Question

Let a be a positive real number such that `int_0^ae^(x-[x])dx` = 10e – 9 where [x] is the greatest integer less than or equal to x. Then, a is equal to ______.

Options

• 10 + log_e3

• 10 – log_e(1 + e)

• 10 + log_e2

• 10 + log_e(1 + e)

Answer 1

Let a be a positive real number such that `int_0^ae^(x-[x])dx` = 10e – 9 where [x] is the greatest integer less than or equal to x. Then, a is equal to `underlinebb(10 + log_e2)`.

Explanation:

Since, a > 0, a∈R

Let n ≤ a < n + 1, n∈W

∴ a = [a] + {a}  .....(i)

Where, [a] is greatest integer factor

{a} is fractional integer factor

∴ [a] = n

Given: `int_0^ae^(x-[x])dx` = 10e – 9

⇒ `int_0^n e^((x))dx + int_n^ae^(x-[x])dx` = 10e – 9

⇒ `nint_0^1e^xdx + int_n^ae^(x - n)dx` = 10e – 9

⇒ n(e – e⁰) + (e^x–n – e^n–n) = 10e – 9

⇒ n = 0 and {a} = In 2

Therefore, from equation (i)

a = [a] + {a}

= 10 + In 2