INVERSE AND GRAPH

then ..........................

dude sinx and cosx are also periodic functions
so how can we find their inverses

for a function to have inverse it must be one one and onto
so we take principal value of these functions (by convention )

so what is the point in arguing that its inverse id defined on all the values of R

please draw the graph and find the inverse of the first function!!!!!!!!!!

THEN I DUNNO THIS UN;

it wud b nice to knoe the soltn

let y=x+sinx
wel take x as 0....
so y=0

since this passes through d origin and its inverse is a mirror image on y=x...... so even tht also pass through 0,0

so we can eliminate option b and c...
now take ∩/2
y=∩/2+1
point is (∩/2,∩/2+1)
find its mirror image on y=x
and try 2 substitue dat point in a and d....one of them will satisfy......

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dude good analysis with options
but what if the question was without options ?????????

but i think it could be done in a better way
what about the graph?????

so answer is approximately a.....

Approximately because...
i got d point as (∩/2+1,∩/2)

and when i subsituted in 1
y=∩/2+1-cos1
cos1 is approx 1
so y=∩/2

so ans is approx. A

ok dude..........

wel if its wrong then it has to do with domains and ranges i guess

akand
abhi tak test nahin dia
or u backed up????????

no i think inverse shud be y=x-sinx

bcos after reflection by y=x sinx will be reflected as -sinx and kalyans graph is correct

Well msp maybe UR right,

coz at 0 here the value is 0, but my inverse gives it 1

So it is y=x-sinx , like U said

I had a doubt about this one,
Now cleared

So,
The graph is as above

and the inverse is y=x-sinx

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The inverse is same..i.e..x+sin x..!