Exams Joint Entrance Examination Mathematics Application of Derivatives Question 5 Question 5 of 212 Medium MCQ
Let α \alpha α and β \beta β respectively be the maximum and the minimum values of the function f ( θ ) = 4 ( sin 4 ( 7 π 2 − θ ) + sin 4 ( 11 π + θ ) ) − 2 ( sin 6 ( 3 π 2 − θ ) + sin 6 ( 9 π − θ ) ) , θ ∈ R f(\theta )=4({\sin }^{4}(\frac{7\pi }{2}-\theta )+{\sin }^{4}(11\pi +\theta ))-2({\sin }^{6}(\frac{3\pi }{2}-\theta )+{\sin }^{6}(9\pi -\theta )),\theta \in R f ( θ ) = 4 ( sin 4 ( 2 7 π − θ ) + sin 4 ( 11 π + θ )) − 2 ( sin 6 ( 2 3 π − θ ) + sin 6 ( 9 π − θ )) , θ ∈ R .
Then α + 2 β \alpha +2\beta α + 2 β is equal to :
Solutions · 1 Post your solution Official solution
f ( θ ) = 4 ( sin − 1 ( 7 π 2 − θ ) + sin 4 ( 11 π + θ ) ) − 2 ( sin 6 ( 3 π 2 − θ ) + sin 6 ( 9 π − θ ) ) = 4 ( cos 4 θ + sin 4 θ cos 2 θ ) − 2 ( cos 2 θ + sin 6 θ ) = 4 ( 1 − 2 sin 2 θ cos 2 θ ) − 2 ( 1 ) ( sin 4 θ + cos 4 θ − sin 2 θ cos 2 θ ) = 4 − 8 sin 2 θ cos 2 θ − 2 ( 1 − 3 sin 2 θ cos 2 θ ) = 2 − 2 sin 2 θ cos 2 θ = 2 − ( sin 2 θ ) 2 2 f ( θ ) m a x = 2 = α f ( θ ) m i n = 2 − 1 2 = 3 2 = β α + 2 β = 2 + 3 = 5 \begin{gathered} f(\theta )=4({\sin }^{-1}(\frac{7\pi }{2}-\theta )+{\sin }^{4}(11\pi +\theta )) \\ \, -2({\sin }^{6}(\frac{3\pi }{2}-\theta )+{\sin }^{6}(9\pi -\theta )) \\ =4({\cos }^{4}\theta +{\sin }^{4}\theta {\cos }^{2}\theta )-2({\cos }^{2}\theta +{\sin }^{6}\theta ) \\ =4(1-2{\sin }^{2}\theta {\cos }^{2}\theta ) \\ \, -2(1)({\sin }^{4}\theta +{\cos }^{4}\theta -{\sin }^{2}\theta {\cos }^{2}\theta ) \\ =4-8{\sin }^{2}\theta {\cos }^{2}\theta -2(1-3{\sin }^{2}\theta {\cos }^{2}\theta ) \\ =2-2{\sin }^{2}\theta {\cos }^{2}\theta \\ =2-\frac{(\sin 2\theta {)}^{2}}{2} \\ f(\theta {)}_{max}=2=\alpha \, f(\theta {)}_{min}=2-\frac{1}{2}=\frac{3}{2}=\beta \\ \alpha +2\beta =2+3=5 \end{gathered} f ( θ ) = 4 ( sin − 1 ( 2 7 π − θ ) + sin 4 ( 11 π + θ )) − 2 ( sin 6 ( 2 3 π − θ ) + sin 6 ( 9 π − θ )) = 4 ( cos 4 θ + sin 4 θ cos 2 θ ) − 2 ( cos 2 θ + sin 6 θ ) = 4 ( 1 − 2 sin 2 θ cos 2 θ ) − 2 ( 1 ) ( sin 4 θ + cos 4 θ − sin 2 θ cos 2 θ ) = 4 − 8 sin 2 θ cos 2 θ − 2 ( 1 − 3 sin 2 θ cos 2 θ ) = 2 − 2 sin 2 θ cos 2 θ = 2 − 2 ( sin 2 θ ) 2 f ( θ ) ma x = 2 = α f ( θ ) min = 2 − 2 1 = 2 3 = β α + 2 β = 2 + 3 = 5 Answer the question to unlock