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Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 11 Maths Textbook Solution.

Answers (1)

Answer:

                Discontinuous 

Hint:

Discontinuous occurs when the both number is equal to zero of the numerator and denominator or the function is undefined at its limit.

 

Solution:

Given,

f(x)=\left(\begin{array}{ll} 1+x^{2}, & \text { if } 0 \leq x \leq 1 \\\\ 2-x, & \text { if } x>1 \end{array}\right)

We observe

[LHL at x=1 ]

                \lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}\left(1+\left(1-h^{2}\right)\right)=\lim _{h \rightarrow 0}\left(2+h^{2}\right)=2

[RHL at x=1 ]

                \begin{gathered} \lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}(2-(1+h))=\lim _{h \rightarrow 0}(1-h)=1 \\\\ \lim _{x \rightarrow 1^{+}} f(x) \neq \lim _{x \rightarrow 1^{-}} f(x) \end{gathered}

Thus, f\left ( x \right )  is discontinuous at x=1 .

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