Question

Let $f\left(x\right)=\{\begin{array}{cc}{x}^2& \text{if}0<x<2\\ 2x-3& \text{if}2\le x<3\\ x+2& \text{if}x\ge 3\end{array}$.
Then

• A. $f\left\{f\left(f\left(\frac{3}{2}\right)\right)\right\}=f\left(\frac{3}{2}\right)$
• B. $1+f\left\{f\left(f\left(\frac{5}{2}\right)\right)\right\}=f\left(\frac{5}{2}\right)$
• C. $f\left\{f\right(0\left)\right\}=f\left(1\right)=1$
• D. none of these

Answer 1

Answer: (A), (B)

The correct options are
A $f\left\{f\left(f\left(\frac{3}{2}\right)\right)\right\}=f\left(\frac{3}{2}\right)$
B $1+f\left\{f\left(f\left(\frac{5}{2}\right)\right)\right\}=f\left(\frac{5}{2}\right)$

consider option A)
$f\left(\frac{3}{2}\right)=\frac{9}{4}$ as $\frac{3}{2}<2$
and $f\left(\frac{9}{4}\right)=2\times \frac{9}{4}-3=\frac{3}{2}$ as $\frac{9}{4}>2$.
Therefore $f\left(f\left(f\left(\frac{3}{2}\right)\right)\right)=f\left(\frac{3}{2}\right)$.
Hence option A is correct answer.
Consider option B)
$f\left(\frac{5}{2}\right)=2\times \frac{5}{2}-3=2$ as $\frac{5}{2}>2$
and $f\left(2\right)=2\times 2-3=1$.
Also, $f\left(1\right)=1$
Hence, $1+f\left\{f\left(f\left(\frac{5}{2}\right)\right)\right\}=2=f\left(\frac{5}{2}\right)$
Hence option B is also a correct answer .
Since the function is not defined for $x=0$ option C can be eliminated.