INVERSE AND GRAPH

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......

[1][1][1]

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

i faced it one of the tests!!!!!!!!!

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

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

Dunno about the direct method for inverse, but the graph can be found like this....

The graph for y=x is....

and for y=sinx is....

So, the graph for y=x+sin x is......

Inverse of this function is its mirror image in y=x which gives...

This I suppose is the graph for y= x+cosx

So, the inverse is y=x+cosx

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..!

SHOW UR WORK
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