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Questions  

f(x)=(1+px)(1px)x,    1x<02x+1x2,    0x1 is continuous in the interval [1,1]then  is equal to 

a
-1
b
-12
c
12
d
1

detailed solution

Correct option is B

f0+0=limh→020+h+10+h−2=−12f0−0=limh→01−ph−1+ph0−h                      =limh→0  −2ph−h1−ph+1+ph=pf(x) is continuous at x=0 in 1,1 iff0+0=f0−0=f0⇒p=−1/2.Hence  (2) is the correct answer.

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