Exams Joint Entrance Examination Mathematics Application of Derivatives Question 4 Question 4 of 212 Medium MCQ
The least value of ( cos 2 θ − 6 sin θ cos θ + 3 sin 2 θ + 2 ) ({\cos }^{2}\theta -6\sin \theta \cos \theta +3{\sin }^{2}\theta +2) ( cos 2 θ − 6 sin θ cos θ + 3 sin 2 θ + 2 ) is
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f ( θ ) = cos 2 θ − 6 sin θ cos θ + 3 sin 2 θ + 2 f ( θ ) = 1 + cos 2 θ 2 − 3 sin 2 θ + 3 ( 1 − cos 2 θ 2 ) + 2 f ( θ ) = 1 2 + cos 2 θ 2 − 3 sin 2 θ + 3 2 − 3 cos 2 θ 2 + 2 f ( θ ) = 4 − cos 2 θ − 3 sin 2 θ f ( θ ) = 4 − ( cos 2 θ + 3 sin 2 θ ) g ( θ ) = cos 2 θ + 3 sin 2 θ \begin{gathered} f(\theta )={\cos }^{2}\theta -6\sin \theta \cos \theta +3{\sin }^{2}\theta +2 \\ f(\theta )=\frac{1+\cos 2\theta }{2}-3\sin 2\theta +3(\frac{1-\cos 2\theta }{2})+2 \\ f(\theta )=\frac{1}{2}+\frac{\cos 2\theta }{2}-3\sin 2\theta +\frac{3}{2}-\frac{3\cos 2\theta }{2}+2 \\ f(\theta )=4-\cos 2\theta -3\sin 2\theta \\ f(\theta )=4-(\cos 2\theta +3\sin 2\theta ) \\ g(\theta )=\cos 2\theta +3\sin 2\theta \end{gathered} f ( θ ) = cos 2 θ − 6 sin θ cos θ + 3 sin 2 θ + 2 f ( θ ) = 2 1 + cos 2 θ − 3 sin 2 θ + 3 ( 2 1 − cos 2 θ ) + 2 f ( θ ) = 2 1 + 2 cos 2 θ − 3 sin 2 θ + 2 3 − 2 3 cos 2 θ + 2 f ( θ ) = 4 − cos 2 θ − 3 sin 2 θ f ( θ ) = 4 − ( cos 2 θ + 3 sin 2 θ ) g ( θ ) = cos 2 θ + 3 sin 2 θ
The expression a cos α + b sin α a\cos \alpha +b\sin \alpha a cos α + b sin α has a range of [ − a 2 + b 2 , a 2 + b 2 ] [-\sqrt{{a}^{2}+{b}^{2}},\sqrt{{a}^{2}+{b}^{2}}] [ − a 2 + b 2 , a 2 + b 2 ] .
− 1 2 + 3 2 ≤ g ( θ ) ≤ 1 2 + 3 2 − 10 ≤ g ( θ ) ≤ 10 \begin{gathered} -\sqrt{{1}^{2}+{3}^{2}}\le g(\theta )\le \sqrt{{1}^{2}+{3}^{2}} \\ -\sqrt{10}\le g(\theta )\le \sqrt{10} \end{gathered} − 1 2 + 3 2 ≤ g ( θ ) ≤ 1 2 + 3 2 − 10 ≤ g ( θ ) ≤ 10
For least value of f ( θ ) f(\theta ) f ( θ )
f ( θ ) = 4 − ( m a x g ( θ ) ) f ( θ ) = 4 − 10 \begin{gathered} f(\theta )=4-(maxg(\theta )) \\ f(\theta )=4-\sqrt{10} \end{gathered} f ( θ ) = 4 − ( ma xg ( θ )) f ( θ ) = 4 − 10
So, least value is 4 − 10 4-\sqrt{10} 4 − 10 Answer the question to unlock