Question

Let a function f defined from $R\to R$ as $f\left(x\right)=\{\begin{array}{cc}2m-x,& x\le 1\\ 4mx+1,& x>1\end{array}$
If function is onto on R, then range of m is

• A. $[-1,\infty )$
• B. $[-1,0)$
• C. {-1}
• D. $(0,\infty )$

Answer 1

The correct option is B $[-1,0)$

Case I: $m=-k,wherek>0$
$\therefore f\left(x\right)=\{\begin{array}{cc}-2k-x,& x\le 1\\ -4kx+1,& x>1\end{array}$
$If-\infty <x\le 1\Rightarrow -1\le -x<\infty \Rightarrow -2k-1\le -2k-x<\infty \Rightarrow -2k-1\le f\left(x\right)<\infty \forall x\le 1$
$If1<x<\infty \Rightarrow -\infty \le -x<-1\Rightarrow -\infty <-4kx<-4k\Rightarrow -\infty <-4kx+1<-4k+1\Rightarrow -\infty <f\left(x\right)<-4k+1\forall x>1$
Since, f(x) is an onto function.
$-4k+1\ge -2k-1\Rightarrow 4m+1\ge 2m-1\Rightarrow 2m\ge -2\Rightarrow m\ge -1\therefore -1\le m<0\left[Asmisnegative\right]$

Case II: $m=k,wherek>0$
$\therefore f\left(x\right)=\{\begin{array}{cc}2k-x,& x\le 1\\ 4kx+1,& x>1\end{array}$
$If-\infty <x\le 1\Rightarrow -1\le -x<\infty \Rightarrow 2k-1\le 2k-x<\infty \Rightarrow 2k-1\le f\left(x\right)<\infty \forall x\le 1...\left[1\right]$
$If1<x<\infty \Rightarrow 4k<4kx<\infty \Rightarrow 4k+1<4kx+1<\infty \Rightarrow 4k+1<f\left(x\right)<\infty \forall x>1...\left[2\right]$

Since, f(x) is an onto function, the union of [1] and [2] should be R. This is not possible as k is positive.
Therefore, m cannot be positive. Case II is not possible.

$Rangeoffis[-1,0)$