Let f (x) = \frac{xcotx}{x + cotx} where x \in \left(\right. 0 , \textrm{ }\textrm{ } \frac{\pi}{2} \left.\right) then

Engineering

Mathematics

Methods to Evaluate Limits

Maxima and Minima

Integration by Substitution

Question

Let f (x) = $\frac{xcotx}{x + cotx}$ where $x \in (0, \frac{π}{2})$ then

$\underset{x \to 0^{+}}{Lim} {(\frac{x}{f (x)})}^{\frac{x}{x - sinx}}$ is equal to e⁶.

f (x) has exactly one point of local maxima in $(0, \frac{π}{2})$ .

$\int_{0}^{\frac{π}{2}} f (x) dx = \frac{π}{2}$

f (x) has exactly one point of local minima in $(0, \frac{π}{2})$ .

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$f (x) = \frac{x}{1 + xtanx}$

now f ' (x) = $\frac{1 + xtanx - x ({xsec}^{2} x + tanx)}{{(1 + xtanx)}^{2}} = \frac{1 - x^{2} \sec^{2} x}{{(1 + xtanx)}^{2}} = \frac{\cos^{2} x - x^{2}}{\cos^{2} x {(1 + xtanx)}^{2}}$

$= \frac{(cosx - x) (cosx + x)}{\cos^{2} x {(1 + xtanx)}^{2}} = (cosx - x) \cdot \underset{+ ve}{\underset{⏟}{\frac{(cosx + x)}{\cos^{2} x {(1 + xtanx)}^{2}}}}$

only one maxima in $(0, \frac{π}{2})$