm2/s

  • 4π cm2/s

  • 2π cm2/s


183.

The equation of displacement of a particle is x(t) = 5t2 - 7t + 3. The acceleration at the moment when its velocity becomes 5 m/sec is

  • 3 m/sec2

  • 7 m/sec2

  • 10 m/sec2

  • 8 m/sec2


184.

The mean value of the function fx = 2ex + 1 on the interval [0, 2] is

  • 2 - loge2e2 + 1

  • 2 + loge2e2 + 1

  • 2 + loge2e2 - 1

  • - 2 + loge2e2 - 1


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185.

The function y = 2x - x2

  • increases in (0, 1) but decreases in (1, 2)

  • decreases in (0, 2)

  • increases m (1, 2) but decreases in (0, 1)

  • increases in (0, 2)


186.

The interval in which the function y = x - 2sinx0  x  2π increases throughout is

  • 5π3, 2π

  • 0, π3

  • π3, 5π3

  • 0, π4


187.

The points of the curve y = x3 + x - 2 at which its tangent are parallel to the straight line y = 4x - 1 are

  • (2, 7), (- 2, - 11)

  • (0, 2), (21/3, 21/3)

  • (- 21/3, - 21/3), (0, - 4)

  • (1, 0), (- 1, - 4)


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188.

The equation of the normal to the curve y = - x + 2 at the point of its intersection with the bisector of the first quadrant is

  • 4x - y + 16 = 0

  • 4x - y = 16

  • 2x - y - 1 = 0

  • 2x - y + 1 = 0


C.

2x - y - 1 = 0

Given, y = - x + 2      ...iAnd bisector of first quadrant is           y = x                     ...iiOn solving Eqs. (i) and (ii), we get           x = 1, 4 From Eq. (i) Points are (1, 1) and (4,0)But(4, 0) notsatJSfy Eq. (n)Point(1, 1) is only point of intersectlon of curve (i) and line(ii)Now, slope of tangent of curve (i) is       dydx = - 12x Slope of tangent at point (1, 1) isdydx1, 1 = - 12 Slope of normal at point (1, 1) is

           = - 1dydx1, 1 = - 1- 12 = 2

 Equation of the normal to the curve at point (1, 1) will be

y - 1 = 2(x - 1)  2x - y - 1 = 0

 


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189.

The angle at which the curve y = x2 and the curve x = 53cost, y = 54sint intersect is

  • tan-1241

  • tan-1412

  • - tan-1241

  • 2tan-1412


190.

The maximum value of the function y = 2tanx - tan2x over 0, π2 is

  • 1

  • 3

  • 2


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