Question

Let $$\begin{gathered} f\left(x\right)=\left\{x^2,0<x<2, \\ 2x-3,2\le x<3 \\ x+2,x\ge 3\right\} \end{gathered}$$

Then, which of the following is correct?

• A. $f\left\{f\left(f\left(\frac{3}{2}\right)\right)\right\}=f\left(\frac{3}{2}\right)$
• B. $f\left(f\left(1\right)\right)=2$
• C. $f\left\{f\left(f\left(\frac{5}{2}\right)\right)\right\}=f\left(\frac{5}{2}\right)$
• D. $1-f\left\{f\left(f\left(\frac{5}{2}\right)\right)\right\}=f\left(\frac{5}{2}\right)$

Answer 1

The correct option is A

$f\left\{f\left(f\left(\frac{3}{2}\right)\right)\right\}=f\left(\frac{3}{2}\right)$

Explanation for the correct option:

Checking for Option (A):

Calculating $f\left(\frac{3}{2}\right)$:

Here f(x) will be defined by: $x^2$.

Therefore, $f\left(\frac{3}{2}\right)=\frac{9}{4}$

Now,

$$\begin{gathered} f\left(f\left(\frac{3}{2}\right)\right)=f\left(\frac{9}{4}\right) \\ =2x-3\left[Here,3>x>2\right] \\ =2\times \frac{9}{4}-3 \\ =\frac{3}{2} \\ \therefore f\left\{f\left(f\left(\frac{3}{2}\right)\right)\right\}=f\left(\frac{3}{2}\right) \end{gathered}$$

Therefore, the correct answer is Option (A).

Explanation for the correct option:

Checking for Option (B):

$$\begin{gathered} f\left(1\right)=x^2 \\ =1 \\ f\left(f\left(1\right)\right)=f\left(1\right) \\ \ne 2 \end{gathered}$$

Hence, Option (B) is incorrect.

Checking for Option (C):

$$\begin{gathered} f\left(\frac{5}{2}\right)=2x-3 \\ =2 \\ f\left(f\left(\frac{5}{2}\right)\right)=f\left(2\right) \\ \ne f\left(\frac{5}{2}\right) \end{gathered}$$

Hence, Option (C) is incorrect.

Checking for Option (D):

$$\begin{gathered} 1-f\left(2\right)=1-2x-3 \\ =-2-2x \\ =-6 \\ \ne f\left(\frac{5}{2}\right) \end{gathered}$$

Hence, Option (D) is incorrect.

Hence, the correct is Option (A).