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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

  • If lim_( xto 2 ) (f(x) -f(2))/( x-2) exist ,then

    A
    ` underset(xto2^(-) )lim f(x) ne underset(xto2^(+) ) lim f(x) `
    B
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    C
    ` underset (xto2) lim f(x) =f(2) `
    D
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  • Given function f(x) ={:{(x^(2),"for " 0 lexlt1),(sqrt(x),"for "1 le x le2):}, "then" int_(0)^(2)f(x)dx is equal to

    A
    ` (4sqrt(2)-1)`
    B
    `1/3(4sqrt(2)-1)`
    C
    ` 1/3 (sqrt(2)-1)`
    D
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