The correct options are
A f(x)=⎧⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪⎩(e1/x+1)(e1/x−1),x<0(1−cosx)x,x>0
C u(x)=⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩sin−12xtan−13x;x∈(0,12]|sinx|x;x<0
D w(x)={log3(x+2);x>2log1/2(x2+5);x<2
We will check by options
Option A:
f(x)=⎧⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪⎩(e1/x+1)(e1/x−1),x<0(1−cosx)x,x>0
LHL=limx→0−f(x)=limh→0f(0−h)
=limh→0(e−1/h+1)(e−1/h−1)
=0+10−1=−1
RHL=limx→0+f(x)=limh→0f(0+h)
=limh→0(1−cosh)h
=limh→02sin2h/2h
=limh→0h42sin2h/2h2h2
=0×1=0
Difference between LHL and RHL =0−(−1)=1
Hence, jump of discontinuity is 1.
Option B:
f(x)=⎧⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪⎩x1/3−1x1/2−1;x>1lnx(x−1);12<x<1
RHL=limx→1+f(x)=limh→0f(1+h)
=limh→0(1+h)1/3−1(1+h)1/2−1
=limh→0(1+h3−19h2+....)−1(1+12h−18h2.....)−1
=limh→0h(13−19h+....)h(12−18h.....)
RHL=23
LHL=limx→1−f(x)=limh→0f(1−h)
=limh→0ln(1−h)1−h−1
limh→0ln(1+(−h))−h
=1 (∵limx→0ln(1+x)x=1)
Difference between LHL and RHL 1−23=13<1
Here, jump of discontinuity for f(x) is not 1 or greater than 1.
Option C:
f(x)=⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩sin−12xtan−13x;x∈(0,12]|sinx|x;x<0
LHL=limx→0−f(x)=limh→0f(0−h)
=limh→0sin|0−h|−h
=limh→0|−sinh|−h
=limh→0sinh−h=−1
RHL=limx→0+f(x)=limh→0f(0+h)
=limh→0sin−12htan−13h
=limh→02hsin−12h2h3htan−13h3h
RHL=23
Difference between LHL and RHL 23−(1)=53>1
Hence, jump of discontinuity for f(x) is greater than 1.
Option D:
f(x)={log3(x+2);x>2log1/2(x2+5);x<2
RHL=limx→2+f(x)=limh→0f(2+h)
=limh→0log3(h+4)
RHL=log34=1.26
LHL=limx→2−f(x)=limh→0f(2−h)
=limh→0log1/2(h2−4h+9)
LHL=log1/29=−0.653
Difference between LHL and RHL 1.26−(−0.653)>1
Hence, jump of discontinuity for f(x) is greater than 1.