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Discuss the continuity of the function \mathrm{f}, where \mathrm{f} is defined by: f(x)= \begin{cases}3, & \text { if } 0 \leq x \leq 1 \\ 4, & \text { if } 1<x<3 \\ 5, & \text { if } 3 \leq x \leq 10\end{cases}
CBSE | Class 12 | Continuity and Differentiability | Exercise 5.1 | Maths | NCERT
Discuss the continuity of the function \mathrm{f}, where \mathrm{f} is defined by: f(x)= \begin{cases}3, & \text { if } 0 \leq x \leq 1 \\ 4, & \text { if } 1<x<3 \\ 5, & \text { if } 3 \leq x \leq 10\end{cases}

Solution: The provided function is f(x)= \begin{cases}3, & \text { if } 0 \leq x \leq 1 \\ 4, & \text { if } 1<x<3 \\ 5, & \text { if } 3 \leq x \leq 10\end{cases}

In interval, 0 \leq x \leq 1, f(x)=3

As a result, f is continuous in this interval.

At x=1,

Left Hand Limit =
\lim _{x \rightarrow 1^{-}} f(x)=3 and R.H.L. =\lim _{x \rightarrow 1^{-}} f(x)=4

Since, L.H.L. \neq R.H.L.

As a result, f(x) is discontinuous at x=1.

At x=3,
Left Hand Limit. =\lim _{x \rightarrow 3} f(x)=4
and Right Hand Limit =\lim _{x \rightarrow-3} f(x)=5

Since, L.H.L. \neq R.H.L. As a result, f(x) is discontinuous at x=3

Hence, f is discontinuous at x=1 and x=3

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