If the interval in which \(f(x) =\displaystyle\frac{x}{4}+\frac{4}{x};\ x \ne 0\) is strictly increasing is \((-\infty,a) \cup (b,\infty)\), then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(a = -b\)
f is strictly increasing where \(f'(x) \gt 0\):
\( \displaystyle f'(x) = \frac{1}{4} - \frac{4}{x^2} \)
\( \displaystyle f'(x) = \frac{x^2 - 16}{4x^2} \)
Since \(4x^2 \gt 0\) for \(x \ne 0\), the sign of f'(x) is the sign of \(x^2 - 16\):
\(f'(x) \gt 0\) when \(x^2 \gt 16\), that is \(x \lt -4\) or \(x \gt 4\)
So f is strictly increasing on \((-\infty, -4) \cup (4, \infty)\), which gives a = −4 and b = 4. Hence \(a = -b\).