If the volume of spherical ball is increasing at the rate of 4&pi

Subject

Mathematics

Class

JEE Class 12

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

1.

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

  • e

  • 1e

  • e2

  • 1e2


2.

If g(x) is the inverse function of f(x) and f'x = 11 +x4, then g'(x) is

  • 1 + [g(x)]4

  • 1 - [g(x)]4

  • 1 + [f(x)]4

  • 11 + g(x)4


3.

The inverse of the matrix 10033052- 1 is

  • - 13- 30031092- 3

  • - 13- 3003- 10- 9- 23

  • - 133003- 10- 9- 23

  • - 13- 300- 3- 10- 9- 23


4.

If the function f(x) = tanπ4 + x1x for x  0 is  = K for x = 0 continuous at x = 0, then K = ?

  • e

  • e- 1

  • e2

  • e- 2


Advertisement
5.

For a invertible matrix A if A(adjA) = 100010 then A =

  • 100

  • - 100

  • 10

  • - 10


6.

If x = f(t) and y = g(t) are differentiable functions of t, then d2ydx2 is

  • f't . g''t - g't . f''tf't3

  • f't . g''t - g't . f''tf't2

  • g't . f''t - f't . g''tf't3

  • g't . f''t + f't . g''tf't3


7.

If α and β are roots of the equation x2 + 5x - 6 = 0, then the value of tan-1α - tan-1β is

  • π2

  • 0

  • π

  • π4


Advertisement

8.

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


A.

43π cm2/s

Let V and r be the volume and radius of spherical ball, respectively.

Volume of spherical ball = 43πr3

       V = 43πr3     ...i 288π = 43πr3   given, V = 288 cm3    288 = 43r3       r3 = 72 × 3 = 8 × 27       r3 = 23 × 33             taking cube roots both sides        r = 6

On differentiating Eq. (i) w.r.t. 't', we get

     dVdt = 4πr2drdt  4π = 4πr2drdt       given dVdt = 4π cubic cm/s    1 = 62drdt      r = 6 drdt = 136Now, surface area of spherical ball, (s) = 4πr2    s = 4πr2

On differentiating both sides, w.r.t. 't', we get

    dsdt = 4 × 2πrdrdt           = 8 × π × 6 × 136           r = 6 and drdt = 136 dsdt = 43π cm2/s


Advertisement
Advertisement
9.

If f(x) = = logsec2xcot2x for x  0= K                         for x = 0 is continuous at x = 0, then K is

  • e- 1

  • 1

  • e

  • 0


10.

If the inverse of the matrix α14- 1231623 does not exist, then the value of α is

  • 1

  • - 1

  • 0

  • - 2


Advertisement