The function f(x) = frac{x}{1 + x tan x} atta | vedclass

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(B) Let $f(x) = \frac{x}{1 + x 	an x}$. To find the maxima of $f(x)$,we can find the minima of its reciprocal $g(x) = \frac{1}{f(x)} = \frac{1 + x

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$x = \cos x$

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(B) Let $f(x) = \frac{x}{1 + x 	an x}$. To find the maxima of $f(x)$,we can find the minima of its reciprocal $g(x) = \frac{1}{f(x)} = \frac{1 + x 	an x}{x} = \frac{1}{x} + 	an x$.Now,differentiate $g(x)$ with respect to $x$:$g'(x) = -\frac{1}{x^2} + \sec^2 x$.For critical points,set $g'(x) = 0$:$-\frac{1}{x^2} + \sec^2 x = 0 \implies \sec^2 x = \frac{1}{x^2} \implies \cos^2 x = x^2 \implies x = \cos x$ (considering the domain where $x > 0$).Now,find the second derivative $g''(x)$:$g''(x) = \frac{2}{x^3} + 2 \sec x (\sec x 	an x) = \frac{2}{x^3} + 2 \sec^2 x 	an x$.At $x = \cos x$,$g''(x) = \frac{2}{\cos^3 x} + 2 \sec^2 x 	an x = 2 \sec^3 x + 2 \sec^2 x 	an x = 2 \sec^2 x (\sec x + 	an x)$.Since $x = \cos x$ lies in the first quadrant,$\sec x > 0$ and $	an x > 0$,so $g''(x) > 0$.Thus,$g(x)$ has a minimum at $x = \cos x$,which implies $f(x)$ has a maximum at $x = \cos x$.

The function $f(x) = \frac{x}{1 + x \tan x}$ attains its maximum value at:

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