Q.

fx=1+px-1-pxx       ; -1x02x+1x-2                              ; 0x1is continuous in [-1,1] then 2p+1 is

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Detailed Solution

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The given function is fx=1+px-1-pxx       ; -1x02x+1x-2                              ; 0x1

Given that the function is continuous in -1,1

Hence, the function is continuous at x=0

So that limx0-fx=f0=limx0+fx

Left hand limit is 

limx0-fx=limx0-1+px-1-pxx =limx0-p21+px+p21-px1 =p

Right hand limit is 

limx0+fx=limx0-2x+1x-2 =-12

Hence, p=-122p+1=0

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fx=1+px-1-pxx       ; -1≤x≤02x+1x-2                              ; 0≤x≤1is continuous in [-1,1] then 2p+1 is