Function \( f(x)=x+\frac{1}{x},x\ne 0 \) is increasing in the interval
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( (-\infty,-1)\cup(1,\infty) \)
\( \displaystyle f'(x)=1-\frac{1}{x^2} \)
\( \displaystyle =\frac{x^2-1}{x^2} \)
\( f'(x)\gt 0 \) when \( x^2\gt 1 \), i.e. x < −1 or x > 1.
So f is increasing on \( (-\infty,-1)\cup(1,\infty) \). On (−1, 0) and (0, 1), \( f'(x)\lt 0 \), so f is decreasing there.
NCERT (2026-27 edition):
Increasing and Decreasing Functions
