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If F ( X ) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 1 − Cos 10 X X 2 , X < 0 a , X = 0 √ X √ 625 + √ X − 25 , X > 0 Then the Value of a So that F (X) May Be Continuous at X = 0, is (A) 25 (B) 50 (C) −25 (D) None of These - Mathematics

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प्रश्न

If \[f\left( x \right) = \begin{cases}\frac{1 - \cos 10x}{x^2} , & x < 0 \\ a , & x = 0 \\ \frac{\sqrt{x}}{\sqrt{625 + \sqrt{x}} - 25}, & x > 0\end{cases}\] then the value of a so that f (x) may be continuous at x = 0, is 

विकल्प

  • 25

  • 50

  • −25

  • none of these

MCQ

उत्तर

If \[f\left( x \right)\]  is continuous at  \[x = 0\] , then 

\[\lim_{x \to 0^-} f\left( x \right) = f\left( 0 \right)\]
\[\Rightarrow \lim_{h \to 0} f\left( - h \right) = f\left( 0 \right)\]

\[\Rightarrow \lim_{h \to 0} \frac{\left( 1 - \cos \left( - 10h \right) \right)}{\left( - h \right)^2} = f\left( 0 \right)\]
\[ \Rightarrow \lim_{h \to 0} \frac{\left( 1 - \cos \left( 10h \right) \right)}{h^2} = f\left( 0 \right)\]
\[ \Rightarrow \lim_{h \to 0} \frac{\left( 2 \sin^2 \left( 5h \right) \right)}{h^2} = a\]
\[ \Rightarrow \lim_{h \to 0} \frac{2 \times 25\left( \sin^2 \left( 5h \right) \right)}{25 h^2} = a\]
\[ \Rightarrow 50 \lim_{h \to 0} \frac{\left( \sin^2 \left( 5h \right) \right)}{\left( 5h \right)^2} = a\]
\[ \Rightarrow 50 \lim_{h \to 0} \left( \frac{\sin \left( 5h \right)}{5h} \right)^2 = a\]
\[ \Rightarrow a = 50\]

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अध्याय 9: Continuity - Exercise 9.4 [पृष्ठ ४६]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 9 Continuity
Exercise 9.4 | Q 34 | पृष्ठ ४६

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