# Mark the correct alternative in each of the following:Let g(x) = 1 + x – [x] and where [x] denotes the greatest integer less than or equal to x. Then for all x, f (g(x)) is equal toA. xB. 1C. f(x)D. g(x)

Given that g(x) = 1 + x – [x] and

where [x] denotes the greatest integer less than or equal to x.

(i) -1 < x <0

g(x) = 1 + x – [x]

g(x) = 1 + x + 1 { [x] = -1}

g(x) = 2 + x

f(g(x))= f(2 + x)

f(g(x))=1 + 2 + x-[2 + x]

f(g(x)) = 3 + x -2 – x

f(g(x)) = 1

(ii) x = 0

f(g(x)) = f(1 + x-[x])

f(g(x)) = 1 + 1 + x –[x] – [1 + x + [x]]

f(g(x)) = 2 + 0 -1

f(g(x)) = 1

(iii) x > 1

f(g(x)) = f(1 + x-[x])

f(g(x)) = f(x>0) = 1

Hence, f(g(x)) = 1 for all cases.

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