Question
Function f(x) = |x| - |x - 1| is monotonically increasing when:
- x < 0
- x > 1
- x < 1
- 0 < x < 1
Solution:
f(x) = |x| - |x - 1|
Case I:
Let x < 0
If x < 0, then |x| = -x
⇒ |x - 1| = -(x - 1)
Now,
f(x) = |x| - |x - 1|
= -x - (-x + 1)
= -x + x - 1
= -1
f'(x) = 0
So, f(x) is not monotonically increasing when x < 0.
Case II:
Let x < 0 < 1
Here,
|x| = x
⇒ |x - 1| = -(x - 1)
Now,
f(x) = |x| - |x - 1|
= x + x -1
= 2x - 1
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\frac{19}{8}$
$\frac{8}{19}$
$\frac{19}{2}$
$\frac{3}{4}$
$(A)$ $P=y+x$
$(B)$ $P=y-x$
$(C)$ $P+Q=1-x+y+y^{\prime}+\left(y^{\prime}\right)^2$
$(D)$ $P-Q=x+y-y^{\prime}-\left(y^{\prime}\right)^2$