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QJEEAdv20Ma2793: JEE Advanced 2020 Mathematics - Class 11 JEEAdv20Ma2793 Solution: Let the function be defined by Suppose the function f has a…

Let the function f:(0,π)→Rf:(0,\pi ) \to R be defined by f(θ)=(sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)4f(\theta ) = {(\sin \theta + \cos \theta )^2} + {(\sin \theta - \cos \theta )^4}

Suppose the function f has a local minimum at θ\theta precisely when θ∈{λ1π,....,λrπ}\theta \in \{ {\lambda _1}\pi ,....,{\lambda _r}\pi \} , where 0 < {\lambda _1} < ...{\lambda _r} < 1 . Then the value of λ1+...+λr{\lambda _1} + ... + {\lambda _r} is .............

Step-by-Step Verified Solution

The given function f : R →\to R be defined by

f(θ)=(sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)4f(\theta ) = {(\sin \theta + \cos \theta )^2} + {(\sin \theta - \cos \theta )^4}

=1+sin⁡2θ+(1−sin⁡2θ)2= 1 + \sin 2\theta + {(1 - \sin 2\theta )^2}

=1+sin⁡2θ+1+sin⁡22θ−2sin⁡2θ= 1 + \sin 2\theta + 1 + {\sin ^2}2\theta - 2\sin 2\theta

=sin⁡22θ−sin⁡2θ+2= {\sin ^2}2\theta - \sin 2\theta + 2

=(sin⁡2θ−12)2+74= {\left( {\sin 2\theta - {1 \over 2}} \right)^2} + {7 \over 4}

The local minimum of function 'f' occurs when

sin⁡2θ=12\sin 2\theta = {1 \over 2}

⇒2θ=π6, 5π6, 13π6, ...\Rightarrow 2\theta = {\pi \over 6},\,{{5\pi } \over 6},\,{{13\pi } \over 6},\,...

⇒θ=π12, 5π12, 13π12, ...\Rightarrow \theta = {\pi \over {12}},\,{{5\pi } \over {12}},\,{{13\pi } \over {12}},\,...

but θ∈{λ1π, λ2π, ..., λrπ}\theta \in \{ {\lambda _1}\pi ,\,{\lambda _2}\pi ,\,...,\,{\lambda _r}\pi \} ,

where 0 < {\lambda _1} < .... < {\lambda _r} < 1 .

∴\therefore θ=π12, 5π12\theta = {\pi \over {12}},\,{{5\pi } \over {12}}

So, λ1+...+λr=112+512=0.50{\lambda _1} + ... + {\lambda _r} = {1 \over {12}} + {5 \over {12}} = 0.50

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