QJEEAdv21Mae22d: JEE Advanced 2021 Mathematics - Class 12 JEEAdv21Mae22d Solution: For any real number x, let [ x ] denote the largest integer…

Question

Verified Solution • 2025–26 Session moderate

For any real number x, let [ x ] denote the largest integer less than or equal to x. If

I = \int\limits_0^{10} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx}

, then the value of 9I is __________.

Accepted Answer

Given,

I = \int_0^{10} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]d} x

.... (i)

{{10x} \over {x + 1}} = 1 \Rightarrow 10x = x + 1 \Rightarrow x = {1 \over 9}

{{10x} \over {x + 1}} = 4 \Rightarrow 10x = 4x + 4 \Rightarrow x = {2 \over 3}

{{10x} \over {x + 1}} = 9 \Rightarrow 10x = 9x + 9 \Rightarrow x = 9

From Eq. (i), we get

I = \int_0^{1/9} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_{1/9}^{2/3} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_{2/3}^9 {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_9^{10} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx} } } }

= \int_0^{1/9} {0.dx + \int_{1/9}^{2/3} {1.dx + \int_{2/3}^9 {2.dx + \int_9^{10} {3.dx} } } }

= 0 + [x]_{1/9}^{2/3} + [2x]_{2/3}^9 + [3x]_9^{10}

= {2 \over 3} - {1 \over 9} + 18 - {4 \over 3} + 30 - 27 = 21 - {2 \over 3} - {1 \over 9} = {{182} \over 9}

Hence, 9I = 182

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