The smallest positive root of the equation ta | vedclass

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(C) The given equation is $	an x - x = 0$,which implies $	an x = x$.To find the roots,we look for the points of intersection of the curves $y = 	an

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$\left(\pi, \frac{3\pi}{2}\right)$

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(C) The given equation is $	an x - x = 0$,which implies $	an x = x$.To find the roots,we look for the points of intersection of the curves $y = 	an x$ and $y = x$.At $x = 0$,both curves intersect at the origin,but we are looking for the smallest positive root.For $x \in \left(0, \frac{\pi}{2}ight)$,$	an x > x$,so there is no intersection in this interval.For $x \in \left(\frac{\pi}{2}, \piight)$,$	an x$ is negative while $x$ is positive,so there is no intersection.For $x \in \left(\pi, \frac{3\pi}{2}ight)$,the curve $y = 	an x$ starts from $-\infty$ at $x = \pi^+$ and increases to $+\infty$ at $x = \frac{3\pi}{2}^-$. Since the line $y = x$ is positive in this interval,the curve $y = 	an x$ must intersect the line $y = x$ at exactly one point.Thus,the smallest positive root lies in the interval $\left(\pi, \frac{3\pi}{2}ight)$.

Molecule type	$\frac{C_P}{C_V}$
$A$. Monoatomic	$I$. $\frac{7}{5}$
$B$. Diatomic rigid molecules	$II$. $\frac{9}{7}$
$C$. Diatomic non-rigid molecules	$III$. $\frac{4}{3}$
$D$. Triatomic rigid molecules	$IV$. $\frac{5}{3}$

The smallest positive root of the equation $\tan x - x = 0$ lies in

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