MHT CET · Maths · Application of Derivatives
The approximate value of \(\log _{10} 99\) is \(\quad\left(\right.\) Given \(\log _{10} \mathrm{e}=0.4343\) )
- A \(1.9657\)
- B \(1.9857\)
- C \(1.1 .9957\)
- D \(1.9757\)
Answer & Solution
Correct Answer
(C) \(1.1 .9957\)
Step-by-step Solution
Detailed explanation
Let \(f(x)=\log _{10} x=\frac{\log _{e} x}{\log _{e} 10}\)
\(f^{\prime}(x)=\frac{1}{x \log 10}\)
Let \(\mathrm{a}=100, \mathrm{~h}=-1\)
\(\therefore f(a)=\log _{10} 100 \quad=\log _{10} 10^{2}=2\)
\(f^{\prime}(a)=\frac{1}{100 \times \log _{e} 10}=\frac{1}{100} \log _{10} e=\frac{1}{100} \times 0.4343\)
We know that
\(f(a-h) \doteqdot f(a)+h f^{\prime}(a)\)
\(\doteqdot 2+(-1) \times \frac{1}{100}(0.4343)=2-0.004343\)\(=1.995657\)
\(\doteqdot 1.9957
\)
\(f^{\prime}(x)=\frac{1}{x \log 10}\)
Let \(\mathrm{a}=100, \mathrm{~h}=-1\)
\(\therefore f(a)=\log _{10} 100 \quad=\log _{10} 10^{2}=2\)
\(f^{\prime}(a)=\frac{1}{100 \times \log _{e} 10}=\frac{1}{100} \log _{10} e=\frac{1}{100} \times 0.4343\)
We know that
\(f(a-h) \doteqdot f(a)+h f^{\prime}(a)\)
\(\doteqdot 2+(-1) \times \frac{1}{100}(0.4343)=2-0.004343\)\(=1.995657\)
\(\doteqdot 1.9957
\)
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