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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 "`

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

Text Solution

Verified by Experts

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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Knowledge Check

  • Let f(x)={:(1+(2x)/(a),","0lexlt1),(ax,","1lexlt2):} if underset(xrarr1)"lim"f(x) is defined then value of 'a' will be -

    A
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    B
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    A
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    B
    continuous but not differentiable in (-3,3)
    C
    continuous as well as differentiable in (-3,3)
    D
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    A
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    B
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    C
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    D
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