If f(x) = begin cases 2 + 2x, & -1 leq x < 0 \\ 1 - x/3, & 0 leq x leq 3 end cases ; g(x) = begin cases - x, & -3 leq x leq 0 \\ x, & 0 < x leq 1 end cases then\: range\: of\: (f\: °\: g)(x) is

Mathematics · Functions

JEE Main 2024 — 29 January, Shift 1 — Question 3

\text{If } f(x) = \begin{cases} 2 + 2x, & -1 \leq x < 0 \\ 1 - \frac{x}{3}, & 0 \leq x \leq 3 \end{cases} \quad ; \quad g(x) = \begin{cases} - x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{cases}

\text{then\: range\: of\: } (f\: \circ\: g)(x) \text{ is}

f(g(x)) = \begin{cases} 2 + 2g(x), & -1 \leq g(x) < 0 \quad \text{(1)} \\ 1 - \frac{g(x)}{3}, & 0 \leq g(x) \leq 3 \quad \text{(2)} \end{cases}

\text{By (1): } x \in \phi

\text{And by (2): } x \in [-3, 0] \text{ and } x \in [0, 1]

Graph of $y = f(x)$ and $y = f(g(x))$ shown

Range of $f(g(x))$ is $\boxed{[0, 1]}$

Answer key and solution verified before publishing.

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