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Write the domain of the following real functions (i) f(x) = (2x + 1)/(x - 9) (ii) p(x) = -5/(4x2 + 1)(iii) g(x) = √(x - 2)(iv) h(x) = x + 6 |
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Answer» (i) f(x) = (2x + 1)/(x - 9) If the denominator vanishes when x = 9 So f(x) is not defined at x = 9 ∴ Domain is x ∈ [R – {9}] (ii) p(x) = -5/(4x2 + 1) p(x) is defined for all values of x. So domain is x ∈ R. (iii) g(x) = √(x - 2) When x < 2 g(x) becomes complex. But given “g” is real valued function. So x > 2 Domain x ∈ (2, α) (iv) h (x) = x + 6 For all values of x, h(x) is defined. Hence domain is x ∈ R. |
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