Question

The function $f\left(x\right)=tanx+\frac{1}{x},\forall x\in \left(0,\frac{\pi }{2}\right)$ has

• A. one local maximum
• B. one local minimum
• C. one local maximum and one minimum
• D. no local maximum or minimum

Answer 1

Answer: (B)

$\frac{dy}{dx}=se{c}^{2}x-\frac{1}{x^2}$
$=\frac{\left(x-cosx\right)\left(x+cosx\right)}{{\left(cos\right)}^{2}x.x^2}$

Hence, at $\alpha$ sign of $\frac{dy}{dx}$ changes from negative to positive
i.e., $y=f\left(x\right)$ has a local minimum