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Continuity — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsContinuity3 Marks
Question
Discuss the continuity of the following function at the point(s) or on the interval indicated against them:
$\begin{aligned}
& =2 x^2-2 x+5, & & \text { for } 0 \leq x \leq 2 \\
& =\frac{1-3 x-x^2}{1-x}, & & \text { for } 2<x<4 \\
& =\frac{x^2-25}{x-5}, & & \text { for } 4 \leq x \leq 7 \text { and } x \neq 5 \\
& =7, & & \text { for } x=5
\end{aligned}$
✓

Answer

The domain of $f$ is $[0,7]$
For $0 \leq x \leq 2, f(x)=2 x^2-2 x+5$, being a polynomial function is continuous.
For $2<\mathrm{x}<4, \mathrm{f}(\mathrm{x})=\frac{1-3 x-x^2}{1-x}$, being a rational function is continuous except at the point where its denominator $1-x=0$, i.e., at the point $x=1$ which does not belong to $(2,4)$.
For $4 \leq x \leq 7, x \neq 5, f(x)=\frac{x^2-25}{x-5}$, being a rational function is continuous except at the point where its denominator $x-5=0$, i.e., at the point $x=5$.
Continuity at $x=5$
$
\begin{aligned}
& f(5)=7 \quad \ldots \text { (Given) } \ldots(1) \\
& \lim _{x \rightarrow 5} f(x)=\lim _{x \rightarrow 5} \frac{x^2-25}{x-5} \\
& =\lim _{x \rightarrow 5} \frac{(x-5)(x+5)}{x-5} \\
& =\lim _{x \rightarrow 5}(x+5) \ldots[\because x \rightarrow 5, x \neq 5 \therefore x-5 \neq 0] \\
& =5+5 \\
& =10 \ldots \text { (2) }
\end{aligned}
$
From (1) and (2)
limx-> f(x) = f(5)
z >5
..f is not continuous at x = 5.
Hence, f is continuous on its domain [0, 7] except at the point x = 5, where it is discontinuous.

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