Mathematics - Functions Question with Solution | TestHub

From the given equation, we have $x^2+(x-f)-4x+3=0$, where $\mathrm{f}=$ F. P. F. such that $0\le f<1$

$\Rightarrow \left(x^2-3x+3\right)-f=0\Rightarrow f=x^2-3x+3$.

But $0\le f<1$. Therefore $0\le x^2-3x+3<1$

Now, solving $x^2-3x+3=0$, we get $x=\frac{3\pm \sqrt{9-12}}{2}=$ imaginary

$\therefore x^2-3x+3\ge 0$ for all $x\in R$.

Again from $x^2-3x+3<1$, we get $x^2-3x+2<0\Rightarrow (x-2)(x-1)<0$

$\Rightarrow 1<x<2$.

Thus the inequality (1) is satisfied if $1<x<2\Rightarrow [x]=1$.

Putting $\{x\}=1$ in the given equation, we get $x^2+1-4x+3=0\Rightarrow x^2-4x+4=0\Rightarrow (x-2{)}^2=0\Rightarrow x=2$.

But $\mathrm{x}=2$ does not satisfy the inequality $1<\mathrm{x}<2$.

Hence no $x$ is available to satisfy the equation. That is, the given equation has no solution.