( |x+2|>3 ) को संख्या रेखा पर दिखाने पर कौन सा रूप बनेगा?
Which form is obtained when ( |x+2|>3 ) is shown on the number line?
Explanation opens after your attempt
Correct Answer
D. ( x<-5 ) या ( x>1 )( x<-5 ) or ( x>1 )
Concept
( |x+2|>3 ) gives ( x+2<-3 ) or ( x+2>3 ). Boundary points stay open in a strict inequality.
Why this answer is correct
The correct answer is D. ( x<-5 ) या ( x>1 ) / ( x<-5 ) or ( x>1 ). ( |x+2|>3 ) gives ( x+2<-3 ) or ( x+2>3 ). Boundary points stay open in a strict inequality.
Exam Tip
( |x+2|>3 ) से ( x+2<-3 ) या ( x+2>3 ) मिलता है। सख्त असमानता में सीमा बिंदु खुले रहते हैं।
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