A) \[(2,{{e}^{2}})\]
B) \[\left( 2,\frac{2}{{{e}^{2}}} \right)\]
C) \[(2,3{{e}^{2}})\]
D) \[\left( 2,\frac{3}{{{e}^{2}}} \right)\]
Correct Answer: C
Solution :
[c] : We have, Now, R.H.L\[=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)\] \[=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,3{{\left( 1+\frac{|\sin x|}{3} \right)}^{\frac{6}{|\sin x|}}}=3{{e}^{2}}\] ...(i) Similarly, L.H.L\[=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=3{{e}^{\alpha }}\] ...(ii) \[\because \]f(x) is continuous at x = 0 \[\therefore \]\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=f(0)\] \[\Rightarrow \]\[3{{e}^{\alpha }}=3{{e}^{2}}=\beta \]. So, \[\alpha =2\] and \[\beta =3{{e}^{2}}\].You need to login to perform this action.
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