JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Definite Integrals

Let a be a positive real numbers such that

\int_{0}^{a} e^{x - [x]} d x = 10 e - 9

where

[x]

is the greatest integer less than or equal to

x

. Then a is equal to :

Answer: (A)

Solution

On each interval

[k, k + 1)

the integrand equals

e^{x - k}

, and

\int_{k}^{k + 1} e^{x - k} d x = e - 1

Writing

a = n + α

with

n

a non-negative integer and

0 \leq α < 1

\int_{0}^{a} e^{x - [x]} d x = n (e - 1) + \int_{n}^{n + α} e^{x - n} d x = n (e - 1) + e^{α} - 1

Comparing with

10 e - 9

, the coefficient of

e

gives

n = 10

, and then

10 e - 10 + e^{α} - 1 = 10 e - 9

leaves

e^{α} = 2

, so

α = lo g_{e} 2

, which lies in

[0, 1)

as required. The value is

a = 10 + lo g_{e} 2

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