Question

Complete values of x satisfying |x+5|+|6-x|=11 is:

A. [-6, 5]
B. [5, 6]
C. [-5, 6]
D. [-6, -5]

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Answer 1

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\: |x + 5| + |6 - x| = 11$

We know, Definition of Modulus function is

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x| -\begin{cases} &\sf{ - x, \: \: if \: x \: < \: 0} \\ &\sf{x \: \: if \: x \: \geqslant \: 0} \end{cases}\end{gathered}\end{gathered}$

So, here breaking points are x = - 5 and x = 6

So, three cases arises.

Case :- 1

$\rm :\longmapsto\:When \: x \: < \: - \: 5$

So, given equation

$\rm :\longmapsto\: |x + 5| + |6 - x| = 11$

reduces to

$\rm :\longmapsto\: - (x + 5) + (6 - x) = 11$

$\rm :\longmapsto\: - x - 5 + 6 - x = 11$

$\rm :\longmapsto\: - 2x + 1 = 11$

$\rm :\longmapsto\: - 2x = 11 - 1$

$\rm :\longmapsto\: - 2x = 10$

$\rm :\longmapsto\: - x = 5$

$\rm :\implies\:x = - 5$

It means there is no solution.

Case :- 2

$\rm :\longmapsto\: When \: - 5 \leqslant x \leqslant 6$

So, given equation

$\rm :\longmapsto\: |x + 5| + |6 - x| = 11$

reduces to

$\rm :\longmapsto\:x + 5 + (6 - x) = 11$

$\rm :\longmapsto\:x + 5 + 6 - x = 11$

$\rm :\longmapsto\:11 = 11$

$\bf\implies \:x \: \in \: [ - 5, \: 6]$

Case :- 3

$\rm :\longmapsto\:When \: x \: > \: 6$

So, given equation

$\rm :\longmapsto\: |x + 5| + |6 - x| = 11$

reduces to

$\rm :\longmapsto\:x + 5 - (6 - x) = 11$

$\rm :\longmapsto\:x + 5 - 6 + x = 11$

$\rm :\longmapsto\:2x - 1 = 11$

$\rm :\longmapsto\:2x = 11 + 1$

$\rm :\longmapsto\:2x = 12$

$\rm :\longmapsto\:x = 6$

It means, there is no solution.

Hence,

The solution set of equation

$\bf :\longmapsto\: |x + 5| + |6 - x| = 11$

is

$\bf\implies \:x \: \in \: [ - 5, \: 6]$

Thus

$\: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf{ \: \: \: \: Option \: C. \: is \: correct \: \: \: \: }}}$

Additional Information :-

$\purple{ \boxed{ \bf{ |x| < y \: \implies \: - y < x < y}}}$

$\purple{ \boxed{ \bf{ |x| \leqslant y \: \implies \: - y \leqslant x \leqslant y}}}$

$\purple{ \boxed{ \bf{ |x| > y \: \implies \: x < - y \: \: or \: \: x > y}}}$

$\purple{ \boxed{ \bf{ |x| \geqslant y \: \implies \: x \leqslant - y \: \: or \: \: x \geqslant y}}}$

$\purple{ \boxed{ \bf{ |x - y| < z \: \implies \: y - z < x < y + z}}}$

$\purple{ \boxed{ \bf{ |x - y| \leqslant z \: \implies \: y - z \leqslant x \leqslant y + z}}}$

Answer 2

Given : |x+5|+|6-x|=11  

To Find :   Values of x satisfying

Solution :

|x+5|+|6-x|=11  

| x |  =  x   if  x ≥ 0

      = - x  if x  <  0

Case 1  :  x  >  6    =>  6  - x   <  0    and x + 5 >  0

=>  x  + 5  -  (6 - x)  = 11

=>  2x - 1  = 11

=> 2x = 12

=> x  =  6

Hence no solution for  x > 6

Case 2  : x  < - 5     => x + 5 <  0      6 - x > 0

Hence

-(x + 5)  + 6 - x  =  11

=> - 2x  + 1 = 11

=> x = - 5  

Hence no solution   for x < - 5

Case 3 :     -5 ≤ x  ≤ 6

x + 5 ≥ 0   ,  6 - x ≥  0

=>  x + 5  + 6 - x  =  11

=> 11  = 11

Satisfied for all values of x

Hence x   ∈ [-5 , 6]

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