DOUBT 1

AIEEE 2015 Guidance › Calculus questions › DOUBT 1

Let f:(4,6)--> (6,infinity) be a funtion defined by f(x) = x +[x/2], where [] is GIF.Then f^-1(x) =

a. x-[x/2]

b.-x-2

c.x-2

d. NOT

How to find the inverse of such funtions involving GIF, fractional part etc..
And soln and explanation for above Q

#Mathematics #Calculus

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7 Answers

3

msp ·2009-05-11 08:42:03

f(f^-1x)=x

u have option use the option to check.

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1

sriraghav ·2009-05-11 08:42:57

What is the proper method??[7]

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11

Mani Pal Singh ·2009-05-11 08:47:09

the basic thing to be kept in finding the inverse of the function is
To find the inverse of f(x) u have to find it mirror image in y=x

So u have to draw the graph of x+[x/2]
i think u can proceed now !!!!

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1

sriraghav ·2009-05-11 08:52:02

?? who pinked when nishant sir is not there...[7]

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1

gordo ·2009-05-11 09:27:12

from (4,6) we have, [x/2]=2
so the function is x+2 in the given domain,
f(x)=x+2
or finverse(x)=x-2,
so C)
cheers!!

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1

sriraghav ·2009-05-11 09:52:50

k...what is the other method than graphical one mani...i also need algebric method

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9

Celestine preetham ·2009-05-11 10:53:26

Q is actually wrong

f-1(x) doesnt exist as range â‰ codomain

so ans is d ( NOT )

pls confirm sri

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