Let f : R → R  be such that f is injective an

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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.


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


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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


A.

AP always

Let the function, f(x) = akx

Which define in f : R  R and injective also.

Now, we have

f(x)f(y) = f(x + y)

 akx . aky = akx + y akx + y = akx + y        f(x), f(y)and f(z)are in GP fy2 = f(x) . f(y)

 a2ky = akx . akz a2ky = ekx + z

On comparing, we get

2ky = k(x + z)

 2y = x + z

 x, y and z are in AP.


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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


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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


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