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

61.

The value of

100011 × 2 + 12 ×3 + 13 ×4 + ... + 1999 × 1000

  • 1000

  • 999

  • 1001

  • 1999


62.

Six positive numbers are in GP, such that their product is 1000. If the fourth term is 1, then the last term is

  • 1000

  • 100

  • 1100

  • 11000


63.

Five numbers are in AP with common difference  0. If the 1st, 3rd and 4th terms are in GP, then

  • the 5th term is always 0.

  • the 1st term is always 0.

  • the middle term is always 0

  • the middle term is always - 2.


Advertisement

64.

The sum of series

11 × 2C025 + 12 × 3C125 + 13 × 4C225 + ... + 126 × 27C2525

is

  • 227 - 126 × 27

  • 227 - 2826 × 27

  • 12226 + 126 × 27

  • 226 - 152


B.

227 - 2826 × 27

Given series is,

11 × 2C025 + 12 × 3C125 + 13 × 4C225 + ... + 126 × 27C25250x1 + x25dx = 0xC025 + C125x + C225x2 + ... + C2525x25dxOn integrating w.r.t. x, taking limits 0 to x, we get1 + x26260x = C025 + C125 . x22 + C225 . x33 + ... + C25 25. x26260x 1261 + x26 - 126 = C025 + C125 . x22 + C25 25. x2626Again, integrating w.r.t. x, taking limits 0 to 1, we get

126011 + x26 - 1dx= 01C025x + C025 . x22 + ... + C2525x2626dx 1261 + x2727 - x01= C025 . x22 +  C125 . x32 × 3 +  ... + C2525x2726 × 2701 12622727 - 1 - 127 = 12C025 + 12 × 3C125 + ... + 126 × 27C2525 11 × 2 . C025 + 12 × 3C125 + 13 × 4C225 + ... + 126 ×27C2525= 227 - 2826 × 27


Advertisement
Advertisement
65.

Let f : R  R  be such that f is injective and f(x) f(y) = f(x + y) for all x, y if f(x), f(y) and f(z) are in GP, then x, y and z are in

  • AP always

  • GP always

  • AP depending on the values of x, y and z

  • GP depending on the values of x, y and z


66.

If P = 1 + 12 × 2 + 13 × 22 + ... and Q = 11 × 2 + 13 × 4 + 15 × 6 + ...,

then

  • P = Q

  • 2P =Q

  • P = 2Q

  • P = 4Q


67.

If x = 1 + 12 × 1! + 14 × 2! + 18 × 3! + ... and y = 1 + x21! + x42! + x63! +... Then, the value of logey is

  • e

  • e2

  • 1

  • 1e


68.

The value of the infinite series

12 + 223! + 12 + 22 + 324! + 12 + 22 + 32 +425! + ... is

  • e

  • 5e

  • 5e6 - 12

  • 5e6


Advertisement
69.

The sum of the series 1 + 12C1n + 13C2n + ... + 1n + 1Cnn is equal to

  • 2n + 1 - 1n + 1

  • 32n - 12n

  • 2n + 1n + 1

  • 2n + 12n


70.

The value of r = 21 + 2 + ... + r - 1r!

  • e

  • 2e

  • e2

  • 3e2


Advertisement