Question

If  \[f\left( x \right) = \begin{cases}a \sin\frac{\pi}{2}\left( x + 1 \right), & x \leq 0 \\ \frac{\tan x - \sin x}{x^3}, & x > 0\end{cases}\] is continuous at x = 0, then a equals

पर्याय

•   \[\frac{1}{2}\]

• \[\frac{1}{3}\]

• \[\frac{1}{4}\]

•   \[\frac{1}{6}\]

Answer 1

\[\frac{1}{2}\]

Given: 

\[f\left( x \right) = \binom{a \sin \frac{\pi}{2}\left( x + 1 \right) , x \leq 0}{\frac{\tan x - \sin x}{x^3} , x > 0}\]

We have
(LHL at x = 0) =  \[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} a \sin \left( \frac{\pi}{2}\left( - h + 1 \right) \right) = a \sin \left( \frac{\pi}{2} \right) = a\]

(RHL at x = 0) =  \[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right) = \lim_{h \to 0} \frac{\tan h - \sin h}{h^3}\]

\[= \lim_{h \to 0} \frac{\frac{\sin h}{\cos h} - \sin h}{h^3}\]
\[ = \lim_{h \to 0} \frac{\frac{\sin h}{\cos h}\left( 1 - \cos h \right)}{h^3}\]
\[ = \lim_{h \to 0} \frac{\left( 1 - \cos h \right) \tan h}{h^3}\]
\[ = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2} \tan h}{4 \times \frac{h^2}{4} \times h}\]
\[ = \frac{2}{4} \lim_{h \to 0} \frac{\sin^2 \frac{h}{2} \tan h}{\frac{h^2}{4} \times h}\]
\[ = \frac{1}{2} \lim_{h \to 0} \left( \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right)^2 \times \lim_{h \to 0} \frac{\tan h }{ h }\]
\[ = \frac{1}{2} \times 1 \times 1\]
\[ = \frac{1}{2}\]

\[\text{ If } f\left( x \right) \text{ is continuous at x = 0, then } \]
\[ \lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right)\]
\[ \Rightarrow a = \frac{1}{2}\]