iTeachChem•2y ago

ITF

Bhai isme koi meri last step ek bar bata do shi hai kya jisme cos^-1 ki inequality solve kri hai

iTeachChem Helper•5/30/24, 6:33 PM

@Apu

iTeachChem Helper•5/30/24, 6:34 PM

Note for OP

to close the thread when your doubt is solved. Mention the user who helped you solve the doubt. This will be added to their stats.

Real potatoOP•5/30/24, 6:36 PM

also do tell me about the graph ig its wrong

Sam•6/2/24, 5:50 PM

Y= cos-¹(2x²-1)
= Cos -¹[cos (2 cos-¹ (x) ) ]

Say 2 cos-¹(x)= t
So we have to first make the graph of cos -¹ ( cos(t) )

Ab according to values of t you'll get your value of y [ straight line mei (x,y) mei se x toh mil gaya y mil jaega straight line ka equation likhne ke liye graphically slope bhi mil jaegi aur ek point bhi.

Sam•6/2/24, 5:53 PM

2 cos-¹(x) belongs to (0,π)
So we'll have 2 cos-¹x as it is

2 cos -¹(x) belongs to (π,2π)

We have the graph
Y-π = -(t-π)
Y+t =2π
Y= 2π- t
So y = 2π- 2 cos-¹(x)

And so on

So

Sam•6/2/24, 5:54 PM

2 cos-¹(x) belongs to (0,π)
Means
Cos-¹(x) belongs to (0,π/2)
Or X≥0

Lly,
2 cos-¹(x) belongs to (π,2π) means ,
Cos-¹(x) belongs to (π/2,π)
Which means x ≤0

Oh please note all the brackets I have used so far are the closed ones

Sam•6/2/24, 5:55 PM

Obv cos-¹(anything) can only Belong to [0,π]
So we'll have to note only in those ranges

SSam Y= cos-¹(2x²-1) = Cos -¹[cos (2 cos-¹ (x) ) ] Say 2 cos-¹(x)= t So we have to ...

Real potatoOP•6/2/24, 6:19 PM

Thanks brother

iTeachChem•6/21/24, 5:23 PM

+solved @Sam crazy effort man

iTeachChem Helper•6/21/24, 5:23 PM

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