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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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(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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