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Help: Critical points

\[f\left( x \right) =\left( { a }^{ 2 }-3a+2 \right) \left( \cos ^{ 2 }{ \frac { x }{ 4 } } -\sin ^{ 2 }{ \frac { x }{ 4 } } \right) +\left( a-1 \right) x+\sin { 1 } \]

The set of all values of \(a\) for which the function above does not posses critical point is \(\text{__________} \).

Note by Akhilesh Prasad
9 months, 3 weeks ago

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What's the answer ?? I'm getting a€(0,4)- {1} ........... I might have missed cases because I'm a expert in doing that.. ;-) Rishabh Cool · 9 months, 3 weeks ago

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@Rishabh Cool The answer given is \((1,\infty)\). Akhilesh Prasad · 9 months, 3 weeks ago

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@Akhilesh Prasad How you solved ?? I found f'(x) and ensured that it does not vanish!! And ultimately got the wrong answer!! Rishabh Cool · 9 months, 3 weeks ago

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@Rishabh Cool Hey were u able to open the file i uploaded. coz it was on google drive Akhilesh Prasad · 9 months, 3 weeks ago

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@Rishabh Cool Finally, found it My solution Akhilesh Prasad · 9 months, 3 weeks ago

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@Akhilesh Prasad And shouldn't where you multiplied inequality by (a-2) cases must be made for a-2>0 and a-2<0?? Rishabh Cool · 9 months, 3 weeks ago

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@Rishabh Cool Considered the cases you told me to, still i am not getting the desired answer. Corrected solution part 1

Corrected solution part 2 Akhilesh Prasad · 9 months, 3 weeks ago

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@Akhilesh Prasad So we both are getting the same answer that is not correct!! ... ?? Rishabh Cool · 9 months, 3 weeks ago

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@Rishabh Cool On a side not, I would wanna ask you are you gonna give JEE this year. Akhilesh Prasad · 9 months, 3 weeks ago

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@Akhilesh Prasad absolutely Yes!! Rishabh Cool · 9 months, 3 weeks ago

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@Akhilesh Prasad Should at the end cases be like:
(1) Right Max value<0
(2) Left Min value>0 Rishabh Cool · 9 months, 3 weeks ago

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@Rishabh Cool It's too time consuming writing the whole answer so i wanted to attach the scan of the hand written answer, but i dunno how to do it, if you could tell me how to do it. Akhilesh Prasad · 9 months, 3 weeks ago

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@Rishabh Cool, I have got a issue in this one too so if you will please, see this one too Akhilesh Prasad · 9 months, 3 weeks ago

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@Rishabh Cool. Would you spare some of your time to please post a solution of this. Akhilesh Prasad · 9 months, 3 weeks ago

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