JEE Advanced 2021 · Paper 2 · Mathematics Q19

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MathematicsInteger+4Question 57 of 57

For any real number $x$ , let $[x]$ denote the largest integer less than or equal to $x$ . If
$I = \int_0^{10}\left[\sqrt{\frac{10x}{x + 1}}\right]dx,$
then the value of $9I$ is ____.

Your answer

Answer and worked solution

Official answer: 182

Where the floor changes. The integrand $\sqrt{\frac{10x}{x+1}}$ increases from 0 to $\sqrt{100/11}\approx3.02$ on $[0,10]$ . It reaches $k$ when $x=\frac{k^2}{10-k^2}$ :

$k=1$ at $x=\frac19$ ;
$k=2$ at $x=\frac23$ ;
$k=3$ at $x=9$ .

Integral.
$I=1\left(\frac23-\frac19\right)+2\left(9-\frac23\right)+3(10-9)=\frac59+\frac{50}3+3=\frac{182}9.$

Value. $9I=182$ .

Topics:Properties of definite integrals Inequalities and modulus

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