A) 0
B) ±1
C) 2
D) \[-2\]
Correct Answer: B
Solution :
[b] \[\because \phi \left( x \right)\]is continuous at \[x=0\] Then \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\phi \left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\phi \left( x \right)=\phi \left( 0 \right).\] Now, \[\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos \lambda x}{x\sin x}=\frac{1}{2}.\] \[\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\frac{\lambda x}{2}}{{{x}^{2}}}.\frac{x}{\sin x}=\frac{1}{2}\] \[\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,2.{{\left( \frac{\sin \frac{\lambda x}{2}}{\frac{\lambda x}{2}} \right)}^{2}}\times \frac{{{\left( \frac{\lambda }{2} \right)}^{2}}}{\left( \frac{\sin x}{x} \right)}=\frac{1}{2}\] \[\Rightarrow 2\times 1\times \frac{{{\lambda }^{2}}}{4}=\frac{1}{2}\Rightarrow {{\lambda }^{2}}=1\] \[\therefore \lambda =\pm 1\] Hence, option [b] is correct.
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