Question

is not continuous at a, then of discontinuity off. Find all points nenſ is said to be discontinuous at a Points of discontinuity of the following is calle continuous at a, and wity of the following functions: a) f(x) = (x2+1 if xs1 1.005"-1.061, 1.0082 1.100, 1.0054 = 1.127, 1.035-* = 0.8, 1.035 1035-* = 0.7.1.07-* = 0.7, 1.07-* = 0.5 =

Answer 1

(a). The function is sediner by ,

.  

f (x) 2-4

Consider the function then g(x) is a polynomial and so continuous everywhere .

g(x) = x2 - 4

Since g(x) is a continuous function
g(r)
is a continuous where
07 ()b
.

Hence $\frac{1}{x^{2}-4}$ is a continuous function where
40
i.e., $x \neq\pm 2$

Hence point of discontinuity of are { 2 , -2 }.

f(r

Answer : { 2 , -2 }

.

(b). The function is given by ,

f(r

f(x)a1
if  $x \leq 1$

   if x> 1

211

For , $x \leq 1$ so f(x) is a polynomial in that region and hence f(x) is continuous there . Similarly f(x) is continuous on the region x > 1. So only point of discontinuity can be at intersection of the two graph i.e., at x=1

So let us check f(x) is continuous at x = 1 or not .

2 lim f() lim -11-1 0 r 1+ r 1+

$\lim_{x\rightarrow 1-}f(x) = \lim_{x\rightarrow 1-} x^{2}+1= 1+1=2$

Since , . So the function f(x) is not continuous at x = 1 .

lim f(x)=lim f(x) r+1+ r+1-

Hence the point of discontinuity of f(x) are {1} .

Answer : { 1 } .