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Continuity and Differentiability
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Solution:
∣x−5∣+∣x+5∣=10
Case-I: x≥5, the equation becomes (x−5)+(x+5)=10 ⇒2x=10⇒x=5 which satisfies the case, therefore accepted.
Case-II: −5<x<5 The above equation becomes −(x−5)+(x+5)=10⇒−x+5+x+5=10 ⇒10=10 which is true. So, the solution is x∈(−5,5) Case-III: x≤−5, The above equation becomes −(x−5)−(x+5)=10⇒−x+5−x−5=10 ⇒−2x=10 ⇒x=−5 which satisfies the above case so, accepted. ∴ final answer is x∈[−5,5]⇒BC