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Question

The approximate value of square root of 25.2 is
  1. 5.01
  2. 5.02
  3. 5.03
  4. 5.04


Solution

The correct option is B 5.02
Let f(x)=x
Now, f(x+δx)f(x)=f(x).δx=δx2x
We may write, 25.2 = 25+0.2
Taking x = 25 and  δx=0.2 we have
f(25.2)f(25)=0.22x
We may write, 25.2 = 25+0.2
Taking x = 25 and δx=0.2, we have
f(25.2)f(25)=0.2225=0.02f(25.2)=f(25)+0.02=25+0.02=5.02(25.2)=5.02

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