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If F ( X ) = | Log 10 X | Then at X = 1 (A) F (X) is Continuous and F' (1+) = Log10 E (B) F (X) is Continuous and F' (1+) = Log10 E (C) F (X) is Continuous and F' (1−) - Mathematics

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प्रश्न

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

विकल्प

  •  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

MCQ

उत्तर

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,

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

 

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अध्याय 9: Continuity - Exercise 9.4 [पृष्ठ ४२]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 9 Continuity
Exercise 9.4 | Q 3 | पृष्ठ ४२

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