Q. 6

If f, g, h are real functions defined by, and h(x) = 2x2 – 3, then find the values of (2f + g – h)(1) and (2f + g – h)(0).

Answer :

Given, and h(x) = 2x3 – 3


We know the square of a real number is never negative.


Clearly, f(x) takes real values only when x + 1 ≥ 0


x ≥ –1


x [–1, ∞)


Thus, domain of f = [–1, ∞)


g(x) is defined for all real values of x, except for 0.


Thus, domain of g = R – {0}


h(x) is defined for all real values of x.


Thus, domain of h = R


We know (2f + g – h)(x) = (2f)(x) + g(x) – h(x)


(2f + g – h)(x) = 2f(x) + g(x) – h(x)




Domain of 2f + g – h = Domain of f Domain of g Domain of h


Domain of 2f + g – h = [–1, ∞) R – {0} R


Domain of 2f + g – h = [–1, ∞) – {0}


i. (2f + g – h)(1)


We have





ii. (2f + g – h)(0)


0 is not in the domain of (2f + g – h)(x).


Hence, (2f + g – h)(0) does not exist.


Thus, and (2f + g – h)(0) does not exist as 0 is not in the domain of (2f + g – h)(x).


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