∫ 7 1 − x 4 g ( x ) e ln ( f ( sin 2 t ) cos 2 t ) d t = 4 1 − x 7 − 7 1 − x 4
For the equation given above, where g ( x ) = 4 1 − x 7 is an invertible function, find f ( 2 1 ) ⋅ f ( 3 1 ) .
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Fabulous solution....!!! I'm Really Foolish since I use Leibniz Theorem ..!! Nice question
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But u said CBSE is foolish. Which means ur are CBSE .. HAHA
good job dude!!
i did the exact thing,, even posted solution like 10 hours ago,, but accidently pressed keep private button,, and i do not know how to undo it,, i can see it but no one else can
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Sorry Mvs Saketh but i can't see your solution That's why I post my solution. I made this question for only one purpose to solve This question By using This Concept...!!
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No i am just telling :) .... either way why did u mention g(x) is an invertible matrix?infact how is it a matrix? Please explain
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@Mvs Saketh – ohh..... Typo in question....Thanks For noticing I fixed that.
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@Deepanshu Gupta – Better would have been if you would have said bijective under certain range.
Exactly same method used, step by step!
yes, the thing why i took a little more time is , i noticed that lower limit is inverse of g ( x ) and wanted to utilise it but i got a easier way and did just like your solution ! good question bro !
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F i r s t t h i n g t h a t s t r i k e i n m i n d t o d o q u e s t i o n i s b y u s i n g L e i b n i t z T h e o r a m . . . B u t P l z a v o i d i t h e r e L e t ′ s d o s o m e d i f f r e n t . . . . : N o t e T h a t : ∫ a b 1 . d x = b − a s o a c c o r d i n g t o q u e s t i o n ⟹ e l n ( f ( s i n 2 t ) c o s 2 t = 1 ⟹ f ( s i n 2 t ) c o s 2 t = 1 ⟹ f ( s i n 2 t ) = c o s 2 t 1 = 1 − s i n 2 t 1 ⟹ f ( x ) = 1 − x 1 ∀ x ∈ [ − 1 , 1 ] ⟶ f ( 1 / 2 ) . f ( 1 / 3 ) = 3 A n s . .