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harsh8686 wrote:If y=|x+5|−|x−5|, then y can take how many integer values?

A. 5
B. 10
C. 11
D. 20
E. 21

$$?\,\,\,:\,\,\,\# \,\,\,{\rm{integer}}\,\,{\rm{values}}\,\,{\rm{for}}\,\,\,y = \left| {x + 5} \right| - \left| {x - 5} \right|$$

$$\left( {\rm{i}} \right)\,\,\,x \le - 5\,\,\,\,\, \Rightarrow \,\,\,\,\,y = - \left( {x + 5} \right) - \left( {5 - x} \right) = - 10\,\,\,\,\,\, \Rightarrow \,\,\,\,\,1\,\,{\rm{integer}}$$

$$\left( {{\rm{ii}}} \right)\,\,\, - 5 < x \le 5\,\,\,\,\, \Rightarrow \,\,\,\,\,y = \left( {x + 5} \right) - \left( {5 - x} \right) = 2x\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{other}}\,\,10 \cdot 2\,\,{\rm{integers}}\,\,\,\left( * \right)\,\,\,$$
$$\left( * \right)\,\,\,x\,\,{\mathop{\rm int}} \,\,\,\left( { - 4 \le x \le 5\,\,\, \Rightarrow \,\,\,10\,\,{\rm{options}}} \right)\,\,\,\,{\rm{or}}\,\,\,\,\,\left\{ \matrix{
\,x \ne {\mathop{\rm int}} \hfill \cr
\,x = {{{\mathop{\rm int}} } \over 2} \hfill \cr} \right.\,\,\,\,\,\,\,\,\left( { - 9 \le \mathop{\rm int}\,{\rm{odd}}\,\, \le 9\,\,\,\,\, \Rightarrow \,\,\,10\,\,{\rm{options}}} \right)$$

$$\left( {{\rm{iii}}} \right)\,\,\,x > 5\,\,\,\,\, \Rightarrow \,\,\,\,\,y = \left( {x + 5} \right) - \left( {x - 5} \right) = 10\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{already}}\,\,{\rm{obtained}}\,\,{\rm{in}}\,\,\left( {{\rm{ii}}} \right)$$

$$? = 1 + 20 = 21$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

harsh8686 wrote:If y=|x+5|−|x−5|, then y can take how many integer values?

A. 5
B. 10
C. 11
D. 20
E. 21

Alternate way:



There are 21 integers between -10 and 10, both of them included (see blue interval).

It´s done!

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

harsh8686 wrote:If y=|x+5|−|x−5|, then y can take how many integer values?

A. 5
B. 10
C. 11
D. 20
E. 21

The CRITICAL POINTS are where the expressions inside the absolute values are equal to 0.

x+5 = 0 when x=-5.
Substituting x=-5 into y = |x+5|-|x-5|, we get:
y = |-5+5| - |-5-5| = 0-10 = -10.

x-5 = 0 when x=5.
Substituting x=5 into y = |x+5|-|x-5|, we get:
y = |5+5| - |5-5| = 10-0 = 10.

The resulting values in blue indicate the range for y.
Thus, y can be equal to any integer value between -10 and 10, inclusive, implying that there are 21 integer options for y.

The correct answer is E.