(B) Given the function $f(x)$ defined in pieces: Step $1$: Calculate $f(-1.75)$. Since $-1.75 \leq -1$, we use $f(x) = x + 2$. $f(-1.75) = -1.75 + 2 =

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(B) Given the function $f(x)$ defined in pieces: Step $1$: Calculate $f(-1.75)$. Since $-1.75 \leq -1$, we use $f(x) = x + 2$. $f(-1.75) = -1.75 + 2 = 0.25$. Step $2$: Calculate $f(0.5)$. Since $-1 $f(0.5) = (0.5)^2 = 0.25$. Step $3$: Calculate $f(1.5)$. Since $1.5 \geq 1$, we use $f(x) = 2 - x$. $f(1.5) = 2 - 1.5 = 0.5$. Step $4$: Sum the values: $f(-1.75) + f(0.5) + f(1.5) = 0.25 + 0.25 + 0.5 = 1$.

Class 12 Mathematics Relations and Functions Type of Functions based on Mapping

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Let $f: R \rightarrow R$ be defined by $f(x) = \begin{cases} x + 2, & x \leq -1 \\ x^2, & -1 < x < 1 \\ 2 - x, & x \geq 1 \end{cases}$. Then the value of $f(-1.75) + f(0.5) + f(1.5)$ is

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$0$
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$1$
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$2$
D
$-1$

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Let $f: R \rightarrow R$ be defined by $f(x) = \begin{cases} x + 2, & x \leq -1 \\ x^2, & -1 < x < 1 \\ 2 - x, & x \geq 1 \end{cases}$. Then the value of $f(-1.75) + f(0.5) + f(1.5)$ is

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