Question

x01, if x112.

Answer 1

The given function is,

f( x )={ x 10 −1   ,  x≤1 x 2          ,  x>1

Consider k be any real number, then the cases will be k<1, k=1 or k>1.

When k<1, then the function is,

f( k )= k 10 −1

The limit of the function is,

lim x→k f( x )= lim x→k ( x 10 −1 ) = k 10 −1

It can be observed that, lim x→k f( x )=f( k ).

Therefore, the function is continuous for all real numbers less than 1.

When k=1, then the function becomes,

f( 1 )= 1 10 −1 =0

The left hand limit of the function is,

LHL= lim x→ 1 − f( x ) = lim x→ 1 − ( x 10 −1 ) =0

The right hand limit of the function is,

RHL= lim x→ 1 + f( x ) = lim x→ 1 + ( x 2 ) =1

It can be observed that, LHL≠RHL.

Therefore, the function is discontinuous at x=1.

When k>1, the function becomes,

f( k )= k 2

The limit of the function is,

lim x→k f( x )= lim x→k ( x 2 ) = k 2

It can be observed that, lim x→k f( x )=f( k ).

Therefore, the function is continuous for all points greater than 1.

Thus, the only point of discontinuity is 1.