Let f(x)={{:(1+(2x)/(a)", "0lexlt1),(ax"...

Let `f(x)={{:(1+(2x)/(a)", "0lexlt1),(ax", "1lexlt2):}."If" lim_(xto1) "f(x) exists, then a is "`

`underset(xto5^(-))f(x)=0`

`underset(xto5^(+))f(x)=1`

`underset(xto5)lim f(x)` does not exist

none of these

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The correct Answer is:

B, C

R.H.L.`=underset(hto0)limf(1+h)=underset(hto0)lim(1+h)=a`
L.H.L.`=underset(hto0)limf(1-h)=underset(hto0)lim{1+(2)/(a)(1-h)}=1+(2)/(a)`
`underset(xto1)limf(x)" exists "implies"R.H.L.=L.H.L. Therefore"`,
`a=1+(2)/(a) " or "a=2,-1`

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