Question
The approximate value of square root of 25.2 is
Answer: Option B
:
B
Let f (x) = √x
Now, f(x+δx)−f(x)=f′(x).δx=δx2√x
We may write, 25.2 = 25 + 0.2
Taking x = 25 and δx=0.2 We have
f(25.2)−f(25)=0.22√25=0.02∴f(25.2)=f(25)+0.02=√25+0.02=5.02⇒√(25.2)=5.02
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:
B
Let f (x) = √x
Now, f(x+δx)−f(x)=f′(x).δx=δx2√x
We may write, 25.2 = 25 + 0.2
Taking x = 25 and δx=0.2 We have
f(25.2)−f(25)=0.22√25=0.02∴f(25.2)=f(25)+0.02=√25+0.02=5.02⇒√(25.2)=5.02
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