Question

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

$\frac{d y}{d x}=sec^{2} x-\frac{1}{x^{2}}$
$=\frac{\left(x - cos x\right) \left(x + cos ⁡ x\right)}{\left(cos\right)^{2} ⁡ x . x^{2}}$

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