Question

If a function, $f\left(x\right)=3-2x$, then find the value of $x$ such that $f\left(x^2\right)=\left[f\right(x){]}^2$.

• A. 1 , 1
• B. -1 , -1
• C. 1 , 2
• D. 1 , -2

Answer 1

The correct option is A 1 , 1

$\text{Given,}f\left(x\right)=3-2x$
$\therefore f\left(x^2\right)=3-2\left(x^2\right)$
and
$\left[f\right(x){]}^2=[3-2x{]}^2=9+4x^2-12x$
$\text{Given,}f\left(x^2\right)=\left[f\right(x){]}^2$
$\therefore 3-2\left(x^2\right)=9+4x^2-12x$
$\Rightarrow 6x^2-12x+6=0\Rightarrow x^2-2x+1=0\Rightarrow (x-1{)}^2=0\Rightarrow x=1$

$\therefore \text{The value of}x\text{is 1}$