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Let `f(x) =x^(2).` Then dom (f ) and range (f ) are respectivelyA. R and RB. `R^(+) " and " R^(+)`C. `R " and " R^(+)`D. `R " and " R-{0}` |
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Answer» Correct Answer - C f(x) `=x^(2) ` is defined for each ` x in R` .So dom (f) =R ` y =x^(2) rArr x = +- sqrt(y)` when `y lt 0` there is no real value of x .So `y ge 0` ` :. ` range `(f) = R^(+)` |
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