JEE Advance - Mathematics (2021 - Paper 2 Online - No. 19)
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 __________.
Answer
182
Explanation
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
$${{10x} \over {x + 1}} = 1 \Rightarrow 10x = x + 1 \Rightarrow x = {1 \over 9}$$
$${{10x} \over {x + 1}} = 4 \Rightarrow 10x = 4x + 4 \Rightarrow x = {2 \over 3}$$
$${{10x} \over {x + 1}} = 9 \Rightarrow 10x = 9x + 9 \Rightarrow x = 9$$
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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