JEE Advance - Mathematics (2020 - Paper 2 Offline - No. 18)
Let the function $$f:(0,\pi ) \to R$$ be defined by $$f(\theta ) = {(\sin \theta + \cos \theta )^2} + {(\sin \theta - \cos \theta )^4}$$
Suppose the function f has a local minimum at $$\theta $$ precisely when $$\theta \in \{ {\lambda _1}\pi ,....,{\lambda _r}\pi \} $$, where $$0 < {\lambda _1} < ...{\lambda _r} < 1$$. Then the value of $${\lambda _1} + ... + {\lambda _r}$$ is .............
Suppose the function f has a local minimum at $$\theta $$ precisely when $$\theta \in \{ {\lambda _1}\pi ,....,{\lambda _r}\pi \} $$, where $$0 < {\lambda _1} < ...{\lambda _r} < 1$$. Then the value of $${\lambda _1} + ... + {\lambda _r}$$ is .............
Answer
0.5
Explanation
The given function f : R $$ \to $$ R be defined by
$$f(\theta ) = {(\sin \theta + \cos \theta )^2} + {(\sin \theta - \cos \theta )^4}$$
$$ = 1 + \sin 2\theta + {(1 - \sin 2\theta )^2}$$
$$ = 1 + \sin 2\theta + 1 + {\sin ^2}2\theta - 2\sin 2\theta $$
$$ = {\sin ^2}2\theta - \sin 2\theta + 2$$
$$ = {\left( {\sin 2\theta - {1 \over 2}} \right)^2} + {7 \over 4}$$
The local minimum of function 'f' occurs when
$$\sin 2\theta = {1 \over 2}$$
$$ \Rightarrow 2\theta = {\pi \over 6},\,{{5\pi } \over 6},\,{{13\pi } \over 6},\,...$$
$$ \Rightarrow \theta = {\pi \over {12}},\,{{5\pi } \over {12}},\,{{13\pi } \over {12}},\,...$$
but $$\theta \in \{ {\lambda _1}\pi ,\,{\lambda _2}\pi ,\,...,\,{\lambda _r}\pi \} $$,
where $$0 < {\lambda _1} < .... < {\lambda _r} < 1$$.
$$ \therefore $$ $$\theta = {\pi \over {12}},\,{{5\pi } \over {12}}$$
So, $${\lambda _1} + ... + {\lambda _r} = {1 \over {12}} + {5 \over {12}} = 0.50$$
$$f(\theta ) = {(\sin \theta + \cos \theta )^2} + {(\sin \theta - \cos \theta )^4}$$
$$ = 1 + \sin 2\theta + {(1 - \sin 2\theta )^2}$$
$$ = 1 + \sin 2\theta + 1 + {\sin ^2}2\theta - 2\sin 2\theta $$
$$ = {\sin ^2}2\theta - \sin 2\theta + 2$$
$$ = {\left( {\sin 2\theta - {1 \over 2}} \right)^2} + {7 \over 4}$$
The local minimum of function 'f' occurs when
$$\sin 2\theta = {1 \over 2}$$
$$ \Rightarrow 2\theta = {\pi \over 6},\,{{5\pi } \over 6},\,{{13\pi } \over 6},\,...$$
$$ \Rightarrow \theta = {\pi \over {12}},\,{{5\pi } \over {12}},\,{{13\pi } \over {12}},\,...$$
but $$\theta \in \{ {\lambda _1}\pi ,\,{\lambda _2}\pi ,\,...,\,{\lambda _r}\pi \} $$,
where $$0 < {\lambda _1} < .... < {\lambda _r} < 1$$.
$$ \therefore $$ $$\theta = {\pi \over {12}},\,{{5\pi } \over {12}}$$
So, $${\lambda _1} + ... + {\lambda _r} = {1 \over {12}} + {5 \over {12}} = 0.50$$
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