The maximum value of \(\sin x.\cos x\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{2}\)
By the double-angle formula \(\sin 2x = 2\sin x\cos x\):
\( \displaystyle \sin x\cos x = \frac{1}{2}\sin 2x \)
Since \(-1 \le \sin 2x \le 1\), the greatest value of \(\sin 2x\) is 1, reached at \(x = \frac{\pi}{4}\). Hence
\( \displaystyle \text{maximum of } \sin x\cos x = \frac{1}{2} \times 1 = \frac{1}{2} \)
NCERT Exemplar:
Application of Derivatives
