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

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

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TopNewestWhat'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 · 1 year, 1 month ago

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– Akhilesh Prasad · 1 year, 1 month ago

The answer given is \((1,\infty)\).Log in to reply

– Rishabh Cool · 1 year, 1 month ago

How you solved ?? I found f'(x) and ensured that it does not vanish!! And ultimately got the wrong answer!!Log in to reply

– Akhilesh Prasad · 1 year, 1 month ago

Hey were u able to open the file i uploaded. coz it was on google driveLog in to reply

My solution – Akhilesh Prasad · 1 year, 1 month ago

Finally, found itLog in to reply

– Rishabh Cool · 1 year, 1 month ago

And shouldn't where you multiplied inequality by (a-2) cases must be made for a-2>0 and a-2<0??Log in to reply

Corrected solution part 1

Considered the cases you told me to, still i am not getting the desired answer.Corrected solution part 2 – Akhilesh Prasad · 1 year, 1 month ago

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– Rishabh Cool · 1 year, 1 month ago

So we both are getting the same answer that is not correct!! ... ??Log in to reply

– Akhilesh Prasad · 1 year, 1 month ago

On a side not, I would wanna ask you are you gonna give JEE this year.Log in to reply

– Rishabh Cool · 1 year, 1 month ago

absolutely Yes!!Log in to reply

(1) Right Max value<0

(2) Left Min value>0 – Rishabh Cool · 1 year, 1 month ago

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– Akhilesh Prasad · 1 year, 1 month ago

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.Log in to reply

@Rishabh Cool, I have got a issue in this one too so if you will please, see this one too – Akhilesh Prasad · 1 year, 1 month ago

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@Rishabh Cool. Would you spare some of your time to please post a solution of this. – Akhilesh Prasad · 1 year, 1 month ago

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