Skip to main content
Back to TRIGONOMETRY RATIOS

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Hard
1 of 1
Application +4 | -1
The extreme values of
4cos⁡(x2)cos⁡(π3+x2)cos⁡(π3−x2)4\cos \left( {{x^2}} \right)\cos \left( {\frac{\mathrm{\pi }}{3} + {x^2}} \right)\cos \left( {\frac{\mathrm{\pi }}{3} - {x^2}} \right) over RR, are
View answer and explanation
Correct option: A
−1,1- 1,1
Option A (-1, 1) is correct because the extreme values are found by simplifying the trigonometric expression and determining the range of the resulting cosine function.

More Questions from TRIGONOMETRY RATIOS

Practice All Chapter MCQs →
If tan⁡α=pq\tan \alpha=\frac{p}{q}, where α=6β\alpha=6 \beta and α\alpha being an acute angle, then 12[pcosec⁡2β−qsec⁡2β]\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta] is equal to The value of cos⁡12∘+cos⁡84∘+cos⁡156∘+cos⁡132∘\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ} is If the △ABC\triangle A B C is acute angled at CC, then The value of the expression sin⁡6  θ+cos⁡6  θ+3sin⁡2  θ  cos⁡2  θ{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta equals sin⁡2A2+sin⁡2B2+sin⁡2C2=\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}= sin⁡2(A/2)+sin⁡2(B/2)−sin⁡2(C/2)=\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →