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If f(x)={(ax^(2)+b",",0 le x lt 1),(x +...

If `f(x)={(ax^(2)+b",",0 le x lt 1),(x +3",",1 lt x le 2), (4",", x =1):}` , then the value of (a, b) for which f(x) cannot be continuous at x = 1 is

A

(2 , 2)

B

(3, 1)

C

(4, 0)

D

(5, 2)

Text Solution

Verified by Experts

The correct Answer is:
D
We have, `lim_(h to 0) f(1-h)=lim_(h to 0) {a(1-h)^(2)+b}=a+b`
`rArr lim_(h to 0) f(1+h) = lim _(h to 0) {(1+h)+3}=4 and f(1)=4`
Since, f(x) is not continuous at x = 1.
` a+b ne 4. `
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