The absolute maximum value of the function \( f(x)=\displaystyle\frac{x}{2}+\displaystyle\frac{2}{x} \) in the interval [−3, 3], is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{13}{6} \)
Note: The given answer is as per the official NTA Answer Key. However, the answer given by NTA appears to be incorrect. The function has no absolute maximum value on [−3, 3], so none of the options is correct, as explained below.
\( f(x)=\displaystyle\frac{x}{2}+\displaystyle\frac{2}{x} \) is not defined at x = 0, which lies inside [−3, 3].
As x comes closer to 0 from the right, \( \displaystyle\frac{2}{x} \) grows without limit:
\( f(1) = 2.5,\quad f(0.1) = 20.05,\quad f(0.01) = 200.005 \)
So there is no largest value of f on the interval.
For reference, \( f'(x) = \displaystyle\frac12 - \displaystyle\frac{2}{x^2} = 0 \) gives x = ±2, with \( f(2) = 2 \) and \( f(-2) = -2 \).
Option (1) \( -2 \) — Incorrect. This is f(−2), the largest value only on the negative side of 0.
Option (2) 2 — Incorrect. This is f(2), the smallest value on the positive side of 0.
Option (3) \( \displaystyle\frac{13}{6} \) — The answer given by NTA, but it seems to be incorrect. It is the value at the end point x = 3. Points near 0 give larger values, for example f(1) = 2.5, which is more than \( \frac{13}{6} \approx 2.17 \).
Option (4) \( \displaystyle\frac{17}{2} \) — Incorrect. f(0.1) = 20.05 is already larger, so 17/2 is not the largest value either.
Advice for the exam: In the real exam, mark the answer that is mathematically correct. If NTA’s provisional answer key marks a different option, challenge it during the answer-key challenge window, giving proper reasoning. NTA does revise its key when a challenge is well justified, so a sound challenge can raise your marks.