UPSC NDA Math Model Paper 1

For x ≠ 0, the given function can be re-written as:
f(x)={

x−x2 2x ;	x≠0
K;	x=0

Since the equation of the function is same for x < 0 and x > 0, we have:

lim

x→0+

f(x)=

lim

x→0−

f(x) =

lim

x→0

1−x

2

=

1−0

2

=

1

2

For the function to be continuous at x = 0, we must have:

lim

x→0

f(x)=f(0)
⇒K=

1

2