Question

Let `f(x)={{:(1+(2x)/(a)", "0lexlt1),(ax", "1lexlt2):}."If" lim_(xto1) "f(x) exists, then a is "`
A. `underset(xto5^(-))f(x)=0`
B. `underset(xto5^(+))f(x)=1`
C. `underset(xto5)lim f(x)` does not exist
D. none of these

Answer 1

Correct Answer - 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`