Thank you for reporting, we will resolve it shortly
Q.
If $|x-5|+|x+5|=10$, then
Continuity and Differentiability
Solution:
$|x - 5| + |x + 5| = 10$
Case-I: $ x \geq 5$, the equation becomes
$(x-5)+(x+5)=10$
$\Rightarrow 2 x =10 \Rightarrow x =5$ which satisfies the case, therefore accepted.
Case-II: $-5< x< 5 $ The above equation becomes
$-(x-5)+(x+5)=10 \Rightarrow -x+5+x+5=10$
$\Rightarrow 10=10 \text { which is true. }$
$\text { So, the solution is } x \in(-5,5)$
$\text { Case-III: } x \leq-5 \text {, The above equation becomes } $
$-( x -5)-( x +5)=10 \Rightarrow - x +5- x -5=10$
$\Rightarrow -2 x =10$
$\Rightarrow x =-5 \text { which satisfies the above case so, accepted. } $
$\therefore \text { final answer is } x \in[-5,5] \Rightarrow B C $