Doubt from functions

Q) 1
@Shwetanshu_2018 @Sneha_2021 @Samarth_2020 and others pls help with Soln.

My trial

too tough to read question @Bhuvanitha_2020 kindly post clear pic
share your approch also with clear pic..
by getting condition its one one function
hence use derivative concept \ge u will get result..

Sry for unclear one

ans may be 2 but not to sure or good approach
@Hari_Shankar sir , @Jagdish_Singh sir please share ur approach

@Bhuvanitha_2020 ans??

Idk i don’t have ans key?
If someone has soln for review test of Allen (26/01) pls post it

It’s one to one function?how Sneha ...but they have given for two values of x we get same y value

condition of one one function
f(x_1)=f(x_2)

But sry Iam not getting it ...for that x1=x2 to be one to one ????isnt here x1 x2 are two diff values

I think it should be C
p=-1,q=-3

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Ok now I got it it’s constant function
f(x)=p+q
So other terms has to be zero
:+1:t2:

how????? whyy other term to be zero logic
@Raunak_2020 what ur approach

I think this is the approach...
@Raunak_2020 pls confirm

They have given for all x1,x2 f(x1)=f(x2) means it has to be constant function...

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if x>tan1 then??
x<tan1 means gif equal to zero doesnt implies p^2-1=0

Here x1 and x2 are diff ...so for all diff values of x we get same value of y so it is constant function.
So p2-1 needs to be zero for function to be continuous and the quadratic has to be zero for function to be constant

Function must be constant for all values of x,hence we must eliminate terms which might make the function vary,that is the coefficients,hence we equate the coefficients to zero

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kk thanks i got it @Bhuvanitha_2020 , @Raunak_2020

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same as @Bhuvanitha_2020 but posting :blush::blush:

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