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If f(x)=\left\{\begin{array}{cc} \alpha+\frac{\sin [x]}{x} & x>0 \\ 2 & x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right] & x<0 \end{array}\right. Where [.] is G.I.F. If f(x) is continuous at x = 0 then \beta-\alpha equal to

Option: 1

1


Option: 2

-1


Option: 3

2


Option: 4

-2


Answers (1)

best_answer

\begin{aligned} & R H L(x=0)=\alpha+0=\alpha \\ & \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 !}-\ldots \ldots \ldots \\ & \operatorname{lt}_{x \rightarrow 0} \frac{\sin x-x}{x^3}=\frac{-1}{6} \\ & L H L=\beta-1 \end{aligned}

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Deependra Verma

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