Question

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 

Options

• 25

• 50

• −25

• none of these

Answer 1

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

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

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

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