Question

If \[f\left( x \right) = \left| \log_{10} x \right|\] then at x = 1

पर्याय

•  f (x) is continuous and f' (1⁺) = log₁₀ e

•  f (x) is continuous and f' (1⁺) = log₁₀ e

•  f (x) is continuous and f' (1⁻) = log₁₀ e

•  f (x) is continuous and f' (1⁻) = −log₁₀ e

Answer 1

f (x) is continuous and 

\[f'\](1⁺) = \[\log_{10} e\]

 f (x) is continuous and 

\[f'\]  (1⁻) = − \[\log_{10} e\]

Given:

\[f\left( x \right) = \left| \log_{10} x \right| = \left| \frac{\log_e x}{\log_e 10} \right| = \left| \left( \log_e x \right) \times \left( \log_{10} e \right) \right| = \left( \log_{10} e \right) \left| \log_e x \right|\]

\[\Rightarrow f'\left( 1^+ \right) = \lim_{h \to 0} \frac{f\left( 1 + h \right) - f\left( 1 \right)}{h} = \lim_{h \to 0} \frac{\left( \log_{10} e \right) \left| \log_e \left( 1 + h \right) \right| - \left( \log_{10} e \right) \left| \log_e 1 \right|}{h} = \left( \log_{10} e \right) \lim_{h \to 0} \frac{\left| \log_e \left( 1 + h \right) \right|}{h} = \left( \log_{10} \left( e \right) \right) \times 1 = \left( \log_{10} e \right)\]

Also,

\[f'\left( 1^- \right) = \lim_{h \to 0} \frac{f\left( 1 - h \right) - f\left( 1 \right)}{h} = \lim_{h \to 0} \frac{\left( \log_{10} e \right) \left| \log_e \left( 1 - h \right) \right| - \left( \log_{10} e \right) \left| \log_e 1 \right|}{h} = - \left( \log_{10} e \right) \lim_{h \to 0} \frac{\left| \log_e \left( 1 - h \right) \right|}{- h} = - \left( \log_{10} e \right) \times 1 = - \left( \log_{10} e \right)\]