1.

Find the domain and range of the following (i) ` f(x)=sqrt(x^(2)-3x+2) " (ii) " f(x)=sqrt(x^(2)-4x+6)`

Answer» (i) Clearly f(x) is defined if `x^(2)-3x+2 ge 0`
`implies (x-1)(x-2) ge 0`
`implies x in (-oo,1] cup [2,oo)`
Now, `f(x)=sqrt(x^(2)-3x+2)`
`=sqrt((x-(3)/(2))^(2)+2-(9)/(4))`
`=sqrt((x-(3)/(2))^(2)-(1)/(4))`
Clearly, `(x-(3)/(2))^(2)-(1)/(4) ge 0`
` :. sqrt((x-(3)/(2))^(2)-(1)/(4)) ge 0`
Therefore, range is `[0,oo).`
Here least value of f(x) occurs when `x-(3)/(2)= +-(1)/(2), ` i.e.,
(ii) `f(x) =sqrt(x^(2)-4x+6)`
`=sqrt((x-2)^(2)+2)`
Clearly `(x-2)^(2) +2 ge 2, AA x in R.`
So, domain of f(x) is R.
Also, `sqrt((x-2)^(2)+2) ge sqrt(2)`
Hence, range is `[ sqrt(2),oo).`


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