The maximum value of \(f(x) = \sin x\cos x\), \(x \in R\), is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{2}\)
Since \(2\sin x\cos x = \sin 2x\),
\( \displaystyle f(x) = \frac{1}{2}\sin 2x \)
Setting the derivative equal to zero:
\( f'(x) = \cos 2x = 0 \), which holds at \( \displaystyle x = \frac{\pi}{4} \)
At \(x = \frac{\pi}{4}\), \(f''(x) = -2\sin 2x = -2 \lt 0\), so it is a maximum:
\( \displaystyle f\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{2} \)
No value can be larger, because \(\sin 2x \le 1\) for every x.
Why the other options are incorrect
Option (1) 1 — Incorrect. This is the maximum of sin 2x; f is half of sin 2x.
Option (3) \(\displaystyle\frac{1}{\sqrt{2}}\) — Incorrect. This is the value of sin x (or cos x) at \(\frac{\pi}{4}\), not of their product.
Option (4) 2 — Incorrect. \(f(x) = \frac{1}{2}\sin 2x\) never exceeds \(\frac{1}{2}\).