Question
The range of the function $$f\left( x \right) = \left| {2x + 1} \right| - 2\left| {x - 1} \right|,\,x\, \in \,R,$$ is :
A.
$$\left[ { - 3,\,3} \right]$$
B.
$$\left[ {0,\,6} \right]$$
C.
$$R$$
D.
none of these
Answer :
$$\left[ { - 3,\,3} \right]$$
Solution :
$$\eqalign{
& {\text{Here, }}f\left( x \right) = - \left( {2x + 1} \right) - 2\left\{ { - \left( {x - 1} \right)} \right\} = - 3,\,x < - \frac{1}{2} \cr
& 2x + 1 - 2\left\{ { - \left( {x - 1} \right)} \right\} = 4x - 1,\, - \frac{1}{2} \leqslant x < 1 \cr
& 2x + 1 - 2\left( {x - 1} \right) = 3,\,x \geqslant 1 \cr} $$
$$f\left( x \right)$$ is the constant $$-3$$ in $$\left( { - \infty ,\, - \frac{1}{2}} \right)$$ and the constant $$3$$ in $$\left[ {1,\, + \infty } \right)$$
In $$ - \frac{1}{2} \leqslant x < 1,\,f'\left( x \right) = 4 > 0$$ and so, $$f\left( x \right)$$ is m.i. in $$\left[ { - \frac{1}{2},\,1} \right)$$
In this interval, $$f\left( x \right)$$ increases from $$-3$$ to $$3 - \in $$
As the function is continuous, range $$ = \left[ { - 3,\,3} \right]$$