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f(x)={(x^2+e^(1/(2-x)))^(-1)k ,x=2, x!=2...

`f(x)={(x^2+e^(1/(2-x)))^(-1)k ,x=2, x!=2` is continuous from right at the point `x=2,` then `k` equals `0` b. `1//4` c. `-1//4` d. none of these

A

0

B

1\4

C

`-1//4`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
`k=underset(xrarr2^(+))(lim)(x^(2)+e^((1)/(2-x)))^(-1)`
`therefore" "k=underset(hrarr0)(lim)[(2+h)^(2)+e^((-1)/(4))]^(-1)`
`=(4+e^(-oo))^(-1)`
`=1//4`
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