The correct option is C x∈(−∞,−15] ∪ [5,∞)
|x+5|≥10 ⋯(A)
Case 1: x+5<0, i.e. x<−5
Then, |x+5|=−x−5
Then, (A) becomes −x−5≥10
⇒x≤−15
But x<−5 ∴x≤−15
⇒x∈(−∞,−15] ⋯(i)
Case 2: x+5≥0, i.e. x≥−5
Then, |x+5|=x+5
Then, (A) becomes x+5≥10
⇒x≥5
But x≥−5 ∴x≥5
⇒x∈[5,∞) ⋯(ii)
From (i) and (ii)
x∈(−∞,−15] ∪ [5,∞)