Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

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

241.

The point on the curve y2 = x the tangent at which makes an angle 45° with X-axis ts

  • 14, 12

  • 12, 14

  • 12, - 12

  • 12, 12


242.

The length of the subtangent to the curve x2y2 = a4 at (- a, a)

  • a2

  • 2a

  • a

  • a3


243.

A stone is thrown vertically upwards and the height x ft reached by the stone in t seconds is given by x = 80 t - 16t2 The stone reaches the maximum height in

  • 2 s

  • 2.5 s

  • 3 s

  • 1.5 s


244.

The maximum value of logxx in 2,  is

  • 1

  • 2e

  • e

  • 1e


Advertisement
245.

The tangent to a given curve y = f(x) is perpendicular to the x-axis, if

  • dydx - 1

  • dxdy = 0

  • dxdy = 1

  • dydx = 0


246.

The minimum value of 27cos2x 81sin2x is

  • - 5

  • 15

  • 1243

  • 127


247.

A stone is thrown vertically upwards from the top of a tower 64 m high according to the law s = 48t - 16t2. The greatest height attained by the stone above ground is

  • 36 m

  • 32 m

  • 100  m

  • 64 m


Advertisement

248.

The length of the subtangent at t on the curve x = at + sint, y = a1 - cost

  • asint

  • 2asint2tant2

  • 2asint2

  • 2asin3t2sect2


A.

asint

Given, x = at + sint, y = a1 - cost dxdt = a1 + acost, dydt = asint dydx = asinta1 + acost = tant2 Length of subtangent = ydydx                                       = a1 - costtant2                                      = 2asint2cost2                                      = asint


Advertisement
Advertisement
249.

A wire of length 20 cm is bent in the form of a sector of a circle. The maximum area that can be enclosed by the wire is

  • 20 sq cm

  • 25 sq cm

  • 10 sq cm

  • 30 sq cm


250.

The condition for the line y = mx + c to be a normal to the parabola y2 = 4ax is

  • c = - 2am - am3

  • c = - am

  • c = am

  • c = 2am + am3


Advertisement