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CGPET 2022 Solved Paper

Section: Mathematics

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Question : 135 of 150

Marks: +1, -0

The function

f(x)=\begin{cases} \frac{\lvert x-3 \rvert}{x-3}, & x \neq 3 \\ 0, & x=3 \end{cases}.

discontinuous at $x=-2$
continuous at $x=-2$
continuous at $x=3$
None of the above

Solution:

We have

f(x)=\begin{cases} \frac{\lvert x-3 \rvert}{x-3}, & x \neq 3 \\ 0, & x=3 \end{cases}.

=\begin{cases} -\frac{x-3}{x-3}, & x<3 \\ 0, & x=3 \\ \frac{x-3}{x-3}, & x>3 \end{cases}

=\begin{cases} -1, & x<3 \\ 0, & x=3 \\ 1, & x>3 \end{cases}

Now, at

x=3

\text{LHL} = -1, \text{RHL} = 1, f(3) = 0

So,

f(x)

is not continuous at

x =3

Again, at

c=-2

f(x) = -1 [\because \text{ when } x<3, f(x) = -1]

which is a polynomial function. So,

f(x)

is

continuous at

x=-2

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