Question

$f\left(x\right)=\frac{x}{1+x\tan x}$ is maximum when $x=$

• A. $sinx$
• B. $cosx$
• C. $tanx$
• D. $cotx$

Answer 1

The correct option is B $cosx$

$f^{\prime }\left(x\right)=\frac{(1+xtanx)-x(xse{c}^{2}x+tanx)}{(1+xtanx{)}^2}$
$=\frac{1-x^{2}se{c}^{2}x}{(1+xtanx{)}^2}$
$f^{\prime }\left(x\right)=0$ when $x=cosx$
$f^{\prime }\left(x\right)<0$ when $x>cosx$
$f^{\prime }\left(x\right)>0$ when $x<cosx$
So $f\left(x\right)$ is minimum at $x=cosx$