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प्रश्न
If \[f\left( x \right) = \begin{cases}\frac{1 - \sin^2 x}{3 \cos^2 x} , & x < \frac{\pi}{2} \\ a , & x = \frac{\pi}{2} \\ \frac{b\left( 1 - \sin x \right)}{\left( \pi - 2x \right)^2}, & x > \frac{\pi}{2}\end{cases}\]. Then, f (x) is continuous at \[x = \frac{\pi}{2}\], if
पर्याय
\[a = \frac{1}{3},\] b = 2
\[a = \frac{1}{3}, b = \frac{8}{3}\]
- \[a = \frac{2}{3}, b = \frac{8}{3}\]
none of these
उत्तर
\[a = \frac{1}{3} , b = \frac{8}{3}\]
Given:
We have
(LHL at x = \[\frac{\pi}{2}\] = \[\lim_{x \to \frac{\pi}{2}^-} f\left( x \right) = \lim_{h \to 0} f\left( \frac{\pi}{2} - h \right)\]
\[= \lim_{h \to 0} \left( \frac{1 - \sin^2 \left( \frac{\pi}{2} - h \right)}{3 \cos^2 \left( \frac{\pi}{2} - h \right)} \right)\]
\[ = \lim_{h \to 0} \left( \frac{1 - \cos^2 h}{3 \sin^2 h} \right)\]
\[ = \frac{1}{3} \lim_{h \to 0} \left( \frac{\sin^2 h}{\sin^2 h} \right)\]
\[ = \frac{1}{3}\]
(RHL at x = \[\frac{\pi}{2}\] = \[\lim_{x \to \frac{\pi}{2}^+} f\left( x \right) = \lim_{h \to 0} f\left( \frac{\pi}{2} + h \right)\]
\[= \lim_{h \to 0} \left( \frac{b\left[ 1 - \sin \left( \frac{\pi}{2} + h \right) \right]}{\left[ \pi - 2\left( \frac{\pi}{2} + h \right) \right]^2} \right)\]
\[ = \lim_{h \to 0} \left( \frac{b\left( 1 - \cos h \right)}{\left[ - 2h \right]^2} \right)\]
\[ = \lim_{h \to 0} \left( \frac{2b \sin^2 \frac{h}{2}}{4 h^2} \right)\]
\[ = \lim_{h \to 0} \left( \frac{2b \sin^2 \frac{h}{2}}{16\frac{h^2}{4}} \right)\]
\[ = \frac{b}{8} \lim_{h \to 0} \left( \frac{\sin\frac{h}{2}}{\frac{h}{2}} \right)^2 \]
\[ = \frac{b}{8} \times 1\]
\[ = \frac{b}{8}\]
Also,
If f(x) is continuous at x = \[\frac{\pi}{2}\], then
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