The function `f(x)=(x^3+x^2-16 x+20)/(x-2)` is not defined for `x=2.` In order to make `f(x)` continuous at `x=2,f(2)` should be defined as
(a)0
(b) 1
(c) 2
(d) 3

Answer

Option `(a)`

Explanation

`\lim _{x \rightarrow 2} f(x)=\lim _{x \rightarrow 2} \frac{x^{3}+x^{2}-16 x+20}{x-2}`
`lim _{x \rightarrow 2} \frac{3 x^{2}+2 x-16}{1}`


(By L Hopital Rule)

`=12+4-16=0`