Master your funtion skills!
Given that f ( x ) = a 0 + a 1 x + a 2 x 2 + … + a n x n , and ∣ f ( x ) ∣ ≤ ∣ e x − 1 − 1 ∣ for all x ≥ 0 . Find the value of y if y ≥ ∣ a 1 + 2 a 2 + 3 a 3 + … + n a n ∣ .
This is a past-year IIT-JEE problem.
The answer is 1.
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1 solution
Haha , as soon as you mentioned that it is an old JEE question , I thought that perhaps 1 is the answer and I went for it !
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Went for 0 and then 1! Same reason! +1 Aditya, if it helps you.
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Cheers ⌣ ¨
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@A Former Brilliant Member – I remember a friend of mine was once taking a JEE mock test paper. Whenever he couldn't solve a question, he went for 0 or 1 and got most of them right!
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@User 123 – Well, I have another story for you . Last year in the NSO first level , there was a friend of mine who hadn't studied and come for the exam . Do you know what he did ? He marked the option C for 37 questions out of 50 and got around 30 of them correct and that stupid (in a friendly way!) got school rank 2 without even studying !!
Seriously , we then had to tolerate his bragging on that topic !
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@A Former Brilliant Member – Wow! He must have been intolerable!
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@User 123 – You bet !
Btw , if you are studying now , sorry for disturbing you .
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@A Former Brilliant Member – No, I'm about to go to bed (well, I have been for the past 20 minutes), and was just whiling away some time. Sorry if I've disturbed you too. Goodnight!
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@User 123 – I am enjoying it out here too ! Well , have sweet dreams :)
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@A Former Brilliant Member – @Azhaghu Roopesh M can you suggest me a good way to prepare for iit-jee
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@Aditya Kumar – Do you currently go to some coaching institute ?
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@A Former Brilliant Member – Yes I do go to a coaching institute
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@Aditya Kumar – That's great! You can get all the basic knowledge of each chapter from your teachers and then you can furnish your skills by solving material from your institute . And yes , there's always Brilliant when you want to go a bit higher with the practice !The Brilliant wiki's are also vary good . And also the peer group here is equally fantastic ,If you want help regarding books , you can always post it as a note and ask .
I'm sure you'll clear JEE if you spend your time wisely .:)
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@A Former Brilliant Member – Thanks.....is solving upto krotov level in physics enough??
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@Aditya Kumar – Who's he ?? Well , a friend of mine who cleared JEE last year and is studying in IITB currently advised me to thoroughly read NCERT and practice my Institute's books well . I can personally advise you to choose any book and solve it completely , I mean it completely to get the idea of questions .
I'll call some of my Physics Genii friends here to help you out :)
@Raghav Vaidyanathan , @Mvs Saketh , @Deepanshu Gupta , @Shashwat Shukla
Guys, help him out !
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@A Former Brilliant Member – krotov is a book tougher than irodov
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@Aditya Kumar – I guess then I'll pass on Krotov !!
Btw don't get discouraged if no one's answering your questions , keep on posting so that people will come to know you are capable of posting good questions .
A personal suggestion will be that you first post your original questions and then post other questions so that people can feel that you can create good questions yourself :)
Good Luck :)
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@A Former Brilliant Member – thanks......good night :)
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@Aditya Kumar – Yeah , good night :)
@A Former Brilliant Member – Thanks. They'll all be about maths! They usually are! Same to you when you go to sleep.
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@User 123 – Hmm, now that you mention it , the only time I had a Math dream is that when I was in Class 7 and in my dream I was in KBC and for the final question , I was asked to prove 1+1=2 and I was stumped there !
I think I still won't be able to prove it . As there were some researchers who had written a 150 page proof !
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@A Former Brilliant Member – Couldn't go to bed yet, thinking I'll ask one doubt as a note and would then go to bed... We can all 'prove' ( 1 + 1 ) = x where x ∈ R , though!
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@User 123 – I'll give it a try . I'll ask some of my friends too visit there too.
There is mistype in 6th line after => I think It should be f ( 1 ) < = 0 !
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i did not understand that line .... could you please elaborate?
H e r e , f ( x ) = a 0 + a 1 x + a 2 x 2 + . . . + a n x n . f ′ ( x ) = a 1 + 2 a 2 x + 3 a 3 x 2 + . . . + n a n x n − 1 f ′ ( 1 ) = a 1 + 2 a 2 + 3 a 3 + . . . + n a n N o w , ∣ f ( x ) ∣ ≤ ∣ ∣ e x − 1 − 1 ∣ ∣ ∀ x ≥ 0 = > ∣ f ( 1 ) ∣ ≤ 0 ! . B u t ∣ f ( 1 ) ∣ ≥ 0 s o ∣ f ( 1 ) ∣ = 0 = > ∣ f ( 1 + h ) ∣ ≤ ∣ ∣ e h − 1 ∣ ∣ ∀ h ≥ − 1 , h = 0 = > ∣ f ( 1 + h ) − 0 ∣ ≤ ∣ ∣ e h − 1 ∣ ∣ ∀ h ≥ − 1 , h = 0 = > ∣ f ( 1 + h ) − f ( 1 ) ∣ ≤ ∣ ∣ e h − 1 ∣ ∣ = > ∣ ∣ ∣ h f ( 1 + h ) − f ( 1 ) ∣ ∣ ∣ ≤ ∣ ∣ ∣ h e h − 1 ∣ ∣ ∣ N o w t a k i n g l i m i t a s h → 0 , h → 0 l im ∣ ∣ ∣ h f ( 1 + h ) − f ( 1 ) ∣ ∣ ∣ ≤ h → 0 l im ∣ ∣ ∣ h e h − 1 ∣ ∣ ∣ f ′ ( 1 ) ≤ 1 ∴ ∣ a 1 + 2 a 2 + 3 a 3 + . . . + n a n ∣ ≤ 1