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If \(f: R \rightarrow R\) is defined by
\(f(x)= \begin{cases}|[x-5]|, & \text { for } x < 5 \\ {[|x-5|],} & \text { for } x \geq 5\end{cases}\)
Then, \((f \circ f)\left(-\frac{7}{2}\right)=\)
(here, \([x]\) is the greatest integer not exceeding \(x\))
Options:
\(f(x)= \begin{cases}|[x-5]|, & \text { for } x < 5 \\ {[|x-5|],} & \text { for } x \geq 5\end{cases}\)
Then, \((f \circ f)\left(-\frac{7}{2}\right)=\)
(here, \([x]\) is the greatest integer not exceeding \(x\))
Solution:
2290 Upvotes
Verified Answer
The correct answer is:
\((f \circ f)\left(\frac{9}{2}\right)\)
\(f \circ f\left(-\frac{7}{2}\right)=f\left[\left[-\frac{7}{2}-5\right]=f(9)=|[19-51]|=4\right.\) By checking option
(A) \(\rightarrow f\left(\mid\left[-\frac{11}{2}-5\right]\right)=6\)
(B) \(\longrightarrow f \circ f\left(-\frac{9}{2}\right)=5\)
(C) \(\rightarrow f o f(3)=3\)
(D) \(\longrightarrow f \circ f\left(\frac{9}{2}\right)=f(\mathrm{l})=4\)
Hence, option (d) is correct.
(A) \(\rightarrow f\left(\mid\left[-\frac{11}{2}-5\right]\right)=6\)
(B) \(\longrightarrow f \circ f\left(-\frac{9}{2}\right)=5\)
(C) \(\rightarrow f o f(3)=3\)
(D) \(\longrightarrow f \circ f\left(\frac{9}{2}\right)=f(\mathrm{l})=4\)
Hence, option (d) is correct.
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