MCQ
જો$D=\begin{vmatrix}{1}&{3\cos\theta}&1\\\sin\theta&1&3\cos\theta\\1&\sin\theta&1\\\end{vmatrix}$ તો $D$ નું મહતમ મૂલ્ય ............. છે.
- A$9$
- B$1$
- ✓$10$
- D$16$
$D=1-3sin \theta cos \theta - 3 cos \theta (sin \theta - 3 cos \theta) +1 (sin^2 \theta-1)$
$=1-3 sin \theta cos \theta -3 sin \theta cos \theta +9 cos^2 \theta + sin^2 \theta -1$
$= (3cos \theta - sin \theta)^2$
$3cos \theta - sin \theta$ નો વિસ્તાર $ \sqrt{ 9+1} = \sqrt 10$ મહતમ મુલ્ય
$(3 cos \theta - sin \theta)^2$ નો વિસ્તાર $[0,10]$
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