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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