Let f : R → R be defined by\(\ f (x) =\begin{cases}x+2\,,&\quad x\le – InterviewSolution

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Let f : R → R be defined by\(\ f (x) =\begin{cases}x+2\,,&\quad x\leq 1\\x^2\,, &\quad -1<x<1\\2-x\,, &\quad x\geq 1\end{cases}\) then the value of f (– 1.75) + f (0.5) + f (1.5) is(a) 0 (b) 1 (c) 2 (d) – 1

Answer»

Answer : (b) 1

f (– 1.75) + f (0.5) + f (1.5)

= (– 1.75 + 2) + (0.5)² + (2 – 1.5)

= 0.25 + 0.25 + 0.5

= 1.

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