The integral $\int_{0}^{1} \frac{1}{7^{\left[\frac{1}{x}\right]}} dx$, where $[.]$ denotes the greatest integer function is equal to

(A) $1+6 \log _{e}\left(\frac{6}{7}\right)$

(B) $1-6 \log _{e}\left(\frac{6}{7}\right)$

(C) $\log _{e}\left(\frac{7}{6}\right)$

(D) $1-7 \log _{e}\left(\frac{6}{7}\right)$

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Definite Integrals

The integral

\int_{0}^{1} \frac{1}{7 ^{[\frac{1}{x}]}} d x

, where

[.]

denotes the greatest integer function is equal to

Answer: (A)

Breaking the integral into intervals where

[\frac{1}{x}]

is constant:

\int_{0}^{1} \frac{1}{7 ^{[1/ x]}} d x = \int_{1/2}^{1} \frac{1}{7} d x + \int_{1/3}^{1/2} \frac{1}{7 ^{2}} d x + \int_{1/4}^{1/3} \frac{1}{7 ^{3}} d x + \dots

= (\frac{1}{7} + \frac{1}{2 \cdot 7 ^{2}} + \frac{1}{3 \cdot 7 ^{3}} + \dots) - (\frac{1}{7 \cdot 2} + \frac{1}{7 ^{2} \cdot 3} + \dots)

Using

ln (1 - x) = - (x + \frac{x ^{2}}{2} + \frac{x ^{3}}{3} + \dots)

= - ln (1 - \frac{1}{7}) - 7 (\frac{1}{7 ^{2} \cdot 2} + \frac{1}{7 ^{3} \cdot 3} + \dots)

= - ln \frac{6}{7} - 7 (- ln (1 - \frac{1}{7}) - \frac{1}{7})

= 6 ln \frac{6}{7} + 1 = 1 + 6 ln (\frac{6}{7})

Answer: (A)

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Single correct option