If $ f: R \rightarrow R $ is defined by $ f(x)=\left\{\begin{array}{cc}\frac{x+2}{x^{2}+3 x+2}, & \text { if } x \in R-\{-1,-2\} \ -1, & \text { if } \quad x=-2 \quad \text { then } f \text { is continuous on the se } \ 0, & \text { if } \quad x=-1\end{array}\right. $

Question

If $ f: R \rightarrow R $ is defined by $ f(x)=\left\{\begin{array}{cc}\frac{x+2}{x^{2}+3 x+2}, & \text { if } x \in R-\{-1,-2\} \ -1, & \text { if } \quad x=-2 \quad \text { then } f \text { is continuous on the se } \ 0, & \text { if } \quad x=-1\end{array}\right. $, $ R $, $ R-\{-2\} $, $ R-\{-1\} $, $ R-\{-1,-2\} $

MARKS App · Accepted Answer

Since, f x is continuous for R , not sure about - 1 , - 2 . Now, we have check continuity at these points. At x = - 2 , LHL = l i m n → 0 - 2 - n + 2 - 2 - n 2 + 3 - 2 - n + 2 = l i m n → 0 - n n 2 + n = - 1 RHL = l i m n → 0 - 2 + n + 2 - 2 + n 2 + 3 - 2 + n + 2 = l i m n → 0 n n 2 - n = - 1 LHL = RHL = f - 2 It is continuous at x = - 2 Now, check for x = - 1 LHL = l i m n → 0 - 1 - n + 2 - 1 - n 2 + 3 - 1 - n + 2 = l i m n → 0 1 - n n 2 - n = ∞ RHL = l i m n → 0 - 1 + n + 2 - 1 + n 2 + 3 - 1 + n + 2 = l i m n → 0 1 + n n 2 + n = ∞ LHL = RHL ≠ f - 1 It is not continuous at x = - 1 The required function is continuous in R - { - 1 }

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