MCQ
$cos^2 \theta - 6sin\theta . cos\theta + 3sin^2\theta + 2$ ની મહત્તમ કિંમત મેળવો.
- ✓$4 + \sqrt{10}$
- B$4 - \sqrt{10}$
- C$4 - \sqrt{2}$
- D$4 + \sqrt{2}$
$cos^2\theta - 6 sin\theta . cos \theta + 3sin^2\theta + 2$
$= 1 - sin^2\theta - 3 sin2\theta + 3sin^2\theta + 2$
$= 1 + 2 sin^2\theta - 3sin2\theta$
$= 4 - \left(cos2\theta + 3sin2\theta\right)$
$= \sqrt{10} \leq cos2\theta + 3sin2\theta \leq \sqrt{10}$
અહીંથી સ્વ પ્રયત્નથી ગણવું.
મહતમ કિમંત $4 + \sqrt{10}$
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