A) 1
B) -1
C) 2
D) -2
Correct Answer: A
Solution :
[a] : R.H.L \[(x=0)=\alpha +0=\alpha \] Now, \[\frac{\sin x-x}{{{x}^{3}}}=\frac{x-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}-.....-x}{{{x}^{3}}}=\frac{-1}{3!}+\frac{{{x}^{2}}}{5!}-.....\] \[\therefore \]\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x-x}{{{x}^{3}}}=\frac{-1}{6}\] L.H.L. \[=\beta -1\] \[f(x)\]is continuous at\[x=0\Rightarrow \beta -1=2=\alpha \] \[\Rightarrow \beta =3,\alpha =2.\]So,\[\beta -\alpha =1\]You need to login to perform this action.
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